From 7c48d8900d905740b0f01f124704ea8239c18202 Mon Sep 17 00:00:00 2001 From: Morten Hjorth-Jensen Date: Thu, 4 Nov 2021 08:47:26 +0100 Subject: [PATCH] update --- doc/pub/week44/html/._week44-bs000.html | 127 +- doc/pub/week44/html/._week44-bs001.html | 127 +- doc/pub/week44/html/._week44-bs002.html | 127 +- doc/pub/week44/html/._week44-bs003.html | 259 +- doc/pub/week44/html/._week44-bs004.html | 157 +- doc/pub/week44/html/._week44-bs005.html | 161 +- doc/pub/week44/html/._week44-bs006.html | 144 +- doc/pub/week44/html/._week44-bs007.html | 152 +- doc/pub/week44/html/._week44-bs008.html | 159 +- doc/pub/week44/html/._week44-bs009.html | 157 +- doc/pub/week44/html/._week44-bs010.html | 157 +- doc/pub/week44/html/._week44-bs011.html | 147 +- doc/pub/week44/html/._week44-bs012.html | 142 +- doc/pub/week44/html/._week44-bs013.html | 139 +- doc/pub/week44/html/._week44-bs014.html | 253 +- doc/pub/week44/html/._week44-bs015.html | 244 +- doc/pub/week44/html/._week44-bs016.html | 187 +- doc/pub/week44/html/._week44-bs017.html | 194 +- doc/pub/week44/html/._week44-bs018.html | 234 +- doc/pub/week44/html/._week44-bs019.html | 272 +- doc/pub/week44/html/._week44-bs020.html | 158 +- doc/pub/week44/html/._week44-bs021.html | 141 +- doc/pub/week44/html/._week44-bs022.html | 131 +- doc/pub/week44/html/._week44-bs023.html | 137 +- doc/pub/week44/html/._week44-bs024.html | 145 +- doc/pub/week44/html/._week44-bs025.html | 149 +- doc/pub/week44/html/._week44-bs026.html | 247 +- doc/pub/week44/html/._week44-bs027.html | 254 +- doc/pub/week44/html/._week44-bs028.html | 156 +- doc/pub/week44/html/._week44-bs029.html | 181 +- doc/pub/week44/html/._week44-bs030.html | 183 +- doc/pub/week44/html/._week44-bs031.html | 165 +- doc/pub/week44/html/._week44-bs032.html | 170 +- doc/pub/week44/html/._week44-bs033.html | 160 +- doc/pub/week44/html/._week44-bs034.html | 158 +- doc/pub/week44/html/._week44-bs035.html | 181 +- doc/pub/week44/html/._week44-bs036.html | 213 +- doc/pub/week44/html/._week44-bs037.html | 158 +- doc/pub/week44/html/._week44-bs038.html | 163 +- doc/pub/week44/html/._week44-bs039.html | 146 +- doc/pub/week44/html/._week44-bs040.html | 171 +- doc/pub/week44/html/._week44-bs041.html | 159 +- doc/pub/week44/html/._week44-bs042.html | 161 +- doc/pub/week44/html/._week44-bs043.html | 185 +- doc/pub/week44/html/._week44-bs044.html | 254 +- doc/pub/week44/html/._week44-bs045.html | 260 +- doc/pub/week44/html/._week44-bs046.html | 236 +- doc/pub/week44/html/._week44-bs047.html | 214 +- doc/pub/week44/html/._week44-bs048.html | 224 +- doc/pub/week44/html/._week44-bs049.html | 199 +- doc/pub/week44/html/._week44-bs050.html | 176 +- doc/pub/week44/html/._week44-bs051.html | 203 +- doc/pub/week44/html/._week44-bs052.html | 245 +- doc/pub/week44/html/._week44-bs053.html | 148 +- doc/pub/week44/html/._week44-bs054.html | 159 +- doc/pub/week44/html/._week44-bs055.html | 153 +- doc/pub/week44/html/._week44-bs056.html | 146 +- doc/pub/week44/html/._week44-bs057.html | 158 +- doc/pub/week44/html/._week44-bs058.html | 183 +- doc/pub/week44/html/._week44-bs059.html | 184 +- doc/pub/week44/html/._week44-bs060.html | 207 +- doc/pub/week44/html/._week44-bs061.html | 277 +- doc/pub/week44/html/._week44-bs062.html | 281 +- doc/pub/week44/html/week44-bs.html | 127 +- doc/pub/week44/html/week44-reveal.html | 133 + doc/pub/week44/html/week44-solarized.html | 137 + doc/pub/week44/html/week44.html | 137 + doc/pub/week44/ipynb/ipynb-week44-src.tar.gz | Bin 294283 -> 294283 bytes doc/pub/week44/ipynb/week44.ipynb | 464 ++- doc/src/week43/programs/diffusion.py~ | 3759 ------------------ doc/src/week43/programs/dill.py~ | 203 - doc/src/week43/programs/ode.py~ | 3759 ------------------ doc/src/week43/programs/odetf.py~ | 250 -- doc/src/week43/programs/poisson.py~ | 3759 ------------------ doc/src/week43/programs/wave.py~ | 3759 ------------------ doc/src/week44/week44.do.txt | 117 + 76 files changed, 6832 insertions(+), 21290 deletions(-) delete mode 100644 doc/src/week43/programs/diffusion.py~ delete mode 100644 doc/src/week43/programs/dill.py~ delete mode 100644 doc/src/week43/programs/ode.py~ delete mode 100644 doc/src/week43/programs/odetf.py~ delete mode 100644 doc/src/week43/programs/poisson.py~ delete mode 100644 doc/src/week43/programs/wave.py~ diff --git a/doc/pub/week44/html/._week44-bs000.html b/doc/pub/week44/html/._week44-bs000.html index a503062e6..9e9a67569 100644 --- a/doc/pub/week44/html/._week44-bs000.html +++ b/doc/pub/week44/html/._week44-bs000.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -365,7 +370,7 @@ MathJax.Hub.Config({
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    • diff --git a/doc/pub/week44/html/._week44-bs001.html b/doc/pub/week44/html/._week44-bs001.html index fcaff62e0..05e2d52ad 100644 --- a/doc/pub/week44/html/._week44-bs001.html +++ b/doc/pub/week44/html/._week44-bs001.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -367,7 +372,7 @@ MathJax.Hub.Config({
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    • diff --git a/doc/pub/week44/html/._week44-bs002.html b/doc/pub/week44/html/._week44-bs002.html index 6114410dd..751a1626c 100644 --- a/doc/pub/week44/html/._week44-bs002.html +++ b/doc/pub/week44/html/._week44-bs002.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -350,7 +355,7 @@ accelerate scientific discovery.
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    • diff --git a/doc/pub/week44/html/._week44-bs003.html b/doc/pub/week44/html/._week44-bs003.html index 072a49115..0b0366499 100644 --- a/doc/pub/week44/html/._week44-bs003.html +++ b/doc/pub/week44/html/._week44-bs003.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,12 +327,138 @@ MathJax.Hub.Config({

     

     

     

    -

    Thursday, Principal Component Analysis

    +

    A short Discussion of Project 2

    -

    For the principal component analysis, -see slides from week 43, in particular from slide 28 and forward +

    For neural networks and regression, should I use a design matrix with information about a polynomial fit or not? +Discuss pros and cons. The example here shows some of these issues.

    + + +
    +
    +
    +
    +
    + 
    """
+Code to test Ridge and NNs using Scikit-Learn only
+"""
+
+import numpy as np
+import pandas as pd
+import matplotlib.pyplot as plt
+from sklearn.model_selection import train_test_split
+from sklearn import linear_model
+from sklearn.neural_network import MLPRegressor
+from sklearn.metrics import accuracy_score
+import seaborn as sns
+
+
+def MSE(y_data,y_model):
+    n = np.size(y_model)
+    return np.sum((y_data-y_model)**2)/n
+# A seed just to ensure that the random numbers are the same for every run.
+# Useful for eventual debugging.
+np.random.seed(315)
+
+n = 100
+x = np.random.rand(n)
+y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)
+
+Maxpolydegree = 5
+X = np.zeros((n,Maxpolydegree-1))
+
+for degree in range(1,Maxpolydegree): #No intercept column
+    X[:,degree-1] = x**(degree)
+
+# We split the data in test and training data
+X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
+
+# Decide which values of lambda to use
+
+nlambdas = 10
+lmbd_vals = np.logspace(-4, 0, nlambdas)
+MSERidgePredict = np.zeros(nlambdas)
+for i in range(nlambdas):
+    lmb = lmbd_vals[i]
+    RegRidge = linear_model.Ridge(lmb)
+    RegRidge.fit(X_train,y_train)
+    ypredictRidge = RegRidge.predict(X_test)
+    MSERidgePredict[i] = MSE(y_test,ypredictRidge)
+
+plt.figure()
+plt.plot(np.log10(lmbd_vals), MSERidgePredict, 'g--', label = 'MSE SL Ridge Test')
+plt.xlabel('log10(lambda)')
+plt.ylabel('MSE')
+plt.legend()
+plt.show()
+
+# Neural Network part
+
+n_hidden_neurons = 50
+epochs = 100
+# store models for later use
+eta_vals = np.logspace(-4, 0, 10)
+# store the models for later use
+DNN_scikit = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)
+test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))
+sns.set()
+for i, eta in enumerate(eta_vals):
+    for j, lmbd in enumerate(lmbd_vals):
+        dnn = MLPRegressor(hidden_layer_sizes=(n_hidden_neurons), activation='logistic',
+                            alpha=lmbd, learning_rate_init=eta, max_iter=epochs)
+        dnn.fit(X_train, y_train)
+        ypredictMLP = dnn.predict(X_test)
+        test_accuracy[i][j] = MSE(ypredictMLP, y_test)
+
+fig, ax = plt.subplots(figsize = (10, 10))
+sns.heatmap(test_accuracy, annot=True, ax=ax, cmap="viridis")
+ax.set_title("Training Accuracy")
+ax.set_ylabel("$\eta$")
+ax.set_xlabel("$\lambda$")
+plt.show()
+
+# Now we redefine our design matrix to include only the x-values and try out our NN
+
+X = np.zeros((n,1))
+X[:,0] = x
+
+# We split the data in test and training data again
+X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
+# Repeat the NN calculation
+DNN_scikit = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)
+test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))
+sns.set()
+for i, eta in enumerate(eta_vals):
+    for j, lmbd in enumerate(lmbd_vals):
+        dnn = MLPRegressor(hidden_layer_sizes=(n_hidden_neurons), activation='logistic',
+                            alpha=lmbd, learning_rate_init=eta, max_iter=epochs)
+        dnn.fit(X_train, y_train)
+        ypredictMLP = dnn.predict(X_test)
+        test_accuracy[i][j] = MSE(ypredictMLP, y_test)
+
+fig, ax = plt.subplots(figsize = (10, 10))
+sns.heatmap(test_accuracy, annot=True, ax=ax, cmap="viridis")
+ax.set_title("Training Accuracy")
+ax.set_ylabel("$\eta$")
+ax.set_xlabel("$\lambda$")
+plt.show()
+
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    + +

    diff --git a/doc/pub/week44/html/._week44-bs004.html b/doc/pub/week44/html/._week44-bs004.html index b6d7730aa..ce201e738 100644 --- a/doc/pub/week44/html/._week44-bs004.html +++ b/doc/pub/week44/html/._week44-bs004.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,34 +327,10 @@ MathJax.Hub.Config({

     

     

     

    -

    A kind of Bird's view on PCA

    +

    Thursday, Principal Component Analysis

    -Why do we maximize variance during Principal Component Analysis? - -

    Variance is a measure of the variability of the data you -have. Potentially the number of components is infinite, so you want to "squeeze" the most -information in each component of the finite set you build. -

    - -

    If, to exaggerate, you were to select a single principal component, -you would want it to account for the most variability possible: hence -the search for maximum variance, so that the one component collects -the most "uniqueness" from the data set. -

    - -

    Maximizing the component vector variances is the same as maximizing -the 'uniqueness' of those vectors. The vectors are as distant -from each other as possible (orthogonal to each other). -

    - -

    Take for example a situation where you have 2 lines that are -orthogonal in a 3D space. You can capture the environment much more -completely with those orthogonal lines than 2 lines that are parallel -(or nearly parallel). When applied to very high dimensional states -using very few vectors, this becomes a much more important -relationship among the vectors to maintain. In a linear algebra sense -you want independent rows to be produced by PCA, otherwise some of -those rows will be redundant. +

    For the principal component analysis, +see slides from week 43, in particular from slide 28 and forward

    @@ -371,7 +352,7 @@ those rows will be redundant.

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    • diff --git a/doc/pub/week44/html/._week44-bs005.html b/doc/pub/week44/html/._week44-bs005.html index 6cb747086..5caf396f5 100644 --- a/doc/pub/week44/html/._week44-bs005.html +++ b/doc/pub/week44/html/._week44-bs005.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,18 +327,34 @@ MathJax.Hub.Config({

     

     

     

    -

    Thursday: Clustering and Unsupervised Learning

    +

    A kind of Bird's view on PCA

    -

    In general terms cluster analysis, or clustering, is the task of grouping a -data-set into different distinct categories based on some measure of equality of -the data. This measure is often referred to as a metric or similarity -measure in the literature (note: sometimes we deal with a dissimilarity -measure instead). Usually, these metrics are formulated as some kind of -distance function between points in a high-dimensional space. +Why do we maximize variance during Principal Component Analysis? + +

    Variance is a measure of the variability of the data you +have. Potentially the number of components is infinite, so you want to "squeeze" the most +information in each component of the finite set you build.

    -

    The simplest, and also the most -common is the Euclidean distance. +

    If, to exaggerate, you were to select a single principal component, +you would want it to account for the most variability possible: hence +the search for maximum variance, so that the one component collects +the most "uniqueness" from the data set. +

    + +

    Maximizing the component vector variances is the same as maximizing +the 'uniqueness' of those vectors. The vectors are as distant +from each other as possible (orthogonal to each other). +

    + +

    Take for example a situation where you have 2 lines that are +orthogonal in a 3D space. You can capture the environment much more +completely with those orthogonal lines than 2 lines that are parallel +(or nearly parallel). When applied to very high dimensional states +using very few vectors, this becomes a much more important +relationship among the vectors to maintain. In a linear algebra sense +you want independent rows to be produced by PCA, otherwise some of +those rows will be redundant.

    @@ -356,7 +377,7 @@ common is the Euclidean distance.

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    • diff --git a/doc/pub/week44/html/._week44-bs006.html b/doc/pub/week44/html/._week44-bs006.html index 69824ad1c..df3b179ce 100644 --- a/doc/pub/week44/html/._week44-bs006.html +++ b/doc/pub/week44/html/._week44-bs006.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,13 +327,18 @@ MathJax.Hub.Config({

     

     

     

    -

    Basic Idea of the \( k \)-means Clustering Algorithm

    +

    Thursday: Clustering and Unsupervised Learning

    -

    The simplest of all clustering algorithms is the k-means algorithm -, sometimes also referred to as Lloyds algorithm. It is the simplest and also -the most common. From its simplicity it obtains both strengths and weaknesses. -These will be discussed in more detail later. The \( k \)-means algorithm is a -centroid based clustering algorithm. +

    In general terms cluster analysis, or clustering, is the task of grouping a +data-set into different distinct categories based on some measure of equality of +the data. This measure is often referred to as a metric or similarity +measure in the literature (note: sometimes we deal with a dissimilarity +measure instead). Usually, these metrics are formulated as some kind of +distance function between points in a high-dimensional space. +

    + +

    The simplest, and also the most +common is the Euclidean distance.

    @@ -352,7 +362,7 @@ These will be discussed in more detail later. The \( k \)-means algorithm is a

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    • diff --git a/doc/pub/week44/html/._week44-bs007.html b/doc/pub/week44/html/._week44-bs007.html index 505fc711f..9b0d441ba 100644 --- a/doc/pub/week44/html/._week44-bs007.html +++ b/doc/pub/week44/html/._week44-bs007.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,28 +327,15 @@ MathJax.Hub.Config({

     

     

     

    -

    The \( k \)-means Algorithm

    +

    Basic Idea of the \( k \)-means Clustering Algorithm

    -

    Assume, we are given \( n \) data points and we wish to split the data into \( K < n \) -different categories, or clusters. We label each cluster by an integer +

    The simplest of all clustering algorithms is the k-means algorithm +, sometimes also referred to as Lloyds algorithm. It is the simplest and also +the most common. From its simplicity it obtains both strengths and weaknesses. +These will be discussed in more detail later. The \( k \)-means algorithm is a +centroid based clustering algorithm.

    -$$ k\in\{1, \cdots, K \}. -$$ - -

    In the basic k-means algorithm each point is assigned to only -one cluster \( k \), and these assignments are non-injective i.e. many-to-one. We -can think of these mappings as an encoder \( k = C(i) \), which assigns the \( i \)-th -data-point \( \bf x_i \) to the \( k \)-th cluster. -

    - -

    \( k \)-means algorithm in words:

    -
      -
    1. We start with guesses / random initializations of our \( k \) cluster centers/centroids
      2. -
    2. For each centroid the points that are most similar are identified
      3. -
    3. Then we move / replace each centroid with a coordinate average of all the points that were assigned to that centroid.
      4. -
    4. Iterate 2-3 until the centroids no longer move (to some tolerance)
      5. -

    diff --git a/doc/pub/week44/html/._week44-bs008.html b/doc/pub/week44/html/._week44-bs008.html index 98b6c480b..c3ef52afe 100644 --- a/doc/pub/week44/html/._week44-bs008.html +++ b/doc/pub/week44/html/._week44-bs008.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,26 +327,28 @@ MathJax.Hub.Config({

     

     

     

    -

    Basic Math of the \( k \)-means Algorithm

    +

    The \( k \)-means Algorithm

    -

    We assume we have \( n \) data-points

    -$$ -\begin{equation}\tag{1} - \boldsymbol{x_i} = \{x_{i, 1}, \cdots, x_{i, p}\}\in\mathbb{R}^p. -\end{equation} -$$ - -

    which we wish to group into \( K < n \) clusters. For our dissimilarity measure we -use the squared Euclidean distance +

    Assume, we are given \( n \) data points and we wish to split the data into \( K < n \) +different categories, or clusters. We label each cluster by an integer

    -$$ -\begin{equation}\tag{2} - d(\boldsymbol{x_i}, \boldsymbol{x_i'}) = \sum_{j=1}^p(x_{ij} - x_{i'j})^2 - = ||\boldsymbol{x_i} - \boldsymbol{x_{i'}}||^2 -\end{equation} + +$$ k\in\{1, \cdots, K \}. $$ +

    In the basic k-means algorithm each point is assigned to only +one cluster \( k \), and these assignments are non-injective i.e. many-to-one. We +can think of these mappings as an encoder \( k = C(i) \), which assigns the \( i \)-th +data-point \( \bf x_i \) to the \( k \)-th cluster. +

    +

    \( k \)-means algorithm in words:

    +
      +
    1. We start with guesses / random initializations of our \( k \) cluster centers/centroids
      2. +
    2. For each centroid the points that are most similar are identified
      3. +
    3. Then we move / replace each centroid with a coordinate average of all the points that were assigned to that centroid.
      4. +
    4. Iterate 2-3 until the centroids no longer move (to some tolerance)
      5. +

    diff --git a/doc/pub/week44/html/._week44-bs009.html b/doc/pub/week44/html/._week44-bs009.html index faea9cbf4..ba6230203 100644 --- a/doc/pub/week44/html/._week44-bs009.html +++ b/doc/pub/week44/html/._week44-bs009.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,29 +327,25 @@ MathJax.Hub.Config({

     

     

     

    -

    Within Cluster Point Scatter

    +

    Basic Math of the \( k \)-means Algorithm

    -

    We define the so called within-cluster point scatter which gives us a -measure of how close each data point assigned to the same cluster tends to be to -the all the others. -

    +

    We assume we have \( n \) data-points

    $$ -\begin{equation}\tag{3} - W(C) = \frac{1}{2}\sum_{k=1}^K\sum_{C(i)=k} - \sum_{C(i')=k}d(\boldsymbol{x_i}, \boldsymbol{x_{i'}}) = - \sum_{k=1}^KN_k\sum_{C(i)=k}||\boldsymbol{x_i} - \boldsymbol{\overline{x_k}}||^2 +\begin{equation}\tag{1} + \boldsymbol{x_i} = \{x_{i, 1}, \cdots, x_{i, p}\}\in\mathbb{R}^p. \end{equation} $$ -

    where \( \boldsymbol{\overline{x_k}} \) is the mean vector associated with the \( k \)-th -cluster, and \( N_k = \sum_{i=1}^nI(C(i) = k) \), where the \( I() \) notation is -similar to the Kronecker delta (Commonly used in statistics, it just means that -when \( i = k \) we have the encoder \( C(i) \)). In other words, the within-cluster -scatter measures the compactness of each cluster with respect to the data points -assigned to each cluster. This is the quantity that the \( k \)-means algorithm aims -to minimize. We refer to this quantity \( W(C) \) as the within cluster scatter -because of its relation to the total scatter. +

    which we wish to group into \( K < n \) clusters. For our dissimilarity measure we +use the squared Euclidean distance

    +$$ +\begin{equation}\tag{2} + d(\boldsymbol{x_i}, \boldsymbol{x_i'}) = \sum_{j=1}^p(x_{ij} - x_{i'j})^2 + = ||\boldsymbol{x_i} - \boldsymbol{x_{i'}}||^2 +\end{equation} +$$ +

    @@ -370,7 +371,7 @@ because of its relation to the total scatter.

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    • diff --git a/doc/pub/week44/html/._week44-bs010.html b/doc/pub/week44/html/._week44-bs010.html index 7dca89891..12e6ed39c 100644 --- a/doc/pub/week44/html/._week44-bs010.html +++ b/doc/pub/week44/html/._week44-bs010.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,24 +327,28 @@ MathJax.Hub.Config({

     

     

     

    -

    More Details

    +

    Within Cluster Point Scatter

    -

    We have

    +

    We define the so called within-cluster point scatter which gives us a +measure of how close each data point assigned to the same cluster tends to be to +the all the others. +

    $$ -\begin{equation}\tag{4} - T = W(C) + B(C) = \frac{1}{2}\sum_{i=1}^n - \sum_{i'=1}^nd(\boldsymbol{x_i}, \boldsymbol{x_{i'}}) - = \frac{1}{2}\sum_{k=1}^K\sum_{C(i)=k} - \Big(\sum_{C(i') = k}d(\boldsymbol{x_i}, \boldsymbol{x_{i'}}) - + \sum_{C(i')\neq k}d(\boldsymbol{x_i}, \boldsymbol{x_{i'}})\Big). +\begin{equation}\tag{3} + W(C) = \frac{1}{2}\sum_{k=1}^K\sum_{C(i)=k} + \sum_{C(i')=k}d(\boldsymbol{x_i}, \boldsymbol{x_{i'}}) = + \sum_{k=1}^KN_k\sum_{C(i)=k}||\boldsymbol{x_i} - \boldsymbol{\overline{x_k}}||^2 \end{equation} $$ -

    This is a quantity that is conserved throughout the \( k \)-means algorithm. It can -be thought of as the total amount of information in the data, and it is composed -of the aforementioned within-cluster scatter and the between-cluster scatter -\( B(C) \). In methods such as principle component analysis the total scatter is not -conserved. +

    where \( \boldsymbol{\overline{x_k}} \) is the mean vector associated with the \( k \)-th +cluster, and \( N_k = \sum_{i=1}^nI(C(i) = k) \), where the \( I() \) notation is +similar to the Kronecker delta (Commonly used in statistics, it just means that +when \( i = k \) we have the encoder \( C(i) \)). In other words, the within-cluster +scatter measures the compactness of each cluster with respect to the data points +assigned to each cluster. This is the quantity that the \( k \)-means algorithm aims +to minimize. We refer to this quantity \( W(C) \) as the within cluster scatter +because of its relation to the total scatter.

    @@ -367,7 +376,7 @@ conserved.

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    • diff --git a/doc/pub/week44/html/._week44-bs011.html b/doc/pub/week44/html/._week44-bs011.html index 417bab534..121ea6376 100644 --- a/doc/pub/week44/html/._week44-bs011.html +++ b/doc/pub/week44/html/._week44-bs011.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,15 +327,25 @@ MathJax.Hub.Config({

     

     

     

    -

    Total Cluster Variance

    -

    Given a cluster mean \( \boldsymbol{m_k} \) we define the total cluster variance

    +

    More Details

    + +

    We have

    $$ -\begin{equation}\tag{5} - \min_{C, \{\boldsymbol{m_k}\}_1^K}\sum_{k=1}^KN_k\sum||\boldsymbol{x_i} - \boldsymbol{m_k}||^2 +\begin{equation}\tag{4} + T = W(C) + B(C) = \frac{1}{2}\sum_{i=1}^n + \sum_{i'=1}^nd(\boldsymbol{x_i}, \boldsymbol{x_{i'}}) + = \frac{1}{2}\sum_{k=1}^K\sum_{C(i)=k} + \Big(\sum_{C(i') = k}d(\boldsymbol{x_i}, \boldsymbol{x_{i'}}) + + \sum_{C(i')\neq k}d(\boldsymbol{x_i}, \boldsymbol{x_{i'}})\Big). \end{equation} $$ -

    Now we have all the pieces necessary to formally revisit the \( k \)-means algorithm.

    +

    This is a quantity that is conserved throughout the \( k \)-means algorithm. It can +be thought of as the total amount of information in the data, and it is composed +of the aforementioned within-cluster scatter and the between-cluster scatter +\( B(C) \). In methods such as principle component analysis the total scatter is not +conserved. +

    @@ -357,7 +372,7 @@ $$

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    • diff --git a/doc/pub/week44/html/._week44-bs012.html b/doc/pub/week44/html/._week44-bs012.html index 3ab23bf81..6af150569 100644 --- a/doc/pub/week44/html/._week44-bs012.html +++ b/doc/pub/week44/html/._week44-bs012.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,15 +327,16 @@ MathJax.Hub.Config({

     

     

     

    -

    The \( k \)-means Clustering Algorithm

    +

    Total Cluster Variance

    +

    Given a cluster mean \( \boldsymbol{m_k} \) we define the total cluster variance

    +$$ +\begin{equation}\tag{5} + \min_{C, \{\boldsymbol{m_k}\}_1^K}\sum_{k=1}^KN_k\sum||\boldsymbol{x_i} - \boldsymbol{m_k}||^2 +\end{equation} +$$ -

    The \( k \)-means clustering algorithm goes as follows

    +

    Now we have all the pieces necessary to formally revisit the \( k \)-means algorithm.

    -
      -
    1. For a given cluster assignment \( C \), and \( k \) cluster means \( \left\{m_1, \cdots, m_k\right\} \). We minimize the total cluster variance with respect to the cluster means \( \{m_k\} \) yielding the means of the currently assigned clusters.
      2. -
    2. Given a current set of \( k \) means \( \{m_k\} \) the total cluster variance is minimized by assigning each observation to the closest (current) cluster mean. That is $$C(i) = \underset{1\leq k\leq K}{\mathrm{argmin}} ||\boldsymbol{x_i} - \boldsymbol{m_k}||^2$$
      3. -
    3. Steps 1 and 2 are repeated until the assignments do not change.
      4. -

    diff --git a/doc/pub/week44/html/._week44-bs013.html b/doc/pub/week44/html/._week44-bs013.html index 5b1fc6b81..97afe5136 100644 --- a/doc/pub/week44/html/._week44-bs013.html +++ b/doc/pub/week44/html/._week44-bs013.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,14 +327,14 @@ MathJax.Hub.Config({

     

     

     

    -

    Summarizing

    +

    The \( k \)-means Clustering Algorithm

    + +

    The \( k \)-means clustering algorithm goes as follows

      -
    1. Before we start we specify a number \( k \) which is the number of clusters we want to try to separate our data into.
      2. -
    2. We initially choose \( k \) random data points in our data as our initial centroids, or means (this is where the name comes from).
      3. -
    3. Assign each data point to their closest centroid, based on the squared Euclidean distance.
      4. -
    4. For each of the \( k \) cluster we update the centroid by calculating new mean values for all the data points in the cluster.
      5. -
    5. Iteratively minimize the within cluster scatter by performing steps (3, 4) until the new assignments stop changing (can be to some tolerance) or until a maximum number of iterations have passed.
      6. +
    6. For a given cluster assignment \( C \), and \( k \) cluster means \( \left\{m_1, \cdots, m_k\right\} \). We minimize the total cluster variance with respect to the cluster means \( \{m_k\} \) yielding the means of the currently assigned clusters.
      7. +
    7. Given a current set of \( k \) means \( \{m_k\} \) the total cluster variance is minimized by assigning each observation to the closest (current) cluster mean. That is $$C(i) = \underset{1\leq k\leq K}{\mathrm{argmin}} ||\boldsymbol{x_i} - \boldsymbol{m_k}||^2$$
      8. +
    8. Steps 1 and 2 are repeated until the assignments do not change.

    @@ -356,7 +361,7 @@ MathJax.Hub.Config({

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    • diff --git a/doc/pub/week44/html/._week44-bs014.html b/doc/pub/week44/html/._week44-bs014.html index 08ec05d41..d3c439731 100644 --- a/doc/pub/week44/html/._week44-bs014.html +++ b/doc/pub/week44/html/._week44-bs014.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,125 +327,15 @@ MathJax.Hub.Config({

     

     

     

    -

    Writing our own Code, the Data Set

    - -

    Let us now program the most basic version of the algorithm using nothing but -Python with numpy arrays. This code is kept intentionally simple to gradually -progress our understanding. There is no vectorization of any kind, and even most -helper functions are not utilized. -

    - -

    We need first a dataset to do our cluster analysis on. In our case -this is a plain vanilla data set using random numbers using a -Gaussian distribution. -

    - - - -
    -
    -
    -
    -
    - 
    import time
-import numpy as np
-import tensorflow as tf
-from matplotlib import image
-import matplotlib.pyplot as plt
-from sklearn.cluster import KMeans
-from IPython.display import display
-
-np.random.seed(2021)
-
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    - -

    Next we define functions, for ease of use later, to generate Gaussians and to -set up our toy data set. -

    - - -
    -
    -
    -
    -
    - 
    def gaussian_points(dim=2, n_points=1000, mean_vector=np.array([0, 0]),
-                    sample_variance=1):
-    """
-    Very simple custom function to generate gaussian distributed point clusters
-    with variable dimension, number of points, means in each direction
-    (must match dim) and sample variance.
-
-    Inputs:
-        dim (int)
-        n_points (int)
-        mean_vector (np.array) (where index 0 is x, index 1 is y etc.)
-        sample_variance (float)
-
-    Returns:
-        data (np.array): with dimensions (dim x n_points)
-    """
-
-    mean_matrix = np.zeros(dim) + mean_vector
-    covariance_matrix = np.eye(dim) * sample_variance
-    data = np.random.multivariate_normal(mean_matrix, covariance_matrix,
-                                    n_points)
-    return data
-
-
-
-def generate_simple_clustering_dataset(dim=2, n_points=1000, plotting=True,
-                                    return_data=True):
-    """
-    Toy model to illustrate k-means clustering
-    """
-
-    data1 = gaussian_points(mean_vector=np.array([5, 5]))
-    data2 = gaussian_points()
-    data3 = gaussian_points(mean_vector=np.array([1, 4.5]))
-    data4 = gaussian_points(mean_vector=np.array([5, 1]))
-    data = np.concatenate((data1, data2, data3, data4), axis=0)
-
-    if plotting:
-        fig, ax = plt.subplots()
-        ax.scatter(data[:, 0], data[:, 1], alpha=0.2)
-        ax.set_title('Toy Model Dataset')
-        plt.show()
-
-
-    if return_data:
-        return data
-
-
-data = generate_simple_clustering_dataset()
-
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    - +

    Summarizing

    +
      +
    1. Before we start we specify a number \( k \) which is the number of clusters we want to try to separate our data into.
      2. +
    2. We initially choose \( k \) random data points in our data as our initial centroids, or means (this is where the name comes from).
      3. +
    3. Assign each data point to their closest centroid, based on the squared Euclidean distance.
      4. +
    4. For each of the \( k \) cluster we update the centroid by calculating new mean values for all the data points in the cluster.
      5. +
    5. Iteratively minimize the within cluster scatter by performing steps (3, 4) until the new assignments stop changing (can be to some tolerance) or until a maximum number of iterations have passed.
      6. +

    diff --git a/doc/pub/week44/html/._week44-bs015.html b/doc/pub/week44/html/._week44-bs015.html index 8237638d3..8bfdcf424 100644 --- a/doc/pub/week44/html/._week44-bs015.html +++ b/doc/pub/week44/html/._week44-bs015.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,10 +327,17 @@ MathJax.Hub.Config({

     

     

     

    -

    Implementing the \( k \)-means Algorithm

    +

    Writing our own Code, the Data Set

    -

    With the above dataset we start -implementing the \( k \)-means algorithm. +

    Let us now program the most basic version of the algorithm using nothing but +Python with numpy arrays. This code is kept intentionally simple to gradually +progress our understanding. There is no vectorization of any kind, and even most +helper functions are not utilized. +

    + +

    We need first a dataset to do our cluster analysis on. In our case +this is a plain vanilla data set using random numbers using a +Gaussian distribution.

    @@ -335,39 +347,89 @@ implementing the \( k \)-means algorithm.
    - 
    n_samples, dimensions = data.shape
-n_clusters = 4
+  
    import time
+import numpy as np
+import tensorflow as tf
+from matplotlib import image
+import matplotlib.pyplot as plt
+from sklearn.cluster import KMeans
+from IPython.display import display
 
-# we randomly initialize our centroids
 np.random.seed(2021)
-centroids = data[np.random.choice(n_samples, n_clusters, replace=False), :]
-distances = np.zeros((n_samples, n_clusters))
+
    
+
    +
    +
    + +
    +
    +
    +
    +
    +
    +
    +
    + -# first we need to calculate the distance to each centroid from our data -for k in range(n_clusters): - for n in range(n_samples): - dist = 0 - for d in range(dimensions): - dist += np.abs(data[n, d] - centroids[k, d])**2 - distances[n, k] = dist +

    Next we define functions, for ease of use later, to generate Gaussians and to +set up our toy data set. +

    -# we initialize an array to keep track of to which cluster each point belongs -# the way we set it up here the index tracks which point and the value which -# cluster the point belongs to -cluster_labels = np.zeros(n_samples, dtype='int') + +
    +
    +
    +
    +
    + 
    def gaussian_points(dim=2, n_points=1000, mean_vector=np.array([0, 0]),
+                    sample_variance=1):
+    """
+    Very simple custom function to generate gaussian distributed point clusters
+    with variable dimension, number of points, means in each direction
+    (must match dim) and sample variance.
 
-# next we loop through our samples and for every point assign it to the cluster
-# to which it has the smallest distance to
-for n in range(n_samples):
-    # tracking variables (all of this is basically just an argmin)
-    smallest = 1e10
-    smallest_row_index = 1e10
-    for k in range(n_clusters):
-        if distances[n, k] < smallest:
-            smallest = distances[n, k]
-            smallest_row_index = k
+    Inputs:
+        dim (int)
+        n_points (int)
+        mean_vector (np.array) (where index 0 is x, index 1 is y etc.)
+        sample_variance (float)
 
-    cluster_labels[n] = smallest_row_index
+    Returns:
+        data (np.array): with dimensions (dim x n_points)
+    """
+
+    mean_matrix = np.zeros(dim) + mean_vector
+    covariance_matrix = np.eye(dim) * sample_variance
+    data = np.random.multivariate_normal(mean_matrix, covariance_matrix,
+                                    n_points)
+    return data
+
+
+
+def generate_simple_clustering_dataset(dim=2, n_points=1000, plotting=True,
+                                    return_data=True):
+    """
+    Toy model to illustrate k-means clustering
+    """
+
+    data1 = gaussian_points(mean_vector=np.array([5, 5]))
+    data2 = gaussian_points()
+    data3 = gaussian_points(mean_vector=np.array([1, 4.5]))
+    data4 = gaussian_points(mean_vector=np.array([5, 1]))
+    data = np.concatenate((data1, data2, data3, data4), axis=0)
+
+    if plotting:
+        fig, ax = plt.subplots()
+        ax.scatter(data[:, 0], data[:, 1], alpha=0.2)
+        ax.set_title('Toy Model Dataset')
+        plt.show()
+
+
+    if return_data:
+        return data
+
+
+data = generate_simple_clustering_dataset()
    @@ -409,7 +471,7 @@ cluster_labels = np24
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    • diff --git a/doc/pub/week44/html/._week44-bs016.html b/doc/pub/week44/html/._week44-bs016.html index 6db2b4390..a0ae49980 100644 --- a/doc/pub/week44/html/._week44-bs016.html +++ b/doc/pub/week44/html/._week44-bs016.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,7 +327,12 @@ MathJax.Hub.Config({

     

     

     

    -

    Plotting

    +

    Implementing the \( k \)-means Algorithm

    + +

    With the above dataset we start +implementing the \( k \)-means algorithm. +

    +
    @@ -330,19 +340,39 @@ MathJax.Hub.Config({
    - 
    fig = plt.figure()
-ax = fig.add_subplot()
-unique_cluster_labels = np.unique(cluster_labels)
-for i in unique_cluster_labels:
-    ax.scatter(data[cluster_labels == i, 0],
-               data[cluster_labels == i, 1],
-               label = i,
-               alpha = 0.2)
-    ax.scatter(centroids[:, 0], centroids[:, 1], c='black')
+  
    n_samples, dimensions = data.shape
+n_clusters = 4
 
-ax.set_title("First Grouping of Points to Centroids")
+# we randomly initialize our centroids
+np.random.seed(2021)
+centroids = data[np.random.choice(n_samples, n_clusters, replace=False), :]
+distances = np.zeros((n_samples, n_clusters))
 
-plt.show()
+# first we need to calculate the distance to each centroid from our data
+for k in range(n_clusters):
+    for n in range(n_samples):
+        dist = 0
+        for d in range(dimensions):
+            dist += np.abs(data[n, d] - centroids[k, d])**2
+            distances[n, k] = dist
+
+# we initialize an array to keep track of to which cluster each point belongs
+# the way we set it up here the index tracks which point and the value which
+# cluster the point belongs to
+cluster_labels = np.zeros(n_samples, dtype='int')
+
+# next we loop through our samples and for every point assign it to the cluster
+# to which it has the smallest distance to
+for n in range(n_samples):
+    # tracking variables (all of this is basically just an argmin)
+    smallest = 1e10
+    smallest_row_index = 1e10
+    for k in range(n_clusters):
+        if distances[n, k] < smallest:
+            smallest = distances[n, k]
+            smallest_row_index = k
+
+    cluster_labels[n] = smallest_row_index
    @@ -358,17 +388,6 @@ plt.show()
    -

    So what do we have so far? We have 'picked' \( k \) centroids at random from our -data points. There are other ways of more intelligently choosing their -initializations, however for our purposes randomly is fine. Then we have -initialized an array 'distances' which holds the information of the distance, -or dissimilarity, of every point to of our centroids. Finally, we have -initialized an array 'cluster_labels' which according to our distances array -holds the information of to which centroid every point is assigned. This was the -first pass of our algorithm. Essentially, all we need to do now is repeat the -distance and assignment steps above until we have reached a desired convergence -or a maximum amount of iterations. -

    @@ -395,7 +414,7 @@ or a maximum amount of iterations.

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    • diff --git a/doc/pub/week44/html/._week44-bs017.html b/doc/pub/week44/html/._week44-bs017.html index ec68a9bdf..606fe7882 100644 --- a/doc/pub/week44/html/._week44-bs017.html +++ b/doc/pub/week44/html/._week44-bs017.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,8 +327,7 @@ MathJax.Hub.Config({

     

     

     

    -

    Continuing

    - +

    Plotting

    @@ -331,50 +335,19 @@ MathJax.Hub.Config({
    - 
    max_iterations = 100
-tolerance = 1e-8
+  
    fig = plt.figure()
+ax = fig.add_subplot()
+unique_cluster_labels = np.unique(cluster_labels)
+for i in unique_cluster_labels:
+    ax.scatter(data[cluster_labels == i, 0],
+               data[cluster_labels == i, 1],
+               label = i,
+               alpha = 0.2)
+    ax.scatter(centroids[:, 0], centroids[:, 1], c='black')
 
-for iteration in range(max_iterations):
-    prev_centroids = centroids.copy()
-    for k in range(n_clusters):
-        # this array will be used to update our centroid positions
-        vector_mean = np.zeros(dimensions)
-        mean_divisor = 0
-        for n in range(n_samples):
-            if cluster_labels[n] == k:
-                vector_mean += data[n, :]
-                mean_divisor += 1
+ax.set_title("First Grouping of Points to Centroids")
 
-        # update according to the k means
-        centroids[k, :] = vector_mean / mean_divisor
-
-    # we find the dissimilarity
-    for k in range(n_clusters):
-        for n in range(n_samples):
-            dist = 0
-            for d in range(dimensions):
-                dist += np.abs(data[n, d] - centroids[k, d])**2
-                distances[n, k] = dist
-
-    # assign each point
-    for n in range(n_samples):
-        smallest = 1e10
-        smallest_row_index = 1e10
-        for k in range(n_clusters):
-            if distances[n, k] < smallest:
-                smallest = distances[n, k]
-                smallest_row_index = k
-
-        cluster_labels[n] = smallest_row_index
-
-    # convergence criteria
-    centroid_difference = np.sum(np.abs(centroids - prev_centroids))
-    if centroid_difference < tolerance:
-        print(f'Converged at iteration {iteration}')
-        break
-
-    elif iteration == max_iterations:
-        print(f'Did not converge in {max_iterations} iterations')
+plt.show()
    @@ -390,6 +363,17 @@ tolerance = 1e-
    +

    So what do we have so far? We have 'picked' \( k \) centroids at random from our +data points. There are other ways of more intelligently choosing their +initializations, however for our purposes randomly is fine. Then we have +initialized an array 'distances' which holds the information of the distance, +or dissimilarity, of every point to of our centroids. Finally, we have +initialized an array 'cluster_labels' which according to our distances array +holds the information of to which centroid every point is assigned. This was the +first pass of our algorithm. Essentially, all we need to do now is repeat the +distance and assignment steps above until we have reached a desired convergence +or a maximum amount of iterations. +

    @@ -416,7 +400,7 @@ tolerance = 1e-

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    • diff --git a/doc/pub/week44/html/._week44-bs018.html b/doc/pub/week44/html/._week44-bs018.html index e3d659802..3ed8ed760 100644 --- a/doc/pub/week44/html/._week44-bs018.html +++ b/doc/pub/week44/html/._week44-bs018.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,10 +327,7 @@ MathJax.Hub.Config({

     

     

     

    -

    Wrapping it up

    -

    We now have a simple , un-optimized \( k \)-means -clustering implementation. Lets plot the final result -

    +

    Continuing

    @@ -334,47 +336,24 @@ clustering implementation. Lets plot the final result
    - 
    fig = plt.figure()
-ax = fig.add_subplot()
-unique_cluster_labels = np.unique(cluster_labels)
-for i in unique_cluster_labels:
-    ax.scatter(data[cluster_labels == i, 0],
-               data[cluster_labels == i, 1],
-               label = i,
-               alpha = 0.2)
-    ax.scatter(centroids[:, 0], centroids[:, 1], c='black')
+  
    max_iterations = 100
+tolerance = 1e-8
 
-ax.set_title("Final Result of K-means Clustering")
+for iteration in range(max_iterations):
+    prev_centroids = centroids.copy()
+    for k in range(n_clusters):
+        # this array will be used to update our centroid positions
+        vector_mean = np.zeros(dimensions)
+        mean_divisor = 0
+        for n in range(n_samples):
+            if cluster_labels[n] == k:
+                vector_mean += data[n, :]
+                mean_divisor += 1
 
-plt.show()
-
    
-
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    - -
    -
    -
    -
    -
    - 
    def naive_kmeans(data, n_clusters=4, max_iterations=100, tolerance=1e-8):
-    start_time = time.time()
-
-    n_samples, dimensions = data.shape
-    n_clusters = 4
-    #np.random.seed(2021)
-    centroids = data[np.random.choice(n_samples, n_clusters, replace=False), :]
-    distances = np.zeros((n_samples, n_clusters))
+        # update according to the k means
+        centroids[k, :] = vector_mean / mean_divisor
 
+    # we find the dissimilarity
     for k in range(n_clusters):
         for n in range(n_samples):
             dist = 0
@@ -382,8 +361,7 @@ plt.show()
                 dist += np.abs(data[n, d] - centroids[k, d])**2
                 distances[n, k] = dist
 
-    cluster_labels = np.zeros(n_samples, dtype='int')
-
+    # assign each point
     for n in range(n_samples):
         smallest = 1e10
         smallest_row_index = 1e10
@@ -394,46 +372,14 @@ plt.show()
 
         cluster_labels[n] = smallest_row_index
 
-    for iteration in range(max_iterations):
-        prev_centroids = centroids.copy()
-        for k in range(n_clusters):
-            vector_mean = np.zeros(dimensions)
-            mean_divisor = 0
-            for n in range(n_samples):
-                if cluster_labels[n] == k:
-                    vector_mean += data[n, :]
-                    mean_divisor += 1
+    # convergence criteria
+    centroid_difference = np.sum(np.abs(centroids - prev_centroids))
+    if centroid_difference < tolerance:
+        print(f'Converged at iteration {iteration}')
+        break
 
-            centroids[k, :] = vector_mean / mean_divisor
-
-        for k in range(n_clusters):
-            for n in range(n_samples):
-                dist = 0
-                for d in range(dimensions):
-                    dist += np.abs(data[n, d] - centroids[k, d])**2
-                    distances[n, k] = dist
-
-        for n in range(n_samples):
-            smallest = 1e10
-            smallest_row_index = 1e10
-            for k in range(n_clusters):
-                if distances[n, k] < smallest:
-                    smallest = distances[n, k]
-                    smallest_row_index = k
-
-            cluster_labels[n] = smallest_row_index
-
-        centroid_difference = np.sum(np.abs(centroids - prev_centroids))
-        if centroid_difference < tolerance:
-            print(f'Converged at iteration {iteration}')
-            print(f'Runtime: {time.time() - start_time} seconds')
-
-            return cluster_labels, centroids
-
-    print(f'Did not converge in {max_iterations} iterations')
-    print(f'Runtime: {time.time() - start_time} seconds')
-
-    return cluster_labels, centroids
+    elif iteration == max_iterations:
+        print(f'Did not converge in {max_iterations} iterations')
    @@ -475,7 +421,7 @@ plt.show()
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    • diff --git a/doc/pub/week44/html/._week44-bs019.html b/doc/pub/week44/html/._week44-bs019.html index 8ac5fb717..ea7120fd6 100644 --- a/doc/pub/week44/html/._week44-bs019.html +++ b/doc/pub/week44/html/._week44-bs019.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,32 +327,133 @@ MathJax.Hub.Config({

     

     

     

    -

    Decision trees, overarching aims

    - -

    We start here with the most basic algorithm, the so-called decision -tree. With this basic algorithm we can in turn build more complex -networks, spanning from homogeneous and heterogenous forests (bagging, -random forests and more) to one of the most popular supervised -algorithms nowadays, the extreme gradient boosting, or just -XGBoost. But let us start with the simplest possible ingredient. +

    Wrapping it up

    +

    We now have a simple , un-optimized \( k \)-means +clustering implementation. Lets plot the final result

    -

    Decision trees are supervised learning algorithms used for both, -classification and regression tasks. -

    -

    The main idea of decision trees -is to find those descriptive features which contain the most -information regarding the target feature and then split the dataset -along the values of these features such that the target feature values -for the resulting underlying datasets are as pure as possible. -

    + +
    +
    +
    +
    +
    + 
    fig = plt.figure()
+ax = fig.add_subplot()
+unique_cluster_labels = np.unique(cluster_labels)
+for i in unique_cluster_labels:
+    ax.scatter(data[cluster_labels == i, 0],
+               data[cluster_labels == i, 1],
+               label = i,
+               alpha = 0.2)
+    ax.scatter(centroids[:, 0], centroids[:, 1], c='black')
+
+ax.set_title("Final Result of K-means Clustering")
+
+plt.show()
+
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    + +
    +
    +
    +
    +
    + 
    def naive_kmeans(data, n_clusters=4, max_iterations=100, tolerance=1e-8):
+    start_time = time.time()
+
+    n_samples, dimensions = data.shape
+    n_clusters = 4
+    #np.random.seed(2021)
+    centroids = data[np.random.choice(n_samples, n_clusters, replace=False), :]
+    distances = np.zeros((n_samples, n_clusters))
+
+    for k in range(n_clusters):
+        for n in range(n_samples):
+            dist = 0
+            for d in range(dimensions):
+                dist += np.abs(data[n, d] - centroids[k, d])**2
+                distances[n, k] = dist
+
+    cluster_labels = np.zeros(n_samples, dtype='int')
+
+    for n in range(n_samples):
+        smallest = 1e10
+        smallest_row_index = 1e10
+        for k in range(n_clusters):
+            if distances[n, k] < smallest:
+                smallest = distances[n, k]
+                smallest_row_index = k
+
+        cluster_labels[n] = smallest_row_index
+
+    for iteration in range(max_iterations):
+        prev_centroids = centroids.copy()
+        for k in range(n_clusters):
+            vector_mean = np.zeros(dimensions)
+            mean_divisor = 0
+            for n in range(n_samples):
+                if cluster_labels[n] == k:
+                    vector_mean += data[n, :]
+                    mean_divisor += 1
+
+            centroids[k, :] = vector_mean / mean_divisor
+
+        for k in range(n_clusters):
+            for n in range(n_samples):
+                dist = 0
+                for d in range(dimensions):
+                    dist += np.abs(data[n, d] - centroids[k, d])**2
+                    distances[n, k] = dist
+
+        for n in range(n_samples):
+            smallest = 1e10
+            smallest_row_index = 1e10
+            for k in range(n_clusters):
+                if distances[n, k] < smallest:
+                    smallest = distances[n, k]
+                    smallest_row_index = k
+
+            cluster_labels[n] = smallest_row_index
+
+        centroid_difference = np.sum(np.abs(centroids - prev_centroids))
+        if centroid_difference < tolerance:
+            print(f'Converged at iteration {iteration}')
+            print(f'Runtime: {time.time() - start_time} seconds')
+
+            return cluster_labels, centroids
+
+    print(f'Did not converge in {max_iterations} iterations')
+    print(f'Runtime: {time.time() - start_time} seconds')
+
+    return cluster_labels, centroids
+
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    -

    The descriptive features which reproduce best the target/output features are normally said -to be the most informative ones. The process of finding the most -informative feature is done until we accomplish a stopping criteria -where we then finally end up in so called leaf nodes. -

    @@ -374,7 +480,7 @@ where we then finally end up in so called leaf nodes.

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    • diff --git a/doc/pub/week44/html/._week44-bs020.html b/doc/pub/week44/html/._week44-bs020.html index 1fce4cb69..fa375ab64 100644 --- a/doc/pub/week44/html/._week44-bs020.html +++ b/doc/pub/week44/html/._week44-bs020.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,18 +327,31 @@ MathJax.Hub.Config({

     

     

     

    -

    Basics of a tree

    +

    Decision trees, overarching aims

    -

    A decision tree is typically divided into a root node, the interior nodes, -and the final leaf nodes or just leaves. These entities are then connected by so-called branches. +

    We start here with the most basic algorithm, the so-called decision +tree. With this basic algorithm we can in turn build more complex +networks, spanning from homogeneous and heterogenous forests (bagging, +random forests and more) to one of the most popular supervised +algorithms nowadays, the extreme gradient boosting, or just +XGBoost. But let us start with the simplest possible ingredient.

    -

    The leaf nodes -contain the predictions we will make for new query instances presented -to our trained model. This is possible since the model has -learned the underlying structure of the training data and hence can, -given some assumptions, make predictions about the target feature value -(class) of unseen query instances. +

    Decision trees are supervised learning algorithms used for both, +classification and regression tasks. +

    + +

    The main idea of decision trees +is to find those descriptive features which contain the most +information regarding the target feature and then split the dataset +along the values of these features such that the target feature values +for the resulting underlying datasets are as pure as possible. +

    + +

    The descriptive features which reproduce best the target/output features are normally said +to be the most informative ones. The process of finding the most +informative feature is done until we accomplish a stopping criteria +where we then finally end up in so called leaf nodes.

    @@ -361,7 +379,7 @@ given some assumptions, make predictions about the target feature value

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    • diff --git a/doc/pub/week44/html/._week44-bs021.html b/doc/pub/week44/html/._week44-bs021.html index 22009764b..4b7ebf18e 100644 --- a/doc/pub/week44/html/._week44-bs021.html +++ b/doc/pub/week44/html/._week44-bs021.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,9 +327,19 @@ MathJax.Hub.Config({

     

     

     

    -

    A Sketch of a Tree, Regression problem

    +

    Basics of a tree

    - +

    A decision tree is typically divided into a root node, the interior nodes, +and the final leaf nodes or just leaves. These entities are then connected by so-called branches. +

    + +

    The leaf nodes +contain the predictions we will make for new query instances presented +to our trained model. This is possible since the model has +learned the underlying structure of the training data and hence can, +given some assumptions, make predictions about the target feature value +(class) of unseen query instances. +

    @@ -351,7 +366,7 @@ MathJax.Hub.Config({

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    • diff --git a/doc/pub/week44/html/._week44-bs022.html b/doc/pub/week44/html/._week44-bs022.html index 3d17dd1c5..c3086fa3a 100644 --- a/doc/pub/week44/html/._week44-bs022.html +++ b/doc/pub/week44/html/._week44-bs022.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,9 +327,9 @@ MathJax.Hub.Config({

     

     

     

    -

    A Sketch of a Tree, Classification problem

    +

    A Sketch of a Tree, Regression problem

    - +

    @@ -351,7 +356,7 @@ MathJax.Hub.Config({

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    • diff --git a/doc/pub/week44/html/._week44-bs023.html b/doc/pub/week44/html/._week44-bs023.html index f8a6a9870..73b13f341 100644 --- a/doc/pub/week44/html/._week44-bs023.html +++ b/doc/pub/week44/html/._week44-bs023.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,15 +327,9 @@ MathJax.Hub.Config({

     

     

     

    -

    A typical Decision Tree with its pertinent Jargon, Classification Problem

    +

    A Sketch of a Tree, Classification problem

    -

    -
    -

    -
    -

    - -

    This tree was produced using the Wisconsin cancer data (discussed here as well, see code examples below) using Scikit-Learn's decision tree classifier. Here we have used the so-called gini index (see below) to split the various branches.

    +

    @@ -357,7 +356,7 @@ MathJax.Hub.Config({

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    • diff --git a/doc/pub/week44/html/._week44-bs024.html b/doc/pub/week44/html/._week44-bs024.html index d09adc34e..0254c0192 100644 --- a/doc/pub/week44/html/._week44-bs024.html +++ b/doc/pub/week44/html/._week44-bs024.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,19 +327,15 @@ MathJax.Hub.Config({

     

     

     

    -

    General Features

    +

    A typical Decision Tree with its pertinent Jargon, Classification Problem

    -

    The overarching approach to decision trees is a top-down approach.

    +

    +
    +

    +
    +

    -
      -
    • A leaf provides the classification of a given instance.
      • -
    • A node specifies a test of some attribute of the instance.
      • -
    • A branch corresponds to a possible values of an attribute.
      • -
    • An instance is classified by starting at the root node of the tree, testing the attribute specified by this node, then moving down the tree branch corresponding to the value of the attribute in the given example.
      • -
    -

    This process is then repeated for the subtree rooted at the new -node. -

    +

    This tree was produced using the Wisconsin cancer data (discussed here as well, see code examples below) using Scikit-Learn's decision tree classifier. Here we have used the so-called gini index (see below) to split the various branches.

    @@ -361,7 +362,7 @@ node.

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    • diff --git a/doc/pub/week44/html/._week44-bs025.html b/doc/pub/week44/html/._week44-bs025.html index a7cfe2d8a..9ff47e0e5 100644 --- a/doc/pub/week44/html/._week44-bs025.html +++ b/doc/pub/week44/html/._week44-bs025.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,20 +327,20 @@ MathJax.Hub.Config({

     

     

     

    -

    How do we set it up?

    +

    General Features

    -

    In simplified terms, the process of training a decision tree and -predicting the target features of query instances is as follows: +

    The overarching approach to decision trees is a top-down approach.

    + +
      +
    • A leaf provides the classification of a given instance.
      • +
    • A node specifies a test of some attribute of the instance.
      • +
    • A branch corresponds to a possible values of an attribute.
      • +
    • An instance is classified by starting at the root node of the tree, testing the attribute specified by this node, then moving down the tree branch corresponding to the value of the attribute in the given example.
      • +
    +

    This process is then repeated for the subtree rooted at the new +node.

    -
      -
    1. Present a dataset containing of a number of training instances characterized by a number of descriptive features and a target feature
      2. -
    2. Train the decision tree model by continuously splitting the target feature along the values of the descriptive features using a measure of information gain during the training process
      3. -
    3. Grow the tree until we accomplish a stopping criteria create leaf nodes which represent the predictions we want to make for new query instances
      4. -
    4. Show query instances to the tree and run down the tree until we arrive at leaf nodes
      5. -
    -

    Then we are essentially done!

    -

      @@ -361,7 +366,7 @@ predicting the target features of query instances is as follows:
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    diff --git a/doc/pub/week44/html/._week44-bs026.html b/doc/pub/week44/html/._week44-bs026.html index b31f290cc..5f0cc0aeb 100644 --- a/doc/pub/week44/html/._week44-bs026.html +++ b/doc/pub/week44/html/._week44-bs026.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,117 +327,19 @@ MathJax.Hub.Config({

     

     

     

    -

    Decision trees and Regression

    +

    How do we set it up?

    - -
    -
    -
    -
    -
    - 
    import numpy as np
-import matplotlib.pyplot as plt
-from sklearn.preprocessing import PolynomialFeatures
-from sklearn.linear_model import LinearRegression
-
-steps=250
-
-distance=0
-x=0
-distance_list=[]
-steps_list=[]
-while x<steps:
-    distance+=np.random.randint(-1,2)
-    distance_list.append(distance)
-    x+=1
-    steps_list.append(x)
-plt.plot(steps_list,distance_list, color='green', label="Random Walk Data")
-
-steps_list=np.asarray(steps_list)
-distance_list=np.asarray(distance_list)
-
-X=steps_list[:,np.newaxis]
-
-#Polynomial fits
-
-#Degree 2
-poly_features=PolynomialFeatures(degree=2, include_bias=False)
-X_poly=poly_features.fit_transform(X)
-
-lin_reg=LinearRegression()
-poly_fit=lin_reg.fit(X_poly,distance_list)
-b=lin_reg.coef_
-c=lin_reg.intercept_
-print ("2nd degree coefficients:")
-print ("zero power: ",c)
-print ("first power: ", b[0])
-print ("second power: ",b[1])
-
-z = np.arange(0, steps, .01)
-z_mod=b[1]*z**2+b[0]*z+c
-
-fit_mod=b[1]*X**2+b[0]*X+c
-plt.plot(z, z_mod, color='r', label="2nd Degree Fit")
-plt.title("Polynomial Regression")
-
-plt.xlabel("Steps")
-plt.ylabel("Distance")
-
-#Degree 10
-poly_features10=PolynomialFeatures(degree=10, include_bias=False)
-X_poly10=poly_features10.fit_transform(X)
-
-poly_fit10=lin_reg.fit(X_poly10,distance_list)
-
-y_plot=poly_fit10.predict(X_poly10)
-plt.plot(X, y_plot, color='black', label="10th Degree Fit")
-
-plt.legend()
-plt.show()
-
-
-#Decision Tree Regression
-from sklearn.tree import DecisionTreeRegressor
-regr_1=DecisionTreeRegressor(max_depth=2)
-regr_2=DecisionTreeRegressor(max_depth=5)
-regr_3=DecisionTreeRegressor(max_depth=7)
-regr_1.fit(X, distance_list)
-regr_2.fit(X, distance_list)
-regr_3.fit(X, distance_list)
-
-X_test = np.arange(0.0, steps, 0.01)[:, np.newaxis]
-y_1 = regr_1.predict(X_test)
-y_2 = regr_2.predict(X_test)
-y_3=regr_3.predict(X_test)
-
-# Plot the results
-plt.figure()
-plt.scatter(X, distance_list, s=2.5, c="black", label="data")
-plt.plot(X_test, y_1, color="red",
-         label="max_depth=2", linewidth=2)
-plt.plot(X_test, y_2, color="green", label="max_depth=5", linewidth=2)
-plt.plot(X_test, y_3, color="m", label="max_depth=7", linewidth=2)
-
-plt.xlabel("Data")
-plt.ylabel("Darget")
-plt.title("Decision Tree Regression")
-plt.legend()
-plt.show()
-
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    +

    In simplified terms, the process of training a decision tree and +predicting the target features of query instances is as follows: +

    +
      +
    1. Present a dataset containing of a number of training instances characterized by a number of descriptive features and a target feature
      2. +
    2. Train the decision tree model by continuously splitting the target feature along the values of the descriptive features using a measure of information gain during the training process
      3. +
    3. Grow the tree until we accomplish a stopping criteria create leaf nodes which represent the predictions we want to make for new query instances
      4. +
    4. Show query instances to the tree and run down the tree until we arrive at leaf nodes
      5. +
    +

    Then we are essentially done!

    @@ -459,7 +366,7 @@ plt.show()

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    • diff --git a/doc/pub/week44/html/._week44-bs027.html b/doc/pub/week44/html/._week44-bs027.html index e2f65cf08..ab2c3a093 100644 --- a/doc/pub/week44/html/._week44-bs027.html +++ b/doc/pub/week44/html/._week44-bs027.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,28 +327,117 @@ MathJax.Hub.Config({

     

     

     

    -

    Building a tree, regression

    +

    Decision trees and Regression

    -

    There are mainly two steps

    -
      -
    1. We split the predictor space (the set of possible values \( x_1,x_2,\dots, x_p \)) into \( J \) distinct and non-non-overlapping regions, \( R_1,R_2,\dots,R_J \).
      2. -
    2. For every observation that falls into the region \( R_j \) , we make the same prediction, which is simply the mean of the response values for the training observations in \( R_j \).
      3. -
    -

    How do we construct the regions \( R_1,\dots,R_J \)? In theory, the -regions could have any shape. However, we choose to divide the -predictor space into high-dimensional rectangles, or boxes, for -simplicity and for ease of interpretation of the resulting predictive -model. The goal is to find boxes \( R_1,\dots,R_J \) that minimize the -MSE, given by -

    + +
    +
    +
    +
    +
    + 
    import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.preprocessing import PolynomialFeatures
+from sklearn.linear_model import LinearRegression
 
-$$
-\sum_{j=1}^J\sum_{i\in R_j}(y_i-\overline{y}_{R_j})^2,
-$$
+steps=250
+
+distance=0
+x=0
+distance_list=[]
+steps_list=[]
+while x<steps:
+    distance+=np.random.randint(-1,2)
+    distance_list.append(distance)
+    x+=1
+    steps_list.append(x)
+plt.plot(steps_list,distance_list, color='green', label="Random Walk Data")
+
+steps_list=np.asarray(steps_list)
+distance_list=np.asarray(distance_list)
+
+X=steps_list[:,np.newaxis]
+
+#Polynomial fits
+
+#Degree 2
+poly_features=PolynomialFeatures(degree=2, include_bias=False)
+X_poly=poly_features.fit_transform(X)
+
+lin_reg=LinearRegression()
+poly_fit=lin_reg.fit(X_poly,distance_list)
+b=lin_reg.coef_
+c=lin_reg.intercept_
+print ("2nd degree coefficients:")
+print ("zero power: ",c)
+print ("first power: ", b[0])
+print ("second power: ",b[1])
+
+z = np.arange(0, steps, .01)
+z_mod=b[1]*z**2+b[0]*z+c
+
+fit_mod=b[1]*X**2+b[0]*X+c
+plt.plot(z, z_mod, color='r', label="2nd Degree Fit")
+plt.title("Polynomial Regression")
+
+plt.xlabel("Steps")
+plt.ylabel("Distance")
+
+#Degree 10
+poly_features10=PolynomialFeatures(degree=10, include_bias=False)
+X_poly10=poly_features10.fit_transform(X)
+
+poly_fit10=lin_reg.fit(X_poly10,distance_list)
+
+y_plot=poly_fit10.predict(X_poly10)
+plt.plot(X, y_plot, color='black', label="10th Degree Fit")
+
+plt.legend()
+plt.show()
+
+
+#Decision Tree Regression
+from sklearn.tree import DecisionTreeRegressor
+regr_1=DecisionTreeRegressor(max_depth=2)
+regr_2=DecisionTreeRegressor(max_depth=5)
+regr_3=DecisionTreeRegressor(max_depth=7)
+regr_1.fit(X, distance_list)
+regr_2.fit(X, distance_list)
+regr_3.fit(X, distance_list)
+
+X_test = np.arange(0.0, steps, 0.01)[:, np.newaxis]
+y_1 = regr_1.predict(X_test)
+y_2 = regr_2.predict(X_test)
+y_3=regr_3.predict(X_test)
+
+# Plot the results
+plt.figure()
+plt.scatter(X, distance_list, s=2.5, c="black", label="data")
+plt.plot(X_test, y_1, color="red",
+         label="max_depth=2", linewidth=2)
+plt.plot(X_test, y_2, color="green", label="max_depth=5", linewidth=2)
+plt.plot(X_test, y_3, color="m", label="max_depth=7", linewidth=2)
+
+plt.xlabel("Data")
+plt.ylabel("Darget")
+plt.title("Decision Tree Regression")
+plt.legend()
+plt.show()
+
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    -

    where \( \overline{y}_{R_j} \) is the mean response for the training observations -within box \( j \). -

    @@ -370,7 +464,7 @@ within box \( j \).

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    • diff --git a/doc/pub/week44/html/._week44-bs028.html b/doc/pub/week44/html/._week44-bs028.html index 346168e4e..ed693496b 100644 --- a/doc/pub/week44/html/._week44-bs028.html +++ b/doc/pub/week44/html/._week44-bs028.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,20 +327,27 @@ MathJax.Hub.Config({

     

     

     

    -

    A top-down approach, recursive binary splitting

    +

    Building a tree, regression

    -

    Unfortunately, it is computationally infeasible to consider every -possible partition of the feature space into \( J \) boxes. The common -strategy is to take a top-down approach +

    There are mainly two steps

    +
      +
    1. We split the predictor space (the set of possible values \( x_1,x_2,\dots, x_p \)) into \( J \) distinct and non-non-overlapping regions, \( R_1,R_2,\dots,R_J \).
      2. +
    2. For every observation that falls into the region \( R_j \) , we make the same prediction, which is simply the mean of the response values for the training observations in \( R_j \).
      3. +
    +

    How do we construct the regions \( R_1,\dots,R_J \)? In theory, the +regions could have any shape. However, we choose to divide the +predictor space into high-dimensional rectangles, or boxes, for +simplicity and for ease of interpretation of the resulting predictive +model. The goal is to find boxes \( R_1,\dots,R_J \) that minimize the +MSE, given by

    -

    The approach is top-down because it begins at the top of the tree (all -observations belong to a single region) and then successively splits -the predictor space; each split is indicated via two new branches -further down on the tree. It is greedy because at each step of the -tree-building process, the best split is made at that particular step, -rather than looking ahead and picking a split that will lead to a -better tree in some future step. +$$ +\sum_{j=1}^J\sum_{i\in R_j}(y_i-\overline{y}_{R_j})^2, +$$ + +

    where \( \overline{y}_{R_j} \) is the mean response for the training observations +within box \( j \).

    @@ -363,7 +375,7 @@ better tree in some future step.

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    • diff --git a/doc/pub/week44/html/._week44-bs029.html b/doc/pub/week44/html/._week44-bs029.html index 4293f1d53..7e5693476 100644 --- a/doc/pub/week44/html/._week44-bs029.html +++ b/doc/pub/week44/html/._week44-bs029.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,52 +327,20 @@ MathJax.Hub.Config({

     

     

     

    -

    Making a tree

    +

    A top-down approach, recursive binary splitting

    -

    In order to implement the recursive binary splitting we start by selecting -the predictor \( x_j \) and a cutpoint \( s \) that splits the predictor space into two regions \( R_1 \) and \( R_2 \) -

    -$$ -\left\{X\vert x_j < s\right\}, -$$ - -

    and

    -$$ -\left\{X\vert x_j \geq s\right\}, -$$ - -

    so that we obtain the lowest MSE, that is

    -$$ -\sum_{i:x_i\in R_j}(y_i-\overline{y}_{R_1})^2+\sum_{i:x_i\in R_2}(y_i-\overline{y}_{R_2})^2, -$$ - -

    which we want to minimize by considering all predictors -\( x_1,x_2,\dots,x_p \). We consider also all possible values of \( s \) for -each predictor. These values could be determined by randomly assigned -numbers or by starting at the midpoint and then proceed till we find -an optimal value. +

    Unfortunately, it is computationally infeasible to consider every +possible partition of the feature space into \( J \) boxes. The common +strategy is to take a top-down approach

    -

    For any \( j \) and \( s \), we define the pair of half-planes where -\( \overline{y}_{R_1} \) is the mean response for the training -observations in \( R_1(j,s) \), and \( \overline{y}_{R_2} \) is the mean -response for the training observations in \( R_2(j,s) \). -

    - -

    Finding the values of \( j \) and \( s \) that minimize the above equation can be -done quite quickly, especially when the number of features \( p \) is not -too large. -

    - -

    Next, we repeat the process, looking -for the best predictor and best cutpoint in order to split the data -further so as to minimize the MSE within each of the resulting -regions. However, this time, instead of splitting the entire predictor -space, we split one of the two previously identified regions. We now -have three regions. Again, we look to split one of these three regions -further, so as to minimize the MSE. The process continues until a -stopping criterion is reached; for instance, we may continue until no -region contains more than five observations. +

    The approach is top-down because it begins at the top of the tree (all +observations belong to a single region) and then successively splits +the predictor space; each split is indicated via two new branches +further down on the tree. It is greedy because at each step of the +tree-building process, the best split is made at that particular step, +rather than looking ahead and picking a split that will lead to a +better tree in some future step.

    @@ -395,7 +368,7 @@ region contains more than five observations.

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    • diff --git a/doc/pub/week44/html/._week44-bs030.html b/doc/pub/week44/html/._week44-bs030.html index 30697795e..77f2ef4b7 100644 --- a/doc/pub/week44/html/._week44-bs030.html +++ b/doc/pub/week44/html/._week44-bs030.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -321,24 +326,54 @@ MathJax.Hub.Config({

     

     

     

    - -

    Pruning the tree

    + +

    Making a tree

    -

    The above procedure is rather straightforward, but leads often to -overfitting and unnecessarily large and complicated trees. The basic -idea is to grow a large tree \( T_0 \) and then prune it back in order to -obtain a subtree. A smaller tree with fewer splits (fewer regions) can -lead to smaller variance and better interpretation at the cost of a -little more bias. +

    In order to implement the recursive binary splitting we start by selecting +the predictor \( x_j \) and a cutpoint \( s \) that splits the predictor space into two regions \( R_1 \) and \( R_2 \) +

    +$$ +\left\{X\vert x_j < s\right\}, +$$ + +

    and

    +$$ +\left\{X\vert x_j \geq s\right\}, +$$ + +

    so that we obtain the lowest MSE, that is

    +$$ +\sum_{i:x_i\in R_j}(y_i-\overline{y}_{R_1})^2+\sum_{i:x_i\in R_2}(y_i-\overline{y}_{R_2})^2, +$$ + +

    which we want to minimize by considering all predictors +\( x_1,x_2,\dots,x_p \). We consider also all possible values of \( s \) for +each predictor. These values could be determined by randomly assigned +numbers or by starting at the midpoint and then proceed till we find +an optimal value.

    -

    The so-called Cost complexity pruning algorithm gives us a -way to do just this. Rather than considering every possible subtree, -we consider a sequence of trees indexed by a nonnegative tuning -parameter \( \alpha \). +

    For any \( j \) and \( s \), we define the pair of half-planes where +\( \overline{y}_{R_1} \) is the mean response for the training +observations in \( R_1(j,s) \), and \( \overline{y}_{R_2} \) is the mean +response for the training observations in \( R_2(j,s) \).

    -

    Read more at the following Scikit-Learn link on pruning.

    +

    Finding the values of \( j \) and \( s \) that minimize the above equation can be +done quite quickly, especially when the number of features \( p \) is not +too large. +

    + +

    Next, we repeat the process, looking +for the best predictor and best cutpoint in order to split the data +further so as to minimize the MSE within each of the resulting +regions. However, this time, instead of splitting the entire predictor +space, we split one of the two previously identified regions. We now +have three regions. Again, we look to split one of these three regions +further, so as to minimize the MSE. The process continues until a +stopping criterion is reached; for instance, we may continue until no +region contains more than five observations. +

    @@ -365,7 +400,7 @@ parameter \( \alpha \).

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    • diff --git a/doc/pub/week44/html/._week44-bs031.html b/doc/pub/week44/html/._week44-bs031.html index 0813c9ac6..8c74e53a5 100644 --- a/doc/pub/week44/html/._week44-bs031.html +++ b/doc/pub/week44/html/._week44-bs031.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -321,36 +326,24 @@ MathJax.Hub.Config({

     

     

     

    - -

    Cost complexity pruning

    + +

    Pruning the tree

    -

    For each value of \( \alpha \) there corresponds a subtree \( T \in T_0 \) such that

    -$$ -\sum_{m=1}^{\overline{T}}\sum_{i:x_i\in R_m}(y_i-\overline{y}_{R_m})^2+\alpha\overline{T}, -$$ - -

    is as small as possible. Here \( \overline{T} \) is -the number of terminal nodes of the tree \( T \) , \( R_m \) is the -rectangle (i.e. the subset of predictor space) corresponding to the \( m \)-th terminal node. +

    The above procedure is rather straightforward, but leads often to +overfitting and unnecessarily large and complicated trees. The basic +idea is to grow a large tree \( T_0 \) and then prune it back in order to +obtain a subtree. A smaller tree with fewer splits (fewer regions) can +lead to smaller variance and better interpretation at the cost of a +little more bias.

    -

    The tuning parameter \( \alpha \) controls a trade-off between the subtree’s -complexity and its fit to the training data. When \( \alpha = 0 \), then the -subtree \( T \) will simply equal \( T_0 \), -because then the above equation just measures the -training error. -However, as \( \alpha \) increases, there is a price to pay for -having a tree with many terminal nodes. The above equation will -tend to be minimized for a smaller subtree. +

    The so-called Cost complexity pruning algorithm gives us a +way to do just this. Rather than considering every possible subtree, +we consider a sequence of trees indexed by a nonnegative tuning +parameter \( \alpha \).

    -

    It turns out that as we increase \( \alpha \) from zero -branches get pruned from the tree in a nested and predictable fashion, -so obtaining the whole sequence of subtrees as a function of \( \alpha \) is -easy. We can select a value of \( \alpha \) using a validation set or using -cross-validation. We then return to the full data set and obtain the -subtree corresponding to \( \alpha \). -

    +

    Read more at the following Scikit-Learn link on pruning.

    @@ -377,7 +370,7 @@ subtree corresponding to \( \alpha \).

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    • diff --git a/doc/pub/week44/html/._week44-bs032.html b/doc/pub/week44/html/._week44-bs032.html index 6a8dd15bf..7fba9f9e6 100644 --- a/doc/pub/week44/html/._week44-bs032.html +++ b/doc/pub/week44/html/._week44-bs032.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,26 +327,35 @@ MathJax.Hub.Config({

     

     

     

    -

    Schematic Regression Procedure

    +

    Cost complexity pruning

    -
    -
    - +

    For each value of \( \alpha \) there corresponds a subtree \( T \in T_0 \) such that

    +$$ +\sum_{m=1}^{\overline{T}}\sum_{i:x_i\in R_m}(y_i-\overline{y}_{R_m})^2+\alpha\overline{T}, +$$ -
      -
    1. Use recursive binary splitting to grow a large tree on the training data, stopping only when each terminal node has fewer than some minimum number of observations.
      2. -
    2. Apply cost complexity pruning to the large tree in order to obtain a sequence of best subtrees, as a function of \( \alpha \).
      3. -
    3. Use for example \( K \)-fold cross-validation to choose \( \alpha \). Divide the training observations into \( K \) folds. For each \( k=1,2,\dots,K \) we:
      4. -
        -
      • repeat steps 1 and 2 on all but the \( k \)-th fold of the training data.
        • -
      • Then we valuate the mean squared prediction error on the data in the left-out \( k \)-th fold, as a function of \( \alpha \).
        • -
      • Finally we average the results for each value of \( \alpha \), and pick \( \alpha \) to minimize the average error.
        • -
      -
    4. Return the subtree from Step 2 that corresponds to the chosen value of \( \alpha \).
      5. -
    -
    -
    +

    is as small as possible. Here \( \overline{T} \) is +the number of terminal nodes of the tree \( T \) , \( R_m \) is the +rectangle (i.e. the subset of predictor space) corresponding to the \( m \)-th terminal node. +

    +

    The tuning parameter \( \alpha \) controls a trade-off between the subtree’s +complexity and its fit to the training data. When \( \alpha = 0 \), then the +subtree \( T \) will simply equal \( T_0 \), +because then the above equation just measures the +training error. +However, as \( \alpha \) increases, there is a price to pay for +having a tree with many terminal nodes. The above equation will +tend to be minimized for a smaller subtree. +

    + +

    It turns out that as we increase \( \alpha \) from zero +branches get pruned from the tree in a nested and predictable fashion, +so obtaining the whole sequence of subtrees as a function of \( \alpha \) is +easy. We can select a value of \( \alpha \) using a validation set or using +cross-validation. We then return to the full data set and obtain the +subtree corresponding to \( \alpha \). +

    @@ -368,7 +382,7 @@ MathJax.Hub.Config({

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    • diff --git a/doc/pub/week44/html/._week44-bs033.html b/doc/pub/week44/html/._week44-bs033.html index 7c3586cda..db2931f59 100644 --- a/doc/pub/week44/html/._week44-bs033.html +++ b/doc/pub/week44/html/._week44-bs033.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,21 +327,26 @@ MathJax.Hub.Config({

     

     

     

    -

    A Classification Tree

    +

    Schematic Regression Procedure

    + +
    +
    + + +
      +
    1. Use recursive binary splitting to grow a large tree on the training data, stopping only when each terminal node has fewer than some minimum number of observations.
      2. +
    2. Apply cost complexity pruning to the large tree in order to obtain a sequence of best subtrees, as a function of \( \alpha \).
      3. +
    3. Use for example \( K \)-fold cross-validation to choose \( \alpha \). Divide the training observations into \( K \) folds. For each \( k=1,2,\dots,K \) we:
      4. +
        +
      • repeat steps 1 and 2 on all but the \( k \)-th fold of the training data.
        • +
      • Then we valuate the mean squared prediction error on the data in the left-out \( k \)-th fold, as a function of \( \alpha \).
        • +
      • Finally we average the results for each value of \( \alpha \), and pick \( \alpha \) to minimize the average error.
        • +
      +
    4. Return the subtree from Step 2 that corresponds to the chosen value of \( \alpha \).
      5. +
    +
    +
    -

    A classification tree is very similar to a regression tree, except -that it is used to predict a qualitative response rather than a -quantitative one. Recall that for a regression tree, the predicted -response for an observation is given by the mean response of the -training observations that belong to the same terminal node. In -contrast, for a classification tree, we predict that each observation -belongs to the most commonly occurring class of training observations -in the region to which it belongs. In interpreting the results of a -classification tree, we are often interested not only in the class -prediction corresponding to a particular terminal node region, but -also in the class proportions among the training observations that -fall into that region. -

    @@ -363,7 +373,7 @@ fall into that region.

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    • diff --git a/doc/pub/week44/html/._week44-bs034.html b/doc/pub/week44/html/._week44-bs034.html index f0fb602e8..fc67cd281 100644 --- a/doc/pub/week44/html/._week44-bs034.html +++ b/doc/pub/week44/html/._week44-bs034.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,25 +327,20 @@ MathJax.Hub.Config({

     

     

     

    -

    Growing a classification tree

    +

    A Classification Tree

    -

    The task of growing a -classification tree is quite similar to the task of growing a -regression tree. Just as in the regression setting, we use recursive -binary splitting to grow a classification tree. However, in the -classification setting, the MSE cannot be used as a criterion for making -the binary splits. A natural alternative to MSE is the classification -error rate. Since we plan to assign an observation in a given region -to the most commonly occurring error rate class of training -observations in that region, the classification error rate is simply -the fraction of the training observations in that region that do not -belong to the most common class. -

    - -

    When building a classification tree, either the Gini index or the -entropy are typically used to evaluate the quality of a particular -split, since these two approaches are more sensitive to node purity -than is the classification error rate. +

    A classification tree is very similar to a regression tree, except +that it is used to predict a qualitative response rather than a +quantitative one. Recall that for a regression tree, the predicted +response for an observation is given by the mean response of the +training observations that belong to the same terminal node. In +contrast, for a classification tree, we predict that each observation +belongs to the most commonly occurring class of training observations +in the region to which it belongs. In interpreting the results of a +classification tree, we are often interested not only in the class +prediction corresponding to a particular terminal node region, but +also in the class proportions among the training observations that +fall into that region.

    @@ -368,7 +368,7 @@ than is the classification error rate.

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    • diff --git a/doc/pub/week44/html/._week44-bs035.html b/doc/pub/week44/html/._week44-bs035.html index ee6d072ba..8d5116f80 100644 --- a/doc/pub/week44/html/._week44-bs035.html +++ b/doc/pub/week44/html/._week44-bs035.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,49 +327,27 @@ MathJax.Hub.Config({

     

     

     

    -

    Classification tree, how to split nodes

    +

    Growing a classification tree

    -

    If our targets are the outcome of a classification process that takes -for example \( k=1,2,\dots,K \) values, the only thing we need to think of -is to set up the splitting criteria for each node. +

    The task of growing a +classification tree is quite similar to the task of growing a +regression tree. Just as in the regression setting, we use recursive +binary splitting to grow a classification tree. However, in the +classification setting, the MSE cannot be used as a criterion for making +the binary splits. A natural alternative to MSE is the classification +error rate. Since we plan to assign an observation in a given region +to the most commonly occurring error rate class of training +observations in that region, the classification error rate is simply +the fraction of the training observations in that region that do not +belong to the most common class.

    -

    We define a PDF \( p_{mk} \) that represents the number of observations of -a class \( k \) in a region \( R_m \) with \( N_m \) observations. We represent -this likelihood function in terms of the proportion \( I(y_i=k) \) of -observations of this class in the region \( R_m \) as +

    When building a classification tree, either the Gini index or the +entropy are typically used to evaluate the quality of a particular +split, since these two approaches are more sensitive to node purity +than is the classification error rate.

    -$$ -p_{mk} = \frac{1}{N_m}\sum_{x_i\in R_m}I(y_i=k). -$$ - -

    We let \( p_{mk} \) represent the majority class of observations in region -\( m \). The three most common ways of splitting a node are given by -

    - -
      -
    • Misclassification error
      • -
    -$$ -p_{mk} = \frac{1}{N_m}\sum_{x_i\in R_m}I(y_i\ne k) = 1-p_{mk}. -$$ - -
      -
    • Gini index \( g \)
      • -
    -$$ -g = \sum_{k=1}^K p_{mk}(1-p_{mk}). -$$ - -
      -
    • Information entropy or just entropy \( s \)
      • -
    -$$ -s = -\sum_{k=1}^K p_{mk}\log{p_{mk}}. -$$ - -

    diff --git a/doc/pub/week44/html/._week44-bs036.html b/doc/pub/week44/html/._week44-bs036.html index 7b410ebee..66f284c67 100644 --- a/doc/pub/week44/html/._week44-bs036.html +++ b/doc/pub/week44/html/._week44-bs036.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,61 +327,47 @@ MathJax.Hub.Config({

     

     

     

    -

    Visualizing the Tree, Classification

    +

    Classification tree, how to split nodes

    - -
    -
    -
    -
    -
    - 
    import os
-from sklearn.datasets import load_breast_cancer
-from sklearn.tree import DecisionTreeClassifier
-from sklearn.model_selection import train_test_split
-from sklearn.metrics import confusion_matrix
-from sklearn.tree import export_graphviz
+
    If our targets are the outcome of a classification process that takes
+for example \( k=1,2,\dots,K \) values, the only thing we need to think of
+is to set up the splitting criteria for each node.
+
    
 
-from IPython.display import Image 
-from pydot import graph_from_dot_data
-import pandas as pd
-import numpy as np
+
    We define a PDF \( p_{mk} \) that represents the number of observations of
+a class \( k \) in a region \( R_m \) with \( N_m \) observations. We represent
+this likelihood function in terms of the proportion \( I(y_i=k) \) of
+observations of this class in the region \( R_m \) as
+
    
 
+$$
+p_{mk} = \frac{1}{N_m}\sum_{x_i\in R_m}I(y_i=k).
+$$
 
-cancer = load_breast_cancer()
-X = pd.DataFrame(cancer.data, columns=cancer.feature_names)
-print(X)
-y = pd.Categorical.from_codes(cancer.target, cancer.target_names)
-y = pd.get_dummies(y)
-print(y)
-X_train, X_test, y_train, y_test = train_test_split(X, y, random_state=1)
-tree_clf = DecisionTreeClassifier(max_depth=5)
-tree_clf.fit(X_train, y_train)
+
    We let \( p_{mk} \) represent the majority class of observations in region
+\( m \). The three most common ways of splitting a node are given by
+
    
 
-export_graphviz(
-    tree_clf,
-    out_file="DataFiles/cancer.dot",
-    feature_names=cancer.feature_names,
-    class_names=cancer.target_names,
-    rounded=True,
-    filled=True
-)
-cmd = 'dot -Tpng DataFiles/cancer.dot -o DataFiles/cancer.png'
-os.system(cmd)
-
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    +
      +
    • Misclassification error
      • +
    +$$ +p_{mk} = \frac{1}{N_m}\sum_{x_i\in R_m}I(y_i\ne k) = 1-p_{mk}. +$$ + +
      +
    • Gini index \( g \)
      • +
    +$$ +g = \sum_{k=1}^K p_{mk}(1-p_{mk}). +$$ + +
      +
    • Information entropy or just entropy \( s \)
      • +
    +$$ +s = -\sum_{k=1}^K p_{mk}\log{p_{mk}}. +$$

    @@ -404,7 +395,7 @@ os.system(cmd)

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    • diff --git a/doc/pub/week44/html/._week44-bs037.html b/doc/pub/week44/html/._week44-bs037.html index 1b223abaa..a77ce957b 100644 --- a/doc/pub/week44/html/._week44-bs037.html +++ b/doc/pub/week44/html/._week44-bs037.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,7 +327,7 @@ MathJax.Hub.Config({

     

     

     

    -

    Visualizing the Tree, The Moons

    +

    Visualizing the Tree, Classification

    @@ -330,29 +335,38 @@ MathJax.Hub.Config({
    - 
    # Common imports
-import numpy as np
-from sklearn.model_selection import  train_test_split 
+  
    import os
+from sklearn.datasets import load_breast_cancer
 from sklearn.tree import DecisionTreeClassifier
-from sklearn.datasets import make_moons
+from sklearn.model_selection import train_test_split
+from sklearn.metrics import confusion_matrix
 from sklearn.tree import export_graphviz
+
+from IPython.display import Image 
 from pydot import graph_from_dot_data
 import pandas as pd
-import os
+import numpy as np
 
-np.random.seed(42)
-X, y = make_moons(n_samples=100, noise=0.25, random_state=53)
-X_train, X_test, y_train, y_test = train_test_split(X,y,random_state=0)
+
+cancer = load_breast_cancer()
+X = pd.DataFrame(cancer.data, columns=cancer.feature_names)
+print(X)
+y = pd.Categorical.from_codes(cancer.target, cancer.target_names)
+y = pd.get_dummies(y)
+print(y)
+X_train, X_test, y_train, y_test = train_test_split(X, y, random_state=1)
 tree_clf = DecisionTreeClassifier(max_depth=5)
 tree_clf.fit(X_train, y_train)
 
 export_graphviz(
     tree_clf,
-    out_file="DataFiles/moons.dot",
+    out_file="DataFiles/cancer.dot",
+    feature_names=cancer.feature_names,
+    class_names=cancer.target_names,
     rounded=True,
     filled=True
 )
-cmd = 'dot -Tpng DataFiles/moons.dot -o DataFiles/moons.png'
+cmd = 'dot -Tpng DataFiles/cancer.dot -o DataFiles/cancer.png'
 os.system(cmd)
    @@ -395,7 +409,7 @@ os.system(cmd)
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  • »
    • diff --git a/doc/pub/week44/html/._week44-bs038.html b/doc/pub/week44/html/._week44-bs038.html index b4086f7f9..b5ca46451 100644 --- a/doc/pub/week44/html/._week44-bs038.html +++ b/doc/pub/week44/html/._week44-bs038.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,10 +327,7 @@ MathJax.Hub.Config({

     

     

     

    -

    Other ways of visualizing the trees

    - -

    Scikit-Learn has also another way to visualize the trees which is very useful, here with the Iris data.

    - +

    Visualizing the Tree, The Moons

    @@ -333,13 +335,30 @@ MathJax.Hub.Config({
    - 
    from sklearn.datasets import load_iris
-from sklearn import tree
-X, y = load_iris(return_X_y=True)
-tree_clf = tree.DecisionTreeClassifier()
-tree_clf = tree_clf.fit(X, y)
-# and then plot the tree
-tree.plot_tree(tree_clf) 
+  
    # Common imports
+import numpy as np
+from sklearn.model_selection import  train_test_split 
+from sklearn.tree import DecisionTreeClassifier
+from sklearn.datasets import make_moons
+from sklearn.tree import export_graphviz
+from pydot import graph_from_dot_data
+import pandas as pd
+import os
+
+np.random.seed(42)
+X, y = make_moons(n_samples=100, noise=0.25, random_state=53)
+X_train, X_test, y_train, y_test = train_test_split(X,y,random_state=0)
+tree_clf = DecisionTreeClassifier(max_depth=5)
+tree_clf.fit(X_train, y_train)
+
+export_graphviz(
+    tree_clf,
+    out_file="DataFiles/moons.dot",
+    rounded=True,
+    filled=True
+)
+cmd = 'dot -Tpng DataFiles/moons.dot -o DataFiles/moons.png'
+os.system(cmd)
    @@ -381,7 +400,7 @@ tree.plot_tree(tree_clf)
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    • -
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  • »
    • diff --git a/doc/pub/week44/html/._week44-bs039.html b/doc/pub/week44/html/._week44-bs039.html index 39785f947..24ec021eb 100644 --- a/doc/pub/week44/html/._week44-bs039.html +++ b/doc/pub/week44/html/._week44-bs039.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,11 +327,9 @@ MathJax.Hub.Config({

     

     

     

    -

    Printing out as text

    +

    Other ways of visualizing the trees

    -

    Alternatively, the tree can also be exported in textual format with the function exporttext. -This method doesn’t require the installation of external libraries and is more compact: -

    +

    Scikit-Learn has also another way to visualize the trees which is very useful, here with the Iris data.

    @@ -336,13 +339,12 @@ This method doesn’t require the installation of external libraries and is 
    from sklearn.datasets import load_iris
-from sklearn.tree import DecisionTreeClassifier
-from sklearn.tree import export_text
-iris = load_iris()
-decision_tree = DecisionTreeClassifier(random_state=0, max_depth=2)
-decision_tree = decision_tree.fit(iris.data, iris.target)
-r = export_text(decision_tree, feature_names=iris['feature_names'])
-print(r)
+from sklearn import tree
+X, y = load_iris(return_X_y=True)
+tree_clf = tree.DecisionTreeClassifier()
+tree_clf = tree_clf.fit(X, y)
+# and then plot the tree
+tree.plot_tree(tree_clf)
    @@ -384,7 +386,7 @@ r = export_text(decision_tree, feature_names
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    • diff --git a/doc/pub/week44/html/._week44-bs040.html b/doc/pub/week44/html/._week44-bs040.html index 55c8a374f..653be736e 100644 --- a/doc/pub/week44/html/._week44-bs040.html +++ b/doc/pub/week44/html/._week44-bs040.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,19 +327,43 @@ MathJax.Hub.Config({

     

     

     

    -

    Algorithms for Setting up Decision Trees

    +

    Printing out as text

    -

    Two algorithms stand out in the set up of decision trees:

    -
      -
    1. The CART (Classification And Regression Tree) algorithm for both classification and regression
      2. -
    2. The ID3 algorithm based on the computation of the information gain for classification
      3. -
    -

    We discuss both algorithms with applications here. The popular library -Scikit-Learn uses the CART algorithm. For classification problems -you can use either the gini index or the entropy to split a tree -in two branches. +

    Alternatively, the tree can also be exported in textual format with the function exporttext. +This method doesn’t require the installation of external libraries and is more compact:

    + + +
    +
    +
    +
    +
    + 
    from sklearn.datasets import load_iris
+from sklearn.tree import DecisionTreeClassifier
+from sklearn.tree import export_text
+iris = load_iris()
+decision_tree = DecisionTreeClassifier(random_state=0, max_depth=2)
+decision_tree = decision_tree.fit(iris.data, iris.target)
+r = export_text(decision_tree, feature_names=iris['feature_names'])
+print(r)
+
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    + +

    diff --git a/doc/pub/week44/html/._week44-bs041.html b/doc/pub/week44/html/._week44-bs041.html index 323d161e2..e5d40e8c7 100644 --- a/doc/pub/week44/html/._week44-bs041.html +++ b/doc/pub/week44/html/._week44-bs041.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,29 +327,17 @@ MathJax.Hub.Config({

     

     

     

    -

    The CART algorithm for Classification

    +

    Algorithms for Setting up Decision Trees

    -

    For classification, the CART algorithm splits the data set in two subsets using a single feature \( k \) and a threshold \( t_k \). -This could be for example a threshold set by a number below a certain circumference of a malign tumor. -

    - -

    How do we find these two quantities? -We search for the pair \( (k,t_k) \) that produces the purest subset using for example the gini factor \( G \). -The cost function it tries to minimize is then -

    -$$ -C(k,t_k) = \frac{m_{\mathrm{left}}}{m}G_{\mathrm{left}}+ \frac{m_{\mathrm{right}}}{m}G_{\mathrm{right}}, -$$ - -

    where \( G_{\mathrm{left/right}} \) measures the impurity of the left/right subset and \( m_{\mathrm{left/right}} \) - is the number of instances in the left/right subset -

    - -

    Once it has successfully split the training set in two, it splits the subsets using the same logic, then the subsubsets -and so on, recursively. It stops recursing once it reaches the maximum depth (defined by the -\( max\_depth \) hyperparameter), or if it cannot find a split that will reduce impurity. A few other -hyperparameters control additional stopping conditions such as the \( min\_samples\_split \), -\( min\_samples\_leaf \), \( min\_weight\_fraction\_leaf \), and \( max\_leaf\_nodes \). +

    Two algorithms stand out in the set up of decision trees:

    +
      +
    1. The CART (Classification And Regression Tree) algorithm for both classification and regression
      2. +
    2. The ID3 algorithm based on the computation of the information gain for classification
      3. +
    +

    We discuss both algorithms with applications here. The popular library +Scikit-Learn uses the CART algorithm. For classification problems +you can use either the gini index or the entropy to split a tree +in two branches.

    @@ -372,7 +365,7 @@ hyperparameters control additional stopping conditions such as the \( min\_sampl

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    • diff --git a/doc/pub/week44/html/._week44-bs042.html b/doc/pub/week44/html/._week44-bs042.html index e97085785..163cc80b3 100644 --- a/doc/pub/week44/html/._week44-bs042.html +++ b/doc/pub/week44/html/._week44-bs042.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,29 +327,29 @@ MathJax.Hub.Config({

     

     

     

    -

    The CART algorithm for Regression

    +

    The CART algorithm for Classification

    -

    The CART algorithm for regression works is similar to the one for classification except that instead of trying to split the -training set in a way that minimizes say the gini or entropy impurity, it now tries to split the training set in a way that minimizes our well-known mean-squared error (MSE). The cost function is now +

    For classification, the CART algorithm splits the data set in two subsets using a single feature \( k \) and a threshold \( t_k \). +This could be for example a threshold set by a number below a certain circumference of a malign tumor. +

    + +

    How do we find these two quantities? +We search for the pair \( (k,t_k) \) that produces the purest subset using for example the gini factor \( G \). +The cost function it tries to minimize is then

    $$ -C(k,t_k) = \frac{m_{\mathrm{left}}}{m}\mathrm{MSE}_{\mathrm{left}}+ \frac{m_{\mathrm{right}}}{m}\mathrm{MSE}_{\mathrm{right}}. +C(k,t_k) = \frac{m_{\mathrm{left}}}{m}G_{\mathrm{left}}+ \frac{m_{\mathrm{right}}}{m}G_{\mathrm{right}}, $$ -

    Here the MSE for a specific node is defined as

    -$$ -\mathrm{MSE}_{\mathrm{node}}=\frac{1}{m_\mathrm{node}}\sum_{i\in \mathrm{node}}(\overline{y}_{\mathrm{node}}-y_i)^2, -$$ +

    where \( G_{\mathrm{left/right}} \) measures the impurity of the left/right subset and \( m_{\mathrm{left/right}} \) + is the number of instances in the left/right subset +

    -

    with

    -$$ -\overline{y}_{\mathrm{node}}=\frac{1}{m_\mathrm{node}}\sum_{i\in \mathrm{node}}y_i, -$$ - -

    the mean value of all observations in a specific node.

    - -

    Without any regularization, the regression task for decision trees, -just like for classification tasks, is prone to overfitting. +

    Once it has successfully split the training set in two, it splits the subsets using the same logic, then the subsubsets +and so on, recursively. It stops recursing once it reaches the maximum depth (defined by the +\( max\_depth \) hyperparameter), or if it cannot find a split that will reduce impurity. A few other +hyperparameters control additional stopping conditions such as the \( min\_samples\_split \), +\( min\_samples\_leaf \), \( min\_weight\_fraction\_leaf \), and \( max\_leaf\_nodes \).

    @@ -372,7 +377,7 @@ just like for classification tasks, is prone to overfitting.

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    • diff --git a/doc/pub/week44/html/._week44-bs043.html b/doc/pub/week44/html/._week44-bs043.html index 4072f3b0a..99ff28b9a 100644 --- a/doc/pub/week44/html/._week44-bs043.html +++ b/doc/pub/week44/html/._week44-bs043.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,46 +327,30 @@ MathJax.Hub.Config({

     

     

     

    -

    Computing the Gini index

    +

    The CART algorithm for Regression

    -

    The example we will look at is a classical one in many Machine -Learning applications. Based on various meteorological features, we -have several so-called attributes which decide whether we at the end -will do some outdoor activity like skiing, going for a bike ride etc -etc. The table here contains the feautures outlook, temperature, -humidity and wind. The target or output is whether we ride -(True=1) or whether we do something else that day (False=0). The -attributes for each feature are then sunny, overcast and rain for the -outlook, hot, cold and mild for temperature, high and normal for -humidity and weak and strong for wind. +

    The CART algorithm for regression works is similar to the one for classification except that instead of trying to split the +training set in a way that minimizes say the gini or entropy impurity, it now tries to split the training set in a way that minimizes our well-known mean-squared error (MSE). The cost function is now

    +$$ +C(k,t_k) = \frac{m_{\mathrm{left}}}{m}\mathrm{MSE}_{\mathrm{left}}+ \frac{m_{\mathrm{right}}}{m}\mathrm{MSE}_{\mathrm{right}}. +$$ -

    The table here summarizes the various attributes and

    -
    -
    - - - - - - - - - - - - - - - - - - - - -
    Day Outlook Temperature Humidity Wind Ride
    1 Sunny Hot High Weak 0
    2 Sunny Hot High Strong 1
    3 Overcast Hot High Weak 1
    4 Rain Mild High Weak 1
    5 Rain Cool Normal Weak 1
    6 Rain Cool Normal Strong 0
    7 Overcast Cool Normal Strong 1
    8 Sunny Mild High Weak 0
    9 Sunny Cool Normal Weak 1
    10 Rain Mild Normal Weak 1
    11 Sunny Mild Normal Strong 1
    12 Overcast Mild High Strong 1
    13 Overcast Hot Normal Weak 1
    14 Rain Mild High Strong 0
    -
    -
    +

    Here the MSE for a specific node is defined as

    +$$ +\mathrm{MSE}_{\mathrm{node}}=\frac{1}{m_\mathrm{node}}\sum_{i\in \mathrm{node}}(\overline{y}_{\mathrm{node}}-y_i)^2, +$$ + +

    with

    +$$ +\overline{y}_{\mathrm{node}}=\frac{1}{m_\mathrm{node}}\sum_{i\in \mathrm{node}}y_i, +$$ + +

    the mean value of all observations in a specific node.

    + +

    Without any regularization, the regression task for decision trees, +just like for classification tasks, is prone to overfitting. +

    @@ -388,7 +377,7 @@ humidity and weak and strong for wind.

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  • »
    • diff --git a/doc/pub/week44/html/._week44-bs044.html b/doc/pub/week44/html/._week44-bs044.html index 940a2c304..54fab72d1 100644 --- a/doc/pub/week44/html/._week44-bs044.html +++ b/doc/pub/week44/html/._week44-bs044.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,97 +327,46 @@ MathJax.Hub.Config({

     

     

     

    -

    Simple Python Code to read in Data and perform Classification

    +

    Computing the Gini index

    +

    The example we will look at is a classical one in many Machine +Learning applications. Based on various meteorological features, we +have several so-called attributes which decide whether we at the end +will do some outdoor activity like skiing, going for a bike ride etc +etc. The table here contains the feautures outlook, temperature, +humidity and wind. The target or output is whether we ride +(True=1) or whether we do something else that day (False=0). The +attributes for each feature are then sunny, overcast and rain for the +outlook, hot, cold and mild for temperature, high and normal for +humidity and weak and strong for wind. +

    - -
    -
    -
    -
    -
    - 
    # Common imports
-import numpy as np
-import pandas as pd
-import matplotlib.pyplot as plt
-from sklearn.tree import DecisionTreeClassifier
-from sklearn.model_selection import train_test_split
-from sklearn.tree import export_graphviz
-from sklearn.preprocessing import StandardScaler, OneHotEncoder
-from sklearn.compose import ColumnTransformer
-from IPython.display import Image 
-from pydot import graph_from_dot_data
-import os
-
-# Where to save the figures and data files
-PROJECT_ROOT_DIR = "Results"
-FIGURE_ID = "Results/FigureFiles"
-DATA_ID = "DataFiles/"
-
-if not os.path.exists(PROJECT_ROOT_DIR):
-    os.mkdir(PROJECT_ROOT_DIR)
-
-if not os.path.exists(FIGURE_ID):
-    os.makedirs(FIGURE_ID)
-
-if not os.path.exists(DATA_ID):
-    os.makedirs(DATA_ID)
-
-def image_path(fig_id):
-    return os.path.join(FIGURE_ID, fig_id)
-
-def data_path(dat_id):
-    return os.path.join(DATA_ID, dat_id)
-
-def save_fig(fig_id):
-    plt.savefig(image_path(fig_id) + ".png", format='png')
-
-infile = open(data_path("rideclass.csv"),'r')
-
-# Read the experimental data with Pandas
-from IPython.display import display
-ridedata = pd.read_csv(infile,names = ('Outlook','Temperature','Humidity','Wind','Ride'))
-ridedata = pd.DataFrame(ridedata)
-
-# Features and targets
-X = ridedata.loc[:, ridedata.columns != 'Ride'].values
-y = ridedata.loc[:, ridedata.columns == 'Ride'].values
-
-# Create the encoder.
-encoder = OneHotEncoder(handle_unknown="ignore")
-# Assume for simplicity all features are categorical.
-encoder.fit(X)    
-# Apply the encoder.
-X = encoder.transform(X)
-print(X)
-# Then do a Classification tree
-tree_clf = DecisionTreeClassifier(max_depth=2)
-tree_clf.fit(X, y)
-print("Train set accuracy with Decision Tree: {:.2f}".format(tree_clf.score(X,y)))
-#transfer to a decision tree graph
-export_graphviz(
-    tree_clf,
-    out_file="DataFiles/ride.dot",
-    rounded=True,
-    filled=True
-)
-cmd = 'dot -Tpng DataFiles/cancer.dot -o DataFiles/cancer.png'
-os.system(cmd)
-
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    - +

    The table here summarizes the various attributes and

    +
    +
    + + + + + + + + + + + + + + + + + + + + +
    Day Outlook Temperature Humidity Wind Ride
    1 Sunny Hot High Weak 0
    2 Sunny Hot High Strong 1
    3 Overcast Hot High Weak 1
    4 Rain Mild High Weak 1
    5 Rain Cool Normal Weak 1
    6 Rain Cool Normal Strong 0
    7 Overcast Cool Normal Strong 1
    8 Sunny Mild High Weak 0
    9 Sunny Cool Normal Weak 1
    10 Rain Mild Normal Weak 1
    11 Sunny Mild Normal Strong 1
    12 Overcast Mild High Strong 1
    13 Overcast Hot Normal Weak 1
    14 Rain Mild High Strong 0
    +
    +

    @@ -439,7 +393,7 @@ os.system(cmd)

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    • diff --git a/doc/pub/week44/html/._week44-bs045.html b/doc/pub/week44/html/._week44-bs045.html index 77ef6b981..1576a78bd 100644 --- a/doc/pub/week44/html/._week44-bs045.html +++ b/doc/pub/week44/html/._week44-bs045.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,15 +327,7 @@ MathJax.Hub.Config({

     

     

     

    -

    Computing the Gini Factor

    - -

    The above functions (gini, entropy and misclassification error) are -important components of the so-called CART algorithm. We will discuss -this algorithm below after we have discussed the information gain -algorithm ID3. -

    - -

    In the example here we have converted all our attributes into numerical values \( 0,1,2 \) etc.

    +

    Simple Python Code to read in Data and perform Classification

    @@ -339,66 +336,73 @@ algorithm ID3.
    - 
    # Split a dataset based on an attribute and an attribute value
-def test_split(index, value, dataset):
-	left, right = list(), list()
-	for row in dataset:
-		if row[index] < value:
-			left.append(row)
-		else:
-			right.append(row)
-	return left, right
- 
-# Calculate the Gini index for a split dataset
-def gini_index(groups, classes):
-	# count all samples at split point
-	n_instances = float(sum([len(group) for group in groups]))
-	# sum weighted Gini index for each group
-	gini = 0.0
-	for group in groups:
-		size = float(len(group))
-		# avoid divide by zero
-		if size == 0:
-			continue
-		score = 0.0
-		# score the group based on the score for each class
-		for class_val in classes:
-			p = [row[-1] for row in group].count(class_val) / size
-			score += p * p
-		# weight the group score by its relative size
-		gini += (1.0 - score) * (size / n_instances)
-	return gini
+  
    # Common imports
+import numpy as np
+import pandas as pd
+import matplotlib.pyplot as plt
+from sklearn.tree import DecisionTreeClassifier
+from sklearn.model_selection import train_test_split
+from sklearn.tree import export_graphviz
+from sklearn.preprocessing import StandardScaler, OneHotEncoder
+from sklearn.compose import ColumnTransformer
+from IPython.display import Image 
+from pydot import graph_from_dot_data
+import os
 
-# Select the best split point for a dataset
-def get_split(dataset):
-	class_values = list(set(row[-1] for row in dataset))
-	b_index, b_value, b_score, b_groups = 999, 999, 999, None
-	for index in range(len(dataset[0])-1):
-		for row in dataset:
-			groups = test_split(index, row[index], dataset)
-			gini = gini_index(groups, class_values)
-			print('X%d < %.3f Gini=%.3f' % ((index+1), row[index], gini))
-			if gini < b_score:
-				b_index, b_value, b_score, b_groups = index, row[index], gini, groups
-	return {'index':b_index, 'value':b_value, 'groups':b_groups}
- 
-dataset = [[0,0,0,0,0],
-            [0,0,0,1,1],
-            [1,0,0,0,1],
-            [2,1,0,0,1],
-            [2,2,1,0,1],
-            [2,2,1,1,0],
-            [1,2,1,1,1],
-            [0,1,0,0,0],
-            [0,2,1,0,1],
-            [2,1,1,0,1],
-            [0,1,1,1,1],
-            [1,1,0,1,1],
-            [1,0,1,0,1],
-            [2,1,0,1,0]]
+# Where to save the figures and data files
+PROJECT_ROOT_DIR = "Results"
+FIGURE_ID = "Results/FigureFiles"
+DATA_ID = "DataFiles/"
 
-split = get_split(dataset)
-print('Split: [X%d < %.3f]' % ((split['index']+1), split['value']))
+if not os.path.exists(PROJECT_ROOT_DIR):
+    os.mkdir(PROJECT_ROOT_DIR)
+
+if not os.path.exists(FIGURE_ID):
+    os.makedirs(FIGURE_ID)
+
+if not os.path.exists(DATA_ID):
+    os.makedirs(DATA_ID)
+
+def image_path(fig_id):
+    return os.path.join(FIGURE_ID, fig_id)
+
+def data_path(dat_id):
+    return os.path.join(DATA_ID, dat_id)
+
+def save_fig(fig_id):
+    plt.savefig(image_path(fig_id) + ".png", format='png')
+
+infile = open(data_path("rideclass.csv"),'r')
+
+# Read the experimental data with Pandas
+from IPython.display import display
+ridedata = pd.read_csv(infile,names = ('Outlook','Temperature','Humidity','Wind','Ride'))
+ridedata = pd.DataFrame(ridedata)
+
+# Features and targets
+X = ridedata.loc[:, ridedata.columns != 'Ride'].values
+y = ridedata.loc[:, ridedata.columns == 'Ride'].values
+
+# Create the encoder.
+encoder = OneHotEncoder(handle_unknown="ignore")
+# Assume for simplicity all features are categorical.
+encoder.fit(X)    
+# Apply the encoder.
+X = encoder.transform(X)
+print(X)
+# Then do a Classification tree
+tree_clf = DecisionTreeClassifier(max_depth=2)
+tree_clf.fit(X, y)
+print("Train set accuracy with Decision Tree: {:.2f}".format(tree_clf.score(X,y)))
+#transfer to a decision tree graph
+export_graphviz(
+    tree_clf,
+    out_file="DataFiles/ride.dot",
+    rounded=True,
+    filled=True
+)
+cmd = 'dot -Tpng DataFiles/cancer.dot -o DataFiles/cancer.png'
+os.system(cmd)
    @@ -440,7 +444,7 @@ split = get_split(dataset)
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    • diff --git a/doc/pub/week44/html/._week44-bs046.html b/doc/pub/week44/html/._week44-bs046.html index 267575516..4ac99e78c 100644 --- a/doc/pub/week44/html/._week44-bs046.html +++ b/doc/pub/week44/html/._week44-bs046.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,37 +327,98 @@ MathJax.Hub.Config({

     

     

     

    -

    Entropy and the ID3 algorithm

    +

    Computing the Gini Factor

    -

    The ID3 algorithm learns decision trees by constructing -them in a top down way, beginning with the question which attribute should be tested at the root of the tree? +

    The above functions (gini, entropy and misclassification error) are +important components of the so-called CART algorithm. We will discuss +this algorithm below after we have discussed the information gain +algorithm ID3.

    -
      -
    1. Each instance attribute is evaluated using a statistical test to determine how well it alone classifies the training examples.
      2. -
    2. The best attribute is selected and used as the test at the root node of the tree.
      3. -
    3. A descendant of the root node is then created for each possible value of this attribute.
      4. -
    4. Training examples are sorted to the appropriate descendant node.
      5. -
    5. The entire process is then repeated using the training examples associated with each descendant node to select the best attribute to test at that point in the tree.
      6. -
    6. This forms a greedy search for an acceptable decision tree, in which the algorithm never backtracks to reconsider earlier choices.
      7. -
    -

    The ID3 algorithm selects which attribute to test at each node in the -tree. -

    +

    In the example here we have converted all our attributes into numerical values \( 0,1,2 \) etc.

    -

    We would like to select the attribute that is most useful for classifying -examples. -

    -

    What is a good quantitative measure of the worth of an attribute?

    + +
    +
    +
    +
    +
    + 
    # Split a dataset based on an attribute and an attribute value
+def test_split(index, value, dataset):
+	left, right = list(), list()
+	for row in dataset:
+		if row[index] < value:
+			left.append(row)
+		else:
+			right.append(row)
+	return left, right
+ 
+# Calculate the Gini index for a split dataset
+def gini_index(groups, classes):
+	# count all samples at split point
+	n_instances = float(sum([len(group) for group in groups]))
+	# sum weighted Gini index for each group
+	gini = 0.0
+	for group in groups:
+		size = float(len(group))
+		# avoid divide by zero
+		if size == 0:
+			continue
+		score = 0.0
+		# score the group based on the score for each class
+		for class_val in classes:
+			p = [row[-1] for row in group].count(class_val) / size
+			score += p * p
+		# weight the group score by its relative size
+		gini += (1.0 - score) * (size / n_instances)
+	return gini
 
-
    Information gain measures how well a given attribute separates the
-training examples according to their target classification.
-
    
+# Select the best split point for a dataset
+def get_split(dataset):
+	class_values = list(set(row[-1] for row in dataset))
+	b_index, b_value, b_score, b_groups = 999, 999, 999, None
+	for index in range(len(dataset[0])-1):
+		for row in dataset:
+			groups = test_split(index, row[index], dataset)
+			gini = gini_index(groups, class_values)
+			print('X%d < %.3f Gini=%.3f' % ((index+1), row[index], gini))
+			if gini < b_score:
+				b_index, b_value, b_score, b_groups = index, row[index], gini, groups
+	return {'index':b_index, 'value':b_value, 'groups':b_groups}
+ 
+dataset = [[0,0,0,0,0],
+            [0,0,0,1,1],
+            [1,0,0,0,1],
+            [2,1,0,0,1],
+            [2,2,1,0,1],
+            [2,2,1,1,0],
+            [1,2,1,1,1],
+            [0,1,0,0,0],
+            [0,2,1,0,1],
+            [2,1,1,0,1],
+            [0,1,1,1,1],
+            [1,1,0,1,1],
+            [1,0,1,0,1],
+            [2,1,0,1,0]]
+
+split = get_split(dataset)
+print('Split: [X%d < %.3f]' % ((split['index']+1), split['value']))
+
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    -

    The ID3 algorithm uses this information gain measure to select among the candidate -attributes at each step while growing the tree. -

    @@ -379,7 +445,7 @@ attributes at each step while growing the tree.

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    • diff --git a/doc/pub/week44/html/._week44-bs047.html b/doc/pub/week44/html/._week44-bs047.html index c28229dc5..4144ab8ba 100644 --- a/doc/pub/week44/html/._week44-bs047.html +++ b/doc/pub/week44/html/._week44-bs047.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,70 +327,37 @@ MathJax.Hub.Config({

     

     

     

    -

    Cancer Data again now with Decision Trees and other Methods

    +

    Entropy and the ID3 algorithm

    - -
    -
    -
    -
    -
    - 
    import matplotlib.pyplot as plt
-import numpy as np
-from sklearn.model_selection import  train_test_split 
-from sklearn.datasets import load_breast_cancer
-from sklearn.svm import SVC
-from sklearn.linear_model import LogisticRegression
-from sklearn.tree import DecisionTreeClassifier
+
    The ID3 algorithm  learns decision trees by constructing
+them in a top down way, beginning with the question which attribute should be tested at the root of the tree?
+
    
 
-# Load the data
-cancer = load_breast_cancer()
+
      
+
    1.  Each instance attribute is evaluated using a statistical test to determine how well it alone classifies the training examples.
      2. 
+
    2.  The best attribute is selected and used as the test at the root node of the tree.
      3. 
+
    3.  A descendant of the root node is then created for each possible value of this attribute.
      4. 
+
    4.  Training examples are sorted to the appropriate descendant node.
      5. 
+
    5.  The entire process is then repeated using the training examples associated with each descendant node to select the best attribute to test at that point in the tree.
      6. 
+
    6.  This forms a greedy search for an acceptable decision tree, in which the algorithm never backtracks to reconsider earlier choices.
      7.  
+
    
+
    The ID3 algorithm selects which attribute to test at each node in the
+tree.
+
    
 
-X_train, X_test, y_train, y_test = train_test_split(cancer.data,cancer.target,random_state=0)
-print(X_train.shape)
-print(X_test.shape)
-# Logistic Regression
-logreg = LogisticRegression(solver='lbfgs')
-logreg.fit(X_train, y_train)
-print("Test set accuracy with Logistic Regression: {:.2f}".format(logreg.score(X_test,y_test)))
-# Support vector machine
-svm = SVC(gamma='auto', C=100)
-svm.fit(X_train, y_train)
-print("Test set accuracy with SVM: {:.2f}".format(svm.score(X_test,y_test)))
-# Decision Trees
-deep_tree_clf = DecisionTreeClassifier(max_depth=None)
-deep_tree_clf.fit(X_train, y_train)
-print("Test set accuracy with Decision Trees: {:.2f}".format(deep_tree_clf.score(X_test,y_test)))
-#now scale the data
-from sklearn.preprocessing import StandardScaler
-scaler = StandardScaler()
-scaler.fit(X_train)
-X_train_scaled = scaler.transform(X_train)
-X_test_scaled = scaler.transform(X_test)
-# Logistic Regression
-logreg.fit(X_train_scaled, y_train)
-print("Test set accuracy Logistic Regression with scaled data: {:.2f}".format(logreg.score(X_test_scaled,y_test)))
-# Support Vector Machine
-svm.fit(X_train_scaled, y_train)
-print("Test set accuracy SVM with scaled data: {:.2f}".format(logreg.score(X_test_scaled,y_test)))
-# Decision Trees
-deep_tree_clf.fit(X_train_scaled, y_train)
-print("Test set accuracy with Decision Trees and scaled data: {:.2f}".format(deep_tree_clf.score(X_test_scaled,y_test)))
-
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    +

    We would like to select the attribute that is most useful for classifying +examples. +

    +

    What is a good quantitative measure of the worth of an attribute?

    + +

    Information gain measures how well a given attribute separates the +training examples according to their target classification. +

    + +

    The ID3 algorithm uses this information gain measure to select among the candidate +attributes at each step while growing the tree. +

    @@ -412,7 +384,7 @@ deep_tree_clf.fit(X_train_scaled, y_train)

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    • diff --git a/doc/pub/week44/html/._week44-bs048.html b/doc/pub/week44/html/._week44-bs048.html index 447786f32..a12212ce1 100644 --- a/doc/pub/week44/html/._week44-bs048.html +++ b/doc/pub/week44/html/._week44-bs048.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,7 +327,7 @@ MathJax.Hub.Config({

     

     

     

    -

    Another example, the moons again

    +

    Cancer Data again now with Decision Trees and other Methods

    @@ -330,70 +335,47 @@ MathJax.Hub.Config({
    - 
    from __future__ import division, print_function, unicode_literals
-
-# Common imports
+  
    import matplotlib.pyplot as plt
 import numpy as np
-import os
-
-# to make this notebook's output stable across runs
-np.random.seed(42)
-
-# To plot pretty figures
-import matplotlib
-import matplotlib.pyplot as plt
-from matplotlib.colors import ListedColormap
-plt.rcParams['axes.labelsize'] = 14
-plt.rcParams['xtick.labelsize'] = 12
-plt.rcParams['ytick.labelsize'] = 12
-
-
+from sklearn.model_selection import  train_test_split 
+from sklearn.datasets import load_breast_cancer
 from sklearn.svm import SVC
-from sklearn import datasets
+from sklearn.linear_model import LogisticRegression
 from sklearn.tree import DecisionTreeClassifier
-from sklearn.datasets import make_moons
-from sklearn.tree import export_graphviz
 
-Xm, ym = make_moons(n_samples=100, noise=0.25, random_state=53)
+# Load the data
+cancer = load_breast_cancer()
 
-deep_tree_clf1 = DecisionTreeClassifier(random_state=42)
-deep_tree_clf2 = DecisionTreeClassifier(min_samples_leaf=4, random_state=42)
-deep_tree_clf1.fit(Xm, ym)
-deep_tree_clf2.fit(Xm, ym)
-
-
-def plot_decision_boundary(clf, X, y, axes=[0, 7.5, 0, 3], iris=True, legend=False, plot_training=True):
-    x1s = np.linspace(axes[0], axes[1], 100)
-    x2s = np.linspace(axes[2], axes[3], 100)
-    x1, x2 = np.meshgrid(x1s, x2s)
-    X_new = np.c_[x1.ravel(), x2.ravel()]
-    y_pred = clf.predict(X_new).reshape(x1.shape)
-    custom_cmap = ListedColormap(['#fafab0','#9898ff','#a0faa0'])
-    plt.contourf(x1, x2, y_pred, alpha=0.3, cmap=custom_cmap)
-    if not iris:
-        custom_cmap2 = ListedColormap(['#7d7d58','#4c4c7f','#507d50'])
-        plt.contour(x1, x2, y_pred, cmap=custom_cmap2, alpha=0.8)
-    if plot_training:
-        plt.plot(X[:, 0][y==0], X[:, 1][y==0], "yo", label="Iris-Setosa")
-        plt.plot(X[:, 0][y==1], X[:, 1][y==1], "bs", label="Iris-Versicolor")
-        plt.plot(X[:, 0][y==2], X[:, 1][y==2], "g^", label="Iris-Virginica")
-        plt.axis(axes)
-    if iris:
-        plt.xlabel("Petal length", fontsize=14)
-        plt.ylabel("Petal width", fontsize=14)
-    else:
-        plt.xlabel(r"$x_1$", fontsize=18)
-        plt.ylabel(r"$x_2$", fontsize=18, rotation=0)
-    if legend:
-        plt.legend(loc="lower right", fontsize=14)
-plt.figure(figsize=(11, 4))
-plt.subplot(121)
-plot_decision_boundary(deep_tree_clf1, Xm, ym, axes=[-1.5, 2.5, -1, 1.5], iris=False)
-plt.title("No restrictions", fontsize=16)
-plt.subplot(122)
-plot_decision_boundary(deep_tree_clf2, Xm, ym, axes=[-1.5, 2.5, -1, 1.5], iris=False)
-plt.title("min_samples_leaf = {}".format(deep_tree_clf2.min_samples_leaf), fontsize=14)
-plt.show()
+X_train, X_test, y_train, y_test = train_test_split(cancer.data,cancer.target,random_state=0)
+print(X_train.shape)
+print(X_test.shape)
+# Logistic Regression
+logreg = LogisticRegression(solver='lbfgs')
+logreg.fit(X_train, y_train)
+print("Test set accuracy with Logistic Regression: {:.2f}".format(logreg.score(X_test,y_test)))
+# Support vector machine
+svm = SVC(gamma='auto', C=100)
+svm.fit(X_train, y_train)
+print("Test set accuracy with SVM: {:.2f}".format(svm.score(X_test,y_test)))
+# Decision Trees
+deep_tree_clf = DecisionTreeClassifier(max_depth=None)
+deep_tree_clf.fit(X_train, y_train)
+print("Test set accuracy with Decision Trees: {:.2f}".format(deep_tree_clf.score(X_test,y_test)))
+#now scale the data
+from sklearn.preprocessing import StandardScaler
+scaler = StandardScaler()
+scaler.fit(X_train)
+X_train_scaled = scaler.transform(X_train)
+X_test_scaled = scaler.transform(X_test)
+# Logistic Regression
+logreg.fit(X_train_scaled, y_train)
+print("Test set accuracy Logistic Regression with scaled data: {:.2f}".format(logreg.score(X_test_scaled,y_test)))
+# Support Vector Machine
+svm.fit(X_train_scaled, y_train)
+print("Test set accuracy SVM with scaled data: {:.2f}".format(logreg.score(X_test_scaled,y_test)))
+# Decision Trees
+deep_tree_clf.fit(X_train_scaled, y_train)
+print("Test set accuracy with Decision Trees and scaled data: {:.2f}".format(deep_tree_clf.score(X_test_scaled,y_test)))
    @@ -435,7 +417,7 @@ plt.show()
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    • diff --git a/doc/pub/week44/html/._week44-bs049.html b/doc/pub/week44/html/._week44-bs049.html index c0813f521..7d1bc14b5 100644 --- a/doc/pub/week44/html/._week44-bs049.html +++ b/doc/pub/week44/html/._week44-bs049.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,7 +327,7 @@ MathJax.Hub.Config({

     

     

     

    -

    Playing around with regions

    +

    Another example, the moons again

    @@ -330,25 +335,69 @@ MathJax.Hub.Config({
    - 
    np.random.seed(6)
-Xs = np.random.rand(100, 2) - 0.5
-ys = (Xs[:, 0] > 0).astype(np.float32) * 2
+  
    from __future__ import division, print_function, unicode_literals
 
-angle = np.pi/4
-rotation_matrix = np.array([[np.cos(angle), -np.sin(angle)], [np.sin(angle), np.cos(angle)]])
-Xsr = Xs.dot(rotation_matrix)
+# Common imports
+import numpy as np
+import os
 
-tree_clf_s = DecisionTreeClassifier(random_state=42)
-tree_clf_s.fit(Xs, ys)
-tree_clf_sr = DecisionTreeClassifier(random_state=42)
-tree_clf_sr.fit(Xsr, ys)
+# to make this notebook's output stable across runs
+np.random.seed(42)
 
+# To plot pretty figures
+import matplotlib
+import matplotlib.pyplot as plt
+from matplotlib.colors import ListedColormap
+plt.rcParams['axes.labelsize'] = 14
+plt.rcParams['xtick.labelsize'] = 12
+plt.rcParams['ytick.labelsize'] = 12
+
+
+from sklearn.svm import SVC
+from sklearn import datasets
+from sklearn.tree import DecisionTreeClassifier
+from sklearn.datasets import make_moons
+from sklearn.tree import export_graphviz
+
+Xm, ym = make_moons(n_samples=100, noise=0.25, random_state=53)
+
+deep_tree_clf1 = DecisionTreeClassifier(random_state=42)
+deep_tree_clf2 = DecisionTreeClassifier(min_samples_leaf=4, random_state=42)
+deep_tree_clf1.fit(Xm, ym)
+deep_tree_clf2.fit(Xm, ym)
+
+
+def plot_decision_boundary(clf, X, y, axes=[0, 7.5, 0, 3], iris=True, legend=False, plot_training=True):
+    x1s = np.linspace(axes[0], axes[1], 100)
+    x2s = np.linspace(axes[2], axes[3], 100)
+    x1, x2 = np.meshgrid(x1s, x2s)
+    X_new = np.c_[x1.ravel(), x2.ravel()]
+    y_pred = clf.predict(X_new).reshape(x1.shape)
+    custom_cmap = ListedColormap(['#fafab0','#9898ff','#a0faa0'])
+    plt.contourf(x1, x2, y_pred, alpha=0.3, cmap=custom_cmap)
+    if not iris:
+        custom_cmap2 = ListedColormap(['#7d7d58','#4c4c7f','#507d50'])
+        plt.contour(x1, x2, y_pred, cmap=custom_cmap2, alpha=0.8)
+    if plot_training:
+        plt.plot(X[:, 0][y==0], X[:, 1][y==0], "yo", label="Iris-Setosa")
+        plt.plot(X[:, 0][y==1], X[:, 1][y==1], "bs", label="Iris-Versicolor")
+        plt.plot(X[:, 0][y==2], X[:, 1][y==2], "g^", label="Iris-Virginica")
+        plt.axis(axes)
+    if iris:
+        plt.xlabel("Petal length", fontsize=14)
+        plt.ylabel("Petal width", fontsize=14)
+    else:
+        plt.xlabel(r"$x_1$", fontsize=18)
+        plt.ylabel(r"$x_2$", fontsize=18, rotation=0)
+    if legend:
+        plt.legend(loc="lower right", fontsize=14)
 plt.figure(figsize=(11, 4))
 plt.subplot(121)
-plot_decision_boundary(tree_clf_s, Xs, ys, axes=[-0.7, 0.7, -0.7, 0.7], iris=False)
+plot_decision_boundary(deep_tree_clf1, Xm, ym, axes=[-1.5, 2.5, -1, 1.5], iris=False)
+plt.title("No restrictions", fontsize=16)
 plt.subplot(122)
-plot_decision_boundary(tree_clf_sr, Xsr, ys, axes=[-0.7, 0.7, -0.7, 0.7], iris=False)
-
+plot_decision_boundary(deep_tree_clf2, Xm, ym, axes=[-1.5, 2.5, -1, 1.5], iris=False)
+plt.title("min_samples_leaf = {}".format(deep_tree_clf2.min_samples_leaf), fontsize=14)
 plt.show()
    @@ -391,7 +440,7 @@ plt.show()
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    • diff --git a/doc/pub/week44/html/._week44-bs050.html b/doc/pub/week44/html/._week44-bs050.html index 66e562fda..9818a8f96 100644 --- a/doc/pub/week44/html/._week44-bs050.html +++ b/doc/pub/week44/html/._week44-bs050.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,7 +327,7 @@ MathJax.Hub.Config({

     

     

     

    -

    Regression trees

    +

    Playing around with regions

    @@ -330,35 +335,26 @@ MathJax.Hub.Config({
    - 
    # Quadratic training set + noise
-np.random.seed(42)
-m = 200
-X = np.random.rand(m, 1)
-y = 4 * (X - 0.5) ** 2
-y = y + np.random.randn(m, 1) / 10
-
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    - -
    -
    -
    -
    -
    - 
    from sklearn.tree import DecisionTreeRegressor
+  
    np.random.seed(6)
+Xs = np.random.rand(100, 2) - 0.5
+ys = (Xs[:, 0] > 0).astype(np.float32) * 2
 
-tree_reg = DecisionTreeRegressor(max_depth=2, random_state=42)
-tree_reg.fit(X, y)
+angle = np.pi/4
+rotation_matrix = np.array([[np.cos(angle), -np.sin(angle)], [np.sin(angle), np.cos(angle)]])
+Xsr = Xs.dot(rotation_matrix)
+
+tree_clf_s = DecisionTreeClassifier(random_state=42)
+tree_clf_s.fit(Xs, ys)
+tree_clf_sr = DecisionTreeClassifier(random_state=42)
+tree_clf_sr.fit(Xsr, ys)
+
+plt.figure(figsize=(11, 4))
+plt.subplot(121)
+plot_decision_boundary(tree_clf_s, Xs, ys, axes=[-0.7, 0.7, -0.7, 0.7], iris=False)
+plt.subplot(122)
+plot_decision_boundary(tree_clf_sr, Xsr, ys, axes=[-0.7, 0.7, -0.7, 0.7], iris=False)
+
+plt.show()
    @@ -400,7 +396,7 @@ tree_reg.fit(X, y)
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    • diff --git a/doc/pub/week44/html/._week44-bs051.html b/doc/pub/week44/html/._week44-bs051.html index eff7106cf..408a13449 100644 --- a/doc/pub/week44/html/._week44-bs051.html +++ b/doc/pub/week44/html/._week44-bs051.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,7 +327,7 @@ MathJax.Hub.Config({

     

     

     

    -

    Final regressor code

    +

    Regression trees

    @@ -330,44 +335,12 @@ MathJax.Hub.Config({
    - 
    from sklearn.tree import DecisionTreeRegressor
-
-tree_reg1 = DecisionTreeRegressor(random_state=42, max_depth=2)
-tree_reg2 = DecisionTreeRegressor(random_state=42, max_depth=3)
-tree_reg1.fit(X, y)
-tree_reg2.fit(X, y)
-
-def plot_regression_predictions(tree_reg, X, y, axes=[0, 1, -0.2, 1], ylabel="$y$"):
-    x1 = np.linspace(axes[0], axes[1], 500).reshape(-1, 1)
-    y_pred = tree_reg.predict(x1)
-    plt.axis(axes)
-    plt.xlabel("$x_1$", fontsize=18)
-    if ylabel:
-        plt.ylabel(ylabel, fontsize=18, rotation=0)
-    plt.plot(X, y, "b.")
-    plt.plot(x1, y_pred, "r.-", linewidth=2, label=r"$\hat{y}$")
-
-plt.figure(figsize=(11, 4))
-plt.subplot(121)
-plot_regression_predictions(tree_reg1, X, y)
-for split, style in ((0.1973, "k-"), (0.0917, "k--"), (0.7718, "k--")):
-    plt.plot([split, split], [-0.2, 1], style, linewidth=2)
-plt.text(0.21, 0.65, "Depth=0", fontsize=15)
-plt.text(0.01, 0.2, "Depth=1", fontsize=13)
-plt.text(0.65, 0.8, "Depth=1", fontsize=13)
-plt.legend(loc="upper center", fontsize=18)
-plt.title("max_depth=2", fontsize=14)
-
-plt.subplot(122)
-plot_regression_predictions(tree_reg2, X, y, ylabel=None)
-for split, style in ((0.1973, "k-"), (0.0917, "k--"), (0.7718, "k--")):
-    plt.plot([split, split], [-0.2, 1], style, linewidth=2)
-for split in (0.0458, 0.1298, 0.2873, 0.9040):
-    plt.plot([split, split], [-0.2, 1], "k:", linewidth=1)
-plt.text(0.3, 0.5, "Depth=2", fontsize=13)
-plt.title("max_depth=3", fontsize=14)
-
-plt.show()
+  
    # Quadratic training set + noise
+np.random.seed(42)
+m = 200
+X = np.random.rand(m, 1)
+y = 4 * (X - 0.5) ** 2
+y = y + np.random.randn(m, 1) / 10
    @@ -387,34 +360,10 @@ plt.show()
    - 
    tree_reg1 = DecisionTreeRegressor(random_state=42)
-tree_reg2 = DecisionTreeRegressor(random_state=42, min_samples_leaf=10)
-tree_reg1.fit(X, y)
-tree_reg2.fit(X, y)
+  
    from sklearn.tree import DecisionTreeRegressor
 
-x1 = np.linspace(0, 1, 500).reshape(-1, 1)
-y_pred1 = tree_reg1.predict(x1)
-y_pred2 = tree_reg2.predict(x1)
-
-plt.figure(figsize=(11, 4))
-
-plt.subplot(121)
-plt.plot(X, y, "b.")
-plt.plot(x1, y_pred1, "r.-", linewidth=2, label=r"$\hat{y}$")
-plt.axis([0, 1, -0.2, 1.1])
-plt.xlabel("$x_1$", fontsize=18)
-plt.ylabel("$y$", fontsize=18, rotation=0)
-plt.legend(loc="upper center", fontsize=18)
-plt.title("No restrictions", fontsize=14)
-
-plt.subplot(122)
-plt.plot(X, y, "b.")
-plt.plot(x1, y_pred2, "r.-", linewidth=2, label=r"$\hat{y}$")
-plt.axis([0, 1, -0.2, 1.1])
-plt.xlabel("$x_1$", fontsize=18)
-plt.title("min_samples_leaf={}".format(tree_reg2.min_samples_leaf), fontsize=14)
-
-plt.show()
+tree_reg = DecisionTreeRegressor(max_depth=2, random_state=42)
+tree_reg.fit(X, y)
    @@ -456,7 +405,7 @@ plt.show()
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    • diff --git a/doc/pub/week44/html/._week44-bs052.html b/doc/pub/week44/html/._week44-bs052.html index b73e7e220..b4bb39471 100644 --- a/doc/pub/week44/html/._week44-bs052.html +++ b/doc/pub/week44/html/._week44-bs052.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,17 +327,115 @@ MathJax.Hub.Config({

     

     

     

    -

    Pros and cons of trees, pros

    +

    Final regressor code

    + + +
    +
    +
    +
    +
    + 
    from sklearn.tree import DecisionTreeRegressor
+
+tree_reg1 = DecisionTreeRegressor(random_state=42, max_depth=2)
+tree_reg2 = DecisionTreeRegressor(random_state=42, max_depth=3)
+tree_reg1.fit(X, y)
+tree_reg2.fit(X, y)
+
+def plot_regression_predictions(tree_reg, X, y, axes=[0, 1, -0.2, 1], ylabel="$y$"):
+    x1 = np.linspace(axes[0], axes[1], 500).reshape(-1, 1)
+    y_pred = tree_reg.predict(x1)
+    plt.axis(axes)
+    plt.xlabel("$x_1$", fontsize=18)
+    if ylabel:
+        plt.ylabel(ylabel, fontsize=18, rotation=0)
+    plt.plot(X, y, "b.")
+    plt.plot(x1, y_pred, "r.-", linewidth=2, label=r"$\hat{y}$")
+
+plt.figure(figsize=(11, 4))
+plt.subplot(121)
+plot_regression_predictions(tree_reg1, X, y)
+for split, style in ((0.1973, "k-"), (0.0917, "k--"), (0.7718, "k--")):
+    plt.plot([split, split], [-0.2, 1], style, linewidth=2)
+plt.text(0.21, 0.65, "Depth=0", fontsize=15)
+plt.text(0.01, 0.2, "Depth=1", fontsize=13)
+plt.text(0.65, 0.8, "Depth=1", fontsize=13)
+plt.legend(loc="upper center", fontsize=18)
+plt.title("max_depth=2", fontsize=14)
+
+plt.subplot(122)
+plot_regression_predictions(tree_reg2, X, y, ylabel=None)
+for split, style in ((0.1973, "k-"), (0.0917, "k--"), (0.7718, "k--")):
+    plt.plot([split, split], [-0.2, 1], style, linewidth=2)
+for split in (0.0458, 0.1298, 0.2873, 0.9040):
+    plt.plot([split, split], [-0.2, 1], "k:", linewidth=1)
+plt.text(0.3, 0.5, "Depth=2", fontsize=13)
+plt.title("max_depth=3", fontsize=14)
+
+plt.show()
+
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    + +
    +
    +
    +
    +
    + 
    tree_reg1 = DecisionTreeRegressor(random_state=42)
+tree_reg2 = DecisionTreeRegressor(random_state=42, min_samples_leaf=10)
+tree_reg1.fit(X, y)
+tree_reg2.fit(X, y)
+
+x1 = np.linspace(0, 1, 500).reshape(-1, 1)
+y_pred1 = tree_reg1.predict(x1)
+y_pred2 = tree_reg2.predict(x1)
+
+plt.figure(figsize=(11, 4))
+
+plt.subplot(121)
+plt.plot(X, y, "b.")
+plt.plot(x1, y_pred1, "r.-", linewidth=2, label=r"$\hat{y}$")
+plt.axis([0, 1, -0.2, 1.1])
+plt.xlabel("$x_1$", fontsize=18)
+plt.ylabel("$y$", fontsize=18, rotation=0)
+plt.legend(loc="upper center", fontsize=18)
+plt.title("No restrictions", fontsize=14)
+
+plt.subplot(122)
+plt.plot(X, y, "b.")
+plt.plot(x1, y_pred2, "r.-", linewidth=2, label=r"$\hat{y}$")
+plt.axis([0, 1, -0.2, 1.1])
+plt.xlabel("$x_1$", fontsize=18)
+plt.title("min_samples_leaf={}".format(tree_reg2.min_samples_leaf), fontsize=14)
+
+plt.show()
+
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    + -
      -
    • White box, easy to interpret model. Some people believe that decision trees more closely mirror human decision-making than do the regression and classification approaches discussed earlier (think of support vector machines)
      • -
    • Trees are very easy to explain to people. In fact, they are even easier to explain than linear regression!
      • -
    • No feature normalization needed
      • -
    • Tree models can handle both continuous and categorical data (Classification and Regression Trees)
      • -
    • Can model nonlinear relationships
      • -
    • Can model interactions between the different descriptive features
      • -
    • Trees can be displayed graphically, and are easily interpreted even by a non-expert (especially if they are small)
      • -

      @@ -358,7 +461,7 @@ MathJax.Hub.Config({
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    • ...
      • -
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      • +
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    • »
    diff --git a/doc/pub/week44/html/._week44-bs053.html b/doc/pub/week44/html/._week44-bs053.html index 8a5cc5c2a..927d92327 100644 --- a/doc/pub/week44/html/._week44-bs053.html +++ b/doc/pub/week44/html/._week44-bs053.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,22 +327,17 @@ MathJax.Hub.Config({

     

     

     

    -

    Disadvantages

    +

    Pros and cons of trees, pros

      -
    • Unfortunately, trees generally do not have the same level of predictive accuracy as some of the other regression and classification approaches
      • -
    • If continuous features are used the tree may become quite large and hence less interpretable
      • -
    • Decision trees are prone to overfit the training data and hence do not well generalize the data if no stopping criteria or improvements like pruning, boosting or bagging are implemented
      • -
    • Small changes in the data may lead to a completely different tree. This issue can be addressed by using ensemble methods like bagging, boosting or random forests
      • -
    • Unbalanced datasets where some target feature values occur much more frequently than others may lead to biased trees since the frequently occurring feature values are preferred over the less frequently occurring ones.
      • -
    • If the number of features is relatively large (high dimensional) and the number of instances is relatively low, the tree might overfit the data
      • -
    • Features with many levels may be preferred over features with less levels since for them it is more easy to split the dataset such that the sub datasets only contain pure target feature values. This issue can be addressed by preferring for instance the information gain ratio as splitting criteria over information gain
      • +
    • White box, easy to interpret model. Some people believe that decision trees more closely mirror human decision-making than do the regression and classification approaches discussed earlier (think of support vector machines)
      • +
    • Trees are very easy to explain to people. In fact, they are even easier to explain than linear regression!
      • +
    • No feature normalization needed
      • +
    • Tree models can handle both continuous and categorical data (Classification and Regression Trees)
      • +
    • Can model nonlinear relationships
      • +
    • Can model interactions between the different descriptive features
      • +
    • Trees can be displayed graphically, and are easily interpreted even by a non-expert (especially if they are small)
    -

    However, by aggregating many decision trees, using methods like -bagging, random forests, and boosting, the predictive performance of -trees can be substantially improved. -

    -

      @@ -362,6 +362,8 @@ trees can be substantially improved.
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      • +
    • ...
      • +
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    • »
    diff --git a/doc/pub/week44/html/._week44-bs054.html b/doc/pub/week44/html/._week44-bs054.html index d700759d6..987c51702 100644 --- a/doc/pub/week44/html/._week44-bs054.html +++ b/doc/pub/week44/html/._week44-bs054.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,29 +327,22 @@ MathJax.Hub.Config({

     

     

     

    -

    Ensemble Methods: From a Single Tree to Many Trees and Extreme Boosting, Meet the Jungle of Methods

    +

    Disadvantages

    -

    As stated above and seen in many of the examples discussed here about -a single decision tree, we often end up overfitting our training -data. This normally means that we have a high variance. Can we reduce -the variance of a statistical learning method? +

      +
    • Unfortunately, trees generally do not have the same level of predictive accuracy as some of the other regression and classification approaches
      • +
    • If continuous features are used the tree may become quite large and hence less interpretable
      • +
    • Decision trees are prone to overfit the training data and hence do not well generalize the data if no stopping criteria or improvements like pruning, boosting or bagging are implemented
      • +
    • Small changes in the data may lead to a completely different tree. This issue can be addressed by using ensemble methods like bagging, boosting or random forests
      • +
    • Unbalanced datasets where some target feature values occur much more frequently than others may lead to biased trees since the frequently occurring feature values are preferred over the less frequently occurring ones.
      • +
    • If the number of features is relatively large (high dimensional) and the number of instances is relatively low, the tree might overfit the data
      • +
    • Features with many levels may be preferred over features with less levels since for them it is more easy to split the dataset such that the sub datasets only contain pure target feature values. This issue can be addressed by preferring for instance the information gain ratio as splitting criteria over information gain
      • +
    +

    However, by aggregating many decision trees, using methods like +bagging, random forests, and boosting, the predictive performance of +trees can be substantially improved.

    -

    This leads us to a set of different methods that can combine different -machine learning algorithms or just use one of them to construct -forests and jungles of trees, homogeneous ones or heterogenous -ones. These methods are recognized by different names which we will -try to explain here. These are -

    - -
      -
    1. Voting classifiers
      2. -
    2. Bagging and Pasting
      3. -
    3. Random forests
      4. -
    4. Boosting methods, from adaptive to Extreme Gradient Boosting (XGBoost)
      5. -
    -

    We discuss these methods here.

    -

      @@ -368,6 +366,7 @@ try to explain here. These are
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    • »
    diff --git a/doc/pub/week44/html/._week44-bs055.html b/doc/pub/week44/html/._week44-bs055.html index 6597f32ab..e2516dd03 100644 --- a/doc/pub/week44/html/._week44-bs055.html +++ b/doc/pub/week44/html/._week44-bs055.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,13 +327,28 @@ MathJax.Hub.Config({

     

     

     

    -

    An Overview of Ensemble Methods

    +

    Ensemble Methods: From a Single Tree to Many Trees and Extreme Boosting, Meet the Jungle of Methods

    -

    -
    -

    -
    -

    +

    As stated above and seen in many of the examples discussed here about +a single decision tree, we often end up overfitting our training +data. This normally means that we have a high variance. Can we reduce +the variance of a statistical learning method? +

    + +

    This leads us to a set of different methods that can combine different +machine learning algorithms or just use one of them to construct +forests and jungles of trees, homogeneous ones or heterogenous +ones. These methods are recognized by different names which we will +try to explain here. These are +

    + +
      +
    1. Voting classifiers
      2. +
    2. Bagging and Pasting
      3. +
    3. Random forests
      4. +
    4. Boosting methods, from adaptive to Extreme Gradient Boosting (XGBoost)
      5. +
    +

    We discuss these methods here.

    @@ -352,6 +372,7 @@ MathJax.Hub.Config({

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  • »
    • diff --git a/doc/pub/week44/html/._week44-bs056.html b/doc/pub/week44/html/._week44-bs056.html index b5e1697f2..27fa50679 100644 --- a/doc/pub/week44/html/._week44-bs056.html +++ b/doc/pub/week44/html/._week44-bs056.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,21 +327,13 @@ MathJax.Hub.Config({

     

     

     

    -

    Bagging

    +

    An Overview of Ensemble Methods

    -

    The plain decision trees suffer from high -variance. This means that if we split the training data into two parts -at random, and fit a decision tree to both halves, the results that we -get could be quite different. In contrast, a procedure with low -variance will yield similar results if applied repeatedly to distinct -data sets; linear regression tends to have low variance, if the ratio -of \( n \) to \( p \) is moderately large. -

    - -

    Bootstrap aggregation, or just bagging, is a -general-purpose procedure for reducing the variance of a statistical -learning method. -

    +

    +
    +

    +
    +

    @@ -359,6 +356,7 @@ learning method.

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  • »
    • diff --git a/doc/pub/week44/html/._week44-bs057.html b/doc/pub/week44/html/._week44-bs057.html index 4becb9352..d8f77547a 100644 --- a/doc/pub/week44/html/._week44-bs057.html +++ b/doc/pub/week44/html/._week44-bs057.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,30 +327,20 @@ MathJax.Hub.Config({

     

     

     

    -

    More bagging

    +

    Bagging

    -

    Bagging typically results in improved accuracy -over prediction using a single tree. Unfortunately, however, it can be -difficult to interpret the resulting model. Recall that one of the -advantages of decision trees is the attractive and easily interpreted -diagram that results. +

    The plain decision trees suffer from high +variance. This means that if we split the training data into two parts +at random, and fit a decision tree to both halves, the results that we +get could be quite different. In contrast, a procedure with low +variance will yield similar results if applied repeatedly to distinct +data sets; linear regression tends to have low variance, if the ratio +of \( n \) to \( p \) is moderately large.

    -

    However, when we bag a large number of trees, it is no longer -possible to represent the resulting statistical learning procedure -using a single tree, and it is no longer clear which variables are -most important to the procedure. Thus, bagging improves prediction -accuracy at the expense of interpretability. Although the collection -of bagged trees is much more difficult to interpret than a single -tree, one can obtain an overall summary of the importance of each -predictor using the MSE (for bagging regression trees) or the Gini -index (for bagging classification trees). In the case of bagging -regression trees, we can record the total amount that the MSE is -decreased due to splits over a given predictor, averaged over all \( B \) possible -trees. A large value indicates an important predictor. Similarly, in -the context of bagging classification trees, we can add up the total -amount that the Gini index is decreased by splits over a given -predictor, averaged over all \( B \) trees. +

    Bootstrap aggregation, or just bagging, is a +general-purpose procedure for reducing the variance of a statistical +learning method.

    @@ -368,6 +363,7 @@ predictor, averaged over all \( B \) trees.

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  • »
    • diff --git a/doc/pub/week44/html/._week44-bs058.html b/doc/pub/week44/html/._week44-bs058.html index f6b8f776c..9d7382b5d 100644 --- a/doc/pub/week44/html/._week44-bs058.html +++ b/doc/pub/week44/html/._week44-bs058.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,42 +327,31 @@ MathJax.Hub.Config({

     

     

     

    -

    Simple Voting Example, head or tail

    +

    More bagging

    - -
    -
    -
    -
    -
    - 
    heads_proba = 0.51
-coin_tosses = (np.random.rand(10000, 10) < heads_proba).astype(np.int32)
-cumulative_heads_ratio = np.cumsum(coin_tosses, axis=0) / np.arange(1, 10001).reshape(-1, 1)
-plt.figure(figsize=(8,3.5))
-plt.plot(cumulative_heads_ratio)
-plt.plot([0, 10000], [0.51, 0.51], "k--", linewidth=2, label="51%")
-plt.plot([0, 10000], [0.5, 0.5], "k-", label="50%")
-plt.xlabel("Number of coin tosses")
-plt.ylabel("Heads ratio")
-plt.legend(loc="lower right")
-plt.axis([0, 10000, 0.42, 0.58])
-save_fig("votingsimple")
-plt.show()
-
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    +

    Bagging typically results in improved accuracy +over prediction using a single tree. Unfortunately, however, it can be +difficult to interpret the resulting model. Recall that one of the +advantages of decision trees is the attractive and easily interpreted +diagram that results. +

    +

    However, when we bag a large number of trees, it is no longer +possible to represent the resulting statistical learning procedure +using a single tree, and it is no longer clear which variables are +most important to the procedure. Thus, bagging improves prediction +accuracy at the expense of interpretability. Although the collection +of bagged trees is much more difficult to interpret than a single +tree, one can obtain an overall summary of the importance of each +predictor using the MSE (for bagging regression trees) or the Gini +index (for bagging classification trees). In the case of bagging +regression trees, we can record the total amount that the MSE is +decreased due to splits over a given predictor, averaged over all \( B \) possible +trees. A large value indicates an important predictor. Similarly, in +the context of bagging classification trees, we can add up the total +amount that the Gini index is decreased by splits over a given +predictor, averaged over all \( B \) trees. +

    @@ -378,6 +372,7 @@ plt.show()

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    • diff --git a/doc/pub/week44/html/._week44-bs059.html b/doc/pub/week44/html/._week44-bs059.html index 8dab5b883..fe5eeac15 100644 --- a/doc/pub/week44/html/._week44-bs059.html +++ b/doc/pub/week44/html/._week44-bs059.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,7 +327,7 @@ MathJax.Hub.Config({

     

     

     

    -

    Using the Voting Classifier

    +

    Simple Voting Example, head or tail

    @@ -330,49 +335,19 @@ MathJax.Hub.Config({
    - 
    from sklearn.model_selection import train_test_split
-from sklearn.datasets import make_moons
-
-X, y = make_moons(n_samples=500, noise=0.30, random_state=42)
-X_train, X_test, y_train, y_test = train_test_split(X, y, random_state=42)
-
-from sklearn.ensemble import RandomForestClassifier
-from sklearn.ensemble import VotingClassifier
-from sklearn.linear_model import LogisticRegression
-from sklearn.svm import SVC
-
-log_clf = LogisticRegression(solver="liblinear", random_state=42)
-rnd_clf = RandomForestClassifier(n_estimators=10, random_state=42)
-svm_clf = SVC(gamma="auto", random_state=42)
-
-voting_clf = VotingClassifier(
-    estimators=[('lr', log_clf), ('rf', rnd_clf), ('svc', svm_clf)],
-    voting='hard')
-
-voting_clf.fit(X_train, y_train)
-
-from sklearn.metrics import accuracy_score
-
-for clf in (log_clf, rnd_clf, svm_clf, voting_clf):
-    clf.fit(X_train, y_train)
-    y_pred = clf.predict(X_test)
-    print(clf.__class__.__name__, accuracy_score(y_test, y_pred))
-
-log_clf = LogisticRegression(solver="liblinear", random_state=42)
-rnd_clf = RandomForestClassifier(n_estimators=10, random_state=42)
-svm_clf = SVC(gamma="auto", probability=True, random_state=42)
-
-voting_clf = VotingClassifier(
-    estimators=[('lr', log_clf), ('rf', rnd_clf), ('svc', svm_clf)],
-    voting='soft')
-voting_clf.fit(X_train, y_train)
-
-from sklearn.metrics import accuracy_score
-
-for clf in (log_clf, rnd_clf, svm_clf, voting_clf):
-    clf.fit(X_train, y_train)
-    y_pred = clf.predict(X_test)
-    print(clf.__class__.__name__, accuracy_score(y_test, y_pred))
+  
    heads_proba = 0.51
+coin_tosses = (np.random.rand(10000, 10) < heads_proba).astype(np.int32)
+cumulative_heads_ratio = np.cumsum(coin_tosses, axis=0) / np.arange(1, 10001).reshape(-1, 1)
+plt.figure(figsize=(8,3.5))
+plt.plot(cumulative_heads_ratio)
+plt.plot([0, 10000], [0.51, 0.51], "k--", linewidth=2, label="51%")
+plt.plot([0, 10000], [0.5, 0.5], "k-", label="50%")
+plt.xlabel("Number of coin tosses")
+plt.ylabel("Heads ratio")
+plt.legend(loc="lower right")
+plt.axis([0, 10000, 0.42, 0.58])
+save_fig("votingsimple")
+plt.show()
    @@ -407,6 +382,7 @@ voting_clf.fit(X_train, y_train)
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    • diff --git a/doc/pub/week44/html/._week44-bs060.html b/doc/pub/week44/html/._week44-bs060.html index ce61f5338..d072ee4a2 100644 --- a/doc/pub/week44/html/._week44-bs060.html +++ b/doc/pub/week44/html/._week44-bs060.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,8 +327,7 @@ MathJax.Hub.Config({

     

     

     

    -

    Please, not the moons again! Voting and Bagging

    - +

    Using the Voting Classifier

    @@ -336,91 +340,39 @@ MathJax.Hub.Config({ X, y = make_moons(n_samples=500, noise=0.30, random_state=42) X_train, X_test, y_train, y_test = train_test_split(X, y, random_state=42) + from sklearn.ensemble import RandomForestClassifier from sklearn.ensemble import VotingClassifier from sklearn.linear_model import LogisticRegression from sklearn.svm import SVC -log_clf = LogisticRegression(random_state=42) -rnd_clf = RandomForestClassifier(random_state=42) -svm_clf = SVC(random_state=42) +log_clf = LogisticRegression(solver="liblinear", random_state=42) +rnd_clf = RandomForestClassifier(n_estimators=10, random_state=42) +svm_clf = SVC(gamma="auto", random_state=42) voting_clf = VotingClassifier( estimators=[('lr', log_clf), ('rf', rnd_clf), ('svc', svm_clf)], voting='hard') + voting_clf.fit(X_train, y_train) - -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    - -
    -
    -
    -
    -
    - 
    from sklearn.metrics import accuracy_score
+
+from sklearn.metrics import accuracy_score
 
 for clf in (log_clf, rnd_clf, svm_clf, voting_clf):
     clf.fit(X_train, y_train)
     y_pred = clf.predict(X_test)
     print(clf.__class__.__name__, accuracy_score(y_test, y_pred))
-
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    - -
    -
    -
    -
    -
    - 
    log_clf = LogisticRegression(random_state=42)
-rnd_clf = RandomForestClassifier(random_state=42)
-svm_clf = SVC(probability=True, random_state=42)
+
+log_clf = LogisticRegression(solver="liblinear", random_state=42)
+rnd_clf = RandomForestClassifier(n_estimators=10, random_state=42)
+svm_clf = SVC(gamma="auto", probability=True, random_state=42)
 
 voting_clf = VotingClassifier(
     estimators=[('lr', log_clf), ('rf', rnd_clf), ('svc', svm_clf)],
     voting='soft')
 voting_clf.fit(X_train, y_train)
-
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    - -
    -
    -
    -
    -
    - 
    from sklearn.metrics import accuracy_score
+
+from sklearn.metrics import accuracy_score
 
 for clf in (log_clf, rnd_clf, svm_clf, voting_clf):
     clf.fit(X_train, y_train)
@@ -459,6 +411,7 @@ voting_clf.fit(X_train, y_train)
   
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    • 
 
 
diff --git a/doc/pub/week44/html/._week44-bs061.html b/doc/pub/week44/html/._week44-bs061.html
index fd91f6a02..32d0f27b4 100644
--- a/doc/pub/week44/html/._week44-bs061.html
+++ b/doc/pub/week44/html/._week44-bs061.html
@@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d
 {'highest level': 2,
  'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'),
               ('Digression First', 2, None, 'digression-first'),
+              ('A short Discussion of Project 2',
+               2,
+               None,
+               'a-short-discussion-of-project-2'),
               ('Thursday, Principal Component Analysis',
                2,
                None,
@@ -251,66 +255,67 @@ MathJax.Hub.Config({
         
       
@@ -322,7 +327,7 @@ MathJax.Hub.Config({
 
     
     
     
     
 
 
-
    Bagging Examples  
    
+
    Please, not the moons again! Voting and Bagging 
    
 
 
 
@@ -331,14 +336,24 @@ MathJax.Hub.Config({
     
    
       
    
         
    
-  
    from sklearn.ensemble import BaggingClassifier
-from sklearn.tree import DecisionTreeClassifier
+  
    from sklearn.model_selection import train_test_split
+from sklearn.datasets import make_moons
 
-bag_clf = BaggingClassifier(
-    DecisionTreeClassifier(random_state=42), n_estimators=500,
-    max_samples=100, bootstrap=True, n_jobs=-1, random_state=42)
-bag_clf.fit(X_train, y_train)
-y_pred = bag_clf.predict(X_test)
+X, y = make_moons(n_samples=500, noise=0.30, random_state=42)
+X_train, X_test, y_train, y_test = train_test_split(X, y, random_state=42)
+from sklearn.ensemble import RandomForestClassifier
+from sklearn.ensemble import VotingClassifier
+from sklearn.linear_model import LogisticRegression
+from sklearn.svm import SVC
+
+log_clf = LogisticRegression(random_state=42)
+rnd_clf = RandomForestClassifier(random_state=42)
+svm_clf = SVC(random_state=42)
+
+voting_clf = VotingClassifier(
+    estimators=[('lr', log_clf), ('rf', rnd_clf), ('svc', svm_clf)],
+    voting='hard')
+voting_clf.fit(X_train, y_train)
 
    
 
    
       
    
@@ -359,76 +374,63 @@ y_pred = bag_clf
         
    
   
    from sklearn.metrics import accuracy_score
-print(accuracy_score(y_test, y_pred))
-
    
-
    
-      
    
-
    -
    -
    -
    -
    -
    -
    -
    -
    -
    - -
    -
    -
    -
    -
    - 
    tree_clf = DecisionTreeClassifier(random_state=42)
-tree_clf.fit(X_train, y_train)
-y_pred_tree = tree_clf.predict(X_test)
-print(accuracy_score(y_test, y_pred_tree))
-
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    -
    - -
    -
    -
    -
    -
    - 
    from matplotlib.colors import ListedColormap
 
-def plot_decision_boundary(clf, X, y, axes=[-1.5, 2.5, -1, 1.5], alpha=0.5, contour=True):
-    x1s = np.linspace(axes[0], axes[1], 100)
-    x2s = np.linspace(axes[2], axes[3], 100)
-    x1, x2 = np.meshgrid(x1s, x2s)
-    X_new = np.c_[x1.ravel(), x2.ravel()]
-    y_pred = clf.predict(X_new).reshape(x1.shape)
-    custom_cmap = ListedColormap(['#fafab0','#9898ff','#a0faa0'])
-    plt.contourf(x1, x2, y_pred, alpha=0.3, cmap=custom_cmap)
-    if contour:
-        custom_cmap2 = ListedColormap(['#7d7d58','#4c4c7f','#507d50'])
-        plt.contour(x1, x2, y_pred, cmap=custom_cmap2, alpha=0.8)
-    plt.plot(X[:, 0][y==0], X[:, 1][y==0], "yo", alpha=alpha)
-    plt.plot(X[:, 0][y==1], X[:, 1][y==1], "bs", alpha=alpha)
-    plt.axis(axes)
-    plt.xlabel(r"$x_1$", fontsize=18)
-    plt.ylabel(r"$x_2$", fontsize=18, rotation=0)
-plt.figure(figsize=(11,4))
-plt.subplot(121)
-plot_decision_boundary(tree_clf, X, y)
-plt.title("Decision Tree", fontsize=14)
-plt.subplot(122)
-plot_decision_boundary(bag_clf, X, y)
-plt.title("Decision Trees with Bagging", fontsize=14)
-save_fig("baggingtree")
-plt.show()
+for clf in (log_clf, rnd_clf, svm_clf, voting_clf):
+    clf.fit(X_train, y_train)
+    y_pred = clf.predict(X_test)
+    print(clf.__class__.__name__, accuracy_score(y_test, y_pred))
+
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    + +
    +
    +
    +
    +
    + 
    log_clf = LogisticRegression(random_state=42)
+rnd_clf = RandomForestClassifier(random_state=42)
+svm_clf = SVC(probability=True, random_state=42)
+
+voting_clf = VotingClassifier(
+    estimators=[('lr', log_clf), ('rf', rnd_clf), ('svc', svm_clf)],
+    voting='soft')
+voting_clf.fit(X_train, y_train)
+
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    + +
    +
    +
    +
    +
    + 
    from sklearn.metrics import accuracy_score
+
+for clf in (log_clf, rnd_clf, svm_clf, voting_clf):
+    clf.fit(X_train, y_train)
+    y_pred = clf.predict(X_test)
+    print(clf.__class__.__name__, accuracy_score(y_test, y_pred))
    @@ -461,6 +463,7 @@ plt.show()
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  • »
    • diff --git a/doc/pub/week44/html/._week44-bs062.html b/doc/pub/week44/html/._week44-bs062.html index bffa54276..371197544 100644 --- a/doc/pub/week44/html/._week44-bs062.html +++ b/doc/pub/week44/html/._week44-bs062.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -322,11 +327,8 @@ MathJax.Hub.Config({

     

     

     

    -

    Making your own Bootstrap: Changing the Level of the Decision Tree

    +

    Bagging Examples

    -

    Let us bring up our good old boostrap example from the linear regression lectures. We change the linerar regression algorithm with -a decision tree wth different depths and perform a bootstrap aggregate (in this case we perform as many bootstraps as data points \( n \)). -

    @@ -334,62 +336,103 @@ a decision tree wth different depths and perform a bootstrap aggregate (in this
    - 
    import matplotlib.pyplot as plt
-import numpy as np
-from sklearn.model_selection import train_test_split
-from sklearn.pipeline import make_pipeline
-from sklearn.utils import resample
-from sklearn.tree import DecisionTreeRegressor
+  
    from sklearn.ensemble import BaggingClassifier
+from sklearn.tree import DecisionTreeClassifier
 
-n = 100
-n_boostraps = 100
-maxdepth = 8
+bag_clf = BaggingClassifier(
+    DecisionTreeClassifier(random_state=42), n_estimators=500,
+    max_samples=100, bootstrap=True, n_jobs=-1, random_state=42)
+bag_clf.fit(X_train, y_train)
+y_pred = bag_clf.predict(X_test)
+
    
+
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    + +
    +
    +
    +
    +
    + 
    from sklearn.metrics import accuracy_score
+print(accuracy_score(y_test, y_pred))
+
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    + +
    +
    +
    +
    +
    + 
    tree_clf = DecisionTreeClassifier(random_state=42)
+tree_clf.fit(X_train, y_train)
+y_pred_tree = tree_clf.predict(X_test)
+print(accuracy_score(y_test, y_pred_tree))
+
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    + +
    +
    +
    +
    +
    + 
    from matplotlib.colors import ListedColormap
 
-# Make data set.
-x = np.linspace(-3, 3, n).reshape(-1, 1)
-y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)+ np.random.normal(0, 0.1, x.shape)
-error = np.zeros(maxdepth)
-bias = np.zeros(maxdepth)
-variance = np.zeros(maxdepth)
-polydegree = np.zeros(maxdepth)
-X_train, X_test, y_train, y_test = train_test_split(x, y, test_size=0.2)
-
-from sklearn.preprocessing import StandardScaler
-scaler = StandardScaler()
-scaler.fit(X_train)
-X_train_scaled = scaler.transform(X_train)
-X_test_scaled = scaler.transform(X_test)
-
-# we produce a simple tree first as benchmark
-simpletree = DecisionTreeRegressor(max_depth=3) 
-simpletree.fit(X_train_scaled, y_train)
-simpleprediction = simpletree.predict(X_test_scaled)
-for degree in range(1,maxdepth):
-    model = DecisionTreeRegressor(max_depth=degree) 
-    y_pred = np.empty((y_test.shape[0], n_boostraps))
-    for i in range(n_boostraps):
-        x_, y_ = resample(X_train_scaled, y_train)
-        model.fit(x_, y_)
-        y_pred[:, i] = model.predict(X_test_scaled)#.ravel()
-
-    polydegree[degree] = degree
-    error[degree] = np.mean( np.mean((y_test - y_pred)**2, axis=1, keepdims=True) )
-    bias[degree] = np.mean( (y_test - np.mean(y_pred, axis=1, keepdims=True))**2 )
-    variance[degree] = np.mean( np.var(y_pred, axis=1, keepdims=True) )
-    print('Polynomial degree:', degree)
-    print('Error:', error[degree])
-    print('Bias^2:', bias[degree])
-    print('Var:', variance[degree])
-    print('{} >= {} + {} = {}'.format(error[degree], bias[degree], variance[degree], bias[degree]+variance[degree]))
- 
-mse_simpletree= np.mean( np.mean((y_test - simpleprediction)**2)
-print(mse_simpletree)
-plt.xlim(1,maxdepth)
-plt.plot(polydegree, error, label='MSE')
-plt.plot(polydegree, bias, label='bias')
-plt.plot(polydegree, variance, label='Variance')
-plt.legend()
-save_fig("baggingboot")
+def plot_decision_boundary(clf, X, y, axes=[-1.5, 2.5, -1, 1.5], alpha=0.5, contour=True):
+    x1s = np.linspace(axes[0], axes[1], 100)
+    x2s = np.linspace(axes[2], axes[3], 100)
+    x1, x2 = np.meshgrid(x1s, x2s)
+    X_new = np.c_[x1.ravel(), x2.ravel()]
+    y_pred = clf.predict(X_new).reshape(x1.shape)
+    custom_cmap = ListedColormap(['#fafab0','#9898ff','#a0faa0'])
+    plt.contourf(x1, x2, y_pred, alpha=0.3, cmap=custom_cmap)
+    if contour:
+        custom_cmap2 = ListedColormap(['#7d7d58','#4c4c7f','#507d50'])
+        plt.contour(x1, x2, y_pred, cmap=custom_cmap2, alpha=0.8)
+    plt.plot(X[:, 0][y==0], X[:, 1][y==0], "yo", alpha=alpha)
+    plt.plot(X[:, 0][y==1], X[:, 1][y==1], "bs", alpha=alpha)
+    plt.axis(axes)
+    plt.xlabel(r"$x_1$", fontsize=18)
+    plt.ylabel(r"$x_2$", fontsize=18, rotation=0)
+plt.figure(figsize=(11,4))
+plt.subplot(121)
+plot_decision_boundary(tree_clf, X, y)
+plt.title("Decision Tree", fontsize=14)
+plt.subplot(122)
+plot_decision_boundary(bag_clf, X, y)
+plt.title("Decision Trees with Bagging", fontsize=14)
+save_fig("baggingtree")
 plt.show()
    @@ -422,6 +465,8 @@ plt.show()
  • 61
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    • +
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    • +
  • »
    • diff --git a/doc/pub/week44/html/week44-bs.html b/doc/pub/week44/html/week44-bs.html index a503062e6..9e9a67569 100644 --- a/doc/pub/week44/html/week44-bs.html +++ b/doc/pub/week44/html/week44-bs.html @@ -38,6 +38,10 @@ doconce format html week44.do.txt --html_style=bootstrap --pygments_html_style=d {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -251,66 +255,67 @@ MathJax.Hub.Config({ @@ -365,7 +370,7 @@ MathJax.Hub.Config({
  • 9
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  • ...
    • -
  • 63
    • +
  • 64
  • »
    • diff --git a/doc/pub/week44/html/week44-reveal.html b/doc/pub/week44/html/week44-reveal.html index 4a3ed90e0..f1682ab6e 100644 --- a/doc/pub/week44/html/week44-reveal.html +++ b/doc/pub/week44/html/week44-reveal.html @@ -236,6 +236,139 @@ accelerate scientific discovery.

    +
    +

    A short Discussion of Project 2

    + +

    For neural networks and regression, should I use a design matrix with information about a polynomial fit or not? +Discuss pros and cons. The example here shows some of these issues. +

    + + + +
    +
    +
    +
    +
    + 
    """
+Code to test Ridge and NNs using Scikit-Learn only
+"""
+
+import numpy as np
+import pandas as pd
+import matplotlib.pyplot as plt
+from sklearn.model_selection import train_test_split
+from sklearn import linear_model
+from sklearn.neural_network import MLPRegressor
+from sklearn.metrics import accuracy_score
+import seaborn as sns
+
+
+def MSE(y_data,y_model):
+    n = np.size(y_model)
+    return np.sum((y_data-y_model)**2)/n
+# A seed just to ensure that the random numbers are the same for every run.
+# Useful for eventual debugging.
+np.random.seed(315)
+
+n = 100
+x = np.random.rand(n)
+y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)
+
+Maxpolydegree = 5
+X = np.zeros((n,Maxpolydegree-1))
+
+for degree in range(1,Maxpolydegree): #No intercept column
+    X[:,degree-1] = x**(degree)
+
+# We split the data in test and training data
+X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
+
+# Decide which values of lambda to use
+
+nlambdas = 10
+lmbd_vals = np.logspace(-4, 0, nlambdas)
+MSERidgePredict = np.zeros(nlambdas)
+for i in range(nlambdas):
+    lmb = lmbd_vals[i]
+    RegRidge = linear_model.Ridge(lmb)
+    RegRidge.fit(X_train,y_train)
+    ypredictRidge = RegRidge.predict(X_test)
+    MSERidgePredict[i] = MSE(y_test,ypredictRidge)
+
+plt.figure()
+plt.plot(np.log10(lmbd_vals), MSERidgePredict, 'g--', label = 'MSE SL Ridge Test')
+plt.xlabel('log10(lambda)')
+plt.ylabel('MSE')
+plt.legend()
+plt.show()
+
+# Neural Network part
+
+n_hidden_neurons = 50
+epochs = 100
+# store models for later use
+eta_vals = np.logspace(-4, 0, 10)
+# store the models for later use
+DNN_scikit = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)
+test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))
+sns.set()
+for i, eta in enumerate(eta_vals):
+    for j, lmbd in enumerate(lmbd_vals):
+        dnn = MLPRegressor(hidden_layer_sizes=(n_hidden_neurons), activation='logistic',
+                            alpha=lmbd, learning_rate_init=eta, max_iter=epochs)
+        dnn.fit(X_train, y_train)
+        ypredictMLP = dnn.predict(X_test)
+        test_accuracy[i][j] = MSE(ypredictMLP, y_test)
+
+fig, ax = plt.subplots(figsize = (10, 10))
+sns.heatmap(test_accuracy, annot=True, ax=ax, cmap="viridis")
+ax.set_title("Training Accuracy")
+ax.set_ylabel("$\eta$")
+ax.set_xlabel("$\lambda$")
+plt.show()
+
+# Now we redefine our design matrix to include only the x-values and try out our NN
+
+X = np.zeros((n,1))
+X[:,0] = x
+
+# We split the data in test and training data again
+X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
+# Repeat the NN calculation
+DNN_scikit = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)
+test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))
+sns.set()
+for i, eta in enumerate(eta_vals):
+    for j, lmbd in enumerate(lmbd_vals):
+        dnn = MLPRegressor(hidden_layer_sizes=(n_hidden_neurons), activation='logistic',
+                            alpha=lmbd, learning_rate_init=eta, max_iter=epochs)
+        dnn.fit(X_train, y_train)
+        ypredictMLP = dnn.predict(X_test)
+        test_accuracy[i][j] = MSE(ypredictMLP, y_test)
+
+fig, ax = plt.subplots(figsize = (10, 10))
+sns.heatmap(test_accuracy, annot=True, ax=ax, cmap="viridis")
+ax.set_title("Training Accuracy")
+ax.set_ylabel("$\eta$")
+ax.set_xlabel("$\lambda$")
+plt.show()
+
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +

    Thursday, Principal Component Analysis

    diff --git a/doc/pub/week44/html/week44-solarized.html b/doc/pub/week44/html/week44-solarized.html index 5576a37cf..372f728d6 100644 --- a/doc/pub/week44/html/week44-solarized.html +++ b/doc/pub/week44/html/week44-solarized.html @@ -65,6 +65,10 @@ div.toc p,a { {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -323,6 +327,139 @@ methods into the real-time experimental data processing loop to accelerate scientific discovery.

    +









    +

    A short Discussion of Project 2

    + +

    For neural networks and regression, should I use a design matrix with information about a polynomial fit or not? +Discuss pros and cons. The example here shows some of these issues. +

    + + + +
    +
    +
    +
    +
    + 
    """
+Code to test Ridge and NNs using Scikit-Learn only
+"""
+
+import numpy as np
+import pandas as pd
+import matplotlib.pyplot as plt
+from sklearn.model_selection import train_test_split
+from sklearn import linear_model
+from sklearn.neural_network import MLPRegressor
+from sklearn.metrics import accuracy_score
+import seaborn as sns
+
+
+def MSE(y_data,y_model):
+    n = np.size(y_model)
+    return np.sum((y_data-y_model)**2)/n
+# A seed just to ensure that the random numbers are the same for every run.
+# Useful for eventual debugging.
+np.random.seed(315)
+
+n = 100
+x = np.random.rand(n)
+y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)
+
+Maxpolydegree = 5
+X = np.zeros((n,Maxpolydegree-1))
+
+for degree in range(1,Maxpolydegree): #No intercept column
+    X[:,degree-1] = x**(degree)
+
+# We split the data in test and training data
+X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
+
+# Decide which values of lambda to use
+
+nlambdas = 10
+lmbd_vals = np.logspace(-4, 0, nlambdas)
+MSERidgePredict = np.zeros(nlambdas)
+for i in range(nlambdas):
+    lmb = lmbd_vals[i]
+    RegRidge = linear_model.Ridge(lmb)
+    RegRidge.fit(X_train,y_train)
+    ypredictRidge = RegRidge.predict(X_test)
+    MSERidgePredict[i] = MSE(y_test,ypredictRidge)
+
+plt.figure()
+plt.plot(np.log10(lmbd_vals), MSERidgePredict, 'g--', label = 'MSE SL Ridge Test')
+plt.xlabel('log10(lambda)')
+plt.ylabel('MSE')
+plt.legend()
+plt.show()
+
+# Neural Network part
+
+n_hidden_neurons = 50
+epochs = 100
+# store models for later use
+eta_vals = np.logspace(-4, 0, 10)
+# store the models for later use
+DNN_scikit = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)
+test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))
+sns.set()
+for i, eta in enumerate(eta_vals):
+    for j, lmbd in enumerate(lmbd_vals):
+        dnn = MLPRegressor(hidden_layer_sizes=(n_hidden_neurons), activation='logistic',
+                            alpha=lmbd, learning_rate_init=eta, max_iter=epochs)
+        dnn.fit(X_train, y_train)
+        ypredictMLP = dnn.predict(X_test)
+        test_accuracy[i][j] = MSE(ypredictMLP, y_test)
+
+fig, ax = plt.subplots(figsize = (10, 10))
+sns.heatmap(test_accuracy, annot=True, ax=ax, cmap="viridis")
+ax.set_title("Training Accuracy")
+ax.set_ylabel("$\eta$")
+ax.set_xlabel("$\lambda$")
+plt.show()
+
+# Now we redefine our design matrix to include only the x-values and try out our NN
+
+X = np.zeros((n,1))
+X[:,0] = x
+
+# We split the data in test and training data again
+X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
+# Repeat the NN calculation
+DNN_scikit = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)
+test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))
+sns.set()
+for i, eta in enumerate(eta_vals):
+    for j, lmbd in enumerate(lmbd_vals):
+        dnn = MLPRegressor(hidden_layer_sizes=(n_hidden_neurons), activation='logistic',
+                            alpha=lmbd, learning_rate_init=eta, max_iter=epochs)
+        dnn.fit(X_train, y_train)
+        ypredictMLP = dnn.predict(X_test)
+        test_accuracy[i][j] = MSE(ypredictMLP, y_test)
+
+fig, ax = plt.subplots(figsize = (10, 10))
+sns.heatmap(test_accuracy, annot=True, ax=ax, cmap="viridis")
+ax.set_title("Training Accuracy")
+ax.set_ylabel("$\eta$")
+ax.set_xlabel("$\lambda$")
+plt.show()
+
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    + +









    Thursday, Principal Component Analysis

    diff --git a/doc/pub/week44/html/week44.html b/doc/pub/week44/html/week44.html index af429774f..1af0a17e0 100644 --- a/doc/pub/week44/html/week44.html +++ b/doc/pub/week44/html/week44.html @@ -142,6 +142,10 @@ div.toc p,a { {'highest level': 2, 'sections': [('Overview of week 44', 2, None, 'overview-of-week-44'), ('Digression First', 2, None, 'digression-first'), + ('A short Discussion of Project 2', + 2, + None, + 'a-short-discussion-of-project-2'), ('Thursday, Principal Component Analysis', 2, None, @@ -400,6 +404,139 @@ methods into the real-time experimental data processing loop to accelerate scientific discovery.

    +









    +

    A short Discussion of Project 2

    + +

    For neural networks and regression, should I use a design matrix with information about a polynomial fit or not? +Discuss pros and cons. The example here shows some of these issues. +

    + + + +
    +
    +
    +
    +
    + 
    """
+Code to test Ridge and NNs using Scikit-Learn only
+"""
+
+import numpy as np
+import pandas as pd
+import matplotlib.pyplot as plt
+from sklearn.model_selection import train_test_split
+from sklearn import linear_model
+from sklearn.neural_network import MLPRegressor
+from sklearn.metrics import accuracy_score
+import seaborn as sns
+
+
+def MSE(y_data,y_model):
+    n = np.size(y_model)
+    return np.sum((y_data-y_model)**2)/n
+# A seed just to ensure that the random numbers are the same for every run.
+# Useful for eventual debugging.
+np.random.seed(315)
+
+n = 100
+x = np.random.rand(n)
+y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)
+
+Maxpolydegree = 5
+X = np.zeros((n,Maxpolydegree-1))
+
+for degree in range(1,Maxpolydegree): #No intercept column
+    X[:,degree-1] = x**(degree)
+
+# We split the data in test and training data
+X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
+
+# Decide which values of lambda to use
+
+nlambdas = 10
+lmbd_vals = np.logspace(-4, 0, nlambdas)
+MSERidgePredict = np.zeros(nlambdas)
+for i in range(nlambdas):
+    lmb = lmbd_vals[i]
+    RegRidge = linear_model.Ridge(lmb)
+    RegRidge.fit(X_train,y_train)
+    ypredictRidge = RegRidge.predict(X_test)
+    MSERidgePredict[i] = MSE(y_test,ypredictRidge)
+
+plt.figure()
+plt.plot(np.log10(lmbd_vals), MSERidgePredict, 'g--', label = 'MSE SL Ridge Test')
+plt.xlabel('log10(lambda)')
+plt.ylabel('MSE')
+plt.legend()
+plt.show()
+
+# Neural Network part
+
+n_hidden_neurons = 50
+epochs = 100
+# store models for later use
+eta_vals = np.logspace(-4, 0, 10)
+# store the models for later use
+DNN_scikit = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)
+test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))
+sns.set()
+for i, eta in enumerate(eta_vals):
+    for j, lmbd in enumerate(lmbd_vals):
+        dnn = MLPRegressor(hidden_layer_sizes=(n_hidden_neurons), activation='logistic',
+                            alpha=lmbd, learning_rate_init=eta, max_iter=epochs)
+        dnn.fit(X_train, y_train)
+        ypredictMLP = dnn.predict(X_test)
+        test_accuracy[i][j] = MSE(ypredictMLP, y_test)
+
+fig, ax = plt.subplots(figsize = (10, 10))
+sns.heatmap(test_accuracy, annot=True, ax=ax, cmap="viridis")
+ax.set_title("Training Accuracy")
+ax.set_ylabel("$\eta$")
+ax.set_xlabel("$\lambda$")
+plt.show()
+
+# Now we redefine our design matrix to include only the x-values and try out our NN
+
+X = np.zeros((n,1))
+X[:,0] = x
+
+# We split the data in test and training data again
+X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
+# Repeat the NN calculation
+DNN_scikit = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)
+test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))
+sns.set()
+for i, eta in enumerate(eta_vals):
+    for j, lmbd in enumerate(lmbd_vals):
+        dnn = MLPRegressor(hidden_layer_sizes=(n_hidden_neurons), activation='logistic',
+                            alpha=lmbd, learning_rate_init=eta, max_iter=epochs)
+        dnn.fit(X_train, y_train)
+        ypredictMLP = dnn.predict(X_test)
+        test_accuracy[i][j] = MSE(ypredictMLP, y_test)
+
+fig, ax = plt.subplots(figsize = (10, 10))
+sns.heatmap(test_accuracy, annot=True, ax=ax, cmap="viridis")
+ax.set_title("Training Accuracy")
+ax.set_ylabel("$\eta$")
+ax.set_xlabel("$\lambda$")
+plt.show()
+
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    +
    + +









    Thursday, Principal Component Analysis

    diff --git a/doc/pub/week44/ipynb/ipynb-week44-src.tar.gz b/doc/pub/week44/ipynb/ipynb-week44-src.tar.gz index f0326d4c216a96bf58ab1a98be2526aa650b0a24..00723e36328fd81e4396afc022787043d199c466 100644 GIT binary patch delta 29 kcmeDFE!h2AkWIdugMoEIBU>vQV=Eg|D;x7xHkQ^}0FEFCYXATM delta 29 kcmeDFE!h2AkWIdugJExXBU>vQV=Eg|D;x7xHkQ^}0GhiA9smFU diff --git a/doc/pub/week44/ipynb/week44.ipynb b/doc/pub/week44/ipynb/week44.ipynb index 148dc25aa..8aeda0d16 100644 --- a/doc/pub/week44/ipynb/week44.ipynb +++ b/doc/pub/week44/ipynb/week44.ipynb @@ -2,7 +2,7 @@ "cells": [ { "cell_type": "markdown", - "id": "cd360e01", + "id": "19d12c31", "metadata": { "editable": true }, @@ -14,7 +14,7 @@ }, { "cell_type": "markdown", - "id": "79e2594a", + "id": "6075cd2c", "metadata": { "editable": true }, @@ -29,7 +29,7 @@ }, { "cell_type": "markdown", - "id": "059c8e58", + "id": "9299817b", "metadata": { "editable": true }, @@ -57,7 +57,7 @@ }, { "cell_type": "markdown", - "id": "5b8367b3", + "id": "3e55fc44", "metadata": { "editable": true }, @@ -75,7 +75,137 @@ }, { "cell_type": "markdown", - "id": "39b2aa0f", + "id": "fba8b629", + "metadata": { + "editable": true + }, + "source": [ + "## A short Discussion of Project 2\n", + "\n", + "For neural networks and regression, should I use a design matrix with information about a polynomial fit or not?\n", + "Discuss pros and cons. The example here shows some of these issues." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "id": "07b52439", + "metadata": { + "collapsed": false, + "editable": true + }, + "outputs": [], + "source": [ + "%matplotlib inline\n", + "\n", + "\"\"\"\n", + "Code to test Ridge and NNs using Scikit-Learn only\n", + "\"\"\"\n", + "\n", + "import numpy as np\n", + "import pandas as pd\n", + "import matplotlib.pyplot as plt\n", + "from sklearn.model_selection import train_test_split\n", + "from sklearn import linear_model\n", + "from sklearn.neural_network import MLPRegressor\n", + "from sklearn.metrics import accuracy_score\n", + "import seaborn as sns\n", + "\n", + "\n", + "def MSE(y_data,y_model):\n", + " n = np.size(y_model)\n", + " return np.sum((y_data-y_model)**2)/n\n", + "# A seed just to ensure that the random numbers are the same for every run.\n", + "# Useful for eventual debugging.\n", + "np.random.seed(315)\n", + "\n", + "n = 100\n", + "x = np.random.rand(n)\n", + "y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)\n", + "\n", + "Maxpolydegree = 5\n", + "X = np.zeros((n,Maxpolydegree-1))\n", + "\n", + "for degree in range(1,Maxpolydegree): #No intercept column\n", + " X[:,degree-1] = x**(degree)\n", + "\n", + "# We split the data in test and training data\n", + "X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)\n", + "\n", + "# Decide which values of lambda to use\n", + "\n", + "nlambdas = 10\n", + "lmbd_vals = np.logspace(-4, 0, nlambdas)\n", + "MSERidgePredict = np.zeros(nlambdas)\n", + "for i in range(nlambdas):\n", + " lmb = lmbd_vals[i]\n", + " RegRidge = linear_model.Ridge(lmb)\n", + " RegRidge.fit(X_train,y_train)\n", + " ypredictRidge = RegRidge.predict(X_test)\n", + " MSERidgePredict[i] = MSE(y_test,ypredictRidge)\n", + "\n", + "plt.figure()\n", + "plt.plot(np.log10(lmbd_vals), MSERidgePredict, 'g--', label = 'MSE SL Ridge Test')\n", + "plt.xlabel('log10(lambda)')\n", + "plt.ylabel('MSE')\n", + "plt.legend()\n", + "plt.show()\n", + "\n", + "# Neural Network part\n", + "\n", + "n_hidden_neurons = 50\n", + "epochs = 100\n", + "# store models for later use\n", + "eta_vals = np.logspace(-4, 0, 10)\n", + "# store the models for later use\n", + "DNN_scikit = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)\n", + "test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))\n", + "sns.set()\n", + "for i, eta in enumerate(eta_vals):\n", + " for j, lmbd in enumerate(lmbd_vals):\n", + " dnn = MLPRegressor(hidden_layer_sizes=(n_hidden_neurons), activation='logistic',\n", + " alpha=lmbd, learning_rate_init=eta, max_iter=epochs)\n", + " dnn.fit(X_train, y_train)\n", + " ypredictMLP = dnn.predict(X_test)\n", + " test_accuracy[i][j] = MSE(ypredictMLP, y_test)\n", + "\n", + "fig, ax = plt.subplots(figsize = (10, 10))\n", + "sns.heatmap(test_accuracy, annot=True, ax=ax, cmap=\"viridis\")\n", + "ax.set_title(\"Training Accuracy\")\n", + "ax.set_ylabel(\"$\\eta$\")\n", + "ax.set_xlabel(\"$\\lambda$\")\n", + "plt.show()\n", + "\n", + "# Now we redefine our design matrix to include only the x-values and try out our NN\n", + "\n", + "X = np.zeros((n,1))\n", + "X[:,0] = x\n", + "\n", + "# We split the data in test and training data again\n", + "X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)\n", + "# Repeat the NN calculation\n", + "DNN_scikit = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)\n", + "test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))\n", + "sns.set()\n", + "for i, eta in enumerate(eta_vals):\n", + " for j, lmbd in enumerate(lmbd_vals):\n", + " dnn = MLPRegressor(hidden_layer_sizes=(n_hidden_neurons), activation='logistic',\n", + " alpha=lmbd, learning_rate_init=eta, max_iter=epochs)\n", + " dnn.fit(X_train, y_train)\n", + " ypredictMLP = dnn.predict(X_test)\n", + " test_accuracy[i][j] = MSE(ypredictMLP, y_test)\n", + "\n", + "fig, ax = plt.subplots(figsize = (10, 10))\n", + "sns.heatmap(test_accuracy, annot=True, ax=ax, cmap=\"viridis\")\n", + "ax.set_title(\"Training Accuracy\")\n", + "ax.set_ylabel(\"$\\eta$\")\n", + "ax.set_xlabel(\"$\\lambda$\")\n", + "plt.show()" + ] + }, + { + "cell_type": "markdown", + "id": "0cb13b9d", "metadata": { "editable": true }, @@ -88,7 +218,7 @@ }, { "cell_type": "markdown", - "id": "33b3b595", + "id": "34f48f2b", "metadata": { "editable": true }, @@ -122,7 +252,7 @@ }, { "cell_type": "markdown", - "id": "4c8e7761", + "id": "ab7ea7a3", "metadata": { "editable": true }, @@ -142,7 +272,7 @@ }, { "cell_type": "markdown", - "id": "b7f04b4c", + "id": "64e7b217", "metadata": { "editable": true }, @@ -158,7 +288,7 @@ }, { "cell_type": "markdown", - "id": "bfabe4ae", + "id": "ffc622bd", "metadata": { "editable": true }, @@ -171,7 +301,7 @@ }, { "cell_type": "markdown", - "id": "065596cf", + "id": "66042bd6", "metadata": { "editable": true }, @@ -183,7 +313,7 @@ }, { "cell_type": "markdown", - "id": "3e8c4e9a", + "id": "6587dbcb", "metadata": { "editable": true }, @@ -205,7 +335,7 @@ }, { "cell_type": "markdown", - "id": "908482ea", + "id": "453c682b", "metadata": { "editable": true }, @@ -217,7 +347,7 @@ }, { "cell_type": "markdown", - "id": "72a1f54a", + "id": "c32c8b64", "metadata": { "editable": true }, @@ -234,7 +364,7 @@ }, { "cell_type": "markdown", - "id": "b21f4a92", + "id": "f6315fee", "metadata": { "editable": true }, @@ -245,7 +375,7 @@ }, { "cell_type": "markdown", - "id": "e8fb7786", + "id": "90c36fbd", "metadata": { "editable": true }, @@ -263,7 +393,7 @@ }, { "cell_type": "markdown", - "id": "aad3ccbb", + "id": "92839267", "metadata": { "editable": true }, @@ -277,7 +407,7 @@ }, { "cell_type": "markdown", - "id": "31e869dc", + "id": "282fcd48", "metadata": { "editable": true }, @@ -296,7 +426,7 @@ }, { "cell_type": "markdown", - "id": "40d43074", + "id": "7c3aebec", "metadata": { "editable": true }, @@ -313,7 +443,7 @@ }, { "cell_type": "markdown", - "id": "ca76119c", + "id": "eb4ae030", "metadata": { "editable": true }, @@ -325,7 +455,7 @@ }, { "cell_type": "markdown", - "id": "95090b1f", + "id": "cdfb7df0", "metadata": { "editable": true }, @@ -346,7 +476,7 @@ }, { "cell_type": "markdown", - "id": "31831318", + "id": "f71a8631", "metadata": { "editable": true }, @@ -360,7 +490,7 @@ }, { "cell_type": "markdown", - "id": "8c678767", + "id": "27ab9c9f", "metadata": { "editable": true }, @@ -371,7 +501,7 @@ }, { "cell_type": "markdown", - "id": "319eadd9", + "id": "4005093f", "metadata": { "editable": true }, @@ -388,7 +518,7 @@ }, { "cell_type": "markdown", - "id": "6b2bc936", + "id": "2a01add6", "metadata": { "editable": true }, @@ -398,7 +528,7 @@ }, { "cell_type": "markdown", - "id": "bb1dccc7", + "id": "0ccb82ad", "metadata": { "editable": true }, @@ -416,7 +546,7 @@ }, { "cell_type": "markdown", - "id": "889e64d9", + "id": "6ff30c97", "metadata": { "editable": true }, @@ -436,7 +566,7 @@ }, { "cell_type": "markdown", - "id": "766af5f3", + "id": "0ee402e9", "metadata": { "editable": true }, @@ -455,16 +585,14 @@ }, { "cell_type": "code", - "execution_count": 1, - "id": "069d4c7d", + "execution_count": 2, + "id": "c1e946bf", "metadata": { "collapsed": false, "editable": true }, "outputs": [], "source": [ - "%matplotlib inline\n", - "\n", "import time\n", "import numpy as np\n", "import tensorflow as tf\n", @@ -478,7 +606,7 @@ }, { "cell_type": "markdown", - "id": "b7d39bb0", + "id": "4b2f1fb0", "metadata": { "editable": true }, @@ -489,8 +617,8 @@ }, { "cell_type": "code", - "execution_count": 2, - "id": "b49760ba", + "execution_count": 3, + "id": "70963327", "metadata": { "collapsed": false, "editable": true @@ -550,7 +678,7 @@ }, { "cell_type": "markdown", - "id": "2226511c", + "id": "23a1347f", "metadata": { "editable": true }, @@ -563,8 +691,8 @@ }, { "cell_type": "code", - "execution_count": 3, - "id": "f55c9436", + "execution_count": 4, + "id": "2d1bc28a", "metadata": { "collapsed": false, "editable": true @@ -609,7 +737,7 @@ }, { "cell_type": "markdown", - "id": "7efbb6fc", + "id": "d66580ed", "metadata": { "editable": true }, @@ -619,8 +747,8 @@ }, { "cell_type": "code", - "execution_count": 4, - "id": "2b41d794", + "execution_count": 5, + "id": "116cfb95", "metadata": { "collapsed": false, "editable": true @@ -644,7 +772,7 @@ }, { "cell_type": "markdown", - "id": "54992d5b", + "id": "ac5de24a", "metadata": { "editable": true }, @@ -663,7 +791,7 @@ }, { "cell_type": "markdown", - "id": "894137b5", + "id": "9db820ca", "metadata": { "editable": true }, @@ -673,8 +801,8 @@ }, { "cell_type": "code", - "execution_count": 5, - "id": "397b0e4d", + "execution_count": 6, + "id": "688a64de", "metadata": { "collapsed": false, "editable": true @@ -730,7 +858,7 @@ }, { "cell_type": "markdown", - "id": "ff4b31c8", + "id": "c671bf91", "metadata": { "editable": true }, @@ -742,8 +870,8 @@ }, { "cell_type": "code", - "execution_count": 6, - "id": "4490b131", + "execution_count": 7, + "id": "544e2cf8", "metadata": { "collapsed": false, "editable": true @@ -767,8 +895,8 @@ }, { "cell_type": "code", - "execution_count": 7, - "id": "47834875", + "execution_count": 8, + "id": "87d8ed8d", "metadata": { "collapsed": false, "editable": true @@ -847,7 +975,7 @@ }, { "cell_type": "markdown", - "id": "4f0b2422", + "id": "d57ceb31", "metadata": { "editable": true }, @@ -878,7 +1006,7 @@ }, { "cell_type": "markdown", - "id": "818ad6d1", + "id": "8ebe69ff", "metadata": { "editable": true }, @@ -898,7 +1026,7 @@ }, { "cell_type": "markdown", - "id": "dcdda417", + "id": "916c4a12", "metadata": { "editable": true }, @@ -910,7 +1038,7 @@ }, { "cell_type": "markdown", - "id": "28eb8b0d", + "id": "648db4a4", "metadata": { "editable": true }, @@ -922,7 +1050,7 @@ }, { "cell_type": "markdown", - "id": "e01a916b", + "id": "a97b3761", "metadata": { "editable": true }, @@ -940,7 +1068,7 @@ }, { "cell_type": "markdown", - "id": "93c7f697", + "id": "c431ae5a", "metadata": { "editable": true }, @@ -963,7 +1091,7 @@ }, { "cell_type": "markdown", - "id": "3b5b070b", + "id": "04e7a1d2", "metadata": { "editable": true }, @@ -986,7 +1114,7 @@ }, { "cell_type": "markdown", - "id": "7cc2bd73", + "id": "5a15d7f0", "metadata": { "editable": true }, @@ -996,8 +1124,8 @@ }, { "cell_type": "code", - "execution_count": 8, - "id": "04fd4471", + "execution_count": 9, + "id": "e4852687", "metadata": { "collapsed": false, "editable": true @@ -1096,7 +1224,7 @@ }, { "cell_type": "markdown", - "id": "386c8882", + "id": "1d0f3998", "metadata": { "editable": true }, @@ -1118,7 +1246,7 @@ }, { "cell_type": "markdown", - "id": "b980b6a4", + "id": "bf91cdbf", "metadata": { "editable": true }, @@ -1130,7 +1258,7 @@ }, { "cell_type": "markdown", - "id": "c863fe52", + "id": "66945981", "metadata": { "editable": true }, @@ -1141,7 +1269,7 @@ }, { "cell_type": "markdown", - "id": "6ba3c8f5", + "id": "6c86adca", "metadata": { "editable": true }, @@ -1163,7 +1291,7 @@ }, { "cell_type": "markdown", - "id": "ae904155", + "id": "42704ae1", "metadata": { "editable": true }, @@ -1176,7 +1304,7 @@ }, { "cell_type": "markdown", - "id": "db635d76", + "id": "599d4583", "metadata": { "editable": true }, @@ -1188,7 +1316,7 @@ }, { "cell_type": "markdown", - "id": "2e74637b", + "id": "68bc1579", "metadata": { "editable": true }, @@ -1198,7 +1326,7 @@ }, { "cell_type": "markdown", - "id": "aaf633fb", + "id": "effaa02c", "metadata": { "editable": true }, @@ -1210,7 +1338,7 @@ }, { "cell_type": "markdown", - "id": "f6e68b1c", + "id": "7bca8dae", "metadata": { "editable": true }, @@ -1220,7 +1348,7 @@ }, { "cell_type": "markdown", - "id": "b11ee68e", + "id": "98f4abf6", "metadata": { "editable": true }, @@ -1232,7 +1360,7 @@ }, { "cell_type": "markdown", - "id": "a3269bff", + "id": "174b8301", "metadata": { "editable": true }, @@ -1265,7 +1393,7 @@ }, { "cell_type": "markdown", - "id": "83af9705", + "id": "c27868e7", "metadata": { "editable": true }, @@ -1289,7 +1417,7 @@ }, { "cell_type": "markdown", - "id": "aa4cb960", + "id": "4e186613", "metadata": { "editable": true }, @@ -1301,7 +1429,7 @@ }, { "cell_type": "markdown", - "id": "de4dc587", + "id": "de14f377", "metadata": { "editable": true }, @@ -1313,7 +1441,7 @@ }, { "cell_type": "markdown", - "id": "6a45fc94", + "id": "d58b7844", "metadata": { "editable": true }, @@ -1341,7 +1469,7 @@ }, { "cell_type": "markdown", - "id": "755c30ad", + "id": "a80689ac", "metadata": { "editable": true }, @@ -1367,7 +1495,7 @@ }, { "cell_type": "markdown", - "id": "c10f79f8", + "id": "9fb59f50", "metadata": { "editable": true }, @@ -1390,7 +1518,7 @@ }, { "cell_type": "markdown", - "id": "f9bb2e6e", + "id": "70a6d2dd", "metadata": { "editable": true }, @@ -1417,7 +1545,7 @@ }, { "cell_type": "markdown", - "id": "311db5f0", + "id": "6e119b7b", "metadata": { "editable": true }, @@ -1436,7 +1564,7 @@ }, { "cell_type": "markdown", - "id": "a2f91713", + "id": "77097ddd", "metadata": { "editable": true }, @@ -1448,7 +1576,7 @@ }, { "cell_type": "markdown", - "id": "ef5ed076", + "id": "4cae3dd5", "metadata": { "editable": true }, @@ -1461,7 +1589,7 @@ }, { "cell_type": "markdown", - "id": "614f9ebe", + "id": "a5e98629", "metadata": { "editable": true }, @@ -1473,7 +1601,7 @@ }, { "cell_type": "markdown", - "id": "0b2fe813", + "id": "aa3ef810", "metadata": { "editable": true }, @@ -1483,7 +1611,7 @@ }, { "cell_type": "markdown", - "id": "b7b63a15", + "id": "c76b96a3", "metadata": { "editable": true }, @@ -1495,7 +1623,7 @@ }, { "cell_type": "markdown", - "id": "d53e92a9", + "id": "4cefe57a", "metadata": { "editable": true }, @@ -1505,7 +1633,7 @@ }, { "cell_type": "markdown", - "id": "60a43e3e", + "id": "c9edea92", "metadata": { "editable": true }, @@ -1517,7 +1645,7 @@ }, { "cell_type": "markdown", - "id": "cd4cabda", + "id": "bd1fb3be", "metadata": { "editable": true }, @@ -1527,8 +1655,8 @@ }, { "cell_type": "code", - "execution_count": 9, - "id": "a26f2c25", + "execution_count": 10, + "id": "5f6ba93c", "metadata": { "collapsed": false, "editable": true @@ -1572,7 +1700,7 @@ }, { "cell_type": "markdown", - "id": "e0f0ef09", + "id": "7e1e2506", "metadata": { "editable": true }, @@ -1582,8 +1710,8 @@ }, { "cell_type": "code", - "execution_count": 10, - "id": "8c53b7c6", + "execution_count": 11, + "id": "02b022b3", "metadata": { "collapsed": false, "editable": true @@ -1618,7 +1746,7 @@ }, { "cell_type": "markdown", - "id": "fa34f6c0", + "id": "61e7bc10", "metadata": { "editable": true }, @@ -1630,8 +1758,8 @@ }, { "cell_type": "code", - "execution_count": 11, - "id": "c6a0e408", + "execution_count": 12, + "id": "7c2b89d8", "metadata": { "collapsed": false, "editable": true @@ -1649,7 +1777,7 @@ }, { "cell_type": "markdown", - "id": "d1bdb68a", + "id": "119e6f80", "metadata": { "editable": true }, @@ -1662,8 +1790,8 @@ }, { "cell_type": "code", - "execution_count": 12, - "id": "ebe595f8", + "execution_count": 13, + "id": "6ee8bc68", "metadata": { "collapsed": false, "editable": true @@ -1682,7 +1810,7 @@ }, { "cell_type": "markdown", - "id": "ae08edff", + "id": "a4b84e02", "metadata": { "editable": true }, @@ -1702,7 +1830,7 @@ }, { "cell_type": "markdown", - "id": "087a0529", + "id": "4f60ad2d", "metadata": { "editable": true }, @@ -1719,7 +1847,7 @@ }, { "cell_type": "markdown", - "id": "d6016c5e", + "id": "24b47ddc", "metadata": { "editable": true }, @@ -1731,7 +1859,7 @@ }, { "cell_type": "markdown", - "id": "b063feb6", + "id": "39464769", "metadata": { "editable": true }, @@ -1748,7 +1876,7 @@ }, { "cell_type": "markdown", - "id": "f07f9c59", + "id": "a3b5165f", "metadata": { "editable": true }, @@ -1761,7 +1889,7 @@ }, { "cell_type": "markdown", - "id": "25bc315d", + "id": "646423be", "metadata": { "editable": true }, @@ -1773,7 +1901,7 @@ }, { "cell_type": "markdown", - "id": "abdd426a", + "id": "869ddef0", "metadata": { "editable": true }, @@ -1783,7 +1911,7 @@ }, { "cell_type": "markdown", - "id": "75254813", + "id": "848e0256", "metadata": { "editable": true }, @@ -1795,7 +1923,7 @@ }, { "cell_type": "markdown", - "id": "f96d7eba", + "id": "96cb2375", "metadata": { "editable": true }, @@ -1805,7 +1933,7 @@ }, { "cell_type": "markdown", - "id": "a59f240b", + "id": "a92647bd", "metadata": { "editable": true }, @@ -1817,7 +1945,7 @@ }, { "cell_type": "markdown", - "id": "8eff4224", + "id": "95cc9640", "metadata": { "editable": true }, @@ -1830,7 +1958,7 @@ }, { "cell_type": "markdown", - "id": "5e94289d", + "id": "6cbf86f3", "metadata": { "editable": true }, @@ -1874,7 +2002,7 @@ }, { "cell_type": "markdown", - "id": "4825caec", + "id": "bcce00cd", "metadata": { "editable": true }, @@ -1884,8 +2012,8 @@ }, { "cell_type": "code", - "execution_count": 13, - "id": "10173cb3", + "execution_count": 14, + "id": "d84c1ee7", "metadata": { "collapsed": false, "editable": true @@ -1963,7 +2091,7 @@ }, { "cell_type": "markdown", - "id": "5a99ca75", + "id": "a7a06011", "metadata": { "editable": true }, @@ -1980,8 +2108,8 @@ }, { "cell_type": "code", - "execution_count": 14, - "id": "59c63ba0", + "execution_count": 15, + "id": "67e927ae", "metadata": { "collapsed": false, "editable": true @@ -2052,7 +2180,7 @@ }, { "cell_type": "markdown", - "id": "2addd397", + "id": "976d8395", "metadata": { "editable": true }, @@ -2091,7 +2219,7 @@ }, { "cell_type": "markdown", - "id": "ee7d4ca0", + "id": "3341e072", "metadata": { "editable": true }, @@ -2101,8 +2229,8 @@ }, { "cell_type": "code", - "execution_count": 15, - "id": "44bffae1", + "execution_count": 16, + "id": "87cdf290", "metadata": { "collapsed": false, "editable": true @@ -2154,7 +2282,7 @@ }, { "cell_type": "markdown", - "id": "88286290", + "id": "f4f316b5", "metadata": { "editable": true }, @@ -2164,8 +2292,8 @@ }, { "cell_type": "code", - "execution_count": 16, - "id": "30773cfa", + "execution_count": 17, + "id": "a78508fe", "metadata": { "collapsed": false, "editable": true @@ -2240,7 +2368,7 @@ }, { "cell_type": "markdown", - "id": "d5620344", + "id": "97bddaf3", "metadata": { "editable": true }, @@ -2250,8 +2378,8 @@ }, { "cell_type": "code", - "execution_count": 17, - "id": "85b1d60f", + "execution_count": 18, + "id": "82dc3b25", "metadata": { "collapsed": false, "editable": true @@ -2282,7 +2410,7 @@ }, { "cell_type": "markdown", - "id": "470b3c72", + "id": "5c4228ee", "metadata": { "editable": true }, @@ -2292,8 +2420,8 @@ }, { "cell_type": "code", - "execution_count": 18, - "id": "263d7105", + "execution_count": 19, + "id": "6483268f", "metadata": { "collapsed": false, "editable": true @@ -2310,8 +2438,8 @@ }, { "cell_type": "code", - "execution_count": 19, - "id": "ae820c0e", + "execution_count": 20, + "id": "7e06c5d3", "metadata": { "collapsed": false, "editable": true @@ -2326,7 +2454,7 @@ }, { "cell_type": "markdown", - "id": "70eaffe8", + "id": "7a690ce3", "metadata": { "editable": true }, @@ -2336,8 +2464,8 @@ }, { "cell_type": "code", - "execution_count": 20, - "id": "24ed3078", + "execution_count": 21, + "id": "e24a9b28", "metadata": { "collapsed": false, "editable": true @@ -2386,8 +2514,8 @@ }, { "cell_type": "code", - "execution_count": 21, - "id": "d9bb3ae2", + "execution_count": 22, + "id": "4612091c", "metadata": { "collapsed": false, "editable": true @@ -2426,7 +2554,7 @@ }, { "cell_type": "markdown", - "id": "e4b2361d", + "id": "0419ea20", "metadata": { "editable": true }, @@ -2450,7 +2578,7 @@ }, { "cell_type": "markdown", - "id": "84042182", + "id": "08b2b371", "metadata": { "editable": true }, @@ -2478,7 +2606,7 @@ }, { "cell_type": "markdown", - "id": "dda7fd42", + "id": "d38b474b", "metadata": { "editable": true }, @@ -2509,7 +2637,7 @@ }, { "cell_type": "markdown", - "id": "ab23b0ce", + "id": "c1f8f3be", "metadata": { "editable": true }, @@ -2525,7 +2653,7 @@ }, { "cell_type": "markdown", - "id": "09ef654d", + "id": "19816c2f", "metadata": { "editable": true }, @@ -2547,7 +2675,7 @@ }, { "cell_type": "markdown", - "id": "040a57a1", + "id": "d04e32ab", "metadata": { "editable": true }, @@ -2579,7 +2707,7 @@ }, { "cell_type": "markdown", - "id": "837a1737", + "id": "abc9e5c3", "metadata": { "editable": true }, @@ -2589,8 +2717,8 @@ }, { "cell_type": "code", - "execution_count": 22, - "id": "ab010a75", + "execution_count": 23, + "id": "7dcbb954", "metadata": { "collapsed": false, "editable": true @@ -2614,7 +2742,7 @@ }, { "cell_type": "markdown", - "id": "cfe1c4d0", + "id": "739626ee", "metadata": { "editable": true }, @@ -2624,8 +2752,8 @@ }, { "cell_type": "code", - "execution_count": 23, - "id": "671cf3dc", + "execution_count": 24, + "id": "8a00e190", "metadata": { "collapsed": false, "editable": true @@ -2679,7 +2807,7 @@ }, { "cell_type": "markdown", - "id": "4df9bb50", + "id": "6682e310", "metadata": { "editable": true }, @@ -2689,8 +2817,8 @@ }, { "cell_type": "code", - "execution_count": 24, - "id": "c9a29b91", + "execution_count": 25, + "id": "ad38c64e", "metadata": { "collapsed": false, "editable": true @@ -2719,8 +2847,8 @@ }, { "cell_type": "code", - "execution_count": 25, - "id": "6c357d4f", + "execution_count": 26, + "id": "11fbf3f1", "metadata": { "collapsed": false, "editable": true @@ -2737,8 +2865,8 @@ }, { "cell_type": "code", - "execution_count": 26, - "id": "5fec2316", + "execution_count": 27, + "id": "e737d148", "metadata": { "collapsed": false, "editable": true @@ -2757,8 +2885,8 @@ }, { "cell_type": "code", - "execution_count": 27, - "id": "401e905d", + "execution_count": 28, + "id": "66ce9dce", "metadata": { "collapsed": false, "editable": true @@ -2775,7 +2903,7 @@ }, { "cell_type": "markdown", - "id": "88c539be", + "id": "48037643", "metadata": { "editable": true }, @@ -2785,8 +2913,8 @@ }, { "cell_type": "code", - "execution_count": 28, - "id": "ca63ff74", + "execution_count": 29, + "id": "239a09a1", "metadata": { "collapsed": false, "editable": true @@ -2805,8 +2933,8 @@ }, { "cell_type": "code", - "execution_count": 29, - "id": "4b4ae266", + "execution_count": 30, + "id": "e146ce02", "metadata": { "collapsed": false, "editable": true @@ -2819,8 +2947,8 @@ }, { "cell_type": "code", - "execution_count": 30, - "id": "0c7094e1", + "execution_count": 31, + "id": "260066c7", "metadata": { "collapsed": false, "editable": true @@ -2835,8 +2963,8 @@ }, { "cell_type": "code", - "execution_count": 31, - "id": "eb5c6446", + "execution_count": 32, + "id": "0a81f1b0", "metadata": { "collapsed": false, "editable": true @@ -2874,7 +3002,7 @@ }, { "cell_type": "markdown", - "id": "0353b2ae", + "id": "97efcdc9", "metadata": { "editable": true }, @@ -2887,8 +3015,8 @@ }, { "cell_type": "code", - "execution_count": 32, - "id": "bee8341b", + "execution_count": 33, + "id": "9c2a48b5", "metadata": { "collapsed": false, "editable": true diff --git a/doc/src/week43/programs/diffusion.py~ b/doc/src/week43/programs/diffusion.py~ deleted file mode 100644 index 2b6c2d667..000000000 --- a/doc/src/week43/programs/diffusion.py~ +++ /dev/null @@ -1,3759 +0,0 @@ -TITLE: Week 43: Solving Differential Equations with Deep Learning and Dimensionality Reduction methods -AUTHOR: Morten Hjorth-Jensen {copyright, 1999-present|CC BY-NC} at Department of Physics, University of Oslo & Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University -DATE: today - -!split - -* Thursday: Wrapping up Recurrent Neural Networks and solving differential equations. -* Friday: Principal Component Analysis and Dimensionality Reduction - -Reading suggestions for both days: "Aurelien Geron's chapters 8 - -!split -===== Recurrent Neural Networks ===== - - -!split -===== Solving ODEs with Deep Learning ===== - - - - - - - -!split -===== Basic ideas of the Principal Component Analysis (PCA) ===== - -The principal component analysis deals with the problem of fitting a -low-dimensional affine subspace $S$ of dimension $d$ much smaller than -the totaldimension $D$ of the problem at hand (our data -set). Mathematically it can be formulated as a statistical problem or -a geometric problem. In our discussion of the theorem for the -classical PCA, we will stay with a statistical approach. This is also -what set the scene historically which for the PCA. - -We have a data set defined by a design/feature matrix $\bm{X}$ (see below for its definition) -* Each data point is determined by $p$ extrinsic (measurement) variables -* We may want to ask the following question: Are there fewer intrinsic variables (say $d << p$) that still approximately describe the data? -* If so, these intrinsic variables may tell us something important and finding these intrinsic variables is what dimension reduction methods do. - - -!split -===== Introducing the Covariance and Correlation functions ===== - -Before we discuss the PCA theorem, we need to remind ourselves about -the definition of the covariance and the correlation function. These are quantities - -Suppose we have defined two vectors -$\hat{x}$ and $\hat{y}$ with $n$ elements each. The covariance matrix $\bm{C}$ is defined as -!bt -\[ -\bm{C}[\bm{x},\bm{y}] = \begin{bmatrix} \mathrm{cov}[\bm{x},\bm{x}] & \mathrm{cov}[\bm{x},\bm{y}] \\ - \mathrm{cov}[\bm{y},\bm{x}] & \mathrm{cov}[\bm{y},\bm{y}] \\ - \end{bmatrix}, -\] -!et -where for example -!bt -\[ -\mathrm{cov}[\bm{x},\bm{y}] =\frac{1}{n} \sum_{i=0}^{n-1}(x_i- \overline{x})(y_i- \overline{y}). -\] -!et -With this definition and recalling that the variance is defined as -!bt -\[ -\mathrm{var}[\bm{x}]=\frac{1}{n} \sum_{i=0}^{n-1}(x_i- \overline{x})^2, -\] -!et -we can rewrite the covariance matrix as -!bt -\[ -\bm{C}[\bm{x},\bm{y}] = \begin{bmatrix} \mathrm{var}[\bm{x}] & \mathrm{cov}[\bm{x},\bm{y}] \\ - \mathrm{cov}[\bm{x},\bm{y}] & \mathrm{var}[\bm{y}] \\ - \end{bmatrix}. -\] -!et - -The covariance takes values between zero and infinity and may thus -lead to problems with loss of numerical precision for particularly -large values. It is common to scale the covariance matrix by -introducing instead the correlation matrix defined via the so-called -correlation function - -!bt -\[ -\mathrm{corr}[\bm{x},\bm{y}]=\frac{\mathrm{cov}[\bm{x},\bm{y}]}{\sqrt{\mathrm{var}[\bm{x}] \mathrm{var}[\bm{y}]}}. -\] -!et - -The correlation function is then given by values $\mathrm{corr}[\bm{x},\bm{y}] -\in [-1,1]$. This avoids eventual problems with too large values. We -can then define the correlation matrix for the two vectors $\bm{x}$ -and $\bm{y}$ as - -!bt -\[ -\bm{K}[\bm{x},\bm{y}] = \begin{bmatrix} 1 & \mathrm{corr}[\bm{x},\bm{y}] \\ - \mathrm{corr}[\bm{y},\bm{x}] & 1 \\ - \end{bmatrix}, -\] -!et - -In the above example this is the function we constructed using _pandas_. - -!split -===== Correlation Function and Design/Feature Matrix ===== - -In our derivation of the various regression algorithms like _Ordinary Least Squares_ or _Ridge regression_ -we defined the design/feature matrix $\bm{X}$ as - -!bt -\[ -\bm{X}=\begin{bmatrix} -x_{0,0} & x_{0,1} & x_{0,2}& \dots & \dots x_{0,p-1}\\ -x_{1,0} & x_{1,1} & x_{1,2}& \dots & \dots x_{1,p-1}\\ -x_{2,0} & x_{2,1} & x_{2,2}& \dots & \dots x_{2,p-1}\\ -\dots & \dots & \dots & \dots \dots & \dots \\ -x_{n-2,0} & x_{n-2,1} & x_{n-2,2}& \dots & \dots x_{n-2,p-1}\\ -x_{n-1,0} & x_{n-1,1} & x_{n-1,2}& \dots & \dots x_{n-1,p-1}\\ -\end{bmatrix}, -\] -!et -with $\bm{X}\in {\mathbb{R}}^{n\times p}$, with the predictors/features $p$ refering to the column numbers and the -entries $n$ being the row elements. -We can rewrite the design/feature matrix in terms of its column vectors as -!bt -\[ -\bm{X}=\begin{bmatrix} \bm{x}_0 & \bm{x}_1 & \bm{x}_2 & \dots & \dots & \bm{x}_{p-1}\end{bmatrix}, -\] -!et -with a given vector -!bt -\[ -\bm{x}_i^T = \begin{bmatrix}x_{0,i} & x_{1,i} & x_{2,i}& \dots & \dots x_{n-1,i}\end{bmatrix}. -\] -!et - -With these definitions, we can now rewrite our $2\times 2$ -correaltion/covariance matrix in terms of a moe general design/feature -matrix $\bm{X}\in {\mathbb{R}}^{n\times p}$. This leads to a $p\times p$ -covariance matrix for the vectors $\bm{x}_i$ with $i=0,1,\dots,p-1$ - -!bt -\[ -\bm{C}[\bm{x}] = \begin{bmatrix} -\mathrm{var}[\bm{x}_0] & \mathrm{cov}[\bm{x}_0,\bm{x}_1] & \mathrm{cov}[\bm{x}_0,\bm{x}_2] & \dots & \dots & \mathrm{cov}[\bm{x}_0,\bm{x}_{p-1}]\\ -\mathrm{cov}[\bm{x}_1,\bm{x}_0] & \mathrm{var}[\bm{x}_1] & \mathrm{cov}[\bm{x}_1,\bm{x}_2] & \dots & \dots & \mathrm{cov}[\bm{x}_1,\bm{x}_{p-1}]\\ -\mathrm{cov}[\bm{x}_2,\bm{x}_0] & \mathrm{cov}[\bm{x}_2,\bm{x}_1] & \mathrm{var}[\bm{x}_2] & \dots & \dots & \mathrm{cov}[\bm{x}_2,\bm{x}_{p-1}]\\ -\dots & \dots & \dots & \dots & \dots & \dots \\ -\dots & \dots & \dots & \dots & \dots & \dots \\ -\mathrm{cov}[\bm{x}_{p-1},\bm{x}_0] & \mathrm{cov}[\bm{x}_{p-1},\bm{x}_1] & \mathrm{cov}[\bm{x}_{p-1},\bm{x}_{2}] & \dots & \dots & \mathrm{var}[\bm{x}_{p-1}]\\ -\end{bmatrix}, -\] -!et -and the correlation matrix -!bt -\[ -\bm{K}[\bm{x}] = \begin{bmatrix} -1 & \mathrm{corr}[\bm{x}_0,\bm{x}_1] & \mathrm{corr}[\bm{x}_0,\bm{x}_2] & \dots & \dots & \mathrm{corr}[\bm{x}_0,\bm{x}_{p-1}]\\ -\mathrm{corr}[\bm{x}_1,\bm{x}_0] & 1 & \mathrm{corr}[\bm{x}_1,\bm{x}_2] & \dots & \dots & \mathrm{corr}[\bm{x}_1,\bm{x}_{p-1}]\\ -\mathrm{corr}[\bm{x}_2,\bm{x}_0] & \mathrm{corr}[\bm{x}_2,\bm{x}_1] & 1 & \dots & \dots & \mathrm{corr}[\bm{x}_2,\bm{x}_{p-1}]\\ -\dots & \dots & \dots & \dots & \dots & \dots \\ -\dots & \dots & \dots & \dots & \dots & \dots \\ -\mathrm{corr}[\bm{x}_{p-1},\bm{x}_0] & \mathrm{corr}[\bm{x}_{p-1},\bm{x}_1] & \mathrm{corr}[\bm{x}_{p-1},\bm{x}_{2}] & \dots & \dots & 1\\ -\end{bmatrix}, -\] -!et - - -!split -===== Covariance Matrix Examples ===== - - -The Numpy function _np.cov_ calculates the covariance elements using -the factor $1/(n-1)$ instead of $1/n$ since it assumes we do not have -the exact mean values. The following simple function uses the -_np.vstack_ function which takes each vector of dimension $1\times n$ -and produces a $2\times n$ matrix $\bm{W}$ - - -!bt -\[ -\bm{W} = \begin{bmatrix} x_0 & y_0 \\ - x_1 & y_1 \\ - x_2 & y_2\\ - \dots & \dots \\ - x_{n-2} & y_{n-2}\\ - x_{n-1} & y_{n-1} & - \end{bmatrix}, -\] -!et - -which in turn is converted into into the $2\times 2$ covariance matrix -$\bm{C}$ via the Numpy function _np.cov()_. We note that we can also calculate -the mean value of each set of samples $\bm{x}$ etc using the Numpy -function _np.mean(x)_. We can also extract the eigenvalues of the -covariance matrix through the _np.linalg.eig()_ function. - -!bc pycod -# Importing various packages -import numpy as np -n = 100 -x = np.random.normal(size=n) -print(np.mean(x)) -y = 4+3*x+np.random.normal(size=n) -print(np.mean(y)) -W = np.vstack((x, y)) -C = np.cov(W) -print(C) -!ec - -!split -===== Correlation Matrix ===== - -The previous example can be converted into the correlation matrix by -simply scaling the matrix elements with the variances. We should also -subtract the mean values for each column. This leads to the following -code which sets up the correlations matrix for the previous example in -a more brute force way. Here we scale the mean values for each column of the design matrix, calculate the relevant mean values and variances and then finally set up the $2\times 2$ correlation matrix (since we have only two vectors). - -!bc pycod -import numpy as np -n = 100 -# define two vectors -x = np.random.random(size=n) -y = 4+3*x+np.random.normal(size=n) -#scaling the x and y vectors -x = x - np.mean(x) -y = y - np.mean(y) -variance_x = np.sum(x@x)/n -variance_y = np.sum(y@y)/n -print(variance_x) -print(variance_y) -cov_xy = np.sum(x@y)/n -cov_xx = np.sum(x@x)/n -cov_yy = np.sum(y@y)/n -C = np.zeros((2,2)) -C[0,0]= cov_xx/variance_x -C[1,1]= cov_yy/variance_y -C[0,1]= cov_xy/np.sqrt(variance_y*variance_x) -C[1,0]= C[0,1] -print(C) -!ec - -We see that the matrix elements along the diagonal are one as they -should be and that the matrix is symmetric. Furthermore, diagonalizing -this matrix we easily see that it is a positive definite matrix. - -The above procedure with _numpy_ can be made more compact if we use _pandas_. - -!split -===== Correlation Matrix with Pandas ===== - -We whow here how we can set up the correlation matrix using _pandas_, as done in this simple code -!bc pycod -import numpy as np -import pandas as pd -n = 10 -x = np.random.normal(size=n) -x = x - np.mean(x) -y = 4+3*x+np.random.normal(size=n) -y = y - np.mean(y) -X = (np.vstack((x, y))).T -print(X) -Xpd = pd.DataFrame(X) -print(Xpd) -correlation_matrix = Xpd.corr() -print(correlation_matrix) -!ec - - -We expand this model to the Franke function discussed above. - -!split -===== Correlation Matrix with Pandas and the Franke function ===== - -!bc pycod -# Common imports -import numpy as np -import pandas as pd - - -def FrankeFunction(x,y): - term1 = 0.75*np.exp(-(0.25*(9*x-2)**2) - 0.25*((9*y-2)**2)) - term2 = 0.75*np.exp(-((9*x+1)**2)/49.0 - 0.1*(9*y+1)) - term3 = 0.5*np.exp(-(9*x-7)**2/4.0 - 0.25*((9*y-3)**2)) - term4 = -0.2*np.exp(-(9*x-4)**2 - (9*y-7)**2) - return term1 + term2 + term3 + term4 - - -def create_X(x, y, n ): - if len(x.shape) > 1: - x = np.ravel(x) - y = np.ravel(y) - - N = len(x) - l = int((n+1)*(n+2)/2) # Number of elements in beta - X = np.ones((N,l)) - - for i in range(1,n+1): - q = int((i)*(i+1)/2) - for k in range(i+1): - X[:,q+k] = (x**(i-k))*(y**k) - - return X - - -# Making meshgrid of datapoints and compute Franke's function -n = 4 -N = 100 -x = np.sort(np.random.uniform(0, 1, N)) -y = np.sort(np.random.uniform(0, 1, N)) -z = FrankeFunction(x, y) -X = create_X(x, y, n=n) - -Xpd = pd.DataFrame(X) -# subtract the mean values and set up the covariance matrix -Xpd = Xpd - Xpd.mean() -covariance_matrix = Xpd.cov() -print(covariance_matrix) -!ec - -We note here that the covariance is zero for the first rows and -columns since all matrix elements in the design matrix were set to one -(we are fitting the function in terms of a polynomial of degree $n$). - -This means that the variance for these elements will be zero and will -cause problems when we set up the correlation matrix. We can simply -drop these elements and construct a correlation -matrix without these elements. - - -!split -===== Rewriting the Covariance and/or Correlation Matrix ===== - -We can rewrite the covariance matrix in a more compact form in terms of the design/feature matrix $\bm{X}$ as -!bt -\[ -\bm{C}[\bm{x}] = \frac{1}{n}\bm{X}\bm{X}^T= \mathbb{E}[\bm{X}\bm{X}^T]. -\] -!et - -To see this let us simply look at a design matrix $\bm{X}\in {\mathbb{R}}^{2\times 2}$ -!bt -\[ -\bm{X}=\begin{bmatrix} -x_{00} & x_{01}\\ -x_{10} & x_{11}\\ -\end{bmatrix}=\begin{bmatrix} -\bm{x}_{0} & \bm{x}_{1}\\ -\end{bmatrix}. -\] -!et - -If we then compute the expectation value -!bt -\[ -\mathbb{E}[\bm{X}\bm{X}^T] = \frac{1}{n}\bm{X}\bm{X}^T=\begin{bmatrix} -x_{00}^2+x_{01}^2 & x_{00}x_{10}+x_{01}x_{11}\\ -x_{10}x_{00}+x_{11}x_{01} & x_{10}^2+x_{11}^2\\ -\end{bmatrix}, -\] -!et -which is just -!bt -\[ -\bm{C}[\bm{x}_0,\bm{x}_1] = \bm{C}[\bm{x}]=\begin{bmatrix} \mathrm{var}[\bm{x}_0] & \mathrm{cov}[\bm{x}_0,\bm{x}_1] \\ - \mathrm{cov}[\bm{x}_1,\bm{x}_0] & \mathrm{var}[\bm{x}_1] \\ - \end{bmatrix}, -\] -!et -where we wrote $$\bm{C}[\bm{x}_0,\bm{x}_1] = \bm{C}[\bm{x}]$$ to indicate that this the covariance of the vectors $\bm{x}$ of the design/feature matrix $\bm{X}$. - -It is easy to generalize this to a matrix $\bm{X}\in {\mathbb{R}}^{n\times p}$. - - -!split -===== Towards the PCA theorem ===== - -We have that the covariance matrix (the correlation matrix involves a simple rescaling) is given as -!bt -\[ -\bm{C}[\bm{x}] = \frac{1}{n}\bm{X}\bm{X}^T= \mathbb{E}[\bm{X}\bm{X}^T]. -\] -!et -Let us now assume that we can perform a series of orthogonal transformations where we employ some orthogonal matrices $\bm{S}$. -These matrices are defined as $\bm{S}\in {\mathbb{R}}^{p\times p}$ and obey the orthogonality requirements $\bm{S}\bm{S}^T=\bm{S}^T\bm{S}=\bm{I}$. The matrix can be written out in terms of the column vectors $\bm{s}_i$ as $\bm{S}=[\bm{s}_0,\bm{s}_1,\dots,\bm{s}_{p-1}]$ and $\bm{s}_i \in {\mathbb{R}}^{p}$. - -Assume also that there is a transformation $\bm{S}\bm{C}[\bm{x}]\bm{S}^T=\bm{C}[\bm{y}]$ such that the new matrix $\bm{C}[\bm{y}]$ is diagonal with elements $[\lambda_0,\lambda_1,\lambda_2,\dots,\lambda_{p-1}]$. - -That is we have -!bt -\[ -\bm{C}[\bm{y}] = \mathbb{E}[\bm{S}\bm{X}\bm{X}^T\bm{S}^T]=\bm{S}\bm{C}[\bm{x}]\bm{S}^T, -\] -!et -since the matrix $\bm{S}$ is not a data dependent matrix. Multiplying with $\bm{S}^T$ from the left we have -!bt -\[ -\bm{S}^T\bm{C}[\bm{y}] = \bm{C}[\bm{x}]\bm{S}^T, -\] -!et -and since $\bm{C}[\bm{y}]$ is diagonal we have for a given eigenvalue $i$ of the covariance matrix that - -!bt -\[ -\bm{S}^T_i\lambda_i = \bm{C}[\bm{x}]\bm{S}^T_i. -\] -!et - -In the derivation of the PCA theorem we will assume that the eigenvalues are ordered in descending order, that is -$\lambda_0 > \lambda_1 > \dots > \lambda_{p-1}$. - - -The eigenvalues tell us then how much we need to stretch the -corresponding eigenvectors. Dimensions with large eigenvalues have -thus large variations (large variance) and define therefore useful -dimensions. The data points are more spread out in the direction of -these eigenvectors. Smaller eigenvalues mean on the other hand that -the corresponding eigenvectors are shrunk accordingly and the data -points are tightly bunched together and there is not much variation in -these specific directions. Hopefully then we could leave it out -dimensions where the eigenvalues are very small. If $p$ is very large, -we could then aim at reducing $p$ to $l << p$ and handle only $l$ -features/predictors. - -!split -===== The Algorithm before the Theorem ===== - -Here's how we would proceed in setting up the algorithm for the PCA, see also discussion below here. -* Set up the datapoints for the design/feature matrix $\bm{X}$ with $\bm{X}\in {\mathbb{R}}^{n\times p}$, with the predictors/features $p$ referring to the column numbers and the entries $n$ being the row elements. -!bt -\[ -\bm{X}=\begin{bmatrix} -x_{0,0} & x_{0,1} & x_{0,2}& \dots & \dots x_{0,p-1}\\ -x_{1,0} & x_{1,1} & x_{1,2}& \dots & \dots x_{1,p-1}\\ -x_{2,0} & x_{2,1} & x_{2,2}& \dots & \dots x_{2,p-1}\\ -\dots & \dots & \dots & \dots \dots & \dots \\ -x_{n-2,0} & x_{n-2,1} & x_{n-2,2}& \dots & \dots x_{n-2,p-1}\\ -x_{n-1,0} & x_{n-1,1} & x_{n-1,2}& \dots & \dots x_{n-1,p-1}\\ -\end{bmatrix}, -\] -!et -* Center the data by subtracting the mean value for each column. This leads to a new matrix $\bm{X}\rightarrow \overline{\bm{X}}$. -* Compute then the covariance/correlation matrix $\mathbb{E}[\overline{\bm{X}}\overline{\bm{X}}^T]$. -* Find the eigenpairs of $\bm{C}$ with eigenvalues $[\lambda_0,\lambda_1,\dots,\lambda_{p-1}]$ and eigenvectors $[\bm{s}_0,\bm{s}_1,\dots,\bm{s}_{p-1}]$. -* Order the eigenvalue (and the eigenvectors accordingly) in order of decreasing eigenvalues. -* Keep only those $l$ eigenvalues larger than a selected threshold value, discarding thus $p-l$ features since we expect small variations in the data here. - - -!split -===== Writing our own PCA code ===== - -We will use a simple example first with two-dimensional data -drawn from a multivariate normal distribution with the following mean and covariance matrix: -!bt -\[ -\mu = (-1,2) \qquad \Sigma = \begin{bmatrix} 4 & 2 \\ -2 & 2 -\end{bmatrix} -\] -!et -Note that the mean refers to each column of data. -We will generate $n = 1000$ points $X = \{ x_1, \ldots, x_N \}$ from -this distribution, and store them in the $1000 \times 2$ matrix $\bm{X}$. - -The following Python code aids in setting up the data and writing out the design matrix. -Note that the function _multivariate_ returns also the covariance discussed above and that it is defined by dividing by $n-1$ instead of $n$. -!bc pycod -import numpy as np -import pandas as pd -import matplotlib.pyplot as plt -from IPython.display import display -n = 10000 -mean = (-1, 2) -cov = [[4, 2], [2, 2]] -X = np.random.multivariate_normal(mean, cov, n) -!ec - -Now we are going to implement the PCA algorithm. We will break it down into various substeps. - -=== Compute the sample mean and center the data === - -The first step of PCA is to compute the sample mean of the data and use it to center the data. Recall that the sample mean is -!bt -\[ -\mu_n = \frac{1}{n} \sum_{i=1}^n x_i -\] -!et -and the mean-centered data $\bar{X} = \{ \bar{x}_1, \ldots, \bar{x}_n \}$ takes the form -!bt -\[ -\bar{x}_i = x_i - \mu_n. -\] -!et -When you are done with these steps, print out $\mu_n$ to verify it is -close to $\mu$ and plot your mean centered data to verify it is -centered at the origin! Compare your code with the functionality from _Scikit-Learn_ discussed above. -The following code elements perform these operations using _pandas_ or using our own functionality for doing so. The latter, using _numpy_ is rather simple through the _mean()_ function. -!bc pycod -df = pd.DataFrame(X) -# Pandas does the centering for us -df = df -df.mean() -# we center it ourselves -X_centered = X - X.mean(axis=0) -!ec - -Alternatively, we could use the functions we discussed -earlier for scaling the data set. That is, we could have used the -_StandardScaler_ function in _Scikit-Learn_, a function which ensures -that for each feature/predictor we study the mean value is zero and -the variance is one (every column in the design/feature matrix). You -would then not get the same results, since we divide by the -variance. The diagonal covariance matrix elements will then be one, -while the non-diagonal ones need to be divided by $2\sqrt{2}$ for our -specific case. - -=== Compute the sample covariance === - -Now we are going to use the mean centered data to compute the sample covariance of the data by using the following equation -!bt -\begin{equation*} -\Sigma_n = \frac{1}{n-1} \sum_{i=1}^n \bar{x}_i^T \bar{x}_i = \frac{1}{n-1} \sum_{i=1}^n (x_i - \mu_n)^T (x_i - \mu_n) -\end{equation*} -!et -where the data points $x_i \in \mathbb{R}^p$ (here in this example $p = 2$) are column vectors and $x^T$ is the transpose of $x$. -We can write our own code or simply use either the functionaly of _numpy_ or that of _pandas_, as follows -!bc pycod -print(df.cov()) -print(np.cov(X_centered.T)) -!ec -Note that the way we define the covariance matrix here has a factor $n-1$ instead of $n$. This is included in the _cov()_ function by _numpy_ and _pandas_. -Our own code here is not very elegant and asks for obvious improvements. It is tailored to this specific $2\times 2$ covariance matrix. -!bc pycod -# extract the relevant columns from the centered design matrix of dim n x 2 -x = X_centered[:,0] -y = X_centered[:,1] -Cov = np.zeros((2,2)) -Cov[0,1] = np.sum(x.T@y)/(n-1.0) -Cov[0,0] = np.sum(x.T@x)/(n-1.0) -Cov[1,1] = np.sum(y.T@y)/(n-1.0) -Cov[1,0]= Cov[0,1] -print("Centered covariance using own code") -print(Cov) -plt.plot(x, y, 'x') -plt.axis('equal') -plt.show() -!ec - -Depending on the number of points $n$, we will get results that are close to the covariance values defined above. -The plot shows how the data are clustered around a line with slope close to one. Is this expected? - -=== Diagonalize the sample covariance matrix to obtain the principal components === - -Now we are ready to solve for the principal components! To do so we -diagonalize the sample covariance matrix $\Sigma$. We can use the -function _np.linalg.eig_ to do so. It will return the eigenvalues and -eigenvectors of $\Sigma$. Once we have these we can perform the -following tasks: - -* We compute the percentage of the total variance captured by the first principal component -* We plot the mean centered data and lines along the first and second principal components -* Then we project the mean centered data onto the first and second principal components, and plot the projected data. -* Finally, we approximate the data as - -!bt -\begin{equation*} -x_i \approx \tilde{x}_i = \mu_n + \langle x_i, v_0 \rangle v_0 -\end{equation*} -!et -where $v_0$ is the first principal component. - -Collecting all these steps we can write our own PCA function and -compare this with the functionality included in _Scikit-Learn_. - -The code here outlines some of the elements we could include in the -analysis. Feel free to extend upon this in order to address the above -questions. - -!bc pycod -# diagonalize and obtain eigenvalues, not necessarily sorted -EigValues, EigVectors = np.linalg.eig(Cov) -# sort eigenvectors and eigenvalues -#permute = EigValues.argsort() -#EigValues = EigValues[permute] -#EigVectors = EigVectors[:,permute] -print("Eigenvalues of Covariance matrix") -for i in range(2): - print(EigValues[i]) -FirstEigvector = EigVectors[:,0] -SecondEigvector = EigVectors[:,1] -print("First eigenvector") -print(FirstEigvector) -print("Second eigenvector") -print(SecondEigvector) -#thereafter we do a PCA with Scikit-learn -from sklearn.decomposition import PCA -pca = PCA(n_components = 2) -X2Dsl = pca.fit_transform(X) -print("Eigenvector of largest eigenvalue") -print(pca.components_.T[:, 0]) - -!ec -This code does not contain all the above elements, but it shows how we can use _Scikit-Learn_ to extract the eigenvector which corresponds to the largest eigenvalue. Try to address the questions we pose before the above code. Try also to change the values of the covariance matrix by making one of the diagonal elements much larger than the other. What do you observe then? - -!split -===== Classical PCA Theorem ===== - -We assume now that we have a design matrix $\bm{X}$ which has been -centered as discussed above. For the sake of simplicity we skip the -overline symbol. The matrix is defined in terms of the various column -vectors $[\bm{x}_0,\bm{x}_1,\dots, \bm{x}_{p-1}]$ each with dimension -$\bm{x}\in {\mathbb{R}}^{n}$. - -We assume also that we have an orthogonal transformation $\bm{W}\in {\mathbb{R}}^{p\times p}$. We define the reconstruction error (which is similar to the mean squared error we have seen before) as -!bt -\[ -J(\bm{W},\bm{Z}) = \frac{1}{n}\sum_i (\bm{x}_i - \overline{\bm{x}}_i)^2, -\] -!et -with $\overline{\bm{x}}_i = \bm{W}\bm{z}_i$, where $\bm{z}_i$ is a row vector with dimension ${\mathbb{R}}^{n}$ of the matrix -$\bm{Z}\in{\mathbb{R}}^{p\times n}$. When doing PCA we want to reduce this dimensionality. - -The PCA theorem states that minimizing the above reconstruction error -corresponds to setting $\bm{W}=\bm{S}$, the orthogonal matrix which -diagonalizes the empirical covariance(correlation) matrix. The optimal -low-dimensional encoding of the data is then given by a set of vectors -$\bm{z}_i$ with at most $l$ vectors, with $l << p$, defined by the -orthogonal projection of the data onto the columns spanned by the -eigenvectors of the covariance(correlations matrix). - -The proof which follows will be updated by mid January 2020. - -!split -===== Proof of the PCA Theorem ===== - -To show the PCA theorem let us start with the assumption that there is one vector $\bm{w}_0$ which corresponds to a solution which minimized the reconstruction error $J$. This is an orthogonal vector. It means that we now approximate the reconstruction error in terms of $\bm{w}_0$ and $\bm{z}_0$ as -!bt -\[ -J(\bm{w}_0,\bm{z}_0)= \frac{1}{n}\sum_i (\bm{x}_i - z_{i0}\bm{w}_0)^2=\frac{1}{n}\sum_i (\bm{x}_i^T\bm{x}_i - 2z_{i0}\bm{w}_0^T\bm{x}_i+z_{i0}^2\bm{w}_0^T\bm{w}_0), -\] -!et -which we can rewrite due to the orthogonality of $\bm{w}_i$ as -!bt -\[ -J(\bm{w}_0,\bm{z}_0)=\frac{1}{n}\sum_i (\bm{x}_i^T\bm{x}_i - 2z_{i0}\bm{w}_0^T\bm{x}_i+z_{i0}^2). -\] -!et -Minimizing $J$ with respect to the unknown parameters $z_{0i}$ we obtain that -!bt -\[ -z_{i0}=\bm{w}_0^T\bm{x}_i, -\] -!et -where the vectors on the rhs are known. - - -!split -===== PCA Proof continued ===== - -We have now found the unknown parameters $z_{i0}$. These correspond to the projected coordinates and we can write -!bt -\[ -J(\bm{w}_0)= \frac{1}{p}\sum_i (\bm{x}_i^T\bm{x}_i - z_{i0}^2)=\mathrm{const}-\frac{1}{n}\sum_i z_{i0}^2. -\] -!et - -We can show that the variance of the projected coordinates defined by $\bm{w}_0^T\bm{x}_i$ are given by -!bt -\[ -\mathrm{var}[\bm{w}_0^T\bm{x}_i] = \frac{1}{n}\sum_i z_{i0}^2, -\] -!et -since the expectation value of -!bt -\[ -\mathbb{E}[\bm{w}_0^T\bm{x}_i] = \mathbb{E}[z_{i0}]= \bm{w}_0^T\mathbb{E}[\bm{x}_i]=0, -\] -!et -where we have used the fact that our data are centered. - -Recalling our definition of the covariance as -!bt -\[ -\bm{C}[\bm{x}] = \frac{1}{n}\bm{X}\bm{X}^T=\mathbb{E}[\bm{X}\bm{X}^T], -\] -!et -we have thus that -!bt -\[ -\mathrm{var}[\bm{w}_0^T\bm{x}_i] = \frac{1}{n}\sum_i z_{i0}^2=\bm{w}_0^T\bm{C}[\bm{x}]\bm{w}_0. -\] -!et - -We are almost there, we have obtained a relation between minimizing -the reconstruction error and the variance and the covariance -matrix. Minimizing the error is equivalent to maximizing the variance -of the projected data. - -!split -===== The final step ===== - -We could trivially maximize the variance of the projection (and -thereby minimize the error in the reconstruction function) by letting -the norm-2 of $\bm{w}_0$ go to infinity. However, this norm since we -want the matrix $\bm{W}$ to be an orthogonal matrix, is constrained by -$\vert\vert \bm{w}_0 \vert\vert_2^2=1$. Imposing this condition via a -Lagrange multiplier we can then in turn maximize - -!bt -\[ -J(\bm{w}_0)= \bm{w}_0^T\bm{C}[\bm{x}]\bm{w}_0+\lambda_0(1-\bm{w}_0^T\bm{w}_0). -\] -!et -Taking the derivative with respect to $\bm{w}_0$ we obtain - -!bt -\[ -\frac{\partial J(\bm{w}_0)}{\partial \bm{w}_0}= 2\bm{C}[\bm{x}]\bm{w}_0-2\lambda_0\bm{w}_0=0, -\] -!et -meaning that -!bt -\[ -\bm{C}[\bm{x}]\bm{w}_0=\lambda_0\bm{w}_0. -\] -!et -_The direction that maximizes the variance (or minimizes the construction error) is an eigenvector of the covariance matrix_! If we left multiply with $\bm{w}_0^T$ we have the variance of the projected data is -!bt -\[ -\bm{w}_0^T\bm{C}[\bm{x}]\bm{w}_0=\lambda_0. -\] -!et - -If we want to maximize the variance (minimize the construction error) -we simply pick the eigenvector of the covariance matrix with the -largest eigenvalue. This establishes the link between the minimization -of the reconstruction function $J$ in terms of an orthogonal matrix -and the maximization of the variance and thereby the covariance of our -observations encoded in the design/feature matrix $\bm{X}$. - -The proof -for the other eigenvectors $\bm{w}_1,\bm{w}_2,\dots$ can be -established by applying the above arguments and using the fact that -our basis of eigenvectors is orthogonal, see "Murphy chapter -12.2":"https://mitpress.mit.edu/books/machine-learning-1". The -discussion in chapter 12.2 of Murphy's text has also a nice link with -the Singular Value Decomposition theorem. For categorical data, see -chapter 12.4 and discussion therein. - -Additional part of the proof for the other eigenvectors will be added by mid January 2020. - -!split -===== Geometric Interpretation and link with Singular Value Decomposition ===== - -This material will be added by mid January 2020. - - -!split -===== Principal Component Analysis ===== - -Principal Component Analysis (PCA) is by far the most popular dimensionality reduction algorithm. -First it identifies the hyperplane that lies closest to the data, and then it projects the data onto it. - -The following Python code uses NumPy’s _svd()_ function to obtain all the principal components of the -training set, then extracts the first two principal components. First we center the data using either _pandas_ or our own code -!bc pycod -import numpy as np -import pandas as pd -from IPython.display import display -np.random.seed(100) -# setting up a 10 x 5 vanilla matrix -rows = 10 -cols = 5 -X = np.random.randn(rows,cols) -df = pd.DataFrame(X) -# Pandas does the centering for us -df = df -df.mean() -display(df) - -# we center it ourselves -X_centered = X - X.mean(axis=0) -# Then check the difference between pandas and our own set up -print(X_centered-df) -#Now we do an SVD -U, s, V = np.linalg.svd(X_centered) -c1 = V.T[:, 0] -c2 = V.T[:, 1] -W2 = V.T[:, :2] -X2D = X_centered.dot(W2) -print(X2D) -!ec - -PCA assumes that the dataset is centered around the origin. Scikit-Learn’s PCA classes take care of centering -the data for you. However, if you implement PCA yourself (as in the preceding example), or if you use other libraries, don’t -forget to center the data first. - -Once you have identified all the principal components, you can reduce the dimensionality of the dataset -down to $d$ dimensions by projecting it onto the hyperplane defined by the first $d$ principal components. -Selecting this hyperplane ensures that the projection will preserve as much variance as possible. -!bc pycod -W2 = V.T[:, :2] -X2D = X_centered.dot(W2) -!ec - -!split -===== PCA and scikit-learn ===== - -Scikit-Learn’s PCA class implements PCA using SVD decomposition just like we did before. The -following code applies PCA to reduce the dimensionality of the dataset down to two dimensions (note -that it automatically takes care of centering the data): -!bc pycod -#thereafter we do a PCA with Scikit-learn -from sklearn.decomposition import PCA -pca = PCA(n_components = 2) -X2D = pca.fit_transform(X) -print(X2D) -!ec -After fitting the PCA transformer to the dataset, you can access the principal components using the -components variable (note that it contains the PCs as horizontal vectors, so, for example, the first -principal component is equal to -!bc pycod -pca.components_.T[:, 0]. -!ec -Another very useful piece of information is the explained variance ratio of each principal component, -available via the $explained\_variance\_ratio$ variable. It indicates the proportion of the dataset’s -variance that lies along the axis of each principal component. - -!split -===== Back to the Cancer Data ===== -We can now repeat the above but applied to real data, in this case our breast cancer data. -Here we compute performance scores on the training data using logistic regression. -!bc pycod -import matplotlib.pyplot as plt -import numpy as np -from sklearn.model_selection import train_test_split -from sklearn.datasets import load_breast_cancer -from sklearn.linear_model import LogisticRegression -cancer = load_breast_cancer() - -X_train, X_test, y_train, y_test = train_test_split(cancer.data,cancer.target,random_state=0) - -logreg = LogisticRegression() -logreg.fit(X_train, y_train) -print("Train set accuracy from Logistic Regression: {:.2f}".format(logreg.score(X_train,y_train))) -# We scale the data -from sklearn.preprocessing import StandardScaler -scaler = StandardScaler() -scaler.fit(X_train) -X_train_scaled = scaler.transform(X_train) -X_test_scaled = scaler.transform(X_test) -# Then perform again a log reg fit -logreg.fit(X_train_scaled, y_train) -print("Train set accuracy scaled data: {:.2f}".format(logreg.score(X_train_scaled,y_train))) -#thereafter we do a PCA with Scikit-learn -from sklearn.decomposition import PCA -pca = PCA(n_components = 2) -X2D_train = pca.fit_transform(X_train_scaled) -# and finally compute the log reg fit and the score on the training data -logreg.fit(X2D_train,y_train) -print("Train set accuracy scaled and PCA data: {:.2f}".format(logreg.score(X2D_train,y_train))) - -!ec - -We see that our training data after the PCA decomposition has a performance similar to the non-scaled data. - -!split -===== More on the PCA ===== - -Instead of arbitrarily choosing the number of dimensions to reduce down to, it is generally preferable to -choose the number of dimensions that add up to a sufficiently large portion of the variance (e.g., 95%). -Unless, of course, you are reducing dimensionality for data visualization — in that case you will -generally want to reduce the dimensionality down to 2 or 3. -The following code computes PCA without reducing dimensionality, then computes the minimum number -of dimensions required to preserve 95% of the training set’s variance: -!bc pycod -pca = PCA() -pca.fit(X) -cumsum = np.cumsum(pca.explained_variance_ratio_) -d = np.argmax(cumsum >= 0.95) + 1 -!ec -You could then set $n\_components=d$ and run PCA again. However, there is a much better option: instead -of specifying the number of principal components you want to preserve, you can set $n\_components$ to be -a float between 0.0 and 1.0, indicating the ratio of variance you wish to preserve: -!bc pycod -pca = PCA(n_components=0.95) -X_reduced = pca.fit_transform(X) -!ec - -!split -===== Incremental PCA ===== - -One problem with the preceding implementation of PCA is that it requires the whole training set to fit in -memory in order for the SVD algorithm to run. Fortunately, Incremental PCA (IPCA) algorithms have -been developed: you can split the training set into mini-batches and feed an IPCA algorithm one minibatch -at a time. This is useful for large training sets, and also to apply PCA online (i.e., on the fly, as new -instances arrive). - -!split -===== Randomized PCA ===== - -Scikit-Learn offers yet another option to perform PCA, called Randomized PCA. This is a stochastic -algorithm that quickly finds an approximation of the first d principal components. Its computational -complexity is $O(m \times d^2)+O(d^3)$, instead of $O(m \times n^2) + O(n^3)$, so it is dramatically faster than the -previous algorithms when $d$ is much smaller than $n$. - - - - -!split -===== Kernel PCA ===== -!bblock - -The kernel trick is a mathematical technique that implicitly maps instances into a -very high-dimensional space (called the feature space), enabling nonlinear classification and regression -with Support Vector Machines. Recall that a linear decision boundary in the high-dimensional feature -space corresponds to a complex nonlinear decision boundary in the original space. -It turns out that the same trick can be applied to PCA, making it possible to perform complex nonlinear -projections for dimensionality reduction. This is called Kernel PCA (kPCA). It is often good at -preserving clusters of instances after projection, or sometimes even unrolling datasets that lie close to a -twisted manifold. -For example, the following code uses Scikit-Learn’s KernelPCA class to perform kPCA with an -!bc pycod -from sklearn.decomposition import KernelPCA -rbf_pca = KernelPCA(n_components = 2, kernel="rbf", gamma=0.04) -X_reduced = rbf_pca.fit_transform(X) -!ec - -!eblock - - -!split -===== LLE ===== - -Locally Linear Embedding (LLE) is another very powerful nonlinear dimensionality reduction -(NLDR) technique. It is a Manifold Learning technique that does not rely on projections like the previous -algorithms. In a nutshell, LLE works by first measuring how each training instance linearly relates to its -closest neighbors (c.n.), and then looking for a low-dimensional representation of the training set where -these local relationships are best preserved (more details shortly). - - - -!split -===== Other techniques ===== - - -There are many other dimensionality reduction techniques, several of which are available in Scikit-Learn. - -Here are some of the most popular: -* _Multidimensional Scaling (MDS)_ reduces dimensionality while trying to preserve the distances between the instances. -* _Isomap_ creates a graph by connecting each instance to its nearest neighbors, then reduces dimensionality while trying to preserve the geodesic distances between the instances. -* _t-Distributed Stochastic Neighbor Embedding_ (t-SNE) reduces dimensionality while trying to keep similar instances close and dissimilar instances apart. It is mostly used for visualization, in particular to visualize clusters of instances in high-dimensional space (e.g., to visualize the MNIST images in 2D). -* Linear Discriminant Analysis (LDA) is actually a classification algorithm, but during training it learns the most discriminative axes between the classes, and these axes can then be used to define a hyperplane onto which to project the data. The benefit is that the projection will keep classes as far apart as possible, so LDA is a good technique to reduce dimensionality before running another classification algorithm such as a Support Vector Machine (SVM) classifier discussed in the SVM lectures. - - -!split -===== Differential equations ===== - -The Universal Approximation Theorem states that a neural network can -approximate any function at a single hidden layer along with one input -and output layer to any given precision. Having this in mind, we will -look closer at whether a neural network manages to solve for a -function in an equation. - - -!split -===== Description of the equation to solve for ===== -A differential equation is a equation where the solution is a function. -The equation describes how the derivatives of the function behaves in a given domain along with some conditions. - -Given a differential equation, it is desirable to know how to -reformulate it into an equation a neural network can solve. Having -decided on which activation functions each layer should use, along -with the number of hidden layers and neurons within each layer, the -changeable parameters of a neural network are the weights and biases -for each neuron in every layer in the net. If a differential equation -is reformulated into an equation where minimization of some parameters -must be done, a neural net could possibly solve this equation. - -A trial solution might be tricky to find in general. Due to the -Universal Approximation Theorem, one could hope that outcome of the -deep neural net might solve a given differential equation, even though -it is used in a simple trial solution. Let us try this idea on some -well-known ordinary differential equations and thereafter try to solve -for functions defined by two variables, giving partial differential -equations. - -!split -===== Ordinary Differential Equations ===== - -An ordinary differential equation (ODE) is an equation involving functions having one variable. - -In general, an ordinary differential equation looks like - -!bt -\begin{equation} \label{ode} -f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right) = 0 -\end{equation} -!et - -where $g(x)$ is the function to find, and $g^{(n)}(x)$ is the $n$-th derivative of $g(x)$. - -The $f\left(x, g(x), g'(x), g''(x), \, \dots \, , g^{(n)}(x)\right)$ is just a way to write that there is an expression involving $x$ and $g(x), \ g'(x), \ g''(x), \, \dots \, , \text{ and } g^{(n)}(x)$ on the left side of the equality sign in (ref{ode}). -The highest order of derivative, that is the value of $n$, determines to the order of the equation. -The equation is referred to as a $n$-th order ODE. -Along with (ref{ode}), some additional conditions of the function $g(x)$ are typically given -for the solution to be unique. - -!split -===== The trial solution ===== - -Let the trial solution $g_t(x)$ be - -!bt -\begin{equation} - g_t(x) = h_1(x) + h_2(x,N(x,P)) -\end{equation} -!et - -where $h_1(x)$ is a function that makes $g_t(x)$ satisfy a given set of conditions, $N(x,P)$ a neural network with weights and biases described by $P$ and $h_2(x, N(x,P))$ some expression involving the neural network. -The role of the function $h_2(x, N(x,P))$, is to ensure that the output from $N(x,P)$ is zero when $g_t(x)$ is evaluated at the values of $x$ where the given conditions must be satisfied. -The function $h_1(x)$ should alone make $g_t(x)$ satisfy the conditions. - -But what about the network $N(x,P)$? -As described previously, an optimization method could be used to minimize the parameters of a neural network, that being its weights and biases, through backward propagation. -For the minimization to be defined, we need to have a cost function at hand to minimize. - -It is given that $f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right)$ should be equal to zero in (ref{ode}). -We can choose to consider the mean squared error as the cost function for an input $x$. -Since we are looking at one input, the cost function is just $f$ squared. -The cost function $c\left(x, P \right)$ can therefore be expressed as - -!bt -c\left(x, P\right) = \big(f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right)\big)^2 -!et - -If $N$ inputs are given as a vector $\vec x$ with elements $x_i$ for $i = 1,\dots,N$, -the cost function becomes - -!bt -\begin{equation} \label{cost} - c\left(\vec x, P\right) = \frac{1}{N} \sum_{i=1}^N \big(f\left(x_i, \, g(x_i), \, g'(x_i), \, g''(x_i), \, \dots \, , \, g^{(n)}(x_i)\right)\big)^2 -\end{equation} -!et - -The neural net should then find some parameters $P$ that minimizes the cost function in -(ref{cost}) for a set of $N$ training samples $x_i$. - -!split -===== Minimizing the cost function using gradient descent and automatic differentiation ===== -To perform the minimization using gradient descent, the gradient of $c\left(\vec x, P\right)$ is needed. -It might happen so that finding an analytical expression of the gradient of $c(\vec x, P)$ from (ref{cost}) gets too messy, depending on which cost function one desires to use. - -Luckily, there exists libraries that makes the job for us through automatic differentiation. -Automatic differentiation is a method of finding the derivatives numerically with very high precision. - -In the forthcoming examples presenting possible usages of Autograd and TensorFlow, -it is shown how one could set up a neural network using gradient descent solving a differential -equation. - -!split -===== Example: Exponential decay and setting up the network using Autograd ===== -An exponential decay of a quantity $g(x)$ is described by the equation - -!bt -\begin{equation} \label{solve_expdec} - g'(x) = -\gamma g(x) -\end{equation} -!et - -with $g(0) = g_0$ for some chosen initial value $g_0$. - -The analytical solution of (ref{solve_expdec}) is - -!bt -\begin{equation} - g(x) = g_0 \exp\left(-\gamma x\right) -\end{equation} -!et - -Having an analytical solution at hand, it is possible to use it to compare how well a neural network finds a solution of (ref{solve_expdec}). - -In this example, a neural network will be implemented using Autograd in order to perform backpropagation. - -!split -===== The function to solve for ===== - -The program will use a neural network to solve - -!bt -\begin{equation} \label{solveode} -g'(x) = -\gamma g(x) -\end{equation} -!et - -where $g(0) = g_0$ with $\gamma$ and $g_0$ being some chosen values. - -In this example, $\gamma = 2$ and $g_0 = 10$. - -!split -===== The trial solution ===== -To begin with, a trial solution $g_t(t)$ must be chosen. A general trial solution for ordinary differential equations could be - -!bt -g_t(x, P) = h_1(x) + h_2(x, N(x, P)) -!et - -with $h_1(x)$ ensuring that $g_t(x)$ satisfies some conditions and $h_2(x,N(x, P))$ an expression involving $x$ and the output from the neural network $N(x,P)$ with $P $ being the collection of the weights and biases for each layer. For now, it is assumed that the network consists of one input layer, one hidden layer, and one output layer. - -In this network, there are no weights and bias at the input layer, so $P = \{ P_{\text{hidden}}, P_{\text{output}} \}$. -If there are $N_{\text{hidden} }$ neurons in the hidden layer, then $P_{\text{hidden}}$ is a $N_{\text{hidden} } \times (1 + N_{\text{input}})$ matrix, given that there are $N_{\text{input}}$ neurons in the input layer. - -The first column in $P_{\text{hidden} }$ represents the bias for each neuron in the hidden layer and the second column represents the weights for each neuron in the hidden layer from the input layer. -If there are $N_{\text{output} }$ neurons in the output layer, then $P_{\text{output}} $ is a $N_{\text{output} } \times (1 + N_{\text{hidden} })$ matrix. - -Its first column represents the bias of each neuron and the remaining columns represents the weights to each neuron. - -It is given that $g(0) = g_0$. The trial solution must fulfill this condition to be a proper solution of (ref{solveode}). A possible way to ensure that $g_t(0, P) = g_0$, is to let $F(N(x,P)) = x \cdot N(x,P)$ and $A(x) = g_0$. This gives the following trial solution: - -!bt -\begin{equation} \label{trial} -g_t(x, P) = g_0 + x \cdot N(x, P) -\end{equation} -!et - -!split -===== Reformulating the problem ===== -We wish that our neural network manages to minimize a given cost function. - -A reformulation of out equation, (ref{solveode}), must therefore be done, -such that it describes the problem a neural network can solve for. - -The neural network must find the set of weights and biases $P$ such that the trial solution in (ref{trial}) satisfies (ref{solveode}). - -The trial solution - -!bt -g_t(x, P) = g_0 + x \cdot N(x, P) -!et - -has been chosen such that it already solves the condition $g(0) = g_0$. What remains, is to find $P$ such that - -!bt -\begin{equation} \label{nnmin} -g_t'(x, P) = - \gamma g_t(x, P) -\end{equation} -!et - -is fulfilled as *best as possible*. - -The left hand side and right hand side of (ref{nnmin}) must be computed separately, and then the neural network must choose weights and biases, contained in $P$, such that the sides are equal as best as possible. -This means that the absolute or squared difference between the sides must be as close to zero, ideally equal to zero. -In this case, the difference squared shows to be an appropriate measurement of how erroneous the trial solution is with respect to $P$ of the neural network. - -This gives the following cost function our neural network must solve for: - -!bt -\min_{P}\Big\{ \big(g_t'(x, P) - ( -\gamma g_t(x, P) \big)^2 \Big\} -!et - -(the notation $\min_{P}\{ f(x, P) \}$ means that we desire to find $P$ that yields the minimum of $f(x, P)$) - -or, in terms of weights and biases for the hidden and output layer in our network: - -!bt -\min_{P_{\text{hidden} }, \ P_{\text{output} }}\Big\{ \big(g_t'(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) - ( -\gamma g_t(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) \big)^2 \Big\} -!et - -for an input value $x$. - -If the neural network evaluates $g_t(x, P)$ at more values for $x$, say $N$ values $x_i$ for $i = 1, \dots, N$, then the *total* error to minimize becomes - -!bt -\begin{equation} \label{min} -\min_{P}\Big\{\frac{1}{N} \sum_{i=1}^N \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 \Big\} -\end{equation} -!et - -Letting $\vec x$ be a vector with elements $x_i$ and $c(\vec x, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2$ denote the cost function, the minimization problem that our network must solve, becomes - -!bt -\min_{P} c(\vec x, P) -!et - -In terms of $P_{\text{hidden} }$ and $P_{\text{output} }$, this could also be expressed as - -$$ -\min_{P_{\text{hidden} }, \ P_{\text{output} }} c(\vec x, \{P_{\text{hidden} }, P_{\text{output} }\}) -$$ - -!split -===== A possible implementation of a neural network using Autograd ===== - -For simplicity, it is assumed that the input is an array $\vec x = (x_1, \dots, x_N)$ with $N$ elements. It is at these points the neural network should find $P$ such that it fulfills (ref{min}). - -First, the neural network must feed forward the inputs. -This means that $\vec x$ must be passed through an input layer, a hidden layer and a output layer. The input layer in this case, does not need to process the data any further. -The input layer will consist of $N_{\text{input} }$ neurons, passing its element to each neuron in the hidden layer. The number of neurons in the hidden layer will be $N_{\text{hidden} }$. - -For the $i$-th in the hidden layer with weight $w_i^{\text{hidden} }$ and bias $b_i^{\text{hidden} }$, the weighting from the $j$-th neuron at the input layer is: - -!bt -\begin{aligned} -z_{i,j}^{\text{hidden}} &= b_i^{\text{hidden}} + w_i^{\text{hidden}}x_j \\ -&= -\begin{pmatrix} -b_i^{\text{hidden}} & w_i^{\text{hidden}} -\end{pmatrix} -\begin{pmatrix} -1 \\ -x_j -\end{pmatrix} -\end{aligned} -!et - -The result after weighting the inputs at the $i$-th hidden neuron can be written as a vector: - -!bt -\begin{aligned} -\vec{z}_{i}^{\text{hidden}} &= \Big( b_i^{\text{hidden}} + w_i^{\text{hidden}}x_1 , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_2, \ \dots \, , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_N\Big) \\ -&= -\begin{pmatrix} - b_i^{\text{hidden}} & w_i^{\text{hidden}} -\end{pmatrix} -\begin{pmatrix} -1 & 1 & \dots & 1 \\ -x_1 & x_2 & \dots & x_N -\end{pmatrix} \\ -&= \vec{p}_{i, \text{hidden}}^T X -\end{aligned} -!et - -The vector $\vec{p}_{i, \text{hidden}}^T$ constitutes each row in $P_{\text{hidden} }$, which contains the weights for the neural network to minimize according to (ref{min}). - -After having found $\vec{z}_{i}^{\text{hidden}} $ for every $i$-th neuron within the hidden layer, the vector will be sent to an activation function $a_i(\vec{z})$. - -In this example, the sigmoid function has been chosen to be the activation function for each hidden neuron: - -!bt -f(z) = \frac{1}{1 + \exp{(-z)}} -!et - -It is possible to use other activations functions for the hidden layer also. - -The output $\vec{x}_i^{\text{hidden} }$from each $i$-th hidden neuron is: - -$$ -\vec{x}_i^{\text{hidden} } = f\big( \vec{z}_{i}^{\text{hidden}} \big) -$$ - -The outputs $\vec{x}_i^{\text{hidden} } $ are then sent to the output layer. - -The output layer consists of one neuron in this case, and combines the output from each of the neurons in the hidden layers. The output layer combines the results from the hidden layer using some weights $ w_i^{\text{output}}$ and biases $b_i^{\text{output}}$. In this case, it is assumes that the number of neurons in the output layer is one. - -The procedure of weighting the output neuron $j$ in the hidden layer to the $i$-th neuron in the output layer is similar as for the hidden layer described previously. - -!bt -\begin{aligned} -z_{1,j}^{\text{output}} & = -\begin{pmatrix} -b_1^{\text{output}} & \vec{w}_1^{\text{output}} -\end{pmatrix} -\begin{pmatrix} -1 \\ -\vec{x}_j^{\text{hidden}} -\end{pmatrix} -\end{aligned} -!et - -Expressing $z_{1,j}^{\text{output}}$ as a vector gives the following way of weighting the inputs from the hidden layer: - -!bt -\vec{z}_{1}^{\text{output}} = -\begin{pmatrix} -b_1^{\text{output}} & \vec{w}_1^{\text{output}} -\end{pmatrix} -\begin{pmatrix} -1 & 1 & \dots & 1 \\ -\vec{x}_1^{\text{hidden}} & \vec{x}_2^{\text{hidden}} & \dots & \vec{x}_N^{\text{hidden}} -\end{pmatrix} -!et - -In this case we seek a continuous range of values since we are approximating a function. This means that after computing $\vec{z}_{1}^{\text{output}}$ the neural network has finished its feed forward step, and $\vec{z}_{1}^{\text{output}}$ is the final output of the network. - -!split -===== Backpropagation using Autograd ===== -The next step is to decide how the parameters should be changed such that they minimize the cost function. - -The chosen cost function for this problem is - -!bt -c(\vec x, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 -!et - -In order to minimize the cost function, an optimization method must be chosen. - -Here, gradient descent with a constant step size has been chosen. - -!split -===== Gradient descent ===== -The idea of the gradient descent algorithm is to update parameters in direction where the cost function decreases goes to a minimum. - -In general, the update of some parameters $\vec \omega$ given a cost function defined by some weights $\vec \omega$, $c(\vec x, \vec \omega)$, goes as follows: - -!bt -\vec \omega_{\text{new} } = \vec \omega - \lambda \nabla_{\vec \omega} c(\vec x, \vec \omega) -!et - -for a number of iterations or until $ \big|\big| \vec \omega_{\text{new} } - \vec \omega \big|\big|$ becomes smaller than some given tolerance. - -The value of $\lambda$ decides how large steps the algorithm must take in the direction of $ \nabla_{\vec \omega} c(\vec x, \vec \omega)$. -The notation $\nabla_{\vec \omega}$ express the gradient with respect to the elements in $\vec \omega$. - -In our case, we have to minimize the cost function $c(\vec x, P)$ with respect to the two sets of weights and biases, that is for the hidden layer $P_{\text{hidden} }$ and for the output layer $P_{\text{output} }$ . - -This means that $P_{\text{hidden} }$ and $P_{\text{output} }$ is updated by - -!bt -\begin{aligned} -P_{\text{hidden},\text{new}} &= P_{\text{hidden}} - \lambda \nabla_{P_{\text{hidden}}} c(\vec x, P) \\ -P_{\text{output},\text{new}} &= P_{\text{output}} - \lambda \nabla_{P_{\text{output}}} c(\vec x, P) -\end{aligned} -!et - -In general, one could risk using a cost function having gradients that are cumbersome to derive analytically. -For our case, the cost functions are just the mean squared error. -One could employ an implementation of the back propagation for this case, but we will emphasis -on how one could use automatic differentiation in order to train the network. - -However, it might be useful to know how automatic differentiation can be used, e.g through Autograd, in order to test an implementation. - -!split -===== The network with one input, hidden, and output layer ===== - -!bc pycod -# Autograd will be used for later, so the numpy wrapper for Autograd must be imported -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -# Assuming one input, hidden, and output layer -def neural_network(params, x): - - # Find the weights (including and biases) for the hidden and output layer. - # Assume that params is a list of parameters for each layer. - # The biases are the first element for each array in params, - # and the weights are the remaning elements in each array in params. - - w_hidden = params[0] - w_output = params[1] - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - ## Hidden layer: - - # Add a row of ones to include bias - x_input = np.concatenate((np.ones((1,num_values)), x_input ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_input) - x_hidden = sigmoid(z_hidden) - - ## Output layer: - - # Include bias: - x_hidden = np.concatenate((np.ones((1,num_values)), x_hidden ), axis = 0) - - z_output = np.matmul(w_output, x_hidden) - x_output = z_output - - return x_output - -# The trial solution using the deep neural network: -def g_trial(x,params, g0 = 10): - return g0 + x*neural_network(params,x) - -# The right side of the ODE: -def g(x, g_trial, gamma = 2): - return -gamma*g_trial - -# The cost function: -def cost_function(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial(x,P) - - # Find the derivative w.r.t x of the neural network - d_net_out = elementwise_grad(neural_network,1)(P,x) - - # Find the derivative w.r.t x of the trial function - d_g_t = elementwise_grad(g_trial,0)(x,P) - - # The right side of the ODE - func = g(x, g_t) - - err_sqr = (d_g_t - func)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum / np.size(err_sqr) - -# Solve the exponential decay ODE using neural network with one input, hidden, and output layer -def solve_ode_neural_network(x, num_neurons_hidden, num_iter, lmb): - ## Set up initial weights and biases - - # For the hidden layer - p0 = npr.randn(num_neurons_hidden, 2 ) - - # For the output layer - p1 = npr.randn(1, num_neurons_hidden + 1 ) # +1 since bias is included - - P = [p0, p1] - - print('Initial cost: %g'%cost_function(P, x)) - - ## Start finding the optimal weights using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_grad = grad(cost_function,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of two arrays; - # one for the gradient w.r.t P_hidden and - # one for the gradient w.r.t P_output - cost_grad = cost_function_grad(P, x) - - P[0] = P[0] - lmb * cost_grad[0] - P[1] = P[1] - lmb * cost_grad[1] - - print('Final cost: %g'%cost_function(P, x)) - - return P - -def g_analytic(x, gamma = 2, g0 = 10): - return g0*np.exp(-gamma*x) - -# Solve the given problem -if __name__ == '__main__': - # Set seed such that the weight are initialized - # with same weights and biases for every run. - npr.seed(15) - - ## Decide the vales of arguments to the function to solve - N = 10 - x = np.linspace(0, 1, N) - - ## Set up the initial parameters - num_hidden_neurons = 10 - num_iter = 10000 - lmb = 0.001 - - # Use the network - P = solve_ode_neural_network(x, num_hidden_neurons, num_iter, lmb) - - # Print the deviation from the trial solution and true solution - res = g_trial(x,P) - res_analytical = g_analytic(x) - - print('Max absolute difference: %g'%np.max(np.abs(res - res_analytical))) - - # Plot the results - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(x, res_analytical) - plt.plot(x, res[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('x') - plt.ylabel('g(x)') - plt.show() -!ec - -!split -===== The network with one input layer, specified number of hidden layers, and one output layer output layer ===== - -It is also possible to extend the construction of our network into a more general one, allowing the network to contain more than one hidden layers. - -The number of neurons within each hidden layer are given as a list of integers in the program below. - -!bc pycod -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -# The neural network with one input layer and one output layer, -# but with number of hidden layers specified by the user. -def deep_neural_network(deep_params, x): - # N_hidden is the number of hidden layers - - N_hidden = np.size(deep_params) - 1 # -1 since params consists of - # parameters to all the hidden - # layers AND the output layer. - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - # Due to multiple hidden layers, define a variable referencing to the - # output of the previous layer: - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output - -# The trial solution using the deep neural network: -def g_trial_deep(x,params, g0 = 10): - return g0 + x*deep_neural_network(params, x) - -# The right side of the ODE: -def g(x, g_trial, gamma = 2): - return -gamma*g_trial - -# The same cost function as before, but calls deep_neural_network instead. -def cost_function_deep(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial_deep(x,P) - - # Find the derivative w.r.t x of the neural network - d_net_out = elementwise_grad(deep_neural_network,1)(P,x) - - # Find the derivative w.r.t x of the trial function - d_g_t = elementwise_grad(g_trial_deep,0)(x,P) - - # The right side of the ODE - func = g(x, g_t) - - err_sqr = (d_g_t - func)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum / np.size(err_sqr) - -# Solve the exponential decay ODE using neural network with one input and one output layer, -# but with specified number of hidden layers from the user. -def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): - # num_hidden_neurons is now a list of number of neurons within each hidden layer - - # The number of elements in the list num_hidden_neurons thus represents - # the number of hidden layers. - - # Find the number of hidden layers: - N_hidden = np.size(num_neurons) - - ## Set up initial weights and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 ) - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: %g'%cost_function_deep(P, x)) - - ## Start finding the optimal weights using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_deep_grad = grad(cost_function_deep,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases - # in the hidden layers and output layers evaluated at x. - cost_deep_grad = cost_function_deep_grad(P, x) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_deep_grad[l] - - print('Final cost: %g'%cost_function_deep(P, x)) - - return P - -def g_analytic(x, gamma = 2, g0 = 10): - return g0*np.exp(-gamma*x) - -# Solve the given problem -if __name__ == '__main__': - npr.seed(15) - - ## Decide the vales of arguments to the function to solve - N = 10 - x = np.linspace(0, 1, N) - - ## Set up the initial parameters - num_hidden_neurons = np.array([10,10]) - num_iter = 10000 - lmb = 0.001 - - P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb) - - res = g_trial_deep(x,P) - res_analytical = g_analytic(x) - - plt.figure(figsize=(10,10)) - - plt.title('Performance of a deep neural network solving an ODE compared to the analytical solution') - plt.plot(x, res_analytical) - plt.plot(x, res[0,:]) - plt.legend(['analytical','dnn']) - plt.ylabel('g(x)') - plt.show() -!ec - -!split -===== Example: Population growth, comparing Autograd, TensorFlow, and Euler's scheme ===== - -A logistic model of population growth assumes that a population converges toward an equilibrium. -The population growth can be modeled by - -!bt -\begin{equation} \label{log} - g'(t) = \alpha g(t)(A - g(t)) -\end{equation} -!et - -where $g(t)$ is the population density at time $t$, $\alpha > 0$ the growth rate and $A > 0$ is the maximum population number in the environment. -Also, at $t = 0$ the population has the size $g(0) = g_0$, where $g_0$ is some chosen constant. - -In this example, similar network as for the exponential decay using Autograd has been used to solve the equation. However, as the implementation might suffer from e.g numerical instability -and high execution time (this might be more apparent in the examples solving PDEs), -a network has been constructed using TensorFlow also. -For comparison, the forward Euler method has been implemented in order to see how the networks performs compared to a numerical scheme. - -!split -===== Setting up the problem ===== - -Here, we will model a population $g(t)$ in an environment having carrying capacity $A$. -The population follows the model - -!bt -\begin{equation} \label{solveode_population} -g'(t) = \alpha g(t)(A - g(t)) -\end{equation} -!et - -where $g(0) = g_0$. - -In this example, we let $\alpha = 2$, $A = 1$, and $g_0 = 1.2$. - -!split -===== The trial solution ===== -We will get a slightly different trial solution, as the boundary conditions are different -compared to the case for exponential decay. - -A possible trial solution satisfying the condition $g(0) = g_0$ could be - -$$ -h_1(t) = g_0 + t \cdot N(t,P) -$$ - -with $N(t,P)$ being the output from the neural network with weights and biases for each layer collected in the set $P$. - -The analytical solution is - -$$ -g(t) = \frac{Ag_0}{g_0 + (A - g_0)\exp(-\alpha A t)} -$$ - -!split -===== The program using Autograd ===== - -The network will be the similar as for the exponential decay example, but with some small modifications for our problem. - -!bc pycod -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -# Function to get the parameters. -# Done such that one can easily change the paramaters after one's liking. -def get_parameters(): - alpha = 2 - A = 1 - g0 = 1.2 - return alpha, A, g0 - -def deep_neural_network(P, x): - # N_hidden is the number of hidden layers - N_hidden = np.size(P) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - # Due to multiple hidden layers, define a variable referencing to the - # output of the previous layer: - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = P[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = P[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output - - -def cost_function_deep(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial_deep(x,P) - - # Find the derivative w.r.t x of the trial function - d_g_t = elementwise_grad(g_trial_deep,0)(x,P) - - # The right side of the ODE - func = f(x, g_t) - - err_sqr = (d_g_t - func)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum / np.size(err_sqr) - -# The right side of the ODE: -def f(x, g_trial): - alpha,A, g0 = get_parameters() - return alpha*g_trial*(A - g_trial) - -# The trial solution using the deep neural network: -def g_trial_deep(x, params): - alpha,A, g0 = get_parameters() - return g0 + x*deep_neural_network(params,x) - -# The analytical solution: -def g_analytic(t): - alpha,A, g0 = get_parameters() - return A*g0/(g0 + (A - g0)*np.exp(-alpha*A*t)) - -def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): - # num_hidden_neurons is now a list of number of neurons within each hidden layer - - # Find the number of hidden layers: - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 ) - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: %g'%cost_function_deep(P, x)) - - ## Start finding the optimal weigths using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_deep_grad = grad(cost_function_deep,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases - # in the hidden layers and output layers evaluated at x. - cost_deep_grad = cost_function_deep_grad(P, x) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_deep_grad[l] - - print('Final cost: %g'%cost_function_deep(P, x)) - - return P - -if __name__ == '__main__': - npr.seed(4155) - - ## Decide the vales of arguments to the function to solve - Nt = 10 - T = 1 - t = np.linspace(0,T, Nt) - - ## Set up the initial parameters - num_hidden_neurons = [100, 50, 25] - num_iter = 1000 - lmb = 1e-3 - - P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb) - - g_dnn_ag = g_trial_deep(t,P) - g_analytical = g_analytic(t) - - # Find the maximum absolute difference between the solutons: - diff_ag = np.max(np.abs(g_dnn_ag - g_analytical)) - print("The max absolute difference between the solutions is: %g"%diff_ag) - - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(t, g_analytical) - plt.plot(t, g_dnn_ag[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('t') - plt.ylabel('g(t)') - - plt.show() -!ec - -!split -===== Using forward Euler to solve the ODE ===== - -A straight-forward way of solving an ODE numerically, is to use Euler's method. - -Euler's method uses Taylor series to approximate the value at a function $f$ at a step $\Delta x$ from $x$: - -$$ -f(x + \Delta x) \approx f(x) + \Delta x f'(x) -$$ - -In our case, using Euler's method to approximate the value of $g$ at a step $\Delta t$ from $t$ yields - -!bt -\begin{aligned} - g(t + \Delta t) &\approx g(t) + \Delta t g'(t) \\ - &= g(t) + \Delta t \big(\alpha g(t)(A - g(t))\big) -\end{aligned} -!et -along with the condition that $g(0) = g_0$. - -Let $t_i = i \cdot \Delta t$ where $\Delta t = \frac{T}{N_t-1}$ where $T$ is the final time our solver must solve for and $N_t$ the number of values for $t \in [0, T]$ for $i = 0, \dots, N_t-1$. - -For $i \geq 1$, we have that -!bt -\begin{aligned} -t_i &= i\Delta t \\ -&= (i - 1)\Delta t + \Delta t \\ -&= t_{i-1} + \Delta t -\end{aligned} -!et - -Now, if $g_i = g(t_i)$ then - -!bt -\begin{equation} - \begin{aligned} - g_i &= g(t_i) \\ - &= g(t_{i-1} + \Delta t) \\ - &\approx g(t_{i-1}) + \Delta t \big(\alpha g(t_{i-1})(A - g(t_{i-1}))\big) \\ - &= g_{i-1} + \Delta t \big(\alpha g_{i-1}(A - g_{i-1})\big) - \end{aligned} -\end{equation} \label{odenum} -!et -for $i \geq 1$ and $g_0 = g(t_0) = g(0) = g_0$. - -Equation (ref{odenum}) could be implemented in the following way, -extending the program that uses the network using Autograd: - -!bc pycod -# Assume that all function definitions from the example program using Autograd -# are located here. - -if __name__ == '__main__': - npr.seed(4155) - - ## Decide the vales of arguments to the function to solve - Nt = 10 - T = 1 - t = np.linspace(0,T, Nt) - - ## Set up the initial parameters - num_hidden_neurons = [100,50,25] - num_iter = 1000 - lmb = 1e-3 - - P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb) - - g_dnn_ag = g_trial_deep(t,P) - g_analytical = g_analytic(t) - - # Find the maximum absolute difference between the solutons: - diff_ag = np.max(np.abs(g_dnn_ag - g_analytical)) - print("The max absolute difference between the solutions is: %g"%diff_ag) - - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(t, g_analytical) - plt.plot(t, g_dnn_ag[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('t') - plt.ylabel('g(t)') - - ## Find an approximation to the funtion using forward Euler - - alpha, A, g0 = get_parameters() - dt = T/(Nt - 1) - - # Perform forward Euler to solve the ODE - g_euler = np.zeros(Nt) - g_euler[0] = g0 - - for i in range(1,Nt): - g_euler[i] = g_euler[i-1] + dt*(alpha*g_euler[i-1]*(A - g_euler[i-1])) - - # Print the errors done by each method - diff1 = np.max(np.abs(g_euler - g_analytical)) - diff2 = np.max(np.abs(g_dnn_ag[0,:] - g_analytical)) - - print('Max absolute difference between Euler method and analytical: %g'%diff1) - print('Max absolute difference between deep neural network and analytical: %g'%diff2) - - # Plot results - plt.figure(figsize=(10,10)) - - plt.plot(t,g_euler) - plt.plot(t,g_analytical) - plt.plot(t,g_dnn_ag[0,:]) - - plt.legend(['euler','analytical','dnn']) - plt.xlabel('Time t') - plt.ylabel('g(t)') - - plt.show() -!ec - -Running the program gives - -!bc -Max absolute difference between Euler method and analytical: 0.011225 -Max absolute difference between deep neural network and analytical: 0.00424909 -!ec - -!split -===== Using TensorFlow to model logistic population growth ===== - -TensorFlow is a library widely used in the machine learning community. -A neural network can be set up in a flexible manner, where various optimization algorithms are implemented and different types of networks can be used, making it easier to experiment on solving differential equations using neural networks. - -!split -===== The general program flow in TensorFlow ===== - -Usually, a program in TensorFlow is divided into two parts; the *construction phase* and the *execution phase*. -In the construction phase, the computational graph that TensorFlow uses to perform its calculations are set up. -In the execution phase, TensorFlow evaluates any procedure that was defined in the construction phase. - -===== Program flow in TensorFlow - Construction phase ===== - -Here, the architecture for the neural network will be set up, along with the cost function and an optimizer class used during training of the network. -Note that TensorFlow uses a different convention for the weighting done in each neuron in each layer within the network than in the implementation using Autograd. -The matrix-vector multiplication between the input from the previous layer and the weighting at the neuron at current layer in the program using Autograd, is the transpose of the convention used in TensorFlow. But it will not affect that much our construction, as TensorFlow takes care of most of the computations. The only thing we have to be aware of, is how the dimensions are for our inputs. - -===== Program flow in TensorFlow - Execution phase ===== - -The computation graph has been defined, and is ready to be evaluated. -In order to get access to the graph, it has to be initialized and be runned within a Session. - -===== The full program modeling logistic population growth using TensorFlow ===== - -!bc pycod -import tensorflow as tf -import numpy as np -import matplotlib.pyplot as plt - -# Just to reset the graph such that it is possible to rerun this in a -# Jupyter cell without resetting the whole kernel. -tf.reset_default_graph() - -# Set a seed to ensure getting the same results from every run -tf.set_random_seed(4155) - -Nt = 10 -T = 1 -t = np.linspace(0,T, Nt) - -## The construction phase - -# Convert the values the trial solution is evaluated at to a tensor. -t_tf = tf.convert_to_tensor(t.reshape(-1,1),dtype=tf.float64) -zeros = tf.reshape(tf.convert_to_tensor(np.zeros(t.shape)),shape=(-1,1)) - -# Define the parameters of the equation -alpha = tf.constant(2.,dtype=tf.float64) -A = tf.constant(1.,dtype=tf.float64) -g0 = tf.constant(1.2,dtype=tf.float64) - -num_iter = 100000 - -# Define the number of neurons at each hidden layer -num_hidden_neurons = [100,50,25] -num_hidden_layers = np.size(num_hidden_neurons) - -# Construct the network. -# tf.name_scope is used to group each step in the construction, -# just for a more organized visualization in TensorBoard -with tf.name_scope('dnn'): - - # Input layer - previous_layer = t_tf - - # Hidden layers - for l in range(num_hidden_layers): - current_layer = tf.layers.dense(previous_layer, num_hidden_neurons[l], name='hidden%d'%(l+1), activation=tf.nn.sigmoid) - previous_layer = current_layer - - # Output layer - dnn_output = tf.layers.dense(previous_layer, 1, name='output') - -# Define the cost function -with tf.name_scope('cost'): - g_trial = g0 + t_tf*dnn_output - d_g_trial = tf.gradients(g_trial,t_tf) - - func = alpha*g_trial*(A - g_trial) - cost = tf.losses.mean_squared_error(zeros, d_g_trial[0] - func) - - -# Choose the method to minimize the cost function, along with a learning rate -learning_rate = 1e-2 -with tf.name_scope('train'): - optimizer = tf.train.GradientDescentOptimizer(learning_rate) - traning_op = optimizer.minimize(cost) - -# Set up a referance to the result from the neural network: -g_dnn_tf = None - -# Define a node that initializes all of the other nodes in the computational graph -# used by TensorFlow: -init = tf.global_variables_initializer() - -## Execution phase - -# Start a session where the graph defined from the construction phase can be evaluated at: -with tf.Session() as sess: - # Initialize the whole graph - init.run() - - # Evaluate the initial cost: - print('Initial cost: %g'%cost.eval()) - - # The training of the network: - for i in range(num_iter): - sess.run(traning_op) - - # If one desires to see how the cost function behaves for each iteration: - #if i % 1000 == 0: - # print(cost.eval()) - - # Training is done, and we have an approximate solution to the ODE - print('Final cost: %g'%cost.eval()) - - # Store the result - g_dnn_tf = g_trial.eval() - -# Compare with analytical solution -def get_parameters(): - alpha = 2 - A = 1 - g0 = 1.2 - return alpha, A, g0 - -def g_analytic(t): - alpha,A, g0 = get_parameters() - return A*g0/(g0 + (A - g0)*np.exp(-alpha*A*t)) - -g_analytical = g_analytic(t) -diff_tf = g_dnn_tf - g_analytical.reshape(-1,1) - -print('\nMax absolute difference between the analytical solution and solution from TensorFlow DNN: %g'%np.max(np.abs(diff_tf))) - -# Plot the result -plt.figure(figsize=(10,10)) - -plt.title('Numerical solutions of the ODE') - -plt.plot(t, g_dnn_tf) -plt.plot(t, g_analytical) - -plt.legend(['dnn, tensorflow', 'exact']) -plt.xlabel('Time t') -plt.ylabel('g(t)') - -plt.show() - -!ec - -!split -===== Example: Solving the one dimensional Poisson equation using Autograd and TensorFlow ===== - -The Poisson equation for $g(x)$ in one dimension is - -!bt -\begin{equation} \label{poisson} - -g''(x) = f(x) -\end{equation} -!et - -where $f(x)$ is a given function for $x \in (0,1)$. - -The conditions that $g(x)$ is chosen to fulfill, are -!bt -\begin{align*} - g(0) &= 0 \\ - g(1) &= 0 -\end{align*} -!et - -This equation can be solved numerically using programs where e.g Autograd and TensorFlow are used. -The results from the networks can then be compared to the analytical solution. -In addition, it could be interesting to see how a typical method for numerically solving second order ODEs compares to the neural networks. - -There exists many different optimization methods implemented in TensorFlow. -In the examples program using TensorFlow, it could also be of interest to see how -the choice of an optimization method affects our results. -In the "TensorFlow documentation about optimizers":"https://www.tensorflow.org/versions/r1.2/api_guides/python/train#Optimizers", a list over available optimization methods are shown. - -!split -===== The specific equation to solve for ===== - -Here, the function $g(x)$ to solve for follows the equation - -!bt --g''(x) = f(x),\qquad x \in (0,1) -!et - -where $f(x)$ is a given function, along with the chosen conditions - -!bt -\begin{aligned} -g(0) = g(1) = 0 -\end{aligned}\label{cond} -!et - -In this example, we consider the case when $f(x) = (3x + x^2)\exp(x)$. - -For this case, a possible trial solution satisfying the conditions could be - -!bt -g_t(x) = x \cdot (1-x) \cdot N(P,x) -!et - -The analytical solution for this problem is - -!bt -g(x) = x(1 - x)\exp(x) -!et - -!split -===== Solving the equation using Autograd ===== - -!bc pycod -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -def deep_neural_network(deep_params, x): - # N_hidden is the number of hidden layers - N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - # Due to multiple hidden layers, define a variable referencing to the - # output of the previous layer: - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output - -def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): - # num_hidden_neurons is now a list of number of neurons within each hidden layer - - # Find the number of hidden layers: - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 ) - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: %g'%cost_function_deep(P, x)) - - ## Start finding the optimal weigths using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_deep_grad = grad(cost_function_deep,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases - # in the hidden layers and output layers evaluated at x. - cost_deep_grad = cost_function_deep_grad(P, x) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_deep_grad[l] - - print('Final cost: %g'%cost_function_deep(P, x)) - - return P - -## Set up the cost function specified for this Poisson equation: - -# The right side of the ODE -def f(x): - return (3*x + x**2)*np.exp(x) - -def cost_function_deep(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial_deep(x,P) - - # Find the derivative w.r.t x of the trial function - d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P) - - right_side = f(x) - - err_sqr = (-d2_g_t - right_side)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum/np.size(err_sqr) - -# The trial solution: -def g_trial_deep(x,P): - return x*(1-x)*deep_neural_network(P,x) - -# The analytic solution; -def g_analytic(x): - return x*(1-x)*np.exp(x) - -if __name__ == '__main__': - npr.seed(4155) - - ## Decide the vales of arguments to the function to solve - Nx = 10 - x = np.linspace(0,1, Nx) - - ## Set up the initial parameters - num_hidden_neurons = [200,100] - num_iter = 1000 - lmb = 1e-3 - - P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb) - - g_dnn_ag = g_trial_deep(x,P) - g_analytical = g_analytic(x) - - # Find the maximum absolute difference between the solutons: - max_diff = np.max(np.abs(g_dnn_ag - g_analytical)) - print("The max absolute difference between the solutions is: %g"%max_diff) - - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(x, g_analytical) - plt.plot(x, g_dnn_ag[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('x') - plt.ylabel('g(x)') - plt.show() -!ec - -!split -===== Comparing with a numerical scheme ===== - -The Poisson equation is possible to solve using Taylor series to approximate the second derivative. - -Using Taylor series, the second derivative can be expressed as - -$$ -g''(x) = \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2} + E_{\Delta x}(x) -$$ - -where $\Delta x$ is a small step size and $E_{\Delta x}(x)$ being the error term. - -Looking away from the error terms gives an approximation to the second derivative: - -!bt -\begin{equation} \label{approx} -g''(x) \approx \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2} -\end{equation} -!et - -If $x_i = i \Delta x = x_{i-1} + \Delta x$ and $g_i = g(x_i)$ for $i = 1,\dots N_x - 2$ with $N_x$ being the number of values for $x$, (ref{approx}) becomes - -!bt -\begin{aligned} -g''(x_i) &\approx \frac{g(x_i + \Delta x) - 2g(x_i) + g(x_i -\Delta x)}{\Delta x^2} \\ -&= \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2} -\end{aligned} -!et - -Since we know from our problem that - -!bt -\begin{aligned} --g''(x) &= f(x) \\ -&= (3x + x^2)\exp(x) -\end{aligned} -!et - -along with the conditions $g(0) = g(1) = 0$, -the following scheme can be used to find an approximate solution for $g(x)$ numerically: - -!bt -\begin{equation} - \begin{aligned} - -\Big( \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2} \Big) &= f(x_i) \\ - -g_{i+1} + 2g_i - g_{i-1} &= \Delta x^2 f(x_i) - \end{aligned} -\end{equation} \label{odesys} -!et - -for $i = 1, \dots, N_x - 2$ where $g_0 = g_{N_x - 1} = 0$ and $f(x_i) = (3x_i + x_i^2)\exp(x_i)$, which is given for our specific problem. - -The equation can be rewritten into a matrix equation: - -!bt -\begin{aligned} -\begin{pmatrix} -2 & -1 & 0 & \dots & 0 \\ --1 & 2 & -1 & \dots & 0 \\ -\vdots & & \ddots & & \vdots \\ -0 & \dots & -1 & 2 & -1 \\ -0 & \dots & 0 & -1 & 2\\ -\end{pmatrix} -\begin{pmatrix} -g_1 \\ -g_2 \\ -\vdots \\ -g_{N_x - 3} \\ -g_{N_x - 2} -\end{pmatrix} -&= -\Delta x^2 -\begin{pmatrix} -f(x_1) \\ -f(x_2) \\ -\vdots \\ -f(x_{N_x - 3}) \\ -f(x_{N_x - 2}) -\end{pmatrix} \\ -A\vec{g} &= \vec{f} -\end{aligned} -!et - -which makes it possible to solve for the vector $\vec{g}$. - -We can then compare the result from this numerical scheme with the output from our network using Autograd: - -!bc pycod -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -def deep_neural_network(deep_params, x): - # N_hidden is the number of hidden layers - N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - # Due to multiple hidden layers, define a variable referencing to the - # output of the previous layer: - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output - -def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): - # num_hidden_neurons is now a list of number of neurons within each hidden layer - - # Find the number of hidden layers: - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 ) - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: %g'%cost_function_deep(P, x)) - - ## Start finding the optimal weigths using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_deep_grad = grad(cost_function_deep,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases - # in the hidden layers and output layers evaluated at x. - cost_deep_grad = cost_function_deep_grad(P, x) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_deep_grad[l] - - print('Final cost: %g'%cost_function_deep(P, x)) - - return P - -## Set up the cost function specified for this Poisson equation: - -# The right side of the ODE -def f(x): - return (3*x + x**2)*np.exp(x) - -def cost_function_deep(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial_deep(x,P) - - # Find the derivative w.r.t x of the trial function - d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P) - - right_side = f(x) - - err_sqr = (-d2_g_t - right_side)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum/np.size(err_sqr) - -# The trial solution: -def g_trial_deep(x,P): - return x*(1-x)*deep_neural_network(P,x) - -# The analytic solution; -def g_analytic(x): - return x*(1-x)*np.exp(x) - -if __name__ == '__main__': - npr.seed(4155) - - ## Decide the vales of arguments to the function to solve - Nx = 10 - x = np.linspace(0,1, Nx) - - ## Set up the initial parameters - num_hidden_neurons = [200,100] - num_iter = 1000 - lmb = 1e-3 - - P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb) - - g_dnn_ag = g_trial_deep(x,P) - g_analytical = g_analytic(x) - - # Find the maximum absolute difference between the solutons: - - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(x, g_analytical) - plt.plot(x, g_dnn_ag[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('x') - plt.ylabel('g(x)') - - ## Perform the computation using the numerical scheme - - dx = 1/(Nx - 1) - - # Set up the matrix A - A = np.zeros((Nx-2,Nx-2)) - - A[0,0] = 2 - A[0,1] = -1 - - for i in range(1,Nx-3): - A[i,i-1] = -1 - A[i,i] = 2 - A[i,i+1] = -1 - - A[Nx - 3, Nx - 4] = -1 - A[Nx - 3, Nx - 3] = 2 - - # Set up the vector f - f_vec = dx**2 * f(x[1:-1]) - - # Solve the equation - g_res = np.linalg.solve(A,f_vec) - - g_vec = np.zeros(Nx) - g_vec[1:-1] = g_res - - # Print the differences between each method - max_diff1 = np.max(np.abs(g_dnn_ag - g_analytical)) - max_diff2 = np.max(np.abs(g_vec - g_analytical)) - print("The max absolute difference between the analytical solution and DNN Autograd: %g"%max_diff1) - print("The max absolute difference between the analytical solution and numerical scheme: %g"%max_diff2) - - # Plot the results - plt.figure(figsize=(10,10)) - - plt.plot(x,g_vec) - plt.plot(x,g_analytical) - plt.plot(x,g_dnn_ag[0,:]) - - plt.legend(['numerical scheme','analytical','dnn']) - plt.show() - -!ec - -The program prints out: -!bc -The max absolute difference between the analytical solution and DNN Autograd: 0.000464088 -The max absolute difference between the analytical solution and numerical scheme: 0.00266858 -!ec - -!split -===== Using gradient descent in TensorFlow to solve Poisson equation ===== -The program follows the similar idea as for the logistic population model. - -What has changed, is what the cost function minimizes and the trial solution. - -!bc pycod -import tensorflow as tf -import numpy as np -import matplotlib.pyplot as plt -## Construction phase - -# Just to reset the graph such that it is possible to rerun this in a -# Jupyter cell without resetting the whole kernel. -tf.reset_default_graph() - -tf.set_random_seed(4155) - -# Convert the values the trial solution is evaluated at to a tensor. -Nx = 10 -x = np.linspace(0,1, Nx) -x_tf = tf.convert_to_tensor(x.reshape(-1,1),dtype=tf.float64) - - -num_iter = 10000 - -# Define the number of neurons at each hidden layer -num_hidden_neurons = [20,10] -num_hidden_layers = np.size(num_hidden_neurons) - -# Construct the network. -# tf.name_scope is used to group each step in the construction, -# just for a more organized visualization in TensorBoard -with tf.name_scope('dnn'): - - # Input layer - previous_layer = x_tf - - # Hidden layers - for l in range(num_hidden_layers): - current_layer = tf.layers.dense(previous_layer, num_hidden_neurons[l], name='hidden%d'%(l+1), activation=tf.nn.sigmoid) - previous_layer = current_layer - - # Output layer - dnn_output = tf.layers.dense(previous_layer, 1, name='output') - -# Define the cost function -with tf.name_scope('cost'): - g_trial = x_tf*(1-x_tf)*dnn_output - d_g_trial = tf.gradients(g_trial,x_tf) - d2_g_trial = tf.gradients(d_g_trial,x_tf) - - right_side = (3*x_tf + x_tf**2)*tf.exp(x_tf) - - err = tf.square( -d2_g_trial[0] - right_side) - cost = tf.reduce_sum(err, name = 'cost') - -# Choose the method to minimize the cost function, along with a learning rate -learning_rate = 1e-2 -with tf.name_scope('train'): - optimizer = tf.train.GradientDescentOptimizer(learning_rate) - traning_op = optimizer.minimize(cost) - -g_dnn_tf = None - -# Define a node that initializes all of the other nodes in the computational graph -# used by TensorFlow: -init = tf.global_variables_initializer() - - -## Execution phase - -# Start a session where the graph defined from the construction phase can be evaluated at: - -with tf.Session() as sess: - # Initialize the whole graph - init.run() - - # Evaluate the initial cost: - print('Initial cost: %g'%cost.eval()) - - # The traning of the network: - for i in range(num_iter): - sess.run(traning_op) - - # Training is done, and we have an approximate solution to the ODE - print('Final cost: %g'%cost.eval()) - - # Store the result - g_dnn_tf = g_trial.eval() - - writer = tf.summary.FileWriter("./output", sess.graph) - writer.close() - -# Evaluate the analytical function to compare with -def g_analytic(x): - return x*(1-x)*np.exp(x) - -g_analytical = g_analytic(x) - -diff_tf = g_dnn_tf - g_analytical.reshape(-1,1) - -print('\nMax absolute difference between the analytical solution and solution from TensorFlow DNN: %g'%np.max(np.abs(diff_tf))) - -# Plot the result -plt.figure(figsize=(10,10)) - -plt.title('Numerical solutions of the ODE') - -plt.plot(x, g_dnn_tf) -plt.plot(x, g_analytical) - -plt.legend(['dnn, tensorflow','exact']) -plt.xlabel('x') -plt.ylabel('g(x)') - -plt.show() - -!ec - -!split -===== Using a different optimization algorithm implemented in TensorFlow to solve Poisson equation ===== - -We can see that the results using GradientDescentOptimizer seems to converge towards the analytical solution. -But there exists many other methods for optimization also, see "the TensorFlow documentation on Optimizers":"https://www.tensorflow.org/versions/r1.2/api_guides/python/train#Optimizers". - -Adam is an optimization algorithm that changes its learning rates accordingly to the function it tries to minimize for every iteration. -The algorithm is described in "this paper":"https://arxiv.org/pdf/1412.6980.pdf". -How much an optimization algorithm has to say for the network to converge, could be interesting to experiment with. -Using the same TensorFlow program as before, the only change to do, is to replace the variable *optimizer*. - -In the program that uses TensorFlow to solve for the Poisson equation, change the line - -!bc -optimizer = tf.train.GradientDescentOptimizer(learning_rate) -!ec - -to - -!bc -optimizer = tf.train.AdamOptimizer(learning_rate) -!ec - - -The program using the Adam optimizer with a different initial learning rate yields indeed an interesting result: -!bc -Max absolute difference between the analytical solution and solution from TensorFlow DNN: 7.11243e-05 -!ec - -!split -===== Partial Differential Equations ===== -A partial differential equation (PDE) has a solution here the function is defined by multiple variables. -The equation may involve all kinds of combinations of which variables the function is differentiated with respect to. - -In general, a partial differential equation for a function $g(x_1,\dots,x_N)$ with $N$ variables may be expressed as - -!bt -\begin{equation} \label{PDE} - f\left(x_1, \, \dots \, , x_N, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1}, \dots , \frac{\partial g(x_1,\dots,x_N) }{\partial x_N}, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(x_1,\dots,x_N) }{\partial x_N^n} \right) = 0 -\end{equation} -!et - -where $f$ is an expression involving all kinds of possible mixed derivatives of $g(x_1,\dots,x_N)$ up to an order $n$. In order for the solution to be unique, some additional conditions must also be given. - -The problem our network must solve for, is similar to the ODE case. -We must have a trial solution $g_t$ at hand. - -For instance, the trial solution could be expressed as -!bt -\begin{align*} - g_t(x_1,\dots,x_N) = h_1(x_1,\dots,x_N) + h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P)) -\end{align*} -!et -where $h_1(x_1,\dots,x_N)$ is a function that ensures $g_t(x_1,\dots,x_N)$ satisfies some given conditions. -The neural network $N(x_1,\dots,x_N,P)$ has weights and biases described by $P$ and $h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P))$ is an expression using the output from the neural network in some way. - -The role of the function $h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P))$, is to ensure that the output of $N(x_1,\dots,x_N,P)$ is zero when $g_t(x_1,\dots,x_N)$ is evaluated at the values of $x_1,\dots,x_N$ where the given conditions must be satisfied. The function $h_1(x_1,\dots,x_N)$ should alone make $g_t(x_1,\dots,x_N)$ satisfy the conditions. - -The network tries then the minimize the cost function following the same ideas as described for the ODE case, but now with more than one variables to consider. -The concept still remains the same; find a set of parameters $P$ such that the expression $f$ in (ref{PDE}) is as close to zero as possible. - -As for the ODE case, the cost function is the mean squared error that the network must try to minimize. The cost function for the network to minimize is - -!bt -\begin{equation*} -c\left(x_1, \dots, x_N, P\right) = \left( f\left(x_1, \, \dots \, , x_N, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1}, \dots , \frac{\partial g(x_1,\dots,x_N) }{\partial x_N}, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(x_1,\dots,x_N) }{\partial x_N^n} \right) \right)^2 -\end{equation*} -!et - -If we let $\vec x = \big( x_1, \dots, x_N \big)$ be an array containing the values for $x_1, \dots, x_N$ respectively, the cost function can be reformulated into the following: -!bt -\begin{equation*} - c\left(\vec{x}, P\right) = f\left( \left( \vec{x}, \frac{\partial g(\vec x) }{\partial x_1}, \dots , \frac{\partial g(\vec x) }{\partial x_N}, \frac{\partial g(\vec x) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(\vec x) }{\partial x_N^n} \right) \right)^2 -\end{equation*} -!et - -If we also have $M$ different sets of values for $x_1, \dots, x_N$, that is $\vec{x}_i = \big(x_1^{(i)}, \dots, x_N^{(i)}\big)$ for $i = 1,\dots,M$ being the rows in matrix $X$, the cost function can be generalized into -!bt -\begin{equation*} -c\left(X, P \right) = \sum_{i=1}^M f\left( \left( \vec{x}_i, \frac{\partial g(\vec{x}_i) }{\partial x_1}, \dots , \frac{\partial g(\vec{x}_i) }{\partial x_N}, \frac{\partial g(\vec{x}_i) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(\vec{x}_i) }{\partial x_N^n} \right) \right)^2 -\end{equation*} -!et - -!split -===== Example: The diffusion equation ===== - -In one spatial dimension, the equation reads -!bt -\begin{equation*} - \frac{\partial g(x,t)}{\partial t} = \frac{\partial^2 g(x,t)}{\partial x^2} -\end{equation*} -!et - -where a possible choice of conditions are -!bt -\begin{align*} -g(0,t) &= 0 ,\qquad t \geq 0 \\ -g(1,t) &= 0, \qquad t \geq 0 \\ -g(x,0) &= u(x),\qquad x\in [0,1] -\end{align*} -!et -with $u(x)$ being some given function. - -!split -===== Defining the problem ===== - -For this case, we want to find $g(x,t)$ such that - -!bt -\begin{equation} - \frac{\partial g(x,t)}{\partial t} = \frac{\partial^2 g(x,t)}{\partial x^2} -\end{equation} \label{diffonedim} -!et - -and - -!bt -\begin{align*} -g(0,t) &= 0 ,\qquad t \geq 0 \\ -g(1,t) &= 0, \qquad t \geq 0 \\ -g(x,0) &= u(x),\qquad x\in [0,1] -\end{align*} -!et -with $u(x) = \sin(\pi x)$. - -First, let us set up the deep neural network. -The deep neural network will follow the same structure as discussed in the examples solving the ODEs. -First, we will look into how Autograd could be used in a network tailored to solve for bivariate functions. - - - -!split -===== Setting up the network using Autograd ===== - -The only change to do here, is to extend our network such that functions of multiple parameters are correctly handled. -In this case we have two variables in our function to solve for, that is time $t$ and position $x$. -The variables will be represented by a one-dimensional array in the program. -The program will evaluate the network at each possible pair $(x,t)$, given an array for the desired $x$-values and $t$-values to approximate the solution at. - -!bc pycod -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -def deep_neural_network(deep_params, x): - # x is now a point and a 1D numpy array; make it a column vector - num_coordinates = np.size(x,0) - x = x.reshape(num_coordinates,-1) - - num_points = np.size(x,1) - - # N_hidden is the number of hidden layers - N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assume that the input layer does nothing to the input x - x_input = x - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output[0][0] -!ec - -!split -===== Setting up the network using Autograd; The trial solution ===== -The cost function must then iterate through the given arrays containing values for $x$ and $t$, defines a point $(x,t)$ the deep neural network and the trial solution is evaluated at, and then finds the Jacobian of the trial solution. - -A possible trial solution for this PDE is - -$$ -g_t(x,t) = h_1(x,t) + x(1-x)tN(x,t,P) -$$ - -with $A(x,t)$ being a function ensuring that $g_t(x,t)$ satisfies our given conditions, and $N(x,t,P)$ being the output from the deep neural network using weights and biases for each layer from $P$. - -To fulfill the conditions, $A(x,t)$ could be: - -$$ -h_1(x,t) = (1-t)\Big(u(x) - \big((1-x)u(0) + x u(1)\big)\Big) = (1-t)u(x) = (1-t)\sin(\pi x) -$$ -since $(0) = u(1) = 0$ and $u(x) = \sin(\pi x)$. - -The Jacobian is used because the program must find the derivative of the trial solution with respect to $x$ and $t$. - -This gives the necessity of computing the Jacobian matrix, as we want to evaluate the gradient with respect to $x$ and $t$ (note that the Jacobian of a scalar-valued multivariate function is simply its gradient). - -In Autograd, the differentiation is by default done with respect to the first input argument of your Python function. Since the points is an array representing $x$ and $t$, the Jacobian is calculated using the values of $x$ and $t$. - -To find the second derivative with respect to $x$ and $t$, the Jacobian can be found for the second time. The result is a Hessian matrix, which is the matrix containing all the possible second order mixed derivatives of $g(x,t)$. - -!bc pycod -# Set up the trial function: -def u(x): - return np.sin(np.pi*x) - -def g_trial(point,P): - x,t = point - return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point) - -# The right side of the ODE: -def f(point): - return 0. - -# The cost function: -def cost_function(P, x, t): - cost_sum = 0 - - g_t_jacobian_func = jacobian(g_trial) - g_t_hessian_func = hessian(g_trial) - - for x_ in x: - for t_ in t: - point = np.array([x_,t_]) - - g_t = g_trial(point,P) - g_t_jacobian = g_t_jacobian_func(point,P) - g_t_hessian = g_t_hessian_func(point,P) - - g_t_dt = g_t_jacobian[1] - g_t_d2x = g_t_hessian[0][0] - - func = f(point) - - err_sqr = ( (g_t_dt - g_t_d2x) - func)**2 - cost_sum += err_sqr - - return cost_sum -!ec - -!split -===== Setting up the network using Autograd; The full program ===== -Having set up the network, along with the trial solution and cost function, we can now see how the deep neural network performs by comparing the results to the analytical solution. - -The analytical solution of our problem is - -$$ -g(x,t) = \exp(-\pi^2 t)\sin(\pi x) -$$ - -A possible way to implement a neural network solving the PDE, is given below. -Be aware, though, that it is fairly slow for the parameters used. -A better result is possible, but requires more iterations, and thus longer time to complete. - -Using only 20 neurons in one hidden layer, the program managed to make the trial solution have the maximum absolute error of 0.0075. The execution time, however, was approximately one day and 14 hours on a computer having Intel i7-7560U 2.4 GHz CPU. - -Indeed, the program below is not optimal in its implementation, but rather serves as an example on how to implement and use a neural network to solve a PDE. -Using TensorFlow in the next example sovling the wave equation, has a much better execution time. - -!bc pycod -import autograd.numpy as np -from autograd import jacobian,hessian,grad -import autograd.numpy.random as npr -from matplotlib import cm -from matplotlib import pyplot as plt -from mpl_toolkits.mplot3d import axes3d - -## Set up the network - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -def deep_neural_network(deep_params, x): - # x is now a point and a 1D numpy array; make it a column vector - num_coordinates = np.size(x,0) - x = x.reshape(num_coordinates,-1) - - num_points = np.size(x,1) - - # N_hidden is the number of hidden layers - N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assume that the input layer does nothing to the input x - x_input = x - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output[0][0] - -## Define the trial solution and cost function -def u(x): - return np.sin(np.pi*x) - -def g_trial(point,P): - x,t = point - return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point) - -# The right side of the ODE: -def f(point): - return 0. - -# The cost function: -def cost_function(P, x, t): - cost_sum = 0 - - g_t_jacobian_func = jacobian(g_trial) - g_t_hessian_func = hessian(g_trial) - - for x_ in x: - for t_ in t: - point = np.array([x_,t_]) - - g_t = g_trial(point,P) - g_t_jacobian = g_t_jacobian_func(point,P) - g_t_hessian = g_t_hessian_func(point,P) - - g_t_dt = g_t_jacobian[1] - g_t_d2x = g_t_hessian[0][0] - - func = f(point) - - err_sqr = ( (g_t_dt - g_t_d2x) - func)**2 - cost_sum += err_sqr - - return cost_sum /( np.size(x)*np.size(t) ) - -## For comparison, define the analytical solution -def g_analytic(point): - x,t = point - return np.exp(-np.pi**2*t)*np.sin(np.pi*x) - -## Set up a function for training the network to solve for the equation -def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb): - ## Set up initial weigths and biases - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: ',cost_function(P, x, t)) - - cost_function_grad = grad(cost_function,0) - - # Let the update be done num_iter times - for i in range(num_iter): - cost_grad = cost_function_grad(P, x , t) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_grad[l] - - print('Final cost: ',cost_function(P, x, t)) - - return P - -if __name__ == '__main__': - ### Use the neural network: - npr.seed(15) - - ## Decide the vales of arguments to the function to solve - Nx = 10; Nt = 10 - x = np.linspace(0, 1, Nx) - t = np.linspace(0,1,Nt) - - ## Set up the parameters for the network - num_hidden_neurons = [100, 25] - num_iter = 250 - lmb = 0.01 - - P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb) - - ## Store the results - g_dnn_ag = np.zeros((Nx, Nt)) - G_analytical = np.zeros((Nx, Nt)) - for i,x_ in enumerate(x): - for j, t_ in enumerate(t): - point = np.array([x_, t_]) - g_dnn_ag[i,j] = g_trial(point,P) - - G_analytical[i,j] = g_analytic(point) - - # Find the map difference between the analytical and the computed solution - diff_ag = np.abs(g_dnn_ag - G_analytical) - print('Max absolute difference between the analytical solution and the network: %g'%np.max(diff_ag)) - - ## Plot the solutions in two dimensions, that being in position and time - - T,X = np.meshgrid(t,x) - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons)) - s = ax.plot_surface(T,X,g_dnn_ag,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Analytical solution') - s = ax.plot_surface(T,X,G_analytical,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Difference') - s = ax.plot_surface(T,X,diff_ag,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - ## Take some slices of the 3D plots just to see the solutions at particular times - indx1 = 0 - indx2 = int(Nt/2) - indx3 = Nt-1 - - t1 = t[indx1] - t2 = t[indx2] - t3 = t[indx3] - - # Slice the results from the DNN - res1 = g_dnn_ag[:,indx1] - res2 = g_dnn_ag[:,indx2] - res3 = g_dnn_ag[:,indx3] - - # Slice the analytical results - res_analytical1 = G_analytical[:,indx1] - res_analytical2 = G_analytical[:,indx2] - res_analytical3 = G_analytical[:,indx3] - - # Plot the slices - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t1) - plt.plot(x, res1) - plt.plot(x,res_analytical1) - plt.legend(['dnn','analytical']) - - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t2) - plt.plot(x, res2) - plt.plot(x,res_analytical2) - plt.legend(['dnn','analytical']) - - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t3) - plt.plot(x, res3) - plt.plot(x,res_analytical3) - plt.legend(['dnn','analytical']) - - plt.show() -!ec - -!split -===== Example: Solving the wave equation using Autograd and TensorFlow ===== - -The wave equation is -!bt -\begin{equation*} - \frac{\partial^2 g(x,t)}{\partial t^2} = c^2\frac{\partial^2 g(x,t)}{\partial x^2} -\end{equation*} -!et - -with $c$ being the specified wave speed. - -Here, the chosen conditions are -!bt -\begin{align*} - g(0,t) &= 0 \\ - g(1,t) &= 0 \\ - g(x,0) &= u(x) \\ - \frac{\partial g(x,t)}{\partial t} \Big |_{t = 0} &= v(x) -\end{align*} -!et -where $\frac{\partial g(x,t)}{\partial t} \Big |_{t = 0}$ means the derivative of $g(x,t)$ with respect to $t$ is evaluated at $t = 0$, and $u(x)$ and $v(x)$ being given functions. - -!split -===== The problem to solve for ===== - -The wave equation to solve for, is - -!bt -\begin{equation} \label{wave} -\frac{\partial^2 g(x,t)}{\partial t^2} = c^2 \frac{\partial^2 g(x,t)}{\partial x^2} -\end{equation} -!et - -where $c$ is the given wave speed. -The chosen conditions for this equation are - -!bt -\begin{aligned} -g(0,t) &= 0, &t \geq 0 \\ -g(1,t) &= 0, &t \geq 0 \\ -g(x,0) &= u(x), &x\in[0,1] \\ -\frac{\partial g(x,t)}{\partial t}\Big |_{t = 0} &= v(x), &x \in [0,1] -\end{aligned} \label{condwave} -!et - -In this example, let $c = 1$ and $u(x) = \sin(\pi x)$ and $v(x) = -\pi\sin(\pi x)$. - - -!split -===== The trial solution ===== -Setting up the network is done in similar matter as for the example of solving the diffusion equation. -The only things we have to change, is the trial solution such that it satisfies the conditions from (ref{condwave}) and the cost function. - -The trial solution becomes slightly different since we have other conditions than in the example of solving the diffusion equation. Here, a possible trial solution $g_t(x,t)$ is - -$$ -g_t(x,t) = h_1(x,t) + x(1-x)t^2N(x,t,P) -$$ - -where - -$$ -h_1(x,t) = (1-t^2)u(x) + tv(x) -$$ - -Note that this trial solution satisfies the conditions only if $u(0) = v(0) = u(1) = v(1) = 0$, which is the case in this example. - -!split -===== The analytical solution ===== - -The analytical solution for our specific problem, is - -$$ -g(x,t) = \sin(\pi x)\cos(\pi t) - \sin(\pi x)\sin(\pi t) -$$ - -!split -===== Solving the wave equation - the full program using Autograd ===== - -!bc pycod -import autograd.numpy as np -from autograd import hessian,grad -import autograd.numpy.random as npr -from matplotlib import cm -from matplotlib import pyplot as plt -from mpl_toolkits.mplot3d import axes3d - -## Set up the trial function: -def u(x): - return np.sin(np.pi*x) - -def v(x): - return -np.pi*np.sin(np.pi*x) - -def h1(point): - x,t = point - return (1 - t**2)*u(x) + t*v(x) - -def g_trial(point,P): - x,t = point - return h1(point) + x*(1-x)*t**2*deep_neural_network(P,point) - -## Define the cost function -def cost_function(P, x, t): - cost_sum = 0 - - g_t_hessian_func = hessian(g_trial) - - for x_ in x: - for t_ in t: - point = np.array([x_,t_]) - - g_t_hessian = g_t_hessian_func(point,P) - - g_t_d2x = g_t_hessian[0][0] - g_t_d2t = g_t_hessian[1][1] - - err_sqr = ( (g_t_d2t - g_t_d2x) )**2 - cost_sum += err_sqr - - return cost_sum / (np.size(t) * np.size(x)) - -## The neural network -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -def deep_neural_network(deep_params, x): - # x is now a point and a 1D numpy array; make it a column vector - num_coordinates = np.size(x,0) - x = x.reshape(num_coordinates,-1) - - num_points = np.size(x,1) - - # N_hidden is the number of hidden layers - N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assume that the input layer does nothing to the input x - x_input = x - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output[0][0] - -## The analytical solution -def g_analytic(point): - x,t = point - return np.sin(np.pi*x)*np.cos(np.pi*t) - np.sin(np.pi*x)*np.sin(np.pi*t) - -def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb): - ## Set up initial weigths and biases - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: ',cost_function(P, x, t)) - - cost_function_grad = grad(cost_function,0) - - # Let the update be done num_iter times - for i in range(num_iter): - cost_grad = cost_function_grad(P, x , t) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_grad[l] - - - print('Final cost: ',cost_function(P, x, t)) - - return P - -if __name__ == '__main__': - ### Use the neural network: - npr.seed(15) - - ## Decide the vales of arguments to the function to solve - Nx = 10; Nt = 10 - x = np.linspace(0, 1, Nx) - t = np.linspace(0,1,Nt) - - ## Set up the parameters for the network - num_hidden_neurons = [50,20] - num_iter = 1000 - lmb = 0.01 - - P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb) - - ## Store the results - res = np.zeros((Nx, Nt)) - res_analytical = np.zeros((Nx, Nt)) - for i,x_ in enumerate(x): - for j, t_ in enumerate(t): - point = np.array([x_, t_]) - res[i,j] = g_trial(point,P) - - res_analytical[i,j] = g_analytic(point) - - diff = np.abs(res - res_analytical) - print("Max difference between analytical and solution from nn: %g"%np.max(diff)) - - ## Plot the solutions in two dimensions, that being in position and time - - T,X = np.meshgrid(t,x) - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons)) - s = ax.plot_surface(T,X,res,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Analytical solution') - s = ax.plot_surface(T,X,res_analytical,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Difference') - s = ax.plot_surface(T,X,diff,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - ## Take some slices of the 3D plots just to see the solutions at particular times - indx1 = 0 - indx2 = int(Nt/2) - indx3 = Nt-1 - - t1 = t[indx1] - t2 = t[indx2] - t3 = t[indx3] - - # Slice the results from the DNN - res1 = res[:,indx1] - res2 = res[:,indx2] - res3 = res[:,indx3] - - # Slice the analytical results - res_analytical1 = res_analytical[:,indx1] - res_analytical2 = res_analytical[:,indx2] - res_analytical3 = res_analytical[:,indx3] - - # Plot the slices - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t1) - plt.plot(x, res1) - plt.plot(x,res_analytical1) - plt.legend(['dnn','analytical']) - - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t2) - plt.plot(x, res2) - plt.plot(x,res_analytical2) - plt.legend(['dnn','analytical']) - - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t3) - plt.plot(x, res3) - plt.plot(x,res_analytical3) - plt.legend(['dnn','analytical']) - - plt.show() -!ec - -!split -===== Solving the wave equation - the full program using TensorFlow ===== -As the program using Autograd is fairly slow, one could hope that using TensorFlow -could make a naive implementation faster, and more numerically robust. - -In addition, having TensorFlow at hand, it could be easier to experiment with different -optimization algorithms, and other constructions of the network. - -The following program solves the given wave equation much faster, - -!bc pycod -import tensorflow as tf -import numpy as np -from matplotlib import cm -from matplotlib import pyplot as plt -from mpl_toolkits.mplot3d import axes3d - -Nx = 10 -x_np = np.linspace(0,1,Nx) - -Nt = 10 -t_np = np.linspace(0,1,Nt) - -X,T = np.meshgrid(x_np, t_np) - -x = X.ravel() -t = T.ravel() - -## The construction phase - -zeros = tf.reshape(tf.convert_to_tensor(np.zeros(x.shape)),shape=(-1,1)) -x = tf.reshape(tf.convert_to_tensor(x),shape=(-1,1)) -t = tf.reshape(tf.convert_to_tensor(t),shape=(-1,1)) - -points = tf.concat([x,t],1) - -num_iter = 100000 -num_hidden_neurons = [90] - -X = tf.convert_to_tensor(X) -T = tf.convert_to_tensor(T) - - -with tf.variable_scope('dnn'): - num_hidden_layers = np.size(num_hidden_neurons) - - previous_layer = points - - for l in range(num_hidden_layers): - current_layer = tf.layers.dense(previous_layer, num_hidden_neurons[l],activation=tf.nn.sigmoid) - previous_layer = current_layer - - dnn_output = tf.layers.dense(previous_layer, 1) - - -def u(x): - return tf.sin(np.pi*x) - -def v(x): - return -np.pi*tf.sin(np.pi*x) - -with tf.name_scope('loss'): - g_trial = (1 - t**2)*u(x) + t*v(x) + x*(1-x)*t**2*dnn_output - - g_trial_d2t = tf.gradients(tf.gradients(g_trial,t),t) - g_trial_d2x = tf.gradients(tf.gradients(g_trial,x),x) - - loss = tf.losses.mean_squared_error(zeros, g_trial_d2t[0] - g_trial_d2x[0]) - -learning_rate = 0.01 -with tf.name_scope('train'): - optimizer = tf.train.GradientDescentOptimizer(learning_rate) - traning_op = optimizer.minimize(loss) - -init = tf.global_variables_initializer() - -g_analytic = tf.sin(np.pi*x)*tf.cos(np.pi*t) - tf.sin(np.pi*x)*tf.sin(np.pi*t) -g_dnn = None - -## The execution phase -with tf.Session() as sess: - init.run() - for i in range(num_iter): - sess.run(traning_op) - - # If one desires to see how the cost function behaves during training - #if i % 100 == 0: - # print(loss.eval()) - - g_analytic = g_analytic.eval() - g_dnn = g_trial.eval() - - -## Compare with the analutical solution -diff = np.abs(g_analytic - g_dnn) -print('Max absolute difference between analytical solution and TensorFlow DNN = ',np.max(diff)) - -G_analytic = g_analytic.reshape((Nt,Nx)) -G_dnn = g_dnn.reshape((Nt,Nx)) - -diff = np.abs(G_analytic - G_dnn) - -# Plot the results - -X,T = np.meshgrid(x_np, t_np) - -fig = plt.figure(figsize=(10,10)) -ax = fig.gca(projection='3d') -ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons)) -s = ax.plot_surface(X,T,G_dnn,linewidth=0,antialiased=False,cmap=cm.viridis) -ax.set_xlabel('Time $t$') -ax.set_ylabel('Position $x$'); - -fig = plt.figure(figsize=(10,10)) -ax = fig.gca(projection='3d') -ax.set_title('Analytical solution') -s = ax.plot_surface(X,T,G_analytic,linewidth=0,antialiased=False,cmap=cm.viridis) -ax.set_xlabel('Time $t$') -ax.set_ylabel('Position $x$'); - -fig = plt.figure(figsize=(10,10)) -ax = fig.gca(projection='3d') -ax.set_title('Difference') -s = ax.plot_surface(X,T,diff,linewidth=0,antialiased=False,cmap=cm.viridis) -ax.set_xlabel('Time $t$') -ax.set_ylabel('Position $x$'); - -## Take some 3D slices - -indx1 = 0 -indx2 = int(Nt/2) -indx3 = Nt-1 - -t1 = t_np[indx1] -t2 = t_np[indx2] -t3 = t_np[indx3] - -# Slice the results from the DNN -res1 = G_dnn[indx1,:] -res2 = G_dnn[indx2,:] -res3 = G_dnn[indx3,:] - -# Slice the analytical results -res_analytical1 = G_analytic[indx1,:] -res_analytical2 = G_analytic[indx2,:] -res_analytical3 = G_analytic[indx3,:] - -# Plot the slices -plt.figure(figsize=(10,10)) -plt.title("Computed solutions at time = %g"%t1) -plt.plot(x_np, res1) -plt.plot(x_np,res_analytical1) -plt.legend(['dnn','analytical']) - -plt.figure(figsize=(10,10)) -plt.title("Computed solutions at time = %g"%t2) -plt.plot(x_np, res2) -plt.plot(x_np,res_analytical2) -plt.legend(['dnn','analytical']) - -plt.figure(figsize=(10,10)) -plt.title("Computed solutions at time = %g"%t3) -plt.plot(x_np, res3) -plt.plot(x_np,res_analytical3) -plt.legend(['dnn','analytical']) - -plt.show() -!ec - -The program manages to find a solution having max absolute difference to the analytical -at approximately 0.0059, by just using some minutes! -It was found, by some testing, that one hidden layer with 90 neurons actually performed well. - -!split -===== Resources ===== - -o "Artificial neural networks for solving ordinary and partial differential equations by I.E. Lagaris et al":"https://pdfs.semanticscholar.org/d061/df393e0e8fbfd0ea24976458b7d42419040d.pdf" -o "Neural networks for solving differential equations by A. Honchar":"https://becominghuman.ai/neural-networks-for-solving-differential-equations-fa230ac5e04c" -o "Solving differential equations using neural networks by M.M Chiaramonte and M. Kiener":"http://cs229.stanford.edu/proj2013/ChiaramonteKiener-SolvingDifferentialEquationsUsingNeuralNetworks.pdf" -o "Introduction to Partial Differential Equations by A. Tveitio, R. Winther":"https://www.springer.com/us/book/9783540225515" - diff --git a/doc/src/week43/programs/dill.py~ b/doc/src/week43/programs/dill.py~ deleted file mode 100644 index c5805a983..000000000 --- a/doc/src/week43/programs/dill.py~ +++ /dev/null @@ -1,203 +0,0 @@ -import numpy as np -import matplotlib.pyplot as plt -import tensorflow as tf -from math import * -import time -import tensorflow.keras -# Different methods from Keras needed to create an RNN -# This is not necessary but it shortened function calls -# that need to be used in the code. -from tensorflow.keras import datasets, layers, models -from tensorflow.keras.layers import Input -from tensorflow.keras import regularizers -from tensorflow.keras import optimizers -from tensorflow.keras.models import Model, Sequential -#from tensorflow.keras.layers.core import Dense, Activation -from tensorflow.keras.layers import Dense, SimpleRNN, LSTM, GRU - -# Define Analytical, Euler-Cromer, and Velocity-Verlet methods of solving -def analytical(k,m,x0,v0,dt,tfinal): - t = np.arange(0,tfinal+dt,dt) - v = -x0 * np.sin(t) + v0 * np.cos(t) - x = x0 * np.cos(t) + v0 * np.sin(t) - K = 1/2 *m*v**2 - U = 1/2 *k*x**2 - return x, v, K, U, t - -def euler_cromer(k,m,x0,v0,dt,tfinal): - n = ceil(tfinal/dt) # Set up arrays - t = np.zeros(n) - v = np.zeros(n) - x = np.zeros(n) - K = np.zeros(n) - U = np.zeros(n) - # Define Initial Conditions - x[0] = x0 - v[0] = v0 - K[0] = 1/2 *m*v0**2 - U[0] = 1/2 *k*x0**2 - - # Integrate using the Euler-Cromer Method - for i in range(n-1): - a = -x[i] - v[i+1] = v[i] + dt*a - x[i+1] = x[i] + dt*v[i+1] - K[i+1] = 1/2 *m*v[i+1]**2 - U[i+1] = 1/2 *k*x[i+1]**2 - t[i+1] = t[i] + dt - return x, v, K, U, t - -def velocity_verlet(k,m,x0,v0,dt,tfinal): - n = ceil(tfinal/dt) - # Set up arrays - t = np.zeros(n) - v = np.zeros(n) - x = np.zeros(n) - K = np.zeros(n) - U = np.zeros(n) - # Define Initial Conditions - x[0] = x0 - v[0] = v0 - K[0] = 1/2 *m*v0**2 - U[0] = 1/2 *k*x0**2 - - # Integrate using the Velocity-Verlet Method - for i in range(n-1): - a = -x[i] - x[i+1] = x[i] + dt*v[i] + dt**2 /2*a - a1 = -x[i+1] - v[i+1] = v[i] + dt/2*(a+a1) - K[i+1] = 1/2 *m*v[i+1]**2 - U[i+1] = 1/2 *k*x[i+1]**2 - t[i+1] = t[i] + dt - return x, v, K, U, t - - -# Analytical Solution to position -def ana_cos(r0,t,k=1,m=1): - w0 = sqrt(k/m) - return r0*np.cos(w0*t) - -# Trial Function for Neural Net -def trial_func(x,y,y0=1): - return x*y + y0 - -# Loss Function for Position and Velocity -def combined_right(trialv,trialy,k=1,m=1): - return -k/m*trialy, trialv - -# Loss Wrapper in order to pass the Loss Function to Neural Net -def con_loss_wrapper(input_tensor): - def con_loss_function(y,y_pred): - trialy = trial_func(input_tensor,y_pred) - trialv = trial_func(input_tensor,y_pred) - righty, rightv = combined_right(trialv,trialy) - - leftv = tf.gradients(trialv,input_tensor)[0] - lefty = tf.gradients(leftv,input_tensor)[0] - - - loss = tf.reduce_mean((tf.math.squared_difference(lefty,righty))) - return loss - return con_loss_function - -# Creates the input data for the Neural Net -def create_input_data(x0=0,xmax=1,num_batch=5,len_batch=15): - input_data = np.linspace(x0,xmax,num_batch*len_batch) - input_data = input_data.reshape(num_batch,len_batch) - - return input_data - -# Creates the Neural Net -def create_net(data,len_batch,lr,epochs,right_side,loss,n_hidden_layer=50): - - input_tensor = Input(shape=(len_batch,)) - - hidden1 = Dense(n_hidden_layer,activation='tanh', - kernel_initializer='random_uniform',bias_initializer='random_uniform')(input_tensor) - hidden2 = Dense(n_hidden_layer,activation='tanh', - kernel_initializer='random_uniform',bias_initializer='random_uniform')(hidden1) - hidden3 = Dense(n_hidden_layer,activation='tanh', - kernel_initializer='random_uniform',bias_initializer='random_uniform')(hidden2) - hidden4 = Dense(n_hidden_layer,activation='tanh', - kernel_initializer='random_uniform',bias_initializer='random_uniform')(hidden3) - output = Dense(len_batch)(hidden4) - - model = Model(input_tensor,output) - - gd = optimizers.SGD(lr=lr) # May need to change first 'lr' to 'learning_rate' depending on TF/Keras version - model.compile(loss=loss(input_tensor),optimizer=gd) - model.fit(data,np.zeros((data.shape[0])),epochs=epochs) - - res = model.predict(data) - - del model - - return res - -# Define Function for Mean Squared Error -def mean_squared_error(analytical, results): - mse = 0 - for i in range(len(analytical)): - mse += (analytical[i] - results[i])**2 - - mse = mse/len(analytical) - return mse - - -# Define Constants -dt = 0.01 -tfinal = 50 -x0 = 1 -v0 = 0 -m = 1 -k = 1 - -num_batch = 1000 -len_batch = 1 - -# Create input data -data = create_input_data(x0=0,xmax=10,num_batch=num_batch,len_batch=len_batch) - -# Create and Run the Neural Net -nn_start_time = time.time() - -velocity = create_net(data,len_batch=len_batch,lr=0.001,n_hidden_layer=50, - epochs=1000,right_side=combined_right,loss=con_loss_wrapper) - -nn_end_time = time.time() - - -# Reshape Neural Net Output for easy graphing and analysis -n = num_batch*len_batch -velocity = velocity.reshape(1,n) -t = data.reshape(1,n)[0] -results_v = trial_func(t,velocity)[0] - -# Create Comparison data from Analytical Solution -analyt = ana_cos(1,t,m=1,k=1) - - - -# Euler-Cromer -ec_start_time = time.time() -ec_x = euler_cromer(k=1,m,x0,v0,dt,tfinal)[0] -ec_end_time = time.time() - -# Velocity-Verlet -vv_start_time = time.time() -vv_x = velocity_verlet(k,m,x0,v0,dt,tfinal)[0] -vv_end_time = time.time() - -# Plot -plt.plot(t,analyt,label='analytical') -plt.plot(t,results_v,label='net') -plt.plot(t, ec_x, label = "Euler-Cromer") -plt.plot(t, vv_x, label = "Velocity-Verlet") -plt.legend(bbox_to_anchor=(1.05, 1), loc='upper left') - -# Find Mean Squared Error -print("The Mean Squared Error of the Neural Net Solution is", mean_squared_error(analyt, results_v), "with a runtime of", nn_end_time-nn_start_time,"seconds.") -print("The Mean Squared Error of the Euler-Cromer is", mean_squared_error(analyt, ec_x), "with a runtime of", ec_end_time-ec_start_time,"seconds.") -print("The Mean Squared Error of the Velocity-Verlet is", mean_squared_error(analyt, vv_x), "with a runtime of", vv_end_time-vv_start_time,"seconds.") - diff --git a/doc/src/week43/programs/ode.py~ b/doc/src/week43/programs/ode.py~ deleted file mode 100644 index 2b6c2d667..000000000 --- a/doc/src/week43/programs/ode.py~ +++ /dev/null @@ -1,3759 +0,0 @@ -TITLE: Week 43: Solving Differential Equations with Deep Learning and Dimensionality Reduction methods -AUTHOR: Morten Hjorth-Jensen {copyright, 1999-present|CC BY-NC} at Department of Physics, University of Oslo & Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University -DATE: today - -!split - -* Thursday: Wrapping up Recurrent Neural Networks and solving differential equations. -* Friday: Principal Component Analysis and Dimensionality Reduction - -Reading suggestions for both days: "Aurelien Geron's chapters 8 - -!split -===== Recurrent Neural Networks ===== - - -!split -===== Solving ODEs with Deep Learning ===== - - - - - - - -!split -===== Basic ideas of the Principal Component Analysis (PCA) ===== - -The principal component analysis deals with the problem of fitting a -low-dimensional affine subspace $S$ of dimension $d$ much smaller than -the totaldimension $D$ of the problem at hand (our data -set). Mathematically it can be formulated as a statistical problem or -a geometric problem. In our discussion of the theorem for the -classical PCA, we will stay with a statistical approach. This is also -what set the scene historically which for the PCA. - -We have a data set defined by a design/feature matrix $\bm{X}$ (see below for its definition) -* Each data point is determined by $p$ extrinsic (measurement) variables -* We may want to ask the following question: Are there fewer intrinsic variables (say $d << p$) that still approximately describe the data? -* If so, these intrinsic variables may tell us something important and finding these intrinsic variables is what dimension reduction methods do. - - -!split -===== Introducing the Covariance and Correlation functions ===== - -Before we discuss the PCA theorem, we need to remind ourselves about -the definition of the covariance and the correlation function. These are quantities - -Suppose we have defined two vectors -$\hat{x}$ and $\hat{y}$ with $n$ elements each. The covariance matrix $\bm{C}$ is defined as -!bt -\[ -\bm{C}[\bm{x},\bm{y}] = \begin{bmatrix} \mathrm{cov}[\bm{x},\bm{x}] & \mathrm{cov}[\bm{x},\bm{y}] \\ - \mathrm{cov}[\bm{y},\bm{x}] & \mathrm{cov}[\bm{y},\bm{y}] \\ - \end{bmatrix}, -\] -!et -where for example -!bt -\[ -\mathrm{cov}[\bm{x},\bm{y}] =\frac{1}{n} \sum_{i=0}^{n-1}(x_i- \overline{x})(y_i- \overline{y}). -\] -!et -With this definition and recalling that the variance is defined as -!bt -\[ -\mathrm{var}[\bm{x}]=\frac{1}{n} \sum_{i=0}^{n-1}(x_i- \overline{x})^2, -\] -!et -we can rewrite the covariance matrix as -!bt -\[ -\bm{C}[\bm{x},\bm{y}] = \begin{bmatrix} \mathrm{var}[\bm{x}] & \mathrm{cov}[\bm{x},\bm{y}] \\ - \mathrm{cov}[\bm{x},\bm{y}] & \mathrm{var}[\bm{y}] \\ - \end{bmatrix}. -\] -!et - -The covariance takes values between zero and infinity and may thus -lead to problems with loss of numerical precision for particularly -large values. It is common to scale the covariance matrix by -introducing instead the correlation matrix defined via the so-called -correlation function - -!bt -\[ -\mathrm{corr}[\bm{x},\bm{y}]=\frac{\mathrm{cov}[\bm{x},\bm{y}]}{\sqrt{\mathrm{var}[\bm{x}] \mathrm{var}[\bm{y}]}}. -\] -!et - -The correlation function is then given by values $\mathrm{corr}[\bm{x},\bm{y}] -\in [-1,1]$. This avoids eventual problems with too large values. We -can then define the correlation matrix for the two vectors $\bm{x}$ -and $\bm{y}$ as - -!bt -\[ -\bm{K}[\bm{x},\bm{y}] = \begin{bmatrix} 1 & \mathrm{corr}[\bm{x},\bm{y}] \\ - \mathrm{corr}[\bm{y},\bm{x}] & 1 \\ - \end{bmatrix}, -\] -!et - -In the above example this is the function we constructed using _pandas_. - -!split -===== Correlation Function and Design/Feature Matrix ===== - -In our derivation of the various regression algorithms like _Ordinary Least Squares_ or _Ridge regression_ -we defined the design/feature matrix $\bm{X}$ as - -!bt -\[ -\bm{X}=\begin{bmatrix} -x_{0,0} & x_{0,1} & x_{0,2}& \dots & \dots x_{0,p-1}\\ -x_{1,0} & x_{1,1} & x_{1,2}& \dots & \dots x_{1,p-1}\\ -x_{2,0} & x_{2,1} & x_{2,2}& \dots & \dots x_{2,p-1}\\ -\dots & \dots & \dots & \dots \dots & \dots \\ -x_{n-2,0} & x_{n-2,1} & x_{n-2,2}& \dots & \dots x_{n-2,p-1}\\ -x_{n-1,0} & x_{n-1,1} & x_{n-1,2}& \dots & \dots x_{n-1,p-1}\\ -\end{bmatrix}, -\] -!et -with $\bm{X}\in {\mathbb{R}}^{n\times p}$, with the predictors/features $p$ refering to the column numbers and the -entries $n$ being the row elements. -We can rewrite the design/feature matrix in terms of its column vectors as -!bt -\[ -\bm{X}=\begin{bmatrix} \bm{x}_0 & \bm{x}_1 & \bm{x}_2 & \dots & \dots & \bm{x}_{p-1}\end{bmatrix}, -\] -!et -with a given vector -!bt -\[ -\bm{x}_i^T = \begin{bmatrix}x_{0,i} & x_{1,i} & x_{2,i}& \dots & \dots x_{n-1,i}\end{bmatrix}. -\] -!et - -With these definitions, we can now rewrite our $2\times 2$ -correaltion/covariance matrix in terms of a moe general design/feature -matrix $\bm{X}\in {\mathbb{R}}^{n\times p}$. This leads to a $p\times p$ -covariance matrix for the vectors $\bm{x}_i$ with $i=0,1,\dots,p-1$ - -!bt -\[ -\bm{C}[\bm{x}] = \begin{bmatrix} -\mathrm{var}[\bm{x}_0] & \mathrm{cov}[\bm{x}_0,\bm{x}_1] & \mathrm{cov}[\bm{x}_0,\bm{x}_2] & \dots & \dots & \mathrm{cov}[\bm{x}_0,\bm{x}_{p-1}]\\ -\mathrm{cov}[\bm{x}_1,\bm{x}_0] & \mathrm{var}[\bm{x}_1] & \mathrm{cov}[\bm{x}_1,\bm{x}_2] & \dots & \dots & \mathrm{cov}[\bm{x}_1,\bm{x}_{p-1}]\\ -\mathrm{cov}[\bm{x}_2,\bm{x}_0] & \mathrm{cov}[\bm{x}_2,\bm{x}_1] & \mathrm{var}[\bm{x}_2] & \dots & \dots & \mathrm{cov}[\bm{x}_2,\bm{x}_{p-1}]\\ -\dots & \dots & \dots & \dots & \dots & \dots \\ -\dots & \dots & \dots & \dots & \dots & \dots \\ -\mathrm{cov}[\bm{x}_{p-1},\bm{x}_0] & \mathrm{cov}[\bm{x}_{p-1},\bm{x}_1] & \mathrm{cov}[\bm{x}_{p-1},\bm{x}_{2}] & \dots & \dots & \mathrm{var}[\bm{x}_{p-1}]\\ -\end{bmatrix}, -\] -!et -and the correlation matrix -!bt -\[ -\bm{K}[\bm{x}] = \begin{bmatrix} -1 & \mathrm{corr}[\bm{x}_0,\bm{x}_1] & \mathrm{corr}[\bm{x}_0,\bm{x}_2] & \dots & \dots & \mathrm{corr}[\bm{x}_0,\bm{x}_{p-1}]\\ -\mathrm{corr}[\bm{x}_1,\bm{x}_0] & 1 & \mathrm{corr}[\bm{x}_1,\bm{x}_2] & \dots & \dots & \mathrm{corr}[\bm{x}_1,\bm{x}_{p-1}]\\ -\mathrm{corr}[\bm{x}_2,\bm{x}_0] & \mathrm{corr}[\bm{x}_2,\bm{x}_1] & 1 & \dots & \dots & \mathrm{corr}[\bm{x}_2,\bm{x}_{p-1}]\\ -\dots & \dots & \dots & \dots & \dots & \dots \\ -\dots & \dots & \dots & \dots & \dots & \dots \\ -\mathrm{corr}[\bm{x}_{p-1},\bm{x}_0] & \mathrm{corr}[\bm{x}_{p-1},\bm{x}_1] & \mathrm{corr}[\bm{x}_{p-1},\bm{x}_{2}] & \dots & \dots & 1\\ -\end{bmatrix}, -\] -!et - - -!split -===== Covariance Matrix Examples ===== - - -The Numpy function _np.cov_ calculates the covariance elements using -the factor $1/(n-1)$ instead of $1/n$ since it assumes we do not have -the exact mean values. The following simple function uses the -_np.vstack_ function which takes each vector of dimension $1\times n$ -and produces a $2\times n$ matrix $\bm{W}$ - - -!bt -\[ -\bm{W} = \begin{bmatrix} x_0 & y_0 \\ - x_1 & y_1 \\ - x_2 & y_2\\ - \dots & \dots \\ - x_{n-2} & y_{n-2}\\ - x_{n-1} & y_{n-1} & - \end{bmatrix}, -\] -!et - -which in turn is converted into into the $2\times 2$ covariance matrix -$\bm{C}$ via the Numpy function _np.cov()_. We note that we can also calculate -the mean value of each set of samples $\bm{x}$ etc using the Numpy -function _np.mean(x)_. We can also extract the eigenvalues of the -covariance matrix through the _np.linalg.eig()_ function. - -!bc pycod -# Importing various packages -import numpy as np -n = 100 -x = np.random.normal(size=n) -print(np.mean(x)) -y = 4+3*x+np.random.normal(size=n) -print(np.mean(y)) -W = np.vstack((x, y)) -C = np.cov(W) -print(C) -!ec - -!split -===== Correlation Matrix ===== - -The previous example can be converted into the correlation matrix by -simply scaling the matrix elements with the variances. We should also -subtract the mean values for each column. This leads to the following -code which sets up the correlations matrix for the previous example in -a more brute force way. Here we scale the mean values for each column of the design matrix, calculate the relevant mean values and variances and then finally set up the $2\times 2$ correlation matrix (since we have only two vectors). - -!bc pycod -import numpy as np -n = 100 -# define two vectors -x = np.random.random(size=n) -y = 4+3*x+np.random.normal(size=n) -#scaling the x and y vectors -x = x - np.mean(x) -y = y - np.mean(y) -variance_x = np.sum(x@x)/n -variance_y = np.sum(y@y)/n -print(variance_x) -print(variance_y) -cov_xy = np.sum(x@y)/n -cov_xx = np.sum(x@x)/n -cov_yy = np.sum(y@y)/n -C = np.zeros((2,2)) -C[0,0]= cov_xx/variance_x -C[1,1]= cov_yy/variance_y -C[0,1]= cov_xy/np.sqrt(variance_y*variance_x) -C[1,0]= C[0,1] -print(C) -!ec - -We see that the matrix elements along the diagonal are one as they -should be and that the matrix is symmetric. Furthermore, diagonalizing -this matrix we easily see that it is a positive definite matrix. - -The above procedure with _numpy_ can be made more compact if we use _pandas_. - -!split -===== Correlation Matrix with Pandas ===== - -We whow here how we can set up the correlation matrix using _pandas_, as done in this simple code -!bc pycod -import numpy as np -import pandas as pd -n = 10 -x = np.random.normal(size=n) -x = x - np.mean(x) -y = 4+3*x+np.random.normal(size=n) -y = y - np.mean(y) -X = (np.vstack((x, y))).T -print(X) -Xpd = pd.DataFrame(X) -print(Xpd) -correlation_matrix = Xpd.corr() -print(correlation_matrix) -!ec - - -We expand this model to the Franke function discussed above. - -!split -===== Correlation Matrix with Pandas and the Franke function ===== - -!bc pycod -# Common imports -import numpy as np -import pandas as pd - - -def FrankeFunction(x,y): - term1 = 0.75*np.exp(-(0.25*(9*x-2)**2) - 0.25*((9*y-2)**2)) - term2 = 0.75*np.exp(-((9*x+1)**2)/49.0 - 0.1*(9*y+1)) - term3 = 0.5*np.exp(-(9*x-7)**2/4.0 - 0.25*((9*y-3)**2)) - term4 = -0.2*np.exp(-(9*x-4)**2 - (9*y-7)**2) - return term1 + term2 + term3 + term4 - - -def create_X(x, y, n ): - if len(x.shape) > 1: - x = np.ravel(x) - y = np.ravel(y) - - N = len(x) - l = int((n+1)*(n+2)/2) # Number of elements in beta - X = np.ones((N,l)) - - for i in range(1,n+1): - q = int((i)*(i+1)/2) - for k in range(i+1): - X[:,q+k] = (x**(i-k))*(y**k) - - return X - - -# Making meshgrid of datapoints and compute Franke's function -n = 4 -N = 100 -x = np.sort(np.random.uniform(0, 1, N)) -y = np.sort(np.random.uniform(0, 1, N)) -z = FrankeFunction(x, y) -X = create_X(x, y, n=n) - -Xpd = pd.DataFrame(X) -# subtract the mean values and set up the covariance matrix -Xpd = Xpd - Xpd.mean() -covariance_matrix = Xpd.cov() -print(covariance_matrix) -!ec - -We note here that the covariance is zero for the first rows and -columns since all matrix elements in the design matrix were set to one -(we are fitting the function in terms of a polynomial of degree $n$). - -This means that the variance for these elements will be zero and will -cause problems when we set up the correlation matrix. We can simply -drop these elements and construct a correlation -matrix without these elements. - - -!split -===== Rewriting the Covariance and/or Correlation Matrix ===== - -We can rewrite the covariance matrix in a more compact form in terms of the design/feature matrix $\bm{X}$ as -!bt -\[ -\bm{C}[\bm{x}] = \frac{1}{n}\bm{X}\bm{X}^T= \mathbb{E}[\bm{X}\bm{X}^T]. -\] -!et - -To see this let us simply look at a design matrix $\bm{X}\in {\mathbb{R}}^{2\times 2}$ -!bt -\[ -\bm{X}=\begin{bmatrix} -x_{00} & x_{01}\\ -x_{10} & x_{11}\\ -\end{bmatrix}=\begin{bmatrix} -\bm{x}_{0} & \bm{x}_{1}\\ -\end{bmatrix}. -\] -!et - -If we then compute the expectation value -!bt -\[ -\mathbb{E}[\bm{X}\bm{X}^T] = \frac{1}{n}\bm{X}\bm{X}^T=\begin{bmatrix} -x_{00}^2+x_{01}^2 & x_{00}x_{10}+x_{01}x_{11}\\ -x_{10}x_{00}+x_{11}x_{01} & x_{10}^2+x_{11}^2\\ -\end{bmatrix}, -\] -!et -which is just -!bt -\[ -\bm{C}[\bm{x}_0,\bm{x}_1] = \bm{C}[\bm{x}]=\begin{bmatrix} \mathrm{var}[\bm{x}_0] & \mathrm{cov}[\bm{x}_0,\bm{x}_1] \\ - \mathrm{cov}[\bm{x}_1,\bm{x}_0] & \mathrm{var}[\bm{x}_1] \\ - \end{bmatrix}, -\] -!et -where we wrote $$\bm{C}[\bm{x}_0,\bm{x}_1] = \bm{C}[\bm{x}]$$ to indicate that this the covariance of the vectors $\bm{x}$ of the design/feature matrix $\bm{X}$. - -It is easy to generalize this to a matrix $\bm{X}\in {\mathbb{R}}^{n\times p}$. - - -!split -===== Towards the PCA theorem ===== - -We have that the covariance matrix (the correlation matrix involves a simple rescaling) is given as -!bt -\[ -\bm{C}[\bm{x}] = \frac{1}{n}\bm{X}\bm{X}^T= \mathbb{E}[\bm{X}\bm{X}^T]. -\] -!et -Let us now assume that we can perform a series of orthogonal transformations where we employ some orthogonal matrices $\bm{S}$. -These matrices are defined as $\bm{S}\in {\mathbb{R}}^{p\times p}$ and obey the orthogonality requirements $\bm{S}\bm{S}^T=\bm{S}^T\bm{S}=\bm{I}$. The matrix can be written out in terms of the column vectors $\bm{s}_i$ as $\bm{S}=[\bm{s}_0,\bm{s}_1,\dots,\bm{s}_{p-1}]$ and $\bm{s}_i \in {\mathbb{R}}^{p}$. - -Assume also that there is a transformation $\bm{S}\bm{C}[\bm{x}]\bm{S}^T=\bm{C}[\bm{y}]$ such that the new matrix $\bm{C}[\bm{y}]$ is diagonal with elements $[\lambda_0,\lambda_1,\lambda_2,\dots,\lambda_{p-1}]$. - -That is we have -!bt -\[ -\bm{C}[\bm{y}] = \mathbb{E}[\bm{S}\bm{X}\bm{X}^T\bm{S}^T]=\bm{S}\bm{C}[\bm{x}]\bm{S}^T, -\] -!et -since the matrix $\bm{S}$ is not a data dependent matrix. Multiplying with $\bm{S}^T$ from the left we have -!bt -\[ -\bm{S}^T\bm{C}[\bm{y}] = \bm{C}[\bm{x}]\bm{S}^T, -\] -!et -and since $\bm{C}[\bm{y}]$ is diagonal we have for a given eigenvalue $i$ of the covariance matrix that - -!bt -\[ -\bm{S}^T_i\lambda_i = \bm{C}[\bm{x}]\bm{S}^T_i. -\] -!et - -In the derivation of the PCA theorem we will assume that the eigenvalues are ordered in descending order, that is -$\lambda_0 > \lambda_1 > \dots > \lambda_{p-1}$. - - -The eigenvalues tell us then how much we need to stretch the -corresponding eigenvectors. Dimensions with large eigenvalues have -thus large variations (large variance) and define therefore useful -dimensions. The data points are more spread out in the direction of -these eigenvectors. Smaller eigenvalues mean on the other hand that -the corresponding eigenvectors are shrunk accordingly and the data -points are tightly bunched together and there is not much variation in -these specific directions. Hopefully then we could leave it out -dimensions where the eigenvalues are very small. If $p$ is very large, -we could then aim at reducing $p$ to $l << p$ and handle only $l$ -features/predictors. - -!split -===== The Algorithm before the Theorem ===== - -Here's how we would proceed in setting up the algorithm for the PCA, see also discussion below here. -* Set up the datapoints for the design/feature matrix $\bm{X}$ with $\bm{X}\in {\mathbb{R}}^{n\times p}$, with the predictors/features $p$ referring to the column numbers and the entries $n$ being the row elements. -!bt -\[ -\bm{X}=\begin{bmatrix} -x_{0,0} & x_{0,1} & x_{0,2}& \dots & \dots x_{0,p-1}\\ -x_{1,0} & x_{1,1} & x_{1,2}& \dots & \dots x_{1,p-1}\\ -x_{2,0} & x_{2,1} & x_{2,2}& \dots & \dots x_{2,p-1}\\ -\dots & \dots & \dots & \dots \dots & \dots \\ -x_{n-2,0} & x_{n-2,1} & x_{n-2,2}& \dots & \dots x_{n-2,p-1}\\ -x_{n-1,0} & x_{n-1,1} & x_{n-1,2}& \dots & \dots x_{n-1,p-1}\\ -\end{bmatrix}, -\] -!et -* Center the data by subtracting the mean value for each column. This leads to a new matrix $\bm{X}\rightarrow \overline{\bm{X}}$. -* Compute then the covariance/correlation matrix $\mathbb{E}[\overline{\bm{X}}\overline{\bm{X}}^T]$. -* Find the eigenpairs of $\bm{C}$ with eigenvalues $[\lambda_0,\lambda_1,\dots,\lambda_{p-1}]$ and eigenvectors $[\bm{s}_0,\bm{s}_1,\dots,\bm{s}_{p-1}]$. -* Order the eigenvalue (and the eigenvectors accordingly) in order of decreasing eigenvalues. -* Keep only those $l$ eigenvalues larger than a selected threshold value, discarding thus $p-l$ features since we expect small variations in the data here. - - -!split -===== Writing our own PCA code ===== - -We will use a simple example first with two-dimensional data -drawn from a multivariate normal distribution with the following mean and covariance matrix: -!bt -\[ -\mu = (-1,2) \qquad \Sigma = \begin{bmatrix} 4 & 2 \\ -2 & 2 -\end{bmatrix} -\] -!et -Note that the mean refers to each column of data. -We will generate $n = 1000$ points $X = \{ x_1, \ldots, x_N \}$ from -this distribution, and store them in the $1000 \times 2$ matrix $\bm{X}$. - -The following Python code aids in setting up the data and writing out the design matrix. -Note that the function _multivariate_ returns also the covariance discussed above and that it is defined by dividing by $n-1$ instead of $n$. -!bc pycod -import numpy as np -import pandas as pd -import matplotlib.pyplot as plt -from IPython.display import display -n = 10000 -mean = (-1, 2) -cov = [[4, 2], [2, 2]] -X = np.random.multivariate_normal(mean, cov, n) -!ec - -Now we are going to implement the PCA algorithm. We will break it down into various substeps. - -=== Compute the sample mean and center the data === - -The first step of PCA is to compute the sample mean of the data and use it to center the data. Recall that the sample mean is -!bt -\[ -\mu_n = \frac{1}{n} \sum_{i=1}^n x_i -\] -!et -and the mean-centered data $\bar{X} = \{ \bar{x}_1, \ldots, \bar{x}_n \}$ takes the form -!bt -\[ -\bar{x}_i = x_i - \mu_n. -\] -!et -When you are done with these steps, print out $\mu_n$ to verify it is -close to $\mu$ and plot your mean centered data to verify it is -centered at the origin! Compare your code with the functionality from _Scikit-Learn_ discussed above. -The following code elements perform these operations using _pandas_ or using our own functionality for doing so. The latter, using _numpy_ is rather simple through the _mean()_ function. -!bc pycod -df = pd.DataFrame(X) -# Pandas does the centering for us -df = df -df.mean() -# we center it ourselves -X_centered = X - X.mean(axis=0) -!ec - -Alternatively, we could use the functions we discussed -earlier for scaling the data set. That is, we could have used the -_StandardScaler_ function in _Scikit-Learn_, a function which ensures -that for each feature/predictor we study the mean value is zero and -the variance is one (every column in the design/feature matrix). You -would then not get the same results, since we divide by the -variance. The diagonal covariance matrix elements will then be one, -while the non-diagonal ones need to be divided by $2\sqrt{2}$ for our -specific case. - -=== Compute the sample covariance === - -Now we are going to use the mean centered data to compute the sample covariance of the data by using the following equation -!bt -\begin{equation*} -\Sigma_n = \frac{1}{n-1} \sum_{i=1}^n \bar{x}_i^T \bar{x}_i = \frac{1}{n-1} \sum_{i=1}^n (x_i - \mu_n)^T (x_i - \mu_n) -\end{equation*} -!et -where the data points $x_i \in \mathbb{R}^p$ (here in this example $p = 2$) are column vectors and $x^T$ is the transpose of $x$. -We can write our own code or simply use either the functionaly of _numpy_ or that of _pandas_, as follows -!bc pycod -print(df.cov()) -print(np.cov(X_centered.T)) -!ec -Note that the way we define the covariance matrix here has a factor $n-1$ instead of $n$. This is included in the _cov()_ function by _numpy_ and _pandas_. -Our own code here is not very elegant and asks for obvious improvements. It is tailored to this specific $2\times 2$ covariance matrix. -!bc pycod -# extract the relevant columns from the centered design matrix of dim n x 2 -x = X_centered[:,0] -y = X_centered[:,1] -Cov = np.zeros((2,2)) -Cov[0,1] = np.sum(x.T@y)/(n-1.0) -Cov[0,0] = np.sum(x.T@x)/(n-1.0) -Cov[1,1] = np.sum(y.T@y)/(n-1.0) -Cov[1,0]= Cov[0,1] -print("Centered covariance using own code") -print(Cov) -plt.plot(x, y, 'x') -plt.axis('equal') -plt.show() -!ec - -Depending on the number of points $n$, we will get results that are close to the covariance values defined above. -The plot shows how the data are clustered around a line with slope close to one. Is this expected? - -=== Diagonalize the sample covariance matrix to obtain the principal components === - -Now we are ready to solve for the principal components! To do so we -diagonalize the sample covariance matrix $\Sigma$. We can use the -function _np.linalg.eig_ to do so. It will return the eigenvalues and -eigenvectors of $\Sigma$. Once we have these we can perform the -following tasks: - -* We compute the percentage of the total variance captured by the first principal component -* We plot the mean centered data and lines along the first and second principal components -* Then we project the mean centered data onto the first and second principal components, and plot the projected data. -* Finally, we approximate the data as - -!bt -\begin{equation*} -x_i \approx \tilde{x}_i = \mu_n + \langle x_i, v_0 \rangle v_0 -\end{equation*} -!et -where $v_0$ is the first principal component. - -Collecting all these steps we can write our own PCA function and -compare this with the functionality included in _Scikit-Learn_. - -The code here outlines some of the elements we could include in the -analysis. Feel free to extend upon this in order to address the above -questions. - -!bc pycod -# diagonalize and obtain eigenvalues, not necessarily sorted -EigValues, EigVectors = np.linalg.eig(Cov) -# sort eigenvectors and eigenvalues -#permute = EigValues.argsort() -#EigValues = EigValues[permute] -#EigVectors = EigVectors[:,permute] -print("Eigenvalues of Covariance matrix") -for i in range(2): - print(EigValues[i]) -FirstEigvector = EigVectors[:,0] -SecondEigvector = EigVectors[:,1] -print("First eigenvector") -print(FirstEigvector) -print("Second eigenvector") -print(SecondEigvector) -#thereafter we do a PCA with Scikit-learn -from sklearn.decomposition import PCA -pca = PCA(n_components = 2) -X2Dsl = pca.fit_transform(X) -print("Eigenvector of largest eigenvalue") -print(pca.components_.T[:, 0]) - -!ec -This code does not contain all the above elements, but it shows how we can use _Scikit-Learn_ to extract the eigenvector which corresponds to the largest eigenvalue. Try to address the questions we pose before the above code. Try also to change the values of the covariance matrix by making one of the diagonal elements much larger than the other. What do you observe then? - -!split -===== Classical PCA Theorem ===== - -We assume now that we have a design matrix $\bm{X}$ which has been -centered as discussed above. For the sake of simplicity we skip the -overline symbol. The matrix is defined in terms of the various column -vectors $[\bm{x}_0,\bm{x}_1,\dots, \bm{x}_{p-1}]$ each with dimension -$\bm{x}\in {\mathbb{R}}^{n}$. - -We assume also that we have an orthogonal transformation $\bm{W}\in {\mathbb{R}}^{p\times p}$. We define the reconstruction error (which is similar to the mean squared error we have seen before) as -!bt -\[ -J(\bm{W},\bm{Z}) = \frac{1}{n}\sum_i (\bm{x}_i - \overline{\bm{x}}_i)^2, -\] -!et -with $\overline{\bm{x}}_i = \bm{W}\bm{z}_i$, where $\bm{z}_i$ is a row vector with dimension ${\mathbb{R}}^{n}$ of the matrix -$\bm{Z}\in{\mathbb{R}}^{p\times n}$. When doing PCA we want to reduce this dimensionality. - -The PCA theorem states that minimizing the above reconstruction error -corresponds to setting $\bm{W}=\bm{S}$, the orthogonal matrix which -diagonalizes the empirical covariance(correlation) matrix. The optimal -low-dimensional encoding of the data is then given by a set of vectors -$\bm{z}_i$ with at most $l$ vectors, with $l << p$, defined by the -orthogonal projection of the data onto the columns spanned by the -eigenvectors of the covariance(correlations matrix). - -The proof which follows will be updated by mid January 2020. - -!split -===== Proof of the PCA Theorem ===== - -To show the PCA theorem let us start with the assumption that there is one vector $\bm{w}_0$ which corresponds to a solution which minimized the reconstruction error $J$. This is an orthogonal vector. It means that we now approximate the reconstruction error in terms of $\bm{w}_0$ and $\bm{z}_0$ as -!bt -\[ -J(\bm{w}_0,\bm{z}_0)= \frac{1}{n}\sum_i (\bm{x}_i - z_{i0}\bm{w}_0)^2=\frac{1}{n}\sum_i (\bm{x}_i^T\bm{x}_i - 2z_{i0}\bm{w}_0^T\bm{x}_i+z_{i0}^2\bm{w}_0^T\bm{w}_0), -\] -!et -which we can rewrite due to the orthogonality of $\bm{w}_i$ as -!bt -\[ -J(\bm{w}_0,\bm{z}_0)=\frac{1}{n}\sum_i (\bm{x}_i^T\bm{x}_i - 2z_{i0}\bm{w}_0^T\bm{x}_i+z_{i0}^2). -\] -!et -Minimizing $J$ with respect to the unknown parameters $z_{0i}$ we obtain that -!bt -\[ -z_{i0}=\bm{w}_0^T\bm{x}_i, -\] -!et -where the vectors on the rhs are known. - - -!split -===== PCA Proof continued ===== - -We have now found the unknown parameters $z_{i0}$. These correspond to the projected coordinates and we can write -!bt -\[ -J(\bm{w}_0)= \frac{1}{p}\sum_i (\bm{x}_i^T\bm{x}_i - z_{i0}^2)=\mathrm{const}-\frac{1}{n}\sum_i z_{i0}^2. -\] -!et - -We can show that the variance of the projected coordinates defined by $\bm{w}_0^T\bm{x}_i$ are given by -!bt -\[ -\mathrm{var}[\bm{w}_0^T\bm{x}_i] = \frac{1}{n}\sum_i z_{i0}^2, -\] -!et -since the expectation value of -!bt -\[ -\mathbb{E}[\bm{w}_0^T\bm{x}_i] = \mathbb{E}[z_{i0}]= \bm{w}_0^T\mathbb{E}[\bm{x}_i]=0, -\] -!et -where we have used the fact that our data are centered. - -Recalling our definition of the covariance as -!bt -\[ -\bm{C}[\bm{x}] = \frac{1}{n}\bm{X}\bm{X}^T=\mathbb{E}[\bm{X}\bm{X}^T], -\] -!et -we have thus that -!bt -\[ -\mathrm{var}[\bm{w}_0^T\bm{x}_i] = \frac{1}{n}\sum_i z_{i0}^2=\bm{w}_0^T\bm{C}[\bm{x}]\bm{w}_0. -\] -!et - -We are almost there, we have obtained a relation between minimizing -the reconstruction error and the variance and the covariance -matrix. Minimizing the error is equivalent to maximizing the variance -of the projected data. - -!split -===== The final step ===== - -We could trivially maximize the variance of the projection (and -thereby minimize the error in the reconstruction function) by letting -the norm-2 of $\bm{w}_0$ go to infinity. However, this norm since we -want the matrix $\bm{W}$ to be an orthogonal matrix, is constrained by -$\vert\vert \bm{w}_0 \vert\vert_2^2=1$. Imposing this condition via a -Lagrange multiplier we can then in turn maximize - -!bt -\[ -J(\bm{w}_0)= \bm{w}_0^T\bm{C}[\bm{x}]\bm{w}_0+\lambda_0(1-\bm{w}_0^T\bm{w}_0). -\] -!et -Taking the derivative with respect to $\bm{w}_0$ we obtain - -!bt -\[ -\frac{\partial J(\bm{w}_0)}{\partial \bm{w}_0}= 2\bm{C}[\bm{x}]\bm{w}_0-2\lambda_0\bm{w}_0=0, -\] -!et -meaning that -!bt -\[ -\bm{C}[\bm{x}]\bm{w}_0=\lambda_0\bm{w}_0. -\] -!et -_The direction that maximizes the variance (or minimizes the construction error) is an eigenvector of the covariance matrix_! If we left multiply with $\bm{w}_0^T$ we have the variance of the projected data is -!bt -\[ -\bm{w}_0^T\bm{C}[\bm{x}]\bm{w}_0=\lambda_0. -\] -!et - -If we want to maximize the variance (minimize the construction error) -we simply pick the eigenvector of the covariance matrix with the -largest eigenvalue. This establishes the link between the minimization -of the reconstruction function $J$ in terms of an orthogonal matrix -and the maximization of the variance and thereby the covariance of our -observations encoded in the design/feature matrix $\bm{X}$. - -The proof -for the other eigenvectors $\bm{w}_1,\bm{w}_2,\dots$ can be -established by applying the above arguments and using the fact that -our basis of eigenvectors is orthogonal, see "Murphy chapter -12.2":"https://mitpress.mit.edu/books/machine-learning-1". The -discussion in chapter 12.2 of Murphy's text has also a nice link with -the Singular Value Decomposition theorem. For categorical data, see -chapter 12.4 and discussion therein. - -Additional part of the proof for the other eigenvectors will be added by mid January 2020. - -!split -===== Geometric Interpretation and link with Singular Value Decomposition ===== - -This material will be added by mid January 2020. - - -!split -===== Principal Component Analysis ===== - -Principal Component Analysis (PCA) is by far the most popular dimensionality reduction algorithm. -First it identifies the hyperplane that lies closest to the data, and then it projects the data onto it. - -The following Python code uses NumPy’s _svd()_ function to obtain all the principal components of the -training set, then extracts the first two principal components. First we center the data using either _pandas_ or our own code -!bc pycod -import numpy as np -import pandas as pd -from IPython.display import display -np.random.seed(100) -# setting up a 10 x 5 vanilla matrix -rows = 10 -cols = 5 -X = np.random.randn(rows,cols) -df = pd.DataFrame(X) -# Pandas does the centering for us -df = df -df.mean() -display(df) - -# we center it ourselves -X_centered = X - X.mean(axis=0) -# Then check the difference between pandas and our own set up -print(X_centered-df) -#Now we do an SVD -U, s, V = np.linalg.svd(X_centered) -c1 = V.T[:, 0] -c2 = V.T[:, 1] -W2 = V.T[:, :2] -X2D = X_centered.dot(W2) -print(X2D) -!ec - -PCA assumes that the dataset is centered around the origin. Scikit-Learn’s PCA classes take care of centering -the data for you. However, if you implement PCA yourself (as in the preceding example), or if you use other libraries, don’t -forget to center the data first. - -Once you have identified all the principal components, you can reduce the dimensionality of the dataset -down to $d$ dimensions by projecting it onto the hyperplane defined by the first $d$ principal components. -Selecting this hyperplane ensures that the projection will preserve as much variance as possible. -!bc pycod -W2 = V.T[:, :2] -X2D = X_centered.dot(W2) -!ec - -!split -===== PCA and scikit-learn ===== - -Scikit-Learn’s PCA class implements PCA using SVD decomposition just like we did before. The -following code applies PCA to reduce the dimensionality of the dataset down to two dimensions (note -that it automatically takes care of centering the data): -!bc pycod -#thereafter we do a PCA with Scikit-learn -from sklearn.decomposition import PCA -pca = PCA(n_components = 2) -X2D = pca.fit_transform(X) -print(X2D) -!ec -After fitting the PCA transformer to the dataset, you can access the principal components using the -components variable (note that it contains the PCs as horizontal vectors, so, for example, the first -principal component is equal to -!bc pycod -pca.components_.T[:, 0]. -!ec -Another very useful piece of information is the explained variance ratio of each principal component, -available via the $explained\_variance\_ratio$ variable. It indicates the proportion of the dataset’s -variance that lies along the axis of each principal component. - -!split -===== Back to the Cancer Data ===== -We can now repeat the above but applied to real data, in this case our breast cancer data. -Here we compute performance scores on the training data using logistic regression. -!bc pycod -import matplotlib.pyplot as plt -import numpy as np -from sklearn.model_selection import train_test_split -from sklearn.datasets import load_breast_cancer -from sklearn.linear_model import LogisticRegression -cancer = load_breast_cancer() - -X_train, X_test, y_train, y_test = train_test_split(cancer.data,cancer.target,random_state=0) - -logreg = LogisticRegression() -logreg.fit(X_train, y_train) -print("Train set accuracy from Logistic Regression: {:.2f}".format(logreg.score(X_train,y_train))) -# We scale the data -from sklearn.preprocessing import StandardScaler -scaler = StandardScaler() -scaler.fit(X_train) -X_train_scaled = scaler.transform(X_train) -X_test_scaled = scaler.transform(X_test) -# Then perform again a log reg fit -logreg.fit(X_train_scaled, y_train) -print("Train set accuracy scaled data: {:.2f}".format(logreg.score(X_train_scaled,y_train))) -#thereafter we do a PCA with Scikit-learn -from sklearn.decomposition import PCA -pca = PCA(n_components = 2) -X2D_train = pca.fit_transform(X_train_scaled) -# and finally compute the log reg fit and the score on the training data -logreg.fit(X2D_train,y_train) -print("Train set accuracy scaled and PCA data: {:.2f}".format(logreg.score(X2D_train,y_train))) - -!ec - -We see that our training data after the PCA decomposition has a performance similar to the non-scaled data. - -!split -===== More on the PCA ===== - -Instead of arbitrarily choosing the number of dimensions to reduce down to, it is generally preferable to -choose the number of dimensions that add up to a sufficiently large portion of the variance (e.g., 95%). -Unless, of course, you are reducing dimensionality for data visualization — in that case you will -generally want to reduce the dimensionality down to 2 or 3. -The following code computes PCA without reducing dimensionality, then computes the minimum number -of dimensions required to preserve 95% of the training set’s variance: -!bc pycod -pca = PCA() -pca.fit(X) -cumsum = np.cumsum(pca.explained_variance_ratio_) -d = np.argmax(cumsum >= 0.95) + 1 -!ec -You could then set $n\_components=d$ and run PCA again. However, there is a much better option: instead -of specifying the number of principal components you want to preserve, you can set $n\_components$ to be -a float between 0.0 and 1.0, indicating the ratio of variance you wish to preserve: -!bc pycod -pca = PCA(n_components=0.95) -X_reduced = pca.fit_transform(X) -!ec - -!split -===== Incremental PCA ===== - -One problem with the preceding implementation of PCA is that it requires the whole training set to fit in -memory in order for the SVD algorithm to run. Fortunately, Incremental PCA (IPCA) algorithms have -been developed: you can split the training set into mini-batches and feed an IPCA algorithm one minibatch -at a time. This is useful for large training sets, and also to apply PCA online (i.e., on the fly, as new -instances arrive). - -!split -===== Randomized PCA ===== - -Scikit-Learn offers yet another option to perform PCA, called Randomized PCA. This is a stochastic -algorithm that quickly finds an approximation of the first d principal components. Its computational -complexity is $O(m \times d^2)+O(d^3)$, instead of $O(m \times n^2) + O(n^3)$, so it is dramatically faster than the -previous algorithms when $d$ is much smaller than $n$. - - - - -!split -===== Kernel PCA ===== -!bblock - -The kernel trick is a mathematical technique that implicitly maps instances into a -very high-dimensional space (called the feature space), enabling nonlinear classification and regression -with Support Vector Machines. Recall that a linear decision boundary in the high-dimensional feature -space corresponds to a complex nonlinear decision boundary in the original space. -It turns out that the same trick can be applied to PCA, making it possible to perform complex nonlinear -projections for dimensionality reduction. This is called Kernel PCA (kPCA). It is often good at -preserving clusters of instances after projection, or sometimes even unrolling datasets that lie close to a -twisted manifold. -For example, the following code uses Scikit-Learn’s KernelPCA class to perform kPCA with an -!bc pycod -from sklearn.decomposition import KernelPCA -rbf_pca = KernelPCA(n_components = 2, kernel="rbf", gamma=0.04) -X_reduced = rbf_pca.fit_transform(X) -!ec - -!eblock - - -!split -===== LLE ===== - -Locally Linear Embedding (LLE) is another very powerful nonlinear dimensionality reduction -(NLDR) technique. It is a Manifold Learning technique that does not rely on projections like the previous -algorithms. In a nutshell, LLE works by first measuring how each training instance linearly relates to its -closest neighbors (c.n.), and then looking for a low-dimensional representation of the training set where -these local relationships are best preserved (more details shortly). - - - -!split -===== Other techniques ===== - - -There are many other dimensionality reduction techniques, several of which are available in Scikit-Learn. - -Here are some of the most popular: -* _Multidimensional Scaling (MDS)_ reduces dimensionality while trying to preserve the distances between the instances. -* _Isomap_ creates a graph by connecting each instance to its nearest neighbors, then reduces dimensionality while trying to preserve the geodesic distances between the instances. -* _t-Distributed Stochastic Neighbor Embedding_ (t-SNE) reduces dimensionality while trying to keep similar instances close and dissimilar instances apart. It is mostly used for visualization, in particular to visualize clusters of instances in high-dimensional space (e.g., to visualize the MNIST images in 2D). -* Linear Discriminant Analysis (LDA) is actually a classification algorithm, but during training it learns the most discriminative axes between the classes, and these axes can then be used to define a hyperplane onto which to project the data. The benefit is that the projection will keep classes as far apart as possible, so LDA is a good technique to reduce dimensionality before running another classification algorithm such as a Support Vector Machine (SVM) classifier discussed in the SVM lectures. - - -!split -===== Differential equations ===== - -The Universal Approximation Theorem states that a neural network can -approximate any function at a single hidden layer along with one input -and output layer to any given precision. Having this in mind, we will -look closer at whether a neural network manages to solve for a -function in an equation. - - -!split -===== Description of the equation to solve for ===== -A differential equation is a equation where the solution is a function. -The equation describes how the derivatives of the function behaves in a given domain along with some conditions. - -Given a differential equation, it is desirable to know how to -reformulate it into an equation a neural network can solve. Having -decided on which activation functions each layer should use, along -with the number of hidden layers and neurons within each layer, the -changeable parameters of a neural network are the weights and biases -for each neuron in every layer in the net. If a differential equation -is reformulated into an equation where minimization of some parameters -must be done, a neural net could possibly solve this equation. - -A trial solution might be tricky to find in general. Due to the -Universal Approximation Theorem, one could hope that outcome of the -deep neural net might solve a given differential equation, even though -it is used in a simple trial solution. Let us try this idea on some -well-known ordinary differential equations and thereafter try to solve -for functions defined by two variables, giving partial differential -equations. - -!split -===== Ordinary Differential Equations ===== - -An ordinary differential equation (ODE) is an equation involving functions having one variable. - -In general, an ordinary differential equation looks like - -!bt -\begin{equation} \label{ode} -f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right) = 0 -\end{equation} -!et - -where $g(x)$ is the function to find, and $g^{(n)}(x)$ is the $n$-th derivative of $g(x)$. - -The $f\left(x, g(x), g'(x), g''(x), \, \dots \, , g^{(n)}(x)\right)$ is just a way to write that there is an expression involving $x$ and $g(x), \ g'(x), \ g''(x), \, \dots \, , \text{ and } g^{(n)}(x)$ on the left side of the equality sign in (ref{ode}). -The highest order of derivative, that is the value of $n$, determines to the order of the equation. -The equation is referred to as a $n$-th order ODE. -Along with (ref{ode}), some additional conditions of the function $g(x)$ are typically given -for the solution to be unique. - -!split -===== The trial solution ===== - -Let the trial solution $g_t(x)$ be - -!bt -\begin{equation} - g_t(x) = h_1(x) + h_2(x,N(x,P)) -\end{equation} -!et - -where $h_1(x)$ is a function that makes $g_t(x)$ satisfy a given set of conditions, $N(x,P)$ a neural network with weights and biases described by $P$ and $h_2(x, N(x,P))$ some expression involving the neural network. -The role of the function $h_2(x, N(x,P))$, is to ensure that the output from $N(x,P)$ is zero when $g_t(x)$ is evaluated at the values of $x$ where the given conditions must be satisfied. -The function $h_1(x)$ should alone make $g_t(x)$ satisfy the conditions. - -But what about the network $N(x,P)$? -As described previously, an optimization method could be used to minimize the parameters of a neural network, that being its weights and biases, through backward propagation. -For the minimization to be defined, we need to have a cost function at hand to minimize. - -It is given that $f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right)$ should be equal to zero in (ref{ode}). -We can choose to consider the mean squared error as the cost function for an input $x$. -Since we are looking at one input, the cost function is just $f$ squared. -The cost function $c\left(x, P \right)$ can therefore be expressed as - -!bt -c\left(x, P\right) = \big(f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right)\big)^2 -!et - -If $N$ inputs are given as a vector $\vec x$ with elements $x_i$ for $i = 1,\dots,N$, -the cost function becomes - -!bt -\begin{equation} \label{cost} - c\left(\vec x, P\right) = \frac{1}{N} \sum_{i=1}^N \big(f\left(x_i, \, g(x_i), \, g'(x_i), \, g''(x_i), \, \dots \, , \, g^{(n)}(x_i)\right)\big)^2 -\end{equation} -!et - -The neural net should then find some parameters $P$ that minimizes the cost function in -(ref{cost}) for a set of $N$ training samples $x_i$. - -!split -===== Minimizing the cost function using gradient descent and automatic differentiation ===== -To perform the minimization using gradient descent, the gradient of $c\left(\vec x, P\right)$ is needed. -It might happen so that finding an analytical expression of the gradient of $c(\vec x, P)$ from (ref{cost}) gets too messy, depending on which cost function one desires to use. - -Luckily, there exists libraries that makes the job for us through automatic differentiation. -Automatic differentiation is a method of finding the derivatives numerically with very high precision. - -In the forthcoming examples presenting possible usages of Autograd and TensorFlow, -it is shown how one could set up a neural network using gradient descent solving a differential -equation. - -!split -===== Example: Exponential decay and setting up the network using Autograd ===== -An exponential decay of a quantity $g(x)$ is described by the equation - -!bt -\begin{equation} \label{solve_expdec} - g'(x) = -\gamma g(x) -\end{equation} -!et - -with $g(0) = g_0$ for some chosen initial value $g_0$. - -The analytical solution of (ref{solve_expdec}) is - -!bt -\begin{equation} - g(x) = g_0 \exp\left(-\gamma x\right) -\end{equation} -!et - -Having an analytical solution at hand, it is possible to use it to compare how well a neural network finds a solution of (ref{solve_expdec}). - -In this example, a neural network will be implemented using Autograd in order to perform backpropagation. - -!split -===== The function to solve for ===== - -The program will use a neural network to solve - -!bt -\begin{equation} \label{solveode} -g'(x) = -\gamma g(x) -\end{equation} -!et - -where $g(0) = g_0$ with $\gamma$ and $g_0$ being some chosen values. - -In this example, $\gamma = 2$ and $g_0 = 10$. - -!split -===== The trial solution ===== -To begin with, a trial solution $g_t(t)$ must be chosen. A general trial solution for ordinary differential equations could be - -!bt -g_t(x, P) = h_1(x) + h_2(x, N(x, P)) -!et - -with $h_1(x)$ ensuring that $g_t(x)$ satisfies some conditions and $h_2(x,N(x, P))$ an expression involving $x$ and the output from the neural network $N(x,P)$ with $P $ being the collection of the weights and biases for each layer. For now, it is assumed that the network consists of one input layer, one hidden layer, and one output layer. - -In this network, there are no weights and bias at the input layer, so $P = \{ P_{\text{hidden}}, P_{\text{output}} \}$. -If there are $N_{\text{hidden} }$ neurons in the hidden layer, then $P_{\text{hidden}}$ is a $N_{\text{hidden} } \times (1 + N_{\text{input}})$ matrix, given that there are $N_{\text{input}}$ neurons in the input layer. - -The first column in $P_{\text{hidden} }$ represents the bias for each neuron in the hidden layer and the second column represents the weights for each neuron in the hidden layer from the input layer. -If there are $N_{\text{output} }$ neurons in the output layer, then $P_{\text{output}} $ is a $N_{\text{output} } \times (1 + N_{\text{hidden} })$ matrix. - -Its first column represents the bias of each neuron and the remaining columns represents the weights to each neuron. - -It is given that $g(0) = g_0$. The trial solution must fulfill this condition to be a proper solution of (ref{solveode}). A possible way to ensure that $g_t(0, P) = g_0$, is to let $F(N(x,P)) = x \cdot N(x,P)$ and $A(x) = g_0$. This gives the following trial solution: - -!bt -\begin{equation} \label{trial} -g_t(x, P) = g_0 + x \cdot N(x, P) -\end{equation} -!et - -!split -===== Reformulating the problem ===== -We wish that our neural network manages to minimize a given cost function. - -A reformulation of out equation, (ref{solveode}), must therefore be done, -such that it describes the problem a neural network can solve for. - -The neural network must find the set of weights and biases $P$ such that the trial solution in (ref{trial}) satisfies (ref{solveode}). - -The trial solution - -!bt -g_t(x, P) = g_0 + x \cdot N(x, P) -!et - -has been chosen such that it already solves the condition $g(0) = g_0$. What remains, is to find $P$ such that - -!bt -\begin{equation} \label{nnmin} -g_t'(x, P) = - \gamma g_t(x, P) -\end{equation} -!et - -is fulfilled as *best as possible*. - -The left hand side and right hand side of (ref{nnmin}) must be computed separately, and then the neural network must choose weights and biases, contained in $P$, such that the sides are equal as best as possible. -This means that the absolute or squared difference between the sides must be as close to zero, ideally equal to zero. -In this case, the difference squared shows to be an appropriate measurement of how erroneous the trial solution is with respect to $P$ of the neural network. - -This gives the following cost function our neural network must solve for: - -!bt -\min_{P}\Big\{ \big(g_t'(x, P) - ( -\gamma g_t(x, P) \big)^2 \Big\} -!et - -(the notation $\min_{P}\{ f(x, P) \}$ means that we desire to find $P$ that yields the minimum of $f(x, P)$) - -or, in terms of weights and biases for the hidden and output layer in our network: - -!bt -\min_{P_{\text{hidden} }, \ P_{\text{output} }}\Big\{ \big(g_t'(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) - ( -\gamma g_t(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) \big)^2 \Big\} -!et - -for an input value $x$. - -If the neural network evaluates $g_t(x, P)$ at more values for $x$, say $N$ values $x_i$ for $i = 1, \dots, N$, then the *total* error to minimize becomes - -!bt -\begin{equation} \label{min} -\min_{P}\Big\{\frac{1}{N} \sum_{i=1}^N \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 \Big\} -\end{equation} -!et - -Letting $\vec x$ be a vector with elements $x_i$ and $c(\vec x, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2$ denote the cost function, the minimization problem that our network must solve, becomes - -!bt -\min_{P} c(\vec x, P) -!et - -In terms of $P_{\text{hidden} }$ and $P_{\text{output} }$, this could also be expressed as - -$$ -\min_{P_{\text{hidden} }, \ P_{\text{output} }} c(\vec x, \{P_{\text{hidden} }, P_{\text{output} }\}) -$$ - -!split -===== A possible implementation of a neural network using Autograd ===== - -For simplicity, it is assumed that the input is an array $\vec x = (x_1, \dots, x_N)$ with $N$ elements. It is at these points the neural network should find $P$ such that it fulfills (ref{min}). - -First, the neural network must feed forward the inputs. -This means that $\vec x$ must be passed through an input layer, a hidden layer and a output layer. The input layer in this case, does not need to process the data any further. -The input layer will consist of $N_{\text{input} }$ neurons, passing its element to each neuron in the hidden layer. The number of neurons in the hidden layer will be $N_{\text{hidden} }$. - -For the $i$-th in the hidden layer with weight $w_i^{\text{hidden} }$ and bias $b_i^{\text{hidden} }$, the weighting from the $j$-th neuron at the input layer is: - -!bt -\begin{aligned} -z_{i,j}^{\text{hidden}} &= b_i^{\text{hidden}} + w_i^{\text{hidden}}x_j \\ -&= -\begin{pmatrix} -b_i^{\text{hidden}} & w_i^{\text{hidden}} -\end{pmatrix} -\begin{pmatrix} -1 \\ -x_j -\end{pmatrix} -\end{aligned} -!et - -The result after weighting the inputs at the $i$-th hidden neuron can be written as a vector: - -!bt -\begin{aligned} -\vec{z}_{i}^{\text{hidden}} &= \Big( b_i^{\text{hidden}} + w_i^{\text{hidden}}x_1 , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_2, \ \dots \, , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_N\Big) \\ -&= -\begin{pmatrix} - b_i^{\text{hidden}} & w_i^{\text{hidden}} -\end{pmatrix} -\begin{pmatrix} -1 & 1 & \dots & 1 \\ -x_1 & x_2 & \dots & x_N -\end{pmatrix} \\ -&= \vec{p}_{i, \text{hidden}}^T X -\end{aligned} -!et - -The vector $\vec{p}_{i, \text{hidden}}^T$ constitutes each row in $P_{\text{hidden} }$, which contains the weights for the neural network to minimize according to (ref{min}). - -After having found $\vec{z}_{i}^{\text{hidden}} $ for every $i$-th neuron within the hidden layer, the vector will be sent to an activation function $a_i(\vec{z})$. - -In this example, the sigmoid function has been chosen to be the activation function for each hidden neuron: - -!bt -f(z) = \frac{1}{1 + \exp{(-z)}} -!et - -It is possible to use other activations functions for the hidden layer also. - -The output $\vec{x}_i^{\text{hidden} }$from each $i$-th hidden neuron is: - -$$ -\vec{x}_i^{\text{hidden} } = f\big( \vec{z}_{i}^{\text{hidden}} \big) -$$ - -The outputs $\vec{x}_i^{\text{hidden} } $ are then sent to the output layer. - -The output layer consists of one neuron in this case, and combines the output from each of the neurons in the hidden layers. The output layer combines the results from the hidden layer using some weights $ w_i^{\text{output}}$ and biases $b_i^{\text{output}}$. In this case, it is assumes that the number of neurons in the output layer is one. - -The procedure of weighting the output neuron $j$ in the hidden layer to the $i$-th neuron in the output layer is similar as for the hidden layer described previously. - -!bt -\begin{aligned} -z_{1,j}^{\text{output}} & = -\begin{pmatrix} -b_1^{\text{output}} & \vec{w}_1^{\text{output}} -\end{pmatrix} -\begin{pmatrix} -1 \\ -\vec{x}_j^{\text{hidden}} -\end{pmatrix} -\end{aligned} -!et - -Expressing $z_{1,j}^{\text{output}}$ as a vector gives the following way of weighting the inputs from the hidden layer: - -!bt -\vec{z}_{1}^{\text{output}} = -\begin{pmatrix} -b_1^{\text{output}} & \vec{w}_1^{\text{output}} -\end{pmatrix} -\begin{pmatrix} -1 & 1 & \dots & 1 \\ -\vec{x}_1^{\text{hidden}} & \vec{x}_2^{\text{hidden}} & \dots & \vec{x}_N^{\text{hidden}} -\end{pmatrix} -!et - -In this case we seek a continuous range of values since we are approximating a function. This means that after computing $\vec{z}_{1}^{\text{output}}$ the neural network has finished its feed forward step, and $\vec{z}_{1}^{\text{output}}$ is the final output of the network. - -!split -===== Backpropagation using Autograd ===== -The next step is to decide how the parameters should be changed such that they minimize the cost function. - -The chosen cost function for this problem is - -!bt -c(\vec x, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 -!et - -In order to minimize the cost function, an optimization method must be chosen. - -Here, gradient descent with a constant step size has been chosen. - -!split -===== Gradient descent ===== -The idea of the gradient descent algorithm is to update parameters in direction where the cost function decreases goes to a minimum. - -In general, the update of some parameters $\vec \omega$ given a cost function defined by some weights $\vec \omega$, $c(\vec x, \vec \omega)$, goes as follows: - -!bt -\vec \omega_{\text{new} } = \vec \omega - \lambda \nabla_{\vec \omega} c(\vec x, \vec \omega) -!et - -for a number of iterations or until $ \big|\big| \vec \omega_{\text{new} } - \vec \omega \big|\big|$ becomes smaller than some given tolerance. - -The value of $\lambda$ decides how large steps the algorithm must take in the direction of $ \nabla_{\vec \omega} c(\vec x, \vec \omega)$. -The notation $\nabla_{\vec \omega}$ express the gradient with respect to the elements in $\vec \omega$. - -In our case, we have to minimize the cost function $c(\vec x, P)$ with respect to the two sets of weights and biases, that is for the hidden layer $P_{\text{hidden} }$ and for the output layer $P_{\text{output} }$ . - -This means that $P_{\text{hidden} }$ and $P_{\text{output} }$ is updated by - -!bt -\begin{aligned} -P_{\text{hidden},\text{new}} &= P_{\text{hidden}} - \lambda \nabla_{P_{\text{hidden}}} c(\vec x, P) \\ -P_{\text{output},\text{new}} &= P_{\text{output}} - \lambda \nabla_{P_{\text{output}}} c(\vec x, P) -\end{aligned} -!et - -In general, one could risk using a cost function having gradients that are cumbersome to derive analytically. -For our case, the cost functions are just the mean squared error. -One could employ an implementation of the back propagation for this case, but we will emphasis -on how one could use automatic differentiation in order to train the network. - -However, it might be useful to know how automatic differentiation can be used, e.g through Autograd, in order to test an implementation. - -!split -===== The network with one input, hidden, and output layer ===== - -!bc pycod -# Autograd will be used for later, so the numpy wrapper for Autograd must be imported -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -# Assuming one input, hidden, and output layer -def neural_network(params, x): - - # Find the weights (including and biases) for the hidden and output layer. - # Assume that params is a list of parameters for each layer. - # The biases are the first element for each array in params, - # and the weights are the remaning elements in each array in params. - - w_hidden = params[0] - w_output = params[1] - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - ## Hidden layer: - - # Add a row of ones to include bias - x_input = np.concatenate((np.ones((1,num_values)), x_input ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_input) - x_hidden = sigmoid(z_hidden) - - ## Output layer: - - # Include bias: - x_hidden = np.concatenate((np.ones((1,num_values)), x_hidden ), axis = 0) - - z_output = np.matmul(w_output, x_hidden) - x_output = z_output - - return x_output - -# The trial solution using the deep neural network: -def g_trial(x,params, g0 = 10): - return g0 + x*neural_network(params,x) - -# The right side of the ODE: -def g(x, g_trial, gamma = 2): - return -gamma*g_trial - -# The cost function: -def cost_function(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial(x,P) - - # Find the derivative w.r.t x of the neural network - d_net_out = elementwise_grad(neural_network,1)(P,x) - - # Find the derivative w.r.t x of the trial function - d_g_t = elementwise_grad(g_trial,0)(x,P) - - # The right side of the ODE - func = g(x, g_t) - - err_sqr = (d_g_t - func)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum / np.size(err_sqr) - -# Solve the exponential decay ODE using neural network with one input, hidden, and output layer -def solve_ode_neural_network(x, num_neurons_hidden, num_iter, lmb): - ## Set up initial weights and biases - - # For the hidden layer - p0 = npr.randn(num_neurons_hidden, 2 ) - - # For the output layer - p1 = npr.randn(1, num_neurons_hidden + 1 ) # +1 since bias is included - - P = [p0, p1] - - print('Initial cost: %g'%cost_function(P, x)) - - ## Start finding the optimal weights using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_grad = grad(cost_function,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of two arrays; - # one for the gradient w.r.t P_hidden and - # one for the gradient w.r.t P_output - cost_grad = cost_function_grad(P, x) - - P[0] = P[0] - lmb * cost_grad[0] - P[1] = P[1] - lmb * cost_grad[1] - - print('Final cost: %g'%cost_function(P, x)) - - return P - -def g_analytic(x, gamma = 2, g0 = 10): - return g0*np.exp(-gamma*x) - -# Solve the given problem -if __name__ == '__main__': - # Set seed such that the weight are initialized - # with same weights and biases for every run. - npr.seed(15) - - ## Decide the vales of arguments to the function to solve - N = 10 - x = np.linspace(0, 1, N) - - ## Set up the initial parameters - num_hidden_neurons = 10 - num_iter = 10000 - lmb = 0.001 - - # Use the network - P = solve_ode_neural_network(x, num_hidden_neurons, num_iter, lmb) - - # Print the deviation from the trial solution and true solution - res = g_trial(x,P) - res_analytical = g_analytic(x) - - print('Max absolute difference: %g'%np.max(np.abs(res - res_analytical))) - - # Plot the results - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(x, res_analytical) - plt.plot(x, res[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('x') - plt.ylabel('g(x)') - plt.show() -!ec - -!split -===== The network with one input layer, specified number of hidden layers, and one output layer output layer ===== - -It is also possible to extend the construction of our network into a more general one, allowing the network to contain more than one hidden layers. - -The number of neurons within each hidden layer are given as a list of integers in the program below. - -!bc pycod -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -# The neural network with one input layer and one output layer, -# but with number of hidden layers specified by the user. -def deep_neural_network(deep_params, x): - # N_hidden is the number of hidden layers - - N_hidden = np.size(deep_params) - 1 # -1 since params consists of - # parameters to all the hidden - # layers AND the output layer. - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - # Due to multiple hidden layers, define a variable referencing to the - # output of the previous layer: - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output - -# The trial solution using the deep neural network: -def g_trial_deep(x,params, g0 = 10): - return g0 + x*deep_neural_network(params, x) - -# The right side of the ODE: -def g(x, g_trial, gamma = 2): - return -gamma*g_trial - -# The same cost function as before, but calls deep_neural_network instead. -def cost_function_deep(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial_deep(x,P) - - # Find the derivative w.r.t x of the neural network - d_net_out = elementwise_grad(deep_neural_network,1)(P,x) - - # Find the derivative w.r.t x of the trial function - d_g_t = elementwise_grad(g_trial_deep,0)(x,P) - - # The right side of the ODE - func = g(x, g_t) - - err_sqr = (d_g_t - func)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum / np.size(err_sqr) - -# Solve the exponential decay ODE using neural network with one input and one output layer, -# but with specified number of hidden layers from the user. -def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): - # num_hidden_neurons is now a list of number of neurons within each hidden layer - - # The number of elements in the list num_hidden_neurons thus represents - # the number of hidden layers. - - # Find the number of hidden layers: - N_hidden = np.size(num_neurons) - - ## Set up initial weights and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 ) - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: %g'%cost_function_deep(P, x)) - - ## Start finding the optimal weights using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_deep_grad = grad(cost_function_deep,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases - # in the hidden layers and output layers evaluated at x. - cost_deep_grad = cost_function_deep_grad(P, x) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_deep_grad[l] - - print('Final cost: %g'%cost_function_deep(P, x)) - - return P - -def g_analytic(x, gamma = 2, g0 = 10): - return g0*np.exp(-gamma*x) - -# Solve the given problem -if __name__ == '__main__': - npr.seed(15) - - ## Decide the vales of arguments to the function to solve - N = 10 - x = np.linspace(0, 1, N) - - ## Set up the initial parameters - num_hidden_neurons = np.array([10,10]) - num_iter = 10000 - lmb = 0.001 - - P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb) - - res = g_trial_deep(x,P) - res_analytical = g_analytic(x) - - plt.figure(figsize=(10,10)) - - plt.title('Performance of a deep neural network solving an ODE compared to the analytical solution') - plt.plot(x, res_analytical) - plt.plot(x, res[0,:]) - plt.legend(['analytical','dnn']) - plt.ylabel('g(x)') - plt.show() -!ec - -!split -===== Example: Population growth, comparing Autograd, TensorFlow, and Euler's scheme ===== - -A logistic model of population growth assumes that a population converges toward an equilibrium. -The population growth can be modeled by - -!bt -\begin{equation} \label{log} - g'(t) = \alpha g(t)(A - g(t)) -\end{equation} -!et - -where $g(t)$ is the population density at time $t$, $\alpha > 0$ the growth rate and $A > 0$ is the maximum population number in the environment. -Also, at $t = 0$ the population has the size $g(0) = g_0$, where $g_0$ is some chosen constant. - -In this example, similar network as for the exponential decay using Autograd has been used to solve the equation. However, as the implementation might suffer from e.g numerical instability -and high execution time (this might be more apparent in the examples solving PDEs), -a network has been constructed using TensorFlow also. -For comparison, the forward Euler method has been implemented in order to see how the networks performs compared to a numerical scheme. - -!split -===== Setting up the problem ===== - -Here, we will model a population $g(t)$ in an environment having carrying capacity $A$. -The population follows the model - -!bt -\begin{equation} \label{solveode_population} -g'(t) = \alpha g(t)(A - g(t)) -\end{equation} -!et - -where $g(0) = g_0$. - -In this example, we let $\alpha = 2$, $A = 1$, and $g_0 = 1.2$. - -!split -===== The trial solution ===== -We will get a slightly different trial solution, as the boundary conditions are different -compared to the case for exponential decay. - -A possible trial solution satisfying the condition $g(0) = g_0$ could be - -$$ -h_1(t) = g_0 + t \cdot N(t,P) -$$ - -with $N(t,P)$ being the output from the neural network with weights and biases for each layer collected in the set $P$. - -The analytical solution is - -$$ -g(t) = \frac{Ag_0}{g_0 + (A - g_0)\exp(-\alpha A t)} -$$ - -!split -===== The program using Autograd ===== - -The network will be the similar as for the exponential decay example, but with some small modifications for our problem. - -!bc pycod -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -# Function to get the parameters. -# Done such that one can easily change the paramaters after one's liking. -def get_parameters(): - alpha = 2 - A = 1 - g0 = 1.2 - return alpha, A, g0 - -def deep_neural_network(P, x): - # N_hidden is the number of hidden layers - N_hidden = np.size(P) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - # Due to multiple hidden layers, define a variable referencing to the - # output of the previous layer: - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = P[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = P[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output - - -def cost_function_deep(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial_deep(x,P) - - # Find the derivative w.r.t x of the trial function - d_g_t = elementwise_grad(g_trial_deep,0)(x,P) - - # The right side of the ODE - func = f(x, g_t) - - err_sqr = (d_g_t - func)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum / np.size(err_sqr) - -# The right side of the ODE: -def f(x, g_trial): - alpha,A, g0 = get_parameters() - return alpha*g_trial*(A - g_trial) - -# The trial solution using the deep neural network: -def g_trial_deep(x, params): - alpha,A, g0 = get_parameters() - return g0 + x*deep_neural_network(params,x) - -# The analytical solution: -def g_analytic(t): - alpha,A, g0 = get_parameters() - return A*g0/(g0 + (A - g0)*np.exp(-alpha*A*t)) - -def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): - # num_hidden_neurons is now a list of number of neurons within each hidden layer - - # Find the number of hidden layers: - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 ) - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: %g'%cost_function_deep(P, x)) - - ## Start finding the optimal weigths using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_deep_grad = grad(cost_function_deep,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases - # in the hidden layers and output layers evaluated at x. - cost_deep_grad = cost_function_deep_grad(P, x) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_deep_grad[l] - - print('Final cost: %g'%cost_function_deep(P, x)) - - return P - -if __name__ == '__main__': - npr.seed(4155) - - ## Decide the vales of arguments to the function to solve - Nt = 10 - T = 1 - t = np.linspace(0,T, Nt) - - ## Set up the initial parameters - num_hidden_neurons = [100, 50, 25] - num_iter = 1000 - lmb = 1e-3 - - P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb) - - g_dnn_ag = g_trial_deep(t,P) - g_analytical = g_analytic(t) - - # Find the maximum absolute difference between the solutons: - diff_ag = np.max(np.abs(g_dnn_ag - g_analytical)) - print("The max absolute difference between the solutions is: %g"%diff_ag) - - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(t, g_analytical) - plt.plot(t, g_dnn_ag[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('t') - plt.ylabel('g(t)') - - plt.show() -!ec - -!split -===== Using forward Euler to solve the ODE ===== - -A straight-forward way of solving an ODE numerically, is to use Euler's method. - -Euler's method uses Taylor series to approximate the value at a function $f$ at a step $\Delta x$ from $x$: - -$$ -f(x + \Delta x) \approx f(x) + \Delta x f'(x) -$$ - -In our case, using Euler's method to approximate the value of $g$ at a step $\Delta t$ from $t$ yields - -!bt -\begin{aligned} - g(t + \Delta t) &\approx g(t) + \Delta t g'(t) \\ - &= g(t) + \Delta t \big(\alpha g(t)(A - g(t))\big) -\end{aligned} -!et -along with the condition that $g(0) = g_0$. - -Let $t_i = i \cdot \Delta t$ where $\Delta t = \frac{T}{N_t-1}$ where $T$ is the final time our solver must solve for and $N_t$ the number of values for $t \in [0, T]$ for $i = 0, \dots, N_t-1$. - -For $i \geq 1$, we have that -!bt -\begin{aligned} -t_i &= i\Delta t \\ -&= (i - 1)\Delta t + \Delta t \\ -&= t_{i-1} + \Delta t -\end{aligned} -!et - -Now, if $g_i = g(t_i)$ then - -!bt -\begin{equation} - \begin{aligned} - g_i &= g(t_i) \\ - &= g(t_{i-1} + \Delta t) \\ - &\approx g(t_{i-1}) + \Delta t \big(\alpha g(t_{i-1})(A - g(t_{i-1}))\big) \\ - &= g_{i-1} + \Delta t \big(\alpha g_{i-1}(A - g_{i-1})\big) - \end{aligned} -\end{equation} \label{odenum} -!et -for $i \geq 1$ and $g_0 = g(t_0) = g(0) = g_0$. - -Equation (ref{odenum}) could be implemented in the following way, -extending the program that uses the network using Autograd: - -!bc pycod -# Assume that all function definitions from the example program using Autograd -# are located here. - -if __name__ == '__main__': - npr.seed(4155) - - ## Decide the vales of arguments to the function to solve - Nt = 10 - T = 1 - t = np.linspace(0,T, Nt) - - ## Set up the initial parameters - num_hidden_neurons = [100,50,25] - num_iter = 1000 - lmb = 1e-3 - - P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb) - - g_dnn_ag = g_trial_deep(t,P) - g_analytical = g_analytic(t) - - # Find the maximum absolute difference between the solutons: - diff_ag = np.max(np.abs(g_dnn_ag - g_analytical)) - print("The max absolute difference between the solutions is: %g"%diff_ag) - - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(t, g_analytical) - plt.plot(t, g_dnn_ag[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('t') - plt.ylabel('g(t)') - - ## Find an approximation to the funtion using forward Euler - - alpha, A, g0 = get_parameters() - dt = T/(Nt - 1) - - # Perform forward Euler to solve the ODE - g_euler = np.zeros(Nt) - g_euler[0] = g0 - - for i in range(1,Nt): - g_euler[i] = g_euler[i-1] + dt*(alpha*g_euler[i-1]*(A - g_euler[i-1])) - - # Print the errors done by each method - diff1 = np.max(np.abs(g_euler - g_analytical)) - diff2 = np.max(np.abs(g_dnn_ag[0,:] - g_analytical)) - - print('Max absolute difference between Euler method and analytical: %g'%diff1) - print('Max absolute difference between deep neural network and analytical: %g'%diff2) - - # Plot results - plt.figure(figsize=(10,10)) - - plt.plot(t,g_euler) - plt.plot(t,g_analytical) - plt.plot(t,g_dnn_ag[0,:]) - - plt.legend(['euler','analytical','dnn']) - plt.xlabel('Time t') - plt.ylabel('g(t)') - - plt.show() -!ec - -Running the program gives - -!bc -Max absolute difference between Euler method and analytical: 0.011225 -Max absolute difference between deep neural network and analytical: 0.00424909 -!ec - -!split -===== Using TensorFlow to model logistic population growth ===== - -TensorFlow is a library widely used in the machine learning community. -A neural network can be set up in a flexible manner, where various optimization algorithms are implemented and different types of networks can be used, making it easier to experiment on solving differential equations using neural networks. - -!split -===== The general program flow in TensorFlow ===== - -Usually, a program in TensorFlow is divided into two parts; the *construction phase* and the *execution phase*. -In the construction phase, the computational graph that TensorFlow uses to perform its calculations are set up. -In the execution phase, TensorFlow evaluates any procedure that was defined in the construction phase. - -===== Program flow in TensorFlow - Construction phase ===== - -Here, the architecture for the neural network will be set up, along with the cost function and an optimizer class used during training of the network. -Note that TensorFlow uses a different convention for the weighting done in each neuron in each layer within the network than in the implementation using Autograd. -The matrix-vector multiplication between the input from the previous layer and the weighting at the neuron at current layer in the program using Autograd, is the transpose of the convention used in TensorFlow. But it will not affect that much our construction, as TensorFlow takes care of most of the computations. The only thing we have to be aware of, is how the dimensions are for our inputs. - -===== Program flow in TensorFlow - Execution phase ===== - -The computation graph has been defined, and is ready to be evaluated. -In order to get access to the graph, it has to be initialized and be runned within a Session. - -===== The full program modeling logistic population growth using TensorFlow ===== - -!bc pycod -import tensorflow as tf -import numpy as np -import matplotlib.pyplot as plt - -# Just to reset the graph such that it is possible to rerun this in a -# Jupyter cell without resetting the whole kernel. -tf.reset_default_graph() - -# Set a seed to ensure getting the same results from every run -tf.set_random_seed(4155) - -Nt = 10 -T = 1 -t = np.linspace(0,T, Nt) - -## The construction phase - -# Convert the values the trial solution is evaluated at to a tensor. -t_tf = tf.convert_to_tensor(t.reshape(-1,1),dtype=tf.float64) -zeros = tf.reshape(tf.convert_to_tensor(np.zeros(t.shape)),shape=(-1,1)) - -# Define the parameters of the equation -alpha = tf.constant(2.,dtype=tf.float64) -A = tf.constant(1.,dtype=tf.float64) -g0 = tf.constant(1.2,dtype=tf.float64) - -num_iter = 100000 - -# Define the number of neurons at each hidden layer -num_hidden_neurons = [100,50,25] -num_hidden_layers = np.size(num_hidden_neurons) - -# Construct the network. -# tf.name_scope is used to group each step in the construction, -# just for a more organized visualization in TensorBoard -with tf.name_scope('dnn'): - - # Input layer - previous_layer = t_tf - - # Hidden layers - for l in range(num_hidden_layers): - current_layer = tf.layers.dense(previous_layer, num_hidden_neurons[l], name='hidden%d'%(l+1), activation=tf.nn.sigmoid) - previous_layer = current_layer - - # Output layer - dnn_output = tf.layers.dense(previous_layer, 1, name='output') - -# Define the cost function -with tf.name_scope('cost'): - g_trial = g0 + t_tf*dnn_output - d_g_trial = tf.gradients(g_trial,t_tf) - - func = alpha*g_trial*(A - g_trial) - cost = tf.losses.mean_squared_error(zeros, d_g_trial[0] - func) - - -# Choose the method to minimize the cost function, along with a learning rate -learning_rate = 1e-2 -with tf.name_scope('train'): - optimizer = tf.train.GradientDescentOptimizer(learning_rate) - traning_op = optimizer.minimize(cost) - -# Set up a referance to the result from the neural network: -g_dnn_tf = None - -# Define a node that initializes all of the other nodes in the computational graph -# used by TensorFlow: -init = tf.global_variables_initializer() - -## Execution phase - -# Start a session where the graph defined from the construction phase can be evaluated at: -with tf.Session() as sess: - # Initialize the whole graph - init.run() - - # Evaluate the initial cost: - print('Initial cost: %g'%cost.eval()) - - # The training of the network: - for i in range(num_iter): - sess.run(traning_op) - - # If one desires to see how the cost function behaves for each iteration: - #if i % 1000 == 0: - # print(cost.eval()) - - # Training is done, and we have an approximate solution to the ODE - print('Final cost: %g'%cost.eval()) - - # Store the result - g_dnn_tf = g_trial.eval() - -# Compare with analytical solution -def get_parameters(): - alpha = 2 - A = 1 - g0 = 1.2 - return alpha, A, g0 - -def g_analytic(t): - alpha,A, g0 = get_parameters() - return A*g0/(g0 + (A - g0)*np.exp(-alpha*A*t)) - -g_analytical = g_analytic(t) -diff_tf = g_dnn_tf - g_analytical.reshape(-1,1) - -print('\nMax absolute difference between the analytical solution and solution from TensorFlow DNN: %g'%np.max(np.abs(diff_tf))) - -# Plot the result -plt.figure(figsize=(10,10)) - -plt.title('Numerical solutions of the ODE') - -plt.plot(t, g_dnn_tf) -plt.plot(t, g_analytical) - -plt.legend(['dnn, tensorflow', 'exact']) -plt.xlabel('Time t') -plt.ylabel('g(t)') - -plt.show() - -!ec - -!split -===== Example: Solving the one dimensional Poisson equation using Autograd and TensorFlow ===== - -The Poisson equation for $g(x)$ in one dimension is - -!bt -\begin{equation} \label{poisson} - -g''(x) = f(x) -\end{equation} -!et - -where $f(x)$ is a given function for $x \in (0,1)$. - -The conditions that $g(x)$ is chosen to fulfill, are -!bt -\begin{align*} - g(0) &= 0 \\ - g(1) &= 0 -\end{align*} -!et - -This equation can be solved numerically using programs where e.g Autograd and TensorFlow are used. -The results from the networks can then be compared to the analytical solution. -In addition, it could be interesting to see how a typical method for numerically solving second order ODEs compares to the neural networks. - -There exists many different optimization methods implemented in TensorFlow. -In the examples program using TensorFlow, it could also be of interest to see how -the choice of an optimization method affects our results. -In the "TensorFlow documentation about optimizers":"https://www.tensorflow.org/versions/r1.2/api_guides/python/train#Optimizers", a list over available optimization methods are shown. - -!split -===== The specific equation to solve for ===== - -Here, the function $g(x)$ to solve for follows the equation - -!bt --g''(x) = f(x),\qquad x \in (0,1) -!et - -where $f(x)$ is a given function, along with the chosen conditions - -!bt -\begin{aligned} -g(0) = g(1) = 0 -\end{aligned}\label{cond} -!et - -In this example, we consider the case when $f(x) = (3x + x^2)\exp(x)$. - -For this case, a possible trial solution satisfying the conditions could be - -!bt -g_t(x) = x \cdot (1-x) \cdot N(P,x) -!et - -The analytical solution for this problem is - -!bt -g(x) = x(1 - x)\exp(x) -!et - -!split -===== Solving the equation using Autograd ===== - -!bc pycod -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -def deep_neural_network(deep_params, x): - # N_hidden is the number of hidden layers - N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - # Due to multiple hidden layers, define a variable referencing to the - # output of the previous layer: - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output - -def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): - # num_hidden_neurons is now a list of number of neurons within each hidden layer - - # Find the number of hidden layers: - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 ) - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: %g'%cost_function_deep(P, x)) - - ## Start finding the optimal weigths using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_deep_grad = grad(cost_function_deep,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases - # in the hidden layers and output layers evaluated at x. - cost_deep_grad = cost_function_deep_grad(P, x) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_deep_grad[l] - - print('Final cost: %g'%cost_function_deep(P, x)) - - return P - -## Set up the cost function specified for this Poisson equation: - -# The right side of the ODE -def f(x): - return (3*x + x**2)*np.exp(x) - -def cost_function_deep(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial_deep(x,P) - - # Find the derivative w.r.t x of the trial function - d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P) - - right_side = f(x) - - err_sqr = (-d2_g_t - right_side)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum/np.size(err_sqr) - -# The trial solution: -def g_trial_deep(x,P): - return x*(1-x)*deep_neural_network(P,x) - -# The analytic solution; -def g_analytic(x): - return x*(1-x)*np.exp(x) - -if __name__ == '__main__': - npr.seed(4155) - - ## Decide the vales of arguments to the function to solve - Nx = 10 - x = np.linspace(0,1, Nx) - - ## Set up the initial parameters - num_hidden_neurons = [200,100] - num_iter = 1000 - lmb = 1e-3 - - P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb) - - g_dnn_ag = g_trial_deep(x,P) - g_analytical = g_analytic(x) - - # Find the maximum absolute difference between the solutons: - max_diff = np.max(np.abs(g_dnn_ag - g_analytical)) - print("The max absolute difference between the solutions is: %g"%max_diff) - - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(x, g_analytical) - plt.plot(x, g_dnn_ag[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('x') - plt.ylabel('g(x)') - plt.show() -!ec - -!split -===== Comparing with a numerical scheme ===== - -The Poisson equation is possible to solve using Taylor series to approximate the second derivative. - -Using Taylor series, the second derivative can be expressed as - -$$ -g''(x) = \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2} + E_{\Delta x}(x) -$$ - -where $\Delta x$ is a small step size and $E_{\Delta x}(x)$ being the error term. - -Looking away from the error terms gives an approximation to the second derivative: - -!bt -\begin{equation} \label{approx} -g''(x) \approx \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2} -\end{equation} -!et - -If $x_i = i \Delta x = x_{i-1} + \Delta x$ and $g_i = g(x_i)$ for $i = 1,\dots N_x - 2$ with $N_x$ being the number of values for $x$, (ref{approx}) becomes - -!bt -\begin{aligned} -g''(x_i) &\approx \frac{g(x_i + \Delta x) - 2g(x_i) + g(x_i -\Delta x)}{\Delta x^2} \\ -&= \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2} -\end{aligned} -!et - -Since we know from our problem that - -!bt -\begin{aligned} --g''(x) &= f(x) \\ -&= (3x + x^2)\exp(x) -\end{aligned} -!et - -along with the conditions $g(0) = g(1) = 0$, -the following scheme can be used to find an approximate solution for $g(x)$ numerically: - -!bt -\begin{equation} - \begin{aligned} - -\Big( \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2} \Big) &= f(x_i) \\ - -g_{i+1} + 2g_i - g_{i-1} &= \Delta x^2 f(x_i) - \end{aligned} -\end{equation} \label{odesys} -!et - -for $i = 1, \dots, N_x - 2$ where $g_0 = g_{N_x - 1} = 0$ and $f(x_i) = (3x_i + x_i^2)\exp(x_i)$, which is given for our specific problem. - -The equation can be rewritten into a matrix equation: - -!bt -\begin{aligned} -\begin{pmatrix} -2 & -1 & 0 & \dots & 0 \\ --1 & 2 & -1 & \dots & 0 \\ -\vdots & & \ddots & & \vdots \\ -0 & \dots & -1 & 2 & -1 \\ -0 & \dots & 0 & -1 & 2\\ -\end{pmatrix} -\begin{pmatrix} -g_1 \\ -g_2 \\ -\vdots \\ -g_{N_x - 3} \\ -g_{N_x - 2} -\end{pmatrix} -&= -\Delta x^2 -\begin{pmatrix} -f(x_1) \\ -f(x_2) \\ -\vdots \\ -f(x_{N_x - 3}) \\ -f(x_{N_x - 2}) -\end{pmatrix} \\ -A\vec{g} &= \vec{f} -\end{aligned} -!et - -which makes it possible to solve for the vector $\vec{g}$. - -We can then compare the result from this numerical scheme with the output from our network using Autograd: - -!bc pycod -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -def deep_neural_network(deep_params, x): - # N_hidden is the number of hidden layers - N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - # Due to multiple hidden layers, define a variable referencing to the - # output of the previous layer: - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output - -def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): - # num_hidden_neurons is now a list of number of neurons within each hidden layer - - # Find the number of hidden layers: - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 ) - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: %g'%cost_function_deep(P, x)) - - ## Start finding the optimal weigths using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_deep_grad = grad(cost_function_deep,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases - # in the hidden layers and output layers evaluated at x. - cost_deep_grad = cost_function_deep_grad(P, x) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_deep_grad[l] - - print('Final cost: %g'%cost_function_deep(P, x)) - - return P - -## Set up the cost function specified for this Poisson equation: - -# The right side of the ODE -def f(x): - return (3*x + x**2)*np.exp(x) - -def cost_function_deep(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial_deep(x,P) - - # Find the derivative w.r.t x of the trial function - d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P) - - right_side = f(x) - - err_sqr = (-d2_g_t - right_side)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum/np.size(err_sqr) - -# The trial solution: -def g_trial_deep(x,P): - return x*(1-x)*deep_neural_network(P,x) - -# The analytic solution; -def g_analytic(x): - return x*(1-x)*np.exp(x) - -if __name__ == '__main__': - npr.seed(4155) - - ## Decide the vales of arguments to the function to solve - Nx = 10 - x = np.linspace(0,1, Nx) - - ## Set up the initial parameters - num_hidden_neurons = [200,100] - num_iter = 1000 - lmb = 1e-3 - - P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb) - - g_dnn_ag = g_trial_deep(x,P) - g_analytical = g_analytic(x) - - # Find the maximum absolute difference between the solutons: - - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(x, g_analytical) - plt.plot(x, g_dnn_ag[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('x') - plt.ylabel('g(x)') - - ## Perform the computation using the numerical scheme - - dx = 1/(Nx - 1) - - # Set up the matrix A - A = np.zeros((Nx-2,Nx-2)) - - A[0,0] = 2 - A[0,1] = -1 - - for i in range(1,Nx-3): - A[i,i-1] = -1 - A[i,i] = 2 - A[i,i+1] = -1 - - A[Nx - 3, Nx - 4] = -1 - A[Nx - 3, Nx - 3] = 2 - - # Set up the vector f - f_vec = dx**2 * f(x[1:-1]) - - # Solve the equation - g_res = np.linalg.solve(A,f_vec) - - g_vec = np.zeros(Nx) - g_vec[1:-1] = g_res - - # Print the differences between each method - max_diff1 = np.max(np.abs(g_dnn_ag - g_analytical)) - max_diff2 = np.max(np.abs(g_vec - g_analytical)) - print("The max absolute difference between the analytical solution and DNN Autograd: %g"%max_diff1) - print("The max absolute difference between the analytical solution and numerical scheme: %g"%max_diff2) - - # Plot the results - plt.figure(figsize=(10,10)) - - plt.plot(x,g_vec) - plt.plot(x,g_analytical) - plt.plot(x,g_dnn_ag[0,:]) - - plt.legend(['numerical scheme','analytical','dnn']) - plt.show() - -!ec - -The program prints out: -!bc -The max absolute difference between the analytical solution and DNN Autograd: 0.000464088 -The max absolute difference between the analytical solution and numerical scheme: 0.00266858 -!ec - -!split -===== Using gradient descent in TensorFlow to solve Poisson equation ===== -The program follows the similar idea as for the logistic population model. - -What has changed, is what the cost function minimizes and the trial solution. - -!bc pycod -import tensorflow as tf -import numpy as np -import matplotlib.pyplot as plt -## Construction phase - -# Just to reset the graph such that it is possible to rerun this in a -# Jupyter cell without resetting the whole kernel. -tf.reset_default_graph() - -tf.set_random_seed(4155) - -# Convert the values the trial solution is evaluated at to a tensor. -Nx = 10 -x = np.linspace(0,1, Nx) -x_tf = tf.convert_to_tensor(x.reshape(-1,1),dtype=tf.float64) - - -num_iter = 10000 - -# Define the number of neurons at each hidden layer -num_hidden_neurons = [20,10] -num_hidden_layers = np.size(num_hidden_neurons) - -# Construct the network. -# tf.name_scope is used to group each step in the construction, -# just for a more organized visualization in TensorBoard -with tf.name_scope('dnn'): - - # Input layer - previous_layer = x_tf - - # Hidden layers - for l in range(num_hidden_layers): - current_layer = tf.layers.dense(previous_layer, num_hidden_neurons[l], name='hidden%d'%(l+1), activation=tf.nn.sigmoid) - previous_layer = current_layer - - # Output layer - dnn_output = tf.layers.dense(previous_layer, 1, name='output') - -# Define the cost function -with tf.name_scope('cost'): - g_trial = x_tf*(1-x_tf)*dnn_output - d_g_trial = tf.gradients(g_trial,x_tf) - d2_g_trial = tf.gradients(d_g_trial,x_tf) - - right_side = (3*x_tf + x_tf**2)*tf.exp(x_tf) - - err = tf.square( -d2_g_trial[0] - right_side) - cost = tf.reduce_sum(err, name = 'cost') - -# Choose the method to minimize the cost function, along with a learning rate -learning_rate = 1e-2 -with tf.name_scope('train'): - optimizer = tf.train.GradientDescentOptimizer(learning_rate) - traning_op = optimizer.minimize(cost) - -g_dnn_tf = None - -# Define a node that initializes all of the other nodes in the computational graph -# used by TensorFlow: -init = tf.global_variables_initializer() - - -## Execution phase - -# Start a session where the graph defined from the construction phase can be evaluated at: - -with tf.Session() as sess: - # Initialize the whole graph - init.run() - - # Evaluate the initial cost: - print('Initial cost: %g'%cost.eval()) - - # The traning of the network: - for i in range(num_iter): - sess.run(traning_op) - - # Training is done, and we have an approximate solution to the ODE - print('Final cost: %g'%cost.eval()) - - # Store the result - g_dnn_tf = g_trial.eval() - - writer = tf.summary.FileWriter("./output", sess.graph) - writer.close() - -# Evaluate the analytical function to compare with -def g_analytic(x): - return x*(1-x)*np.exp(x) - -g_analytical = g_analytic(x) - -diff_tf = g_dnn_tf - g_analytical.reshape(-1,1) - -print('\nMax absolute difference between the analytical solution and solution from TensorFlow DNN: %g'%np.max(np.abs(diff_tf))) - -# Plot the result -plt.figure(figsize=(10,10)) - -plt.title('Numerical solutions of the ODE') - -plt.plot(x, g_dnn_tf) -plt.plot(x, g_analytical) - -plt.legend(['dnn, tensorflow','exact']) -plt.xlabel('x') -plt.ylabel('g(x)') - -plt.show() - -!ec - -!split -===== Using a different optimization algorithm implemented in TensorFlow to solve Poisson equation ===== - -We can see that the results using GradientDescentOptimizer seems to converge towards the analytical solution. -But there exists many other methods for optimization also, see "the TensorFlow documentation on Optimizers":"https://www.tensorflow.org/versions/r1.2/api_guides/python/train#Optimizers". - -Adam is an optimization algorithm that changes its learning rates accordingly to the function it tries to minimize for every iteration. -The algorithm is described in "this paper":"https://arxiv.org/pdf/1412.6980.pdf". -How much an optimization algorithm has to say for the network to converge, could be interesting to experiment with. -Using the same TensorFlow program as before, the only change to do, is to replace the variable *optimizer*. - -In the program that uses TensorFlow to solve for the Poisson equation, change the line - -!bc -optimizer = tf.train.GradientDescentOptimizer(learning_rate) -!ec - -to - -!bc -optimizer = tf.train.AdamOptimizer(learning_rate) -!ec - - -The program using the Adam optimizer with a different initial learning rate yields indeed an interesting result: -!bc -Max absolute difference between the analytical solution and solution from TensorFlow DNN: 7.11243e-05 -!ec - -!split -===== Partial Differential Equations ===== -A partial differential equation (PDE) has a solution here the function is defined by multiple variables. -The equation may involve all kinds of combinations of which variables the function is differentiated with respect to. - -In general, a partial differential equation for a function $g(x_1,\dots,x_N)$ with $N$ variables may be expressed as - -!bt -\begin{equation} \label{PDE} - f\left(x_1, \, \dots \, , x_N, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1}, \dots , \frac{\partial g(x_1,\dots,x_N) }{\partial x_N}, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(x_1,\dots,x_N) }{\partial x_N^n} \right) = 0 -\end{equation} -!et - -where $f$ is an expression involving all kinds of possible mixed derivatives of $g(x_1,\dots,x_N)$ up to an order $n$. In order for the solution to be unique, some additional conditions must also be given. - -The problem our network must solve for, is similar to the ODE case. -We must have a trial solution $g_t$ at hand. - -For instance, the trial solution could be expressed as -!bt -\begin{align*} - g_t(x_1,\dots,x_N) = h_1(x_1,\dots,x_N) + h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P)) -\end{align*} -!et -where $h_1(x_1,\dots,x_N)$ is a function that ensures $g_t(x_1,\dots,x_N)$ satisfies some given conditions. -The neural network $N(x_1,\dots,x_N,P)$ has weights and biases described by $P$ and $h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P))$ is an expression using the output from the neural network in some way. - -The role of the function $h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P))$, is to ensure that the output of $N(x_1,\dots,x_N,P)$ is zero when $g_t(x_1,\dots,x_N)$ is evaluated at the values of $x_1,\dots,x_N$ where the given conditions must be satisfied. The function $h_1(x_1,\dots,x_N)$ should alone make $g_t(x_1,\dots,x_N)$ satisfy the conditions. - -The network tries then the minimize the cost function following the same ideas as described for the ODE case, but now with more than one variables to consider. -The concept still remains the same; find a set of parameters $P$ such that the expression $f$ in (ref{PDE}) is as close to zero as possible. - -As for the ODE case, the cost function is the mean squared error that the network must try to minimize. The cost function for the network to minimize is - -!bt -\begin{equation*} -c\left(x_1, \dots, x_N, P\right) = \left( f\left(x_1, \, \dots \, , x_N, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1}, \dots , \frac{\partial g(x_1,\dots,x_N) }{\partial x_N}, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(x_1,\dots,x_N) }{\partial x_N^n} \right) \right)^2 -\end{equation*} -!et - -If we let $\vec x = \big( x_1, \dots, x_N \big)$ be an array containing the values for $x_1, \dots, x_N$ respectively, the cost function can be reformulated into the following: -!bt -\begin{equation*} - c\left(\vec{x}, P\right) = f\left( \left( \vec{x}, \frac{\partial g(\vec x) }{\partial x_1}, \dots , \frac{\partial g(\vec x) }{\partial x_N}, \frac{\partial g(\vec x) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(\vec x) }{\partial x_N^n} \right) \right)^2 -\end{equation*} -!et - -If we also have $M$ different sets of values for $x_1, \dots, x_N$, that is $\vec{x}_i = \big(x_1^{(i)}, \dots, x_N^{(i)}\big)$ for $i = 1,\dots,M$ being the rows in matrix $X$, the cost function can be generalized into -!bt -\begin{equation*} -c\left(X, P \right) = \sum_{i=1}^M f\left( \left( \vec{x}_i, \frac{\partial g(\vec{x}_i) }{\partial x_1}, \dots , \frac{\partial g(\vec{x}_i) }{\partial x_N}, \frac{\partial g(\vec{x}_i) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(\vec{x}_i) }{\partial x_N^n} \right) \right)^2 -\end{equation*} -!et - -!split -===== Example: The diffusion equation ===== - -In one spatial dimension, the equation reads -!bt -\begin{equation*} - \frac{\partial g(x,t)}{\partial t} = \frac{\partial^2 g(x,t)}{\partial x^2} -\end{equation*} -!et - -where a possible choice of conditions are -!bt -\begin{align*} -g(0,t) &= 0 ,\qquad t \geq 0 \\ -g(1,t) &= 0, \qquad t \geq 0 \\ -g(x,0) &= u(x),\qquad x\in [0,1] -\end{align*} -!et -with $u(x)$ being some given function. - -!split -===== Defining the problem ===== - -For this case, we want to find $g(x,t)$ such that - -!bt -\begin{equation} - \frac{\partial g(x,t)}{\partial t} = \frac{\partial^2 g(x,t)}{\partial x^2} -\end{equation} \label{diffonedim} -!et - -and - -!bt -\begin{align*} -g(0,t) &= 0 ,\qquad t \geq 0 \\ -g(1,t) &= 0, \qquad t \geq 0 \\ -g(x,0) &= u(x),\qquad x\in [0,1] -\end{align*} -!et -with $u(x) = \sin(\pi x)$. - -First, let us set up the deep neural network. -The deep neural network will follow the same structure as discussed in the examples solving the ODEs. -First, we will look into how Autograd could be used in a network tailored to solve for bivariate functions. - - - -!split -===== Setting up the network using Autograd ===== - -The only change to do here, is to extend our network such that functions of multiple parameters are correctly handled. -In this case we have two variables in our function to solve for, that is time $t$ and position $x$. -The variables will be represented by a one-dimensional array in the program. -The program will evaluate the network at each possible pair $(x,t)$, given an array for the desired $x$-values and $t$-values to approximate the solution at. - -!bc pycod -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -def deep_neural_network(deep_params, x): - # x is now a point and a 1D numpy array; make it a column vector - num_coordinates = np.size(x,0) - x = x.reshape(num_coordinates,-1) - - num_points = np.size(x,1) - - # N_hidden is the number of hidden layers - N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assume that the input layer does nothing to the input x - x_input = x - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output[0][0] -!ec - -!split -===== Setting up the network using Autograd; The trial solution ===== -The cost function must then iterate through the given arrays containing values for $x$ and $t$, defines a point $(x,t)$ the deep neural network and the trial solution is evaluated at, and then finds the Jacobian of the trial solution. - -A possible trial solution for this PDE is - -$$ -g_t(x,t) = h_1(x,t) + x(1-x)tN(x,t,P) -$$ - -with $A(x,t)$ being a function ensuring that $g_t(x,t)$ satisfies our given conditions, and $N(x,t,P)$ being the output from the deep neural network using weights and biases for each layer from $P$. - -To fulfill the conditions, $A(x,t)$ could be: - -$$ -h_1(x,t) = (1-t)\Big(u(x) - \big((1-x)u(0) + x u(1)\big)\Big) = (1-t)u(x) = (1-t)\sin(\pi x) -$$ -since $(0) = u(1) = 0$ and $u(x) = \sin(\pi x)$. - -The Jacobian is used because the program must find the derivative of the trial solution with respect to $x$ and $t$. - -This gives the necessity of computing the Jacobian matrix, as we want to evaluate the gradient with respect to $x$ and $t$ (note that the Jacobian of a scalar-valued multivariate function is simply its gradient). - -In Autograd, the differentiation is by default done with respect to the first input argument of your Python function. Since the points is an array representing $x$ and $t$, the Jacobian is calculated using the values of $x$ and $t$. - -To find the second derivative with respect to $x$ and $t$, the Jacobian can be found for the second time. The result is a Hessian matrix, which is the matrix containing all the possible second order mixed derivatives of $g(x,t)$. - -!bc pycod -# Set up the trial function: -def u(x): - return np.sin(np.pi*x) - -def g_trial(point,P): - x,t = point - return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point) - -# The right side of the ODE: -def f(point): - return 0. - -# The cost function: -def cost_function(P, x, t): - cost_sum = 0 - - g_t_jacobian_func = jacobian(g_trial) - g_t_hessian_func = hessian(g_trial) - - for x_ in x: - for t_ in t: - point = np.array([x_,t_]) - - g_t = g_trial(point,P) - g_t_jacobian = g_t_jacobian_func(point,P) - g_t_hessian = g_t_hessian_func(point,P) - - g_t_dt = g_t_jacobian[1] - g_t_d2x = g_t_hessian[0][0] - - func = f(point) - - err_sqr = ( (g_t_dt - g_t_d2x) - func)**2 - cost_sum += err_sqr - - return cost_sum -!ec - -!split -===== Setting up the network using Autograd; The full program ===== -Having set up the network, along with the trial solution and cost function, we can now see how the deep neural network performs by comparing the results to the analytical solution. - -The analytical solution of our problem is - -$$ -g(x,t) = \exp(-\pi^2 t)\sin(\pi x) -$$ - -A possible way to implement a neural network solving the PDE, is given below. -Be aware, though, that it is fairly slow for the parameters used. -A better result is possible, but requires more iterations, and thus longer time to complete. - -Using only 20 neurons in one hidden layer, the program managed to make the trial solution have the maximum absolute error of 0.0075. The execution time, however, was approximately one day and 14 hours on a computer having Intel i7-7560U 2.4 GHz CPU. - -Indeed, the program below is not optimal in its implementation, but rather serves as an example on how to implement and use a neural network to solve a PDE. -Using TensorFlow in the next example sovling the wave equation, has a much better execution time. - -!bc pycod -import autograd.numpy as np -from autograd import jacobian,hessian,grad -import autograd.numpy.random as npr -from matplotlib import cm -from matplotlib import pyplot as plt -from mpl_toolkits.mplot3d import axes3d - -## Set up the network - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -def deep_neural_network(deep_params, x): - # x is now a point and a 1D numpy array; make it a column vector - num_coordinates = np.size(x,0) - x = x.reshape(num_coordinates,-1) - - num_points = np.size(x,1) - - # N_hidden is the number of hidden layers - N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assume that the input layer does nothing to the input x - x_input = x - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output[0][0] - -## Define the trial solution and cost function -def u(x): - return np.sin(np.pi*x) - -def g_trial(point,P): - x,t = point - return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point) - -# The right side of the ODE: -def f(point): - return 0. - -# The cost function: -def cost_function(P, x, t): - cost_sum = 0 - - g_t_jacobian_func = jacobian(g_trial) - g_t_hessian_func = hessian(g_trial) - - for x_ in x: - for t_ in t: - point = np.array([x_,t_]) - - g_t = g_trial(point,P) - g_t_jacobian = g_t_jacobian_func(point,P) - g_t_hessian = g_t_hessian_func(point,P) - - g_t_dt = g_t_jacobian[1] - g_t_d2x = g_t_hessian[0][0] - - func = f(point) - - err_sqr = ( (g_t_dt - g_t_d2x) - func)**2 - cost_sum += err_sqr - - return cost_sum /( np.size(x)*np.size(t) ) - -## For comparison, define the analytical solution -def g_analytic(point): - x,t = point - return np.exp(-np.pi**2*t)*np.sin(np.pi*x) - -## Set up a function for training the network to solve for the equation -def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb): - ## Set up initial weigths and biases - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: ',cost_function(P, x, t)) - - cost_function_grad = grad(cost_function,0) - - # Let the update be done num_iter times - for i in range(num_iter): - cost_grad = cost_function_grad(P, x , t) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_grad[l] - - print('Final cost: ',cost_function(P, x, t)) - - return P - -if __name__ == '__main__': - ### Use the neural network: - npr.seed(15) - - ## Decide the vales of arguments to the function to solve - Nx = 10; Nt = 10 - x = np.linspace(0, 1, Nx) - t = np.linspace(0,1,Nt) - - ## Set up the parameters for the network - num_hidden_neurons = [100, 25] - num_iter = 250 - lmb = 0.01 - - P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb) - - ## Store the results - g_dnn_ag = np.zeros((Nx, Nt)) - G_analytical = np.zeros((Nx, Nt)) - for i,x_ in enumerate(x): - for j, t_ in enumerate(t): - point = np.array([x_, t_]) - g_dnn_ag[i,j] = g_trial(point,P) - - G_analytical[i,j] = g_analytic(point) - - # Find the map difference between the analytical and the computed solution - diff_ag = np.abs(g_dnn_ag - G_analytical) - print('Max absolute difference between the analytical solution and the network: %g'%np.max(diff_ag)) - - ## Plot the solutions in two dimensions, that being in position and time - - T,X = np.meshgrid(t,x) - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons)) - s = ax.plot_surface(T,X,g_dnn_ag,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Analytical solution') - s = ax.plot_surface(T,X,G_analytical,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Difference') - s = ax.plot_surface(T,X,diff_ag,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - ## Take some slices of the 3D plots just to see the solutions at particular times - indx1 = 0 - indx2 = int(Nt/2) - indx3 = Nt-1 - - t1 = t[indx1] - t2 = t[indx2] - t3 = t[indx3] - - # Slice the results from the DNN - res1 = g_dnn_ag[:,indx1] - res2 = g_dnn_ag[:,indx2] - res3 = g_dnn_ag[:,indx3] - - # Slice the analytical results - res_analytical1 = G_analytical[:,indx1] - res_analytical2 = G_analytical[:,indx2] - res_analytical3 = G_analytical[:,indx3] - - # Plot the slices - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t1) - plt.plot(x, res1) - plt.plot(x,res_analytical1) - plt.legend(['dnn','analytical']) - - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t2) - plt.plot(x, res2) - plt.plot(x,res_analytical2) - plt.legend(['dnn','analytical']) - - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t3) - plt.plot(x, res3) - plt.plot(x,res_analytical3) - plt.legend(['dnn','analytical']) - - plt.show() -!ec - -!split -===== Example: Solving the wave equation using Autograd and TensorFlow ===== - -The wave equation is -!bt -\begin{equation*} - \frac{\partial^2 g(x,t)}{\partial t^2} = c^2\frac{\partial^2 g(x,t)}{\partial x^2} -\end{equation*} -!et - -with $c$ being the specified wave speed. - -Here, the chosen conditions are -!bt -\begin{align*} - g(0,t) &= 0 \\ - g(1,t) &= 0 \\ - g(x,0) &= u(x) \\ - \frac{\partial g(x,t)}{\partial t} \Big |_{t = 0} &= v(x) -\end{align*} -!et -where $\frac{\partial g(x,t)}{\partial t} \Big |_{t = 0}$ means the derivative of $g(x,t)$ with respect to $t$ is evaluated at $t = 0$, and $u(x)$ and $v(x)$ being given functions. - -!split -===== The problem to solve for ===== - -The wave equation to solve for, is - -!bt -\begin{equation} \label{wave} -\frac{\partial^2 g(x,t)}{\partial t^2} = c^2 \frac{\partial^2 g(x,t)}{\partial x^2} -\end{equation} -!et - -where $c$ is the given wave speed. -The chosen conditions for this equation are - -!bt -\begin{aligned} -g(0,t) &= 0, &t \geq 0 \\ -g(1,t) &= 0, &t \geq 0 \\ -g(x,0) &= u(x), &x\in[0,1] \\ -\frac{\partial g(x,t)}{\partial t}\Big |_{t = 0} &= v(x), &x \in [0,1] -\end{aligned} \label{condwave} -!et - -In this example, let $c = 1$ and $u(x) = \sin(\pi x)$ and $v(x) = -\pi\sin(\pi x)$. - - -!split -===== The trial solution ===== -Setting up the network is done in similar matter as for the example of solving the diffusion equation. -The only things we have to change, is the trial solution such that it satisfies the conditions from (ref{condwave}) and the cost function. - -The trial solution becomes slightly different since we have other conditions than in the example of solving the diffusion equation. Here, a possible trial solution $g_t(x,t)$ is - -$$ -g_t(x,t) = h_1(x,t) + x(1-x)t^2N(x,t,P) -$$ - -where - -$$ -h_1(x,t) = (1-t^2)u(x) + tv(x) -$$ - -Note that this trial solution satisfies the conditions only if $u(0) = v(0) = u(1) = v(1) = 0$, which is the case in this example. - -!split -===== The analytical solution ===== - -The analytical solution for our specific problem, is - -$$ -g(x,t) = \sin(\pi x)\cos(\pi t) - \sin(\pi x)\sin(\pi t) -$$ - -!split -===== Solving the wave equation - the full program using Autograd ===== - -!bc pycod -import autograd.numpy as np -from autograd import hessian,grad -import autograd.numpy.random as npr -from matplotlib import cm -from matplotlib import pyplot as plt -from mpl_toolkits.mplot3d import axes3d - -## Set up the trial function: -def u(x): - return np.sin(np.pi*x) - -def v(x): - return -np.pi*np.sin(np.pi*x) - -def h1(point): - x,t = point - return (1 - t**2)*u(x) + t*v(x) - -def g_trial(point,P): - x,t = point - return h1(point) + x*(1-x)*t**2*deep_neural_network(P,point) - -## Define the cost function -def cost_function(P, x, t): - cost_sum = 0 - - g_t_hessian_func = hessian(g_trial) - - for x_ in x: - for t_ in t: - point = np.array([x_,t_]) - - g_t_hessian = g_t_hessian_func(point,P) - - g_t_d2x = g_t_hessian[0][0] - g_t_d2t = g_t_hessian[1][1] - - err_sqr = ( (g_t_d2t - g_t_d2x) )**2 - cost_sum += err_sqr - - return cost_sum / (np.size(t) * np.size(x)) - -## The neural network -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -def deep_neural_network(deep_params, x): - # x is now a point and a 1D numpy array; make it a column vector - num_coordinates = np.size(x,0) - x = x.reshape(num_coordinates,-1) - - num_points = np.size(x,1) - - # N_hidden is the number of hidden layers - N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assume that the input layer does nothing to the input x - x_input = x - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output[0][0] - -## The analytical solution -def g_analytic(point): - x,t = point - return np.sin(np.pi*x)*np.cos(np.pi*t) - np.sin(np.pi*x)*np.sin(np.pi*t) - -def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb): - ## Set up initial weigths and biases - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: ',cost_function(P, x, t)) - - cost_function_grad = grad(cost_function,0) - - # Let the update be done num_iter times - for i in range(num_iter): - cost_grad = cost_function_grad(P, x , t) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_grad[l] - - - print('Final cost: ',cost_function(P, x, t)) - - return P - -if __name__ == '__main__': - ### Use the neural network: - npr.seed(15) - - ## Decide the vales of arguments to the function to solve - Nx = 10; Nt = 10 - x = np.linspace(0, 1, Nx) - t = np.linspace(0,1,Nt) - - ## Set up the parameters for the network - num_hidden_neurons = [50,20] - num_iter = 1000 - lmb = 0.01 - - P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb) - - ## Store the results - res = np.zeros((Nx, Nt)) - res_analytical = np.zeros((Nx, Nt)) - for i,x_ in enumerate(x): - for j, t_ in enumerate(t): - point = np.array([x_, t_]) - res[i,j] = g_trial(point,P) - - res_analytical[i,j] = g_analytic(point) - - diff = np.abs(res - res_analytical) - print("Max difference between analytical and solution from nn: %g"%np.max(diff)) - - ## Plot the solutions in two dimensions, that being in position and time - - T,X = np.meshgrid(t,x) - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons)) - s = ax.plot_surface(T,X,res,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Analytical solution') - s = ax.plot_surface(T,X,res_analytical,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Difference') - s = ax.plot_surface(T,X,diff,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - ## Take some slices of the 3D plots just to see the solutions at particular times - indx1 = 0 - indx2 = int(Nt/2) - indx3 = Nt-1 - - t1 = t[indx1] - t2 = t[indx2] - t3 = t[indx3] - - # Slice the results from the DNN - res1 = res[:,indx1] - res2 = res[:,indx2] - res3 = res[:,indx3] - - # Slice the analytical results - res_analytical1 = res_analytical[:,indx1] - res_analytical2 = res_analytical[:,indx2] - res_analytical3 = res_analytical[:,indx3] - - # Plot the slices - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t1) - plt.plot(x, res1) - plt.plot(x,res_analytical1) - plt.legend(['dnn','analytical']) - - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t2) - plt.plot(x, res2) - plt.plot(x,res_analytical2) - plt.legend(['dnn','analytical']) - - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t3) - plt.plot(x, res3) - plt.plot(x,res_analytical3) - plt.legend(['dnn','analytical']) - - plt.show() -!ec - -!split -===== Solving the wave equation - the full program using TensorFlow ===== -As the program using Autograd is fairly slow, one could hope that using TensorFlow -could make a naive implementation faster, and more numerically robust. - -In addition, having TensorFlow at hand, it could be easier to experiment with different -optimization algorithms, and other constructions of the network. - -The following program solves the given wave equation much faster, - -!bc pycod -import tensorflow as tf -import numpy as np -from matplotlib import cm -from matplotlib import pyplot as plt -from mpl_toolkits.mplot3d import axes3d - -Nx = 10 -x_np = np.linspace(0,1,Nx) - -Nt = 10 -t_np = np.linspace(0,1,Nt) - -X,T = np.meshgrid(x_np, t_np) - -x = X.ravel() -t = T.ravel() - -## The construction phase - -zeros = tf.reshape(tf.convert_to_tensor(np.zeros(x.shape)),shape=(-1,1)) -x = tf.reshape(tf.convert_to_tensor(x),shape=(-1,1)) -t = tf.reshape(tf.convert_to_tensor(t),shape=(-1,1)) - -points = tf.concat([x,t],1) - -num_iter = 100000 -num_hidden_neurons = [90] - -X = tf.convert_to_tensor(X) -T = tf.convert_to_tensor(T) - - -with tf.variable_scope('dnn'): - num_hidden_layers = np.size(num_hidden_neurons) - - previous_layer = points - - for l in range(num_hidden_layers): - current_layer = tf.layers.dense(previous_layer, num_hidden_neurons[l],activation=tf.nn.sigmoid) - previous_layer = current_layer - - dnn_output = tf.layers.dense(previous_layer, 1) - - -def u(x): - return tf.sin(np.pi*x) - -def v(x): - return -np.pi*tf.sin(np.pi*x) - -with tf.name_scope('loss'): - g_trial = (1 - t**2)*u(x) + t*v(x) + x*(1-x)*t**2*dnn_output - - g_trial_d2t = tf.gradients(tf.gradients(g_trial,t),t) - g_trial_d2x = tf.gradients(tf.gradients(g_trial,x),x) - - loss = tf.losses.mean_squared_error(zeros, g_trial_d2t[0] - g_trial_d2x[0]) - -learning_rate = 0.01 -with tf.name_scope('train'): - optimizer = tf.train.GradientDescentOptimizer(learning_rate) - traning_op = optimizer.minimize(loss) - -init = tf.global_variables_initializer() - -g_analytic = tf.sin(np.pi*x)*tf.cos(np.pi*t) - tf.sin(np.pi*x)*tf.sin(np.pi*t) -g_dnn = None - -## The execution phase -with tf.Session() as sess: - init.run() - for i in range(num_iter): - sess.run(traning_op) - - # If one desires to see how the cost function behaves during training - #if i % 100 == 0: - # print(loss.eval()) - - g_analytic = g_analytic.eval() - g_dnn = g_trial.eval() - - -## Compare with the analutical solution -diff = np.abs(g_analytic - g_dnn) -print('Max absolute difference between analytical solution and TensorFlow DNN = ',np.max(diff)) - -G_analytic = g_analytic.reshape((Nt,Nx)) -G_dnn = g_dnn.reshape((Nt,Nx)) - -diff = np.abs(G_analytic - G_dnn) - -# Plot the results - -X,T = np.meshgrid(x_np, t_np) - -fig = plt.figure(figsize=(10,10)) -ax = fig.gca(projection='3d') -ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons)) -s = ax.plot_surface(X,T,G_dnn,linewidth=0,antialiased=False,cmap=cm.viridis) -ax.set_xlabel('Time $t$') -ax.set_ylabel('Position $x$'); - -fig = plt.figure(figsize=(10,10)) -ax = fig.gca(projection='3d') -ax.set_title('Analytical solution') -s = ax.plot_surface(X,T,G_analytic,linewidth=0,antialiased=False,cmap=cm.viridis) -ax.set_xlabel('Time $t$') -ax.set_ylabel('Position $x$'); - -fig = plt.figure(figsize=(10,10)) -ax = fig.gca(projection='3d') -ax.set_title('Difference') -s = ax.plot_surface(X,T,diff,linewidth=0,antialiased=False,cmap=cm.viridis) -ax.set_xlabel('Time $t$') -ax.set_ylabel('Position $x$'); - -## Take some 3D slices - -indx1 = 0 -indx2 = int(Nt/2) -indx3 = Nt-1 - -t1 = t_np[indx1] -t2 = t_np[indx2] -t3 = t_np[indx3] - -# Slice the results from the DNN -res1 = G_dnn[indx1,:] -res2 = G_dnn[indx2,:] -res3 = G_dnn[indx3,:] - -# Slice the analytical results -res_analytical1 = G_analytic[indx1,:] -res_analytical2 = G_analytic[indx2,:] -res_analytical3 = G_analytic[indx3,:] - -# Plot the slices -plt.figure(figsize=(10,10)) -plt.title("Computed solutions at time = %g"%t1) -plt.plot(x_np, res1) -plt.plot(x_np,res_analytical1) -plt.legend(['dnn','analytical']) - -plt.figure(figsize=(10,10)) -plt.title("Computed solutions at time = %g"%t2) -plt.plot(x_np, res2) -plt.plot(x_np,res_analytical2) -plt.legend(['dnn','analytical']) - -plt.figure(figsize=(10,10)) -plt.title("Computed solutions at time = %g"%t3) -plt.plot(x_np, res3) -plt.plot(x_np,res_analytical3) -plt.legend(['dnn','analytical']) - -plt.show() -!ec - -The program manages to find a solution having max absolute difference to the analytical -at approximately 0.0059, by just using some minutes! -It was found, by some testing, that one hidden layer with 90 neurons actually performed well. - -!split -===== Resources ===== - -o "Artificial neural networks for solving ordinary and partial differential equations by I.E. Lagaris et al":"https://pdfs.semanticscholar.org/d061/df393e0e8fbfd0ea24976458b7d42419040d.pdf" -o "Neural networks for solving differential equations by A. Honchar":"https://becominghuman.ai/neural-networks-for-solving-differential-equations-fa230ac5e04c" -o "Solving differential equations using neural networks by M.M Chiaramonte and M. Kiener":"http://cs229.stanford.edu/proj2013/ChiaramonteKiener-SolvingDifferentialEquationsUsingNeuralNetworks.pdf" -o "Introduction to Partial Differential Equations by A. Tveitio, R. Winther":"https://www.springer.com/us/book/9783540225515" - diff --git a/doc/src/week43/programs/odetf.py~ b/doc/src/week43/programs/odetf.py~ deleted file mode 100644 index d974171c6..000000000 --- a/doc/src/week43/programs/odetf.py~ +++ /dev/null @@ -1,250 +0,0 @@ -# For matrices and calculations -import numpy as np -# For machine learning (backend for keras) -import tensorflow as tf -# User-friendly machine learning library -# Front end for TensorFlow -import tensorflow.keras -# Different methods from Keras needed to create an RNN -# This is not necessary but it shortened function calls -# that need to be used in the code. -from tensorflow.keras import datasets, layers, models -from tensorflow.keras.layers import Input -from tensorflow.keras import regularizers -from tensorflow.keras.models import Model, Sequential -#from tensorflow.keras.layers.core import Dense, Activation -from tensorflow.keras.layers import Dense, SimpleRNN, LSTM, GRU -# For timing the code -from timeit import default_timer as timer -# For plotting -import matplotlib.pyplot as plt - - -# Define Analytical, Euler-Cromer, and Velocity-Verlet methods of solving -def analytical(k,m,x0,v0,dt,tfinal): - t = np.arange(0,tfinal+dt,dt) - v = -x0 * np.sin(t) + v0 * np.cos(t) - x = x0 * np.cos(t) + v0 * np.sin(t) - K = 1/2 *m*v**2 - U = 1/2 *k*x**2 - return x, v, K, U, t - -def euler_cromer(k,m,x0,v0,dt,tfinal): - n = np.ceil(tfinal/dt)\ - # Set up arrays - t = np.zeros(n) - v = np.zeros(n) - x = np.zeros(n) - K = np.zeros(n) - U = np.zeros(n) - # Define Initial Conditions - x[0] = x0 - v[0] = v0 - K[0] = 1/2 *m*v0**2 - U[0] = 1/2 *k*x0**2 - - # Integrate using the Euler-Cromer Method - for i in range(n-1): - a = -x[i] - v[i+1] = v[i] + dt*a - x[i+1] = x[i] + dt*v[i+1] - K[i+1] = 1/2 *m*v[i+1]**2 - U[i+1] = 1/2 *k*x[i+1]**2 - t[i+1] = t[i] + dt - return x, v, K, U, t - -def velocity_verlet(k,m,x0,v0,dt,tfinal): - n = np.ceil(tfinal/dt) - # Set up arrays - t = np.zeros(n) - v = np.zeros(n) - x = np.zeros(n) - K = np.zeros(n) - U = np.zeros(n) - # Define Initial Conditions - x[0] = x0 - v[0] = v0 - K[0] = 1/2 *m*v0**2 - U[0] = 1/2 *k*x0**2 - - # Integrate using the Velocity-Verlet Method - for i in range(n-1): - a = -x[i] - x[i+1] = x[i] + dt*v[i] + dt**2 /2*a - a1 = -x[i+1] - v[i+1] = v[i] + dt/2*(a+a1) - K[i+1] = 1/2 *m*v[i+1]**2 - U[i+1] = 1/2 *k*x[i+1]**2 - t[i+1] = t[i] + dt - return x, v, K, U, t - -# Define Constants -dt = 0.01 -tfinal = 50 -x0 = 1 -v0 = 0 -m = 1 -k = 1 - -# Call the Integration Function -ax, av, aK, aU, at = analytical(k,m,x0,v0,dt,tfinal) -ecx, ecv, ecK, ecU, ect = euler_cromer(k,m,x0,v0,dt,tfinal) -vvx, vvv, vvK, vvU,vvt = velocity_verlet(k,m,x0,v0,dt,tfinal) - -# Plots -fig, axes = plt.subplots(1,3, figsize = (15,5)) -fig.suptitle("System of an Undamped Spring", fontsize=14, y = 1.05) - -axes[0].plot(at, ax, label = "Analytical Method") -axes[0].plot(ect, ecx, label = "Euler-Cromer Method") -axes[0].plot(vvt, vvx, label = "Velocity-Verlet Method") -axes[0].set_title("Position as a Function of Dimensionless Time") -axes[0].set_xlabel("Dimensionless Time") -axes[0].set_ylabel("Position") - -axes[1].plot(at, av, label = "Analytical Method") -axes[1].plot(ect, ecv, label = "Euler-Cromer Method") -axes[1].plot(vvt, vvv, label = "Velocity-Verlet Method") -axes[1].set_title("Velocity as a Function of Dimensionless Time") -axes[1].set_xlabel("Dimensionless Time") -axes[1].set_ylabel("Velocity") - -axes[2].plot(at, aU+aK, label = "Analytical Method") -axes[2].plot(ect, ecU+ecK, label = "Euler-Cromer Method") -axes[2].plot(vvt, vvU+vvK, label = "Velocity-Verlet Method") -axes[2].set_title("Energy as a Function of Dimensionless Time") -axes[2].set_xlabel("Dimensionless Time") -axes[2].set_ylabel("Total Energy") - -plt.legend(bbox_to_anchor=(1.05, 1), loc='upper left') - -plt.tight_layout() - - -def damp(gamma,m,x0,v0,DeltaT,tfinal): - n = np.ceil(tfinal/DeltaT) - # Set up arrays - t = np.zeros(n) - v = np.zeros(n) - r = np.zeros(n) - # Define Initial Conditions - r[0] = x0 - v[0] = v0 - # Integrate over using the Velocity Verlet Method - for i in range(n-1): - a = -r[i] - 2*gamma*v[i] - r[i+1] = r[i] + DeltaT*v[i] + DeltaT**2 /2 *a - a1 = -r[i+1] - 2*gamma*v[i] - v[i+1] = v[i] + DeltaT/2*(a+a1) - t[i+1] = t[i] + DeltaT - - return t,r,v - -# Define Constants -under_gamma = 0.1 -crit_gamma = 1 -over_gamma = 2 -m = 1 -x0 = 1 -v0 = 0 -dt = 0.1 -tfinal = 50 - -# Call the Integration Function -t, under_r, under_v = damp(under_gamma,m,x0,v0,dt,tfinal) -t, crit_r, crit_v = damp(crit_gamma,m,x0,v0,dt,tfinal) -t, over_r, over_v = damp(over_gamma,m,x0,v0,dt,tfinal) - - -# Plots -fig, axes = plt.subplots(1,2, figsize = (15,5)) -fig.suptitle("System of a Damped Spring", fontsize=14, y = 1.05) - -axes[0].plot(t, under_r, label = "Under Damping") -axes[0].plot(t, crit_r, label = "Critical Damping") -axes[0].plot(t, over_r, label = "Over Damping") -axes[0].set_title("Position as a Function of Dimensionless Time") -axes[0].set_xlabel("Dimensionless Time") -axes[0].set_ylabel("Position") - -axes[1].plot(t, under_v, label = "Under Damping") -axes[1].plot(t, crit_v, label = "Critical Damping") -axes[1].plot(t, over_v, label = "Over Damping") -axes[1].set_title("Velocity as a Function of Dimensionless Time") -axes[1].set_xlabel("Dimensionless Time") -axes[1].set_ylabel("Velocity") - -plt.legend(bbox_to_anchor=(1.05, 1), loc='upper left') - -plt.tight_layout() - - -def forced(gamma,r0,v0,F0,omega,d,DeltaT,tfinal): - n = np.ceil(tfinal/DeltaT) - # Set up arrays - t = np.zeros(n) - v = np.zeros(n) - r = np.zeros(n) - # Define Initial Conditions - r[0] = r0 - v[0] = v0 - - # Integrate using the 4th-Order RK Method - for i in range(n-1): - t[i+1] = t[i] + DeltaT - - Force = (-r[i] - 2*gamma*v[i] - F0*np.cos(omega*t[i]-d))*m - k1x = DeltaT*v[i] - k1v = DeltaT*Force - - vv = v[i]+k1v*0.5*DeltaT - rr = r[i]+k1x*0.5*DeltaT - Force = (-rr - 2*gamma*vv - F0*np.cos(omega*(t[i]+DeltaT*0.5)-d))*m - k2x = DeltaT*vv - k2v = DeltaT*Force - - vv = v[i]+k2v*0.5*DeltaT - rr = r[i]+k2x*0.5*DeltaT - Force = (-rr - 2*gamma*vv - F0*np.cos(omega*(t[i]+DeltaT*0.5)-d))*m - k3x = DeltaT*vv - k3v = DeltaT*Force - - vv = v[i]+k3v*DeltaT - rr = r[i]+k3x*DeltaT - Force = (-rr - 2*gamma*vv - F0*np.cos(omega*(t[i]+DeltaT*0.5)-d))*m - k4x = DeltaT*vv - k4v = DeltaT*Force - - r[i+1] = r[i]+(k1x+2*k2x+2*k3x+k4x)/6. - v[i+1] = v[i]+(k1v+2*k2v+2*k3v+k4v)/6. - - return t,r,v - -# Define Constants -gamma = 0.1 -r0 = 1 -v0 = 0 -F0 = 3 -omega = 3 -d = 0 -dt = 0.1 -tfinal = 100 - -# Call the Integration Function -t, r, v = forced(gamma,r0,v0,F0,omega,d,dt,tfinal) - -# Plots -fig, axes = plt.subplots(1,2, figsize = (15,5)) -fig.suptitle("System of a Damped Spring with a Driving Force", fontsize=14, y = 1.05) - -axes[0].plot(t, r) -axes[0].set_title("Position as a Function of Dimensionless Time") -axes[0].set_xlabel("Dimensionless Time") -axes[0].set_ylabel("Position") - -axes[1].plot(t, v) -axes[1].set_title("Velocity as a Function of Dimensionless Time") -axes[1].set_xlabel("Dimensionless Time") -axes[1].set_ylabel("Velocity") - -plt.tight_layout() diff --git a/doc/src/week43/programs/poisson.py~ b/doc/src/week43/programs/poisson.py~ deleted file mode 100644 index 2b6c2d667..000000000 --- a/doc/src/week43/programs/poisson.py~ +++ /dev/null @@ -1,3759 +0,0 @@ -TITLE: Week 43: Solving Differential Equations with Deep Learning and Dimensionality Reduction methods -AUTHOR: Morten Hjorth-Jensen {copyright, 1999-present|CC BY-NC} at Department of Physics, University of Oslo & Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University -DATE: today - -!split - -* Thursday: Wrapping up Recurrent Neural Networks and solving differential equations. -* Friday: Principal Component Analysis and Dimensionality Reduction - -Reading suggestions for both days: "Aurelien Geron's chapters 8 - -!split -===== Recurrent Neural Networks ===== - - -!split -===== Solving ODEs with Deep Learning ===== - - - - - - - -!split -===== Basic ideas of the Principal Component Analysis (PCA) ===== - -The principal component analysis deals with the problem of fitting a -low-dimensional affine subspace $S$ of dimension $d$ much smaller than -the totaldimension $D$ of the problem at hand (our data -set). Mathematically it can be formulated as a statistical problem or -a geometric problem. In our discussion of the theorem for the -classical PCA, we will stay with a statistical approach. This is also -what set the scene historically which for the PCA. - -We have a data set defined by a design/feature matrix $\bm{X}$ (see below for its definition) -* Each data point is determined by $p$ extrinsic (measurement) variables -* We may want to ask the following question: Are there fewer intrinsic variables (say $d << p$) that still approximately describe the data? -* If so, these intrinsic variables may tell us something important and finding these intrinsic variables is what dimension reduction methods do. - - -!split -===== Introducing the Covariance and Correlation functions ===== - -Before we discuss the PCA theorem, we need to remind ourselves about -the definition of the covariance and the correlation function. These are quantities - -Suppose we have defined two vectors -$\hat{x}$ and $\hat{y}$ with $n$ elements each. The covariance matrix $\bm{C}$ is defined as -!bt -\[ -\bm{C}[\bm{x},\bm{y}] = \begin{bmatrix} \mathrm{cov}[\bm{x},\bm{x}] & \mathrm{cov}[\bm{x},\bm{y}] \\ - \mathrm{cov}[\bm{y},\bm{x}] & \mathrm{cov}[\bm{y},\bm{y}] \\ - \end{bmatrix}, -\] -!et -where for example -!bt -\[ -\mathrm{cov}[\bm{x},\bm{y}] =\frac{1}{n} \sum_{i=0}^{n-1}(x_i- \overline{x})(y_i- \overline{y}). -\] -!et -With this definition and recalling that the variance is defined as -!bt -\[ -\mathrm{var}[\bm{x}]=\frac{1}{n} \sum_{i=0}^{n-1}(x_i- \overline{x})^2, -\] -!et -we can rewrite the covariance matrix as -!bt -\[ -\bm{C}[\bm{x},\bm{y}] = \begin{bmatrix} \mathrm{var}[\bm{x}] & \mathrm{cov}[\bm{x},\bm{y}] \\ - \mathrm{cov}[\bm{x},\bm{y}] & \mathrm{var}[\bm{y}] \\ - \end{bmatrix}. -\] -!et - -The covariance takes values between zero and infinity and may thus -lead to problems with loss of numerical precision for particularly -large values. It is common to scale the covariance matrix by -introducing instead the correlation matrix defined via the so-called -correlation function - -!bt -\[ -\mathrm{corr}[\bm{x},\bm{y}]=\frac{\mathrm{cov}[\bm{x},\bm{y}]}{\sqrt{\mathrm{var}[\bm{x}] \mathrm{var}[\bm{y}]}}. -\] -!et - -The correlation function is then given by values $\mathrm{corr}[\bm{x},\bm{y}] -\in [-1,1]$. This avoids eventual problems with too large values. We -can then define the correlation matrix for the two vectors $\bm{x}$ -and $\bm{y}$ as - -!bt -\[ -\bm{K}[\bm{x},\bm{y}] = \begin{bmatrix} 1 & \mathrm{corr}[\bm{x},\bm{y}] \\ - \mathrm{corr}[\bm{y},\bm{x}] & 1 \\ - \end{bmatrix}, -\] -!et - -In the above example this is the function we constructed using _pandas_. - -!split -===== Correlation Function and Design/Feature Matrix ===== - -In our derivation of the various regression algorithms like _Ordinary Least Squares_ or _Ridge regression_ -we defined the design/feature matrix $\bm{X}$ as - -!bt -\[ -\bm{X}=\begin{bmatrix} -x_{0,0} & x_{0,1} & x_{0,2}& \dots & \dots x_{0,p-1}\\ -x_{1,0} & x_{1,1} & x_{1,2}& \dots & \dots x_{1,p-1}\\ -x_{2,0} & x_{2,1} & x_{2,2}& \dots & \dots x_{2,p-1}\\ -\dots & \dots & \dots & \dots \dots & \dots \\ -x_{n-2,0} & x_{n-2,1} & x_{n-2,2}& \dots & \dots x_{n-2,p-1}\\ -x_{n-1,0} & x_{n-1,1} & x_{n-1,2}& \dots & \dots x_{n-1,p-1}\\ -\end{bmatrix}, -\] -!et -with $\bm{X}\in {\mathbb{R}}^{n\times p}$, with the predictors/features $p$ refering to the column numbers and the -entries $n$ being the row elements. -We can rewrite the design/feature matrix in terms of its column vectors as -!bt -\[ -\bm{X}=\begin{bmatrix} \bm{x}_0 & \bm{x}_1 & \bm{x}_2 & \dots & \dots & \bm{x}_{p-1}\end{bmatrix}, -\] -!et -with a given vector -!bt -\[ -\bm{x}_i^T = \begin{bmatrix}x_{0,i} & x_{1,i} & x_{2,i}& \dots & \dots x_{n-1,i}\end{bmatrix}. -\] -!et - -With these definitions, we can now rewrite our $2\times 2$ -correaltion/covariance matrix in terms of a moe general design/feature -matrix $\bm{X}\in {\mathbb{R}}^{n\times p}$. This leads to a $p\times p$ -covariance matrix for the vectors $\bm{x}_i$ with $i=0,1,\dots,p-1$ - -!bt -\[ -\bm{C}[\bm{x}] = \begin{bmatrix} -\mathrm{var}[\bm{x}_0] & \mathrm{cov}[\bm{x}_0,\bm{x}_1] & \mathrm{cov}[\bm{x}_0,\bm{x}_2] & \dots & \dots & \mathrm{cov}[\bm{x}_0,\bm{x}_{p-1}]\\ -\mathrm{cov}[\bm{x}_1,\bm{x}_0] & \mathrm{var}[\bm{x}_1] & \mathrm{cov}[\bm{x}_1,\bm{x}_2] & \dots & \dots & \mathrm{cov}[\bm{x}_1,\bm{x}_{p-1}]\\ -\mathrm{cov}[\bm{x}_2,\bm{x}_0] & \mathrm{cov}[\bm{x}_2,\bm{x}_1] & \mathrm{var}[\bm{x}_2] & \dots & \dots & \mathrm{cov}[\bm{x}_2,\bm{x}_{p-1}]\\ -\dots & \dots & \dots & \dots & \dots & \dots \\ -\dots & \dots & \dots & \dots & \dots & \dots \\ -\mathrm{cov}[\bm{x}_{p-1},\bm{x}_0] & \mathrm{cov}[\bm{x}_{p-1},\bm{x}_1] & \mathrm{cov}[\bm{x}_{p-1},\bm{x}_{2}] & \dots & \dots & \mathrm{var}[\bm{x}_{p-1}]\\ -\end{bmatrix}, -\] -!et -and the correlation matrix -!bt -\[ -\bm{K}[\bm{x}] = \begin{bmatrix} -1 & \mathrm{corr}[\bm{x}_0,\bm{x}_1] & \mathrm{corr}[\bm{x}_0,\bm{x}_2] & \dots & \dots & \mathrm{corr}[\bm{x}_0,\bm{x}_{p-1}]\\ -\mathrm{corr}[\bm{x}_1,\bm{x}_0] & 1 & \mathrm{corr}[\bm{x}_1,\bm{x}_2] & \dots & \dots & \mathrm{corr}[\bm{x}_1,\bm{x}_{p-1}]\\ -\mathrm{corr}[\bm{x}_2,\bm{x}_0] & \mathrm{corr}[\bm{x}_2,\bm{x}_1] & 1 & \dots & \dots & \mathrm{corr}[\bm{x}_2,\bm{x}_{p-1}]\\ -\dots & \dots & \dots & \dots & \dots & \dots \\ -\dots & \dots & \dots & \dots & \dots & \dots \\ -\mathrm{corr}[\bm{x}_{p-1},\bm{x}_0] & \mathrm{corr}[\bm{x}_{p-1},\bm{x}_1] & \mathrm{corr}[\bm{x}_{p-1},\bm{x}_{2}] & \dots & \dots & 1\\ -\end{bmatrix}, -\] -!et - - -!split -===== Covariance Matrix Examples ===== - - -The Numpy function _np.cov_ calculates the covariance elements using -the factor $1/(n-1)$ instead of $1/n$ since it assumes we do not have -the exact mean values. The following simple function uses the -_np.vstack_ function which takes each vector of dimension $1\times n$ -and produces a $2\times n$ matrix $\bm{W}$ - - -!bt -\[ -\bm{W} = \begin{bmatrix} x_0 & y_0 \\ - x_1 & y_1 \\ - x_2 & y_2\\ - \dots & \dots \\ - x_{n-2} & y_{n-2}\\ - x_{n-1} & y_{n-1} & - \end{bmatrix}, -\] -!et - -which in turn is converted into into the $2\times 2$ covariance matrix -$\bm{C}$ via the Numpy function _np.cov()_. We note that we can also calculate -the mean value of each set of samples $\bm{x}$ etc using the Numpy -function _np.mean(x)_. We can also extract the eigenvalues of the -covariance matrix through the _np.linalg.eig()_ function. - -!bc pycod -# Importing various packages -import numpy as np -n = 100 -x = np.random.normal(size=n) -print(np.mean(x)) -y = 4+3*x+np.random.normal(size=n) -print(np.mean(y)) -W = np.vstack((x, y)) -C = np.cov(W) -print(C) -!ec - -!split -===== Correlation Matrix ===== - -The previous example can be converted into the correlation matrix by -simply scaling the matrix elements with the variances. We should also -subtract the mean values for each column. This leads to the following -code which sets up the correlations matrix for the previous example in -a more brute force way. Here we scale the mean values for each column of the design matrix, calculate the relevant mean values and variances and then finally set up the $2\times 2$ correlation matrix (since we have only two vectors). - -!bc pycod -import numpy as np -n = 100 -# define two vectors -x = np.random.random(size=n) -y = 4+3*x+np.random.normal(size=n) -#scaling the x and y vectors -x = x - np.mean(x) -y = y - np.mean(y) -variance_x = np.sum(x@x)/n -variance_y = np.sum(y@y)/n -print(variance_x) -print(variance_y) -cov_xy = np.sum(x@y)/n -cov_xx = np.sum(x@x)/n -cov_yy = np.sum(y@y)/n -C = np.zeros((2,2)) -C[0,0]= cov_xx/variance_x -C[1,1]= cov_yy/variance_y -C[0,1]= cov_xy/np.sqrt(variance_y*variance_x) -C[1,0]= C[0,1] -print(C) -!ec - -We see that the matrix elements along the diagonal are one as they -should be and that the matrix is symmetric. Furthermore, diagonalizing -this matrix we easily see that it is a positive definite matrix. - -The above procedure with _numpy_ can be made more compact if we use _pandas_. - -!split -===== Correlation Matrix with Pandas ===== - -We whow here how we can set up the correlation matrix using _pandas_, as done in this simple code -!bc pycod -import numpy as np -import pandas as pd -n = 10 -x = np.random.normal(size=n) -x = x - np.mean(x) -y = 4+3*x+np.random.normal(size=n) -y = y - np.mean(y) -X = (np.vstack((x, y))).T -print(X) -Xpd = pd.DataFrame(X) -print(Xpd) -correlation_matrix = Xpd.corr() -print(correlation_matrix) -!ec - - -We expand this model to the Franke function discussed above. - -!split -===== Correlation Matrix with Pandas and the Franke function ===== - -!bc pycod -# Common imports -import numpy as np -import pandas as pd - - -def FrankeFunction(x,y): - term1 = 0.75*np.exp(-(0.25*(9*x-2)**2) - 0.25*((9*y-2)**2)) - term2 = 0.75*np.exp(-((9*x+1)**2)/49.0 - 0.1*(9*y+1)) - term3 = 0.5*np.exp(-(9*x-7)**2/4.0 - 0.25*((9*y-3)**2)) - term4 = -0.2*np.exp(-(9*x-4)**2 - (9*y-7)**2) - return term1 + term2 + term3 + term4 - - -def create_X(x, y, n ): - if len(x.shape) > 1: - x = np.ravel(x) - y = np.ravel(y) - - N = len(x) - l = int((n+1)*(n+2)/2) # Number of elements in beta - X = np.ones((N,l)) - - for i in range(1,n+1): - q = int((i)*(i+1)/2) - for k in range(i+1): - X[:,q+k] = (x**(i-k))*(y**k) - - return X - - -# Making meshgrid of datapoints and compute Franke's function -n = 4 -N = 100 -x = np.sort(np.random.uniform(0, 1, N)) -y = np.sort(np.random.uniform(0, 1, N)) -z = FrankeFunction(x, y) -X = create_X(x, y, n=n) - -Xpd = pd.DataFrame(X) -# subtract the mean values and set up the covariance matrix -Xpd = Xpd - Xpd.mean() -covariance_matrix = Xpd.cov() -print(covariance_matrix) -!ec - -We note here that the covariance is zero for the first rows and -columns since all matrix elements in the design matrix were set to one -(we are fitting the function in terms of a polynomial of degree $n$). - -This means that the variance for these elements will be zero and will -cause problems when we set up the correlation matrix. We can simply -drop these elements and construct a correlation -matrix without these elements. - - -!split -===== Rewriting the Covariance and/or Correlation Matrix ===== - -We can rewrite the covariance matrix in a more compact form in terms of the design/feature matrix $\bm{X}$ as -!bt -\[ -\bm{C}[\bm{x}] = \frac{1}{n}\bm{X}\bm{X}^T= \mathbb{E}[\bm{X}\bm{X}^T]. -\] -!et - -To see this let us simply look at a design matrix $\bm{X}\in {\mathbb{R}}^{2\times 2}$ -!bt -\[ -\bm{X}=\begin{bmatrix} -x_{00} & x_{01}\\ -x_{10} & x_{11}\\ -\end{bmatrix}=\begin{bmatrix} -\bm{x}_{0} & \bm{x}_{1}\\ -\end{bmatrix}. -\] -!et - -If we then compute the expectation value -!bt -\[ -\mathbb{E}[\bm{X}\bm{X}^T] = \frac{1}{n}\bm{X}\bm{X}^T=\begin{bmatrix} -x_{00}^2+x_{01}^2 & x_{00}x_{10}+x_{01}x_{11}\\ -x_{10}x_{00}+x_{11}x_{01} & x_{10}^2+x_{11}^2\\ -\end{bmatrix}, -\] -!et -which is just -!bt -\[ -\bm{C}[\bm{x}_0,\bm{x}_1] = \bm{C}[\bm{x}]=\begin{bmatrix} \mathrm{var}[\bm{x}_0] & \mathrm{cov}[\bm{x}_0,\bm{x}_1] \\ - \mathrm{cov}[\bm{x}_1,\bm{x}_0] & \mathrm{var}[\bm{x}_1] \\ - \end{bmatrix}, -\] -!et -where we wrote $$\bm{C}[\bm{x}_0,\bm{x}_1] = \bm{C}[\bm{x}]$$ to indicate that this the covariance of the vectors $\bm{x}$ of the design/feature matrix $\bm{X}$. - -It is easy to generalize this to a matrix $\bm{X}\in {\mathbb{R}}^{n\times p}$. - - -!split -===== Towards the PCA theorem ===== - -We have that the covariance matrix (the correlation matrix involves a simple rescaling) is given as -!bt -\[ -\bm{C}[\bm{x}] = \frac{1}{n}\bm{X}\bm{X}^T= \mathbb{E}[\bm{X}\bm{X}^T]. -\] -!et -Let us now assume that we can perform a series of orthogonal transformations where we employ some orthogonal matrices $\bm{S}$. -These matrices are defined as $\bm{S}\in {\mathbb{R}}^{p\times p}$ and obey the orthogonality requirements $\bm{S}\bm{S}^T=\bm{S}^T\bm{S}=\bm{I}$. The matrix can be written out in terms of the column vectors $\bm{s}_i$ as $\bm{S}=[\bm{s}_0,\bm{s}_1,\dots,\bm{s}_{p-1}]$ and $\bm{s}_i \in {\mathbb{R}}^{p}$. - -Assume also that there is a transformation $\bm{S}\bm{C}[\bm{x}]\bm{S}^T=\bm{C}[\bm{y}]$ such that the new matrix $\bm{C}[\bm{y}]$ is diagonal with elements $[\lambda_0,\lambda_1,\lambda_2,\dots,\lambda_{p-1}]$. - -That is we have -!bt -\[ -\bm{C}[\bm{y}] = \mathbb{E}[\bm{S}\bm{X}\bm{X}^T\bm{S}^T]=\bm{S}\bm{C}[\bm{x}]\bm{S}^T, -\] -!et -since the matrix $\bm{S}$ is not a data dependent matrix. Multiplying with $\bm{S}^T$ from the left we have -!bt -\[ -\bm{S}^T\bm{C}[\bm{y}] = \bm{C}[\bm{x}]\bm{S}^T, -\] -!et -and since $\bm{C}[\bm{y}]$ is diagonal we have for a given eigenvalue $i$ of the covariance matrix that - -!bt -\[ -\bm{S}^T_i\lambda_i = \bm{C}[\bm{x}]\bm{S}^T_i. -\] -!et - -In the derivation of the PCA theorem we will assume that the eigenvalues are ordered in descending order, that is -$\lambda_0 > \lambda_1 > \dots > \lambda_{p-1}$. - - -The eigenvalues tell us then how much we need to stretch the -corresponding eigenvectors. Dimensions with large eigenvalues have -thus large variations (large variance) and define therefore useful -dimensions. The data points are more spread out in the direction of -these eigenvectors. Smaller eigenvalues mean on the other hand that -the corresponding eigenvectors are shrunk accordingly and the data -points are tightly bunched together and there is not much variation in -these specific directions. Hopefully then we could leave it out -dimensions where the eigenvalues are very small. If $p$ is very large, -we could then aim at reducing $p$ to $l << p$ and handle only $l$ -features/predictors. - -!split -===== The Algorithm before the Theorem ===== - -Here's how we would proceed in setting up the algorithm for the PCA, see also discussion below here. -* Set up the datapoints for the design/feature matrix $\bm{X}$ with $\bm{X}\in {\mathbb{R}}^{n\times p}$, with the predictors/features $p$ referring to the column numbers and the entries $n$ being the row elements. -!bt -\[ -\bm{X}=\begin{bmatrix} -x_{0,0} & x_{0,1} & x_{0,2}& \dots & \dots x_{0,p-1}\\ -x_{1,0} & x_{1,1} & x_{1,2}& \dots & \dots x_{1,p-1}\\ -x_{2,0} & x_{2,1} & x_{2,2}& \dots & \dots x_{2,p-1}\\ -\dots & \dots & \dots & \dots \dots & \dots \\ -x_{n-2,0} & x_{n-2,1} & x_{n-2,2}& \dots & \dots x_{n-2,p-1}\\ -x_{n-1,0} & x_{n-1,1} & x_{n-1,2}& \dots & \dots x_{n-1,p-1}\\ -\end{bmatrix}, -\] -!et -* Center the data by subtracting the mean value for each column. This leads to a new matrix $\bm{X}\rightarrow \overline{\bm{X}}$. -* Compute then the covariance/correlation matrix $\mathbb{E}[\overline{\bm{X}}\overline{\bm{X}}^T]$. -* Find the eigenpairs of $\bm{C}$ with eigenvalues $[\lambda_0,\lambda_1,\dots,\lambda_{p-1}]$ and eigenvectors $[\bm{s}_0,\bm{s}_1,\dots,\bm{s}_{p-1}]$. -* Order the eigenvalue (and the eigenvectors accordingly) in order of decreasing eigenvalues. -* Keep only those $l$ eigenvalues larger than a selected threshold value, discarding thus $p-l$ features since we expect small variations in the data here. - - -!split -===== Writing our own PCA code ===== - -We will use a simple example first with two-dimensional data -drawn from a multivariate normal distribution with the following mean and covariance matrix: -!bt -\[ -\mu = (-1,2) \qquad \Sigma = \begin{bmatrix} 4 & 2 \\ -2 & 2 -\end{bmatrix} -\] -!et -Note that the mean refers to each column of data. -We will generate $n = 1000$ points $X = \{ x_1, \ldots, x_N \}$ from -this distribution, and store them in the $1000 \times 2$ matrix $\bm{X}$. - -The following Python code aids in setting up the data and writing out the design matrix. -Note that the function _multivariate_ returns also the covariance discussed above and that it is defined by dividing by $n-1$ instead of $n$. -!bc pycod -import numpy as np -import pandas as pd -import matplotlib.pyplot as plt -from IPython.display import display -n = 10000 -mean = (-1, 2) -cov = [[4, 2], [2, 2]] -X = np.random.multivariate_normal(mean, cov, n) -!ec - -Now we are going to implement the PCA algorithm. We will break it down into various substeps. - -=== Compute the sample mean and center the data === - -The first step of PCA is to compute the sample mean of the data and use it to center the data. Recall that the sample mean is -!bt -\[ -\mu_n = \frac{1}{n} \sum_{i=1}^n x_i -\] -!et -and the mean-centered data $\bar{X} = \{ \bar{x}_1, \ldots, \bar{x}_n \}$ takes the form -!bt -\[ -\bar{x}_i = x_i - \mu_n. -\] -!et -When you are done with these steps, print out $\mu_n$ to verify it is -close to $\mu$ and plot your mean centered data to verify it is -centered at the origin! Compare your code with the functionality from _Scikit-Learn_ discussed above. -The following code elements perform these operations using _pandas_ or using our own functionality for doing so. The latter, using _numpy_ is rather simple through the _mean()_ function. -!bc pycod -df = pd.DataFrame(X) -# Pandas does the centering for us -df = df -df.mean() -# we center it ourselves -X_centered = X - X.mean(axis=0) -!ec - -Alternatively, we could use the functions we discussed -earlier for scaling the data set. That is, we could have used the -_StandardScaler_ function in _Scikit-Learn_, a function which ensures -that for each feature/predictor we study the mean value is zero and -the variance is one (every column in the design/feature matrix). You -would then not get the same results, since we divide by the -variance. The diagonal covariance matrix elements will then be one, -while the non-diagonal ones need to be divided by $2\sqrt{2}$ for our -specific case. - -=== Compute the sample covariance === - -Now we are going to use the mean centered data to compute the sample covariance of the data by using the following equation -!bt -\begin{equation*} -\Sigma_n = \frac{1}{n-1} \sum_{i=1}^n \bar{x}_i^T \bar{x}_i = \frac{1}{n-1} \sum_{i=1}^n (x_i - \mu_n)^T (x_i - \mu_n) -\end{equation*} -!et -where the data points $x_i \in \mathbb{R}^p$ (here in this example $p = 2$) are column vectors and $x^T$ is the transpose of $x$. -We can write our own code or simply use either the functionaly of _numpy_ or that of _pandas_, as follows -!bc pycod -print(df.cov()) -print(np.cov(X_centered.T)) -!ec -Note that the way we define the covariance matrix here has a factor $n-1$ instead of $n$. This is included in the _cov()_ function by _numpy_ and _pandas_. -Our own code here is not very elegant and asks for obvious improvements. It is tailored to this specific $2\times 2$ covariance matrix. -!bc pycod -# extract the relevant columns from the centered design matrix of dim n x 2 -x = X_centered[:,0] -y = X_centered[:,1] -Cov = np.zeros((2,2)) -Cov[0,1] = np.sum(x.T@y)/(n-1.0) -Cov[0,0] = np.sum(x.T@x)/(n-1.0) -Cov[1,1] = np.sum(y.T@y)/(n-1.0) -Cov[1,0]= Cov[0,1] -print("Centered covariance using own code") -print(Cov) -plt.plot(x, y, 'x') -plt.axis('equal') -plt.show() -!ec - -Depending on the number of points $n$, we will get results that are close to the covariance values defined above. -The plot shows how the data are clustered around a line with slope close to one. Is this expected? - -=== Diagonalize the sample covariance matrix to obtain the principal components === - -Now we are ready to solve for the principal components! To do so we -diagonalize the sample covariance matrix $\Sigma$. We can use the -function _np.linalg.eig_ to do so. It will return the eigenvalues and -eigenvectors of $\Sigma$. Once we have these we can perform the -following tasks: - -* We compute the percentage of the total variance captured by the first principal component -* We plot the mean centered data and lines along the first and second principal components -* Then we project the mean centered data onto the first and second principal components, and plot the projected data. -* Finally, we approximate the data as - -!bt -\begin{equation*} -x_i \approx \tilde{x}_i = \mu_n + \langle x_i, v_0 \rangle v_0 -\end{equation*} -!et -where $v_0$ is the first principal component. - -Collecting all these steps we can write our own PCA function and -compare this with the functionality included in _Scikit-Learn_. - -The code here outlines some of the elements we could include in the -analysis. Feel free to extend upon this in order to address the above -questions. - -!bc pycod -# diagonalize and obtain eigenvalues, not necessarily sorted -EigValues, EigVectors = np.linalg.eig(Cov) -# sort eigenvectors and eigenvalues -#permute = EigValues.argsort() -#EigValues = EigValues[permute] -#EigVectors = EigVectors[:,permute] -print("Eigenvalues of Covariance matrix") -for i in range(2): - print(EigValues[i]) -FirstEigvector = EigVectors[:,0] -SecondEigvector = EigVectors[:,1] -print("First eigenvector") -print(FirstEigvector) -print("Second eigenvector") -print(SecondEigvector) -#thereafter we do a PCA with Scikit-learn -from sklearn.decomposition import PCA -pca = PCA(n_components = 2) -X2Dsl = pca.fit_transform(X) -print("Eigenvector of largest eigenvalue") -print(pca.components_.T[:, 0]) - -!ec -This code does not contain all the above elements, but it shows how we can use _Scikit-Learn_ to extract the eigenvector which corresponds to the largest eigenvalue. Try to address the questions we pose before the above code. Try also to change the values of the covariance matrix by making one of the diagonal elements much larger than the other. What do you observe then? - -!split -===== Classical PCA Theorem ===== - -We assume now that we have a design matrix $\bm{X}$ which has been -centered as discussed above. For the sake of simplicity we skip the -overline symbol. The matrix is defined in terms of the various column -vectors $[\bm{x}_0,\bm{x}_1,\dots, \bm{x}_{p-1}]$ each with dimension -$\bm{x}\in {\mathbb{R}}^{n}$. - -We assume also that we have an orthogonal transformation $\bm{W}\in {\mathbb{R}}^{p\times p}$. We define the reconstruction error (which is similar to the mean squared error we have seen before) as -!bt -\[ -J(\bm{W},\bm{Z}) = \frac{1}{n}\sum_i (\bm{x}_i - \overline{\bm{x}}_i)^2, -\] -!et -with $\overline{\bm{x}}_i = \bm{W}\bm{z}_i$, where $\bm{z}_i$ is a row vector with dimension ${\mathbb{R}}^{n}$ of the matrix -$\bm{Z}\in{\mathbb{R}}^{p\times n}$. When doing PCA we want to reduce this dimensionality. - -The PCA theorem states that minimizing the above reconstruction error -corresponds to setting $\bm{W}=\bm{S}$, the orthogonal matrix which -diagonalizes the empirical covariance(correlation) matrix. The optimal -low-dimensional encoding of the data is then given by a set of vectors -$\bm{z}_i$ with at most $l$ vectors, with $l << p$, defined by the -orthogonal projection of the data onto the columns spanned by the -eigenvectors of the covariance(correlations matrix). - -The proof which follows will be updated by mid January 2020. - -!split -===== Proof of the PCA Theorem ===== - -To show the PCA theorem let us start with the assumption that there is one vector $\bm{w}_0$ which corresponds to a solution which minimized the reconstruction error $J$. This is an orthogonal vector. It means that we now approximate the reconstruction error in terms of $\bm{w}_0$ and $\bm{z}_0$ as -!bt -\[ -J(\bm{w}_0,\bm{z}_0)= \frac{1}{n}\sum_i (\bm{x}_i - z_{i0}\bm{w}_0)^2=\frac{1}{n}\sum_i (\bm{x}_i^T\bm{x}_i - 2z_{i0}\bm{w}_0^T\bm{x}_i+z_{i0}^2\bm{w}_0^T\bm{w}_0), -\] -!et -which we can rewrite due to the orthogonality of $\bm{w}_i$ as -!bt -\[ -J(\bm{w}_0,\bm{z}_0)=\frac{1}{n}\sum_i (\bm{x}_i^T\bm{x}_i - 2z_{i0}\bm{w}_0^T\bm{x}_i+z_{i0}^2). -\] -!et -Minimizing $J$ with respect to the unknown parameters $z_{0i}$ we obtain that -!bt -\[ -z_{i0}=\bm{w}_0^T\bm{x}_i, -\] -!et -where the vectors on the rhs are known. - - -!split -===== PCA Proof continued ===== - -We have now found the unknown parameters $z_{i0}$. These correspond to the projected coordinates and we can write -!bt -\[ -J(\bm{w}_0)= \frac{1}{p}\sum_i (\bm{x}_i^T\bm{x}_i - z_{i0}^2)=\mathrm{const}-\frac{1}{n}\sum_i z_{i0}^2. -\] -!et - -We can show that the variance of the projected coordinates defined by $\bm{w}_0^T\bm{x}_i$ are given by -!bt -\[ -\mathrm{var}[\bm{w}_0^T\bm{x}_i] = \frac{1}{n}\sum_i z_{i0}^2, -\] -!et -since the expectation value of -!bt -\[ -\mathbb{E}[\bm{w}_0^T\bm{x}_i] = \mathbb{E}[z_{i0}]= \bm{w}_0^T\mathbb{E}[\bm{x}_i]=0, -\] -!et -where we have used the fact that our data are centered. - -Recalling our definition of the covariance as -!bt -\[ -\bm{C}[\bm{x}] = \frac{1}{n}\bm{X}\bm{X}^T=\mathbb{E}[\bm{X}\bm{X}^T], -\] -!et -we have thus that -!bt -\[ -\mathrm{var}[\bm{w}_0^T\bm{x}_i] = \frac{1}{n}\sum_i z_{i0}^2=\bm{w}_0^T\bm{C}[\bm{x}]\bm{w}_0. -\] -!et - -We are almost there, we have obtained a relation between minimizing -the reconstruction error and the variance and the covariance -matrix. Minimizing the error is equivalent to maximizing the variance -of the projected data. - -!split -===== The final step ===== - -We could trivially maximize the variance of the projection (and -thereby minimize the error in the reconstruction function) by letting -the norm-2 of $\bm{w}_0$ go to infinity. However, this norm since we -want the matrix $\bm{W}$ to be an orthogonal matrix, is constrained by -$\vert\vert \bm{w}_0 \vert\vert_2^2=1$. Imposing this condition via a -Lagrange multiplier we can then in turn maximize - -!bt -\[ -J(\bm{w}_0)= \bm{w}_0^T\bm{C}[\bm{x}]\bm{w}_0+\lambda_0(1-\bm{w}_0^T\bm{w}_0). -\] -!et -Taking the derivative with respect to $\bm{w}_0$ we obtain - -!bt -\[ -\frac{\partial J(\bm{w}_0)}{\partial \bm{w}_0}= 2\bm{C}[\bm{x}]\bm{w}_0-2\lambda_0\bm{w}_0=0, -\] -!et -meaning that -!bt -\[ -\bm{C}[\bm{x}]\bm{w}_0=\lambda_0\bm{w}_0. -\] -!et -_The direction that maximizes the variance (or minimizes the construction error) is an eigenvector of the covariance matrix_! If we left multiply with $\bm{w}_0^T$ we have the variance of the projected data is -!bt -\[ -\bm{w}_0^T\bm{C}[\bm{x}]\bm{w}_0=\lambda_0. -\] -!et - -If we want to maximize the variance (minimize the construction error) -we simply pick the eigenvector of the covariance matrix with the -largest eigenvalue. This establishes the link between the minimization -of the reconstruction function $J$ in terms of an orthogonal matrix -and the maximization of the variance and thereby the covariance of our -observations encoded in the design/feature matrix $\bm{X}$. - -The proof -for the other eigenvectors $\bm{w}_1,\bm{w}_2,\dots$ can be -established by applying the above arguments and using the fact that -our basis of eigenvectors is orthogonal, see "Murphy chapter -12.2":"https://mitpress.mit.edu/books/machine-learning-1". The -discussion in chapter 12.2 of Murphy's text has also a nice link with -the Singular Value Decomposition theorem. For categorical data, see -chapter 12.4 and discussion therein. - -Additional part of the proof for the other eigenvectors will be added by mid January 2020. - -!split -===== Geometric Interpretation and link with Singular Value Decomposition ===== - -This material will be added by mid January 2020. - - -!split -===== Principal Component Analysis ===== - -Principal Component Analysis (PCA) is by far the most popular dimensionality reduction algorithm. -First it identifies the hyperplane that lies closest to the data, and then it projects the data onto it. - -The following Python code uses NumPy’s _svd()_ function to obtain all the principal components of the -training set, then extracts the first two principal components. First we center the data using either _pandas_ or our own code -!bc pycod -import numpy as np -import pandas as pd -from IPython.display import display -np.random.seed(100) -# setting up a 10 x 5 vanilla matrix -rows = 10 -cols = 5 -X = np.random.randn(rows,cols) -df = pd.DataFrame(X) -# Pandas does the centering for us -df = df -df.mean() -display(df) - -# we center it ourselves -X_centered = X - X.mean(axis=0) -# Then check the difference between pandas and our own set up -print(X_centered-df) -#Now we do an SVD -U, s, V = np.linalg.svd(X_centered) -c1 = V.T[:, 0] -c2 = V.T[:, 1] -W2 = V.T[:, :2] -X2D = X_centered.dot(W2) -print(X2D) -!ec - -PCA assumes that the dataset is centered around the origin. Scikit-Learn’s PCA classes take care of centering -the data for you. However, if you implement PCA yourself (as in the preceding example), or if you use other libraries, don’t -forget to center the data first. - -Once you have identified all the principal components, you can reduce the dimensionality of the dataset -down to $d$ dimensions by projecting it onto the hyperplane defined by the first $d$ principal components. -Selecting this hyperplane ensures that the projection will preserve as much variance as possible. -!bc pycod -W2 = V.T[:, :2] -X2D = X_centered.dot(W2) -!ec - -!split -===== PCA and scikit-learn ===== - -Scikit-Learn’s PCA class implements PCA using SVD decomposition just like we did before. The -following code applies PCA to reduce the dimensionality of the dataset down to two dimensions (note -that it automatically takes care of centering the data): -!bc pycod -#thereafter we do a PCA with Scikit-learn -from sklearn.decomposition import PCA -pca = PCA(n_components = 2) -X2D = pca.fit_transform(X) -print(X2D) -!ec -After fitting the PCA transformer to the dataset, you can access the principal components using the -components variable (note that it contains the PCs as horizontal vectors, so, for example, the first -principal component is equal to -!bc pycod -pca.components_.T[:, 0]. -!ec -Another very useful piece of information is the explained variance ratio of each principal component, -available via the $explained\_variance\_ratio$ variable. It indicates the proportion of the dataset’s -variance that lies along the axis of each principal component. - -!split -===== Back to the Cancer Data ===== -We can now repeat the above but applied to real data, in this case our breast cancer data. -Here we compute performance scores on the training data using logistic regression. -!bc pycod -import matplotlib.pyplot as plt -import numpy as np -from sklearn.model_selection import train_test_split -from sklearn.datasets import load_breast_cancer -from sklearn.linear_model import LogisticRegression -cancer = load_breast_cancer() - -X_train, X_test, y_train, y_test = train_test_split(cancer.data,cancer.target,random_state=0) - -logreg = LogisticRegression() -logreg.fit(X_train, y_train) -print("Train set accuracy from Logistic Regression: {:.2f}".format(logreg.score(X_train,y_train))) -# We scale the data -from sklearn.preprocessing import StandardScaler -scaler = StandardScaler() -scaler.fit(X_train) -X_train_scaled = scaler.transform(X_train) -X_test_scaled = scaler.transform(X_test) -# Then perform again a log reg fit -logreg.fit(X_train_scaled, y_train) -print("Train set accuracy scaled data: {:.2f}".format(logreg.score(X_train_scaled,y_train))) -#thereafter we do a PCA with Scikit-learn -from sklearn.decomposition import PCA -pca = PCA(n_components = 2) -X2D_train = pca.fit_transform(X_train_scaled) -# and finally compute the log reg fit and the score on the training data -logreg.fit(X2D_train,y_train) -print("Train set accuracy scaled and PCA data: {:.2f}".format(logreg.score(X2D_train,y_train))) - -!ec - -We see that our training data after the PCA decomposition has a performance similar to the non-scaled data. - -!split -===== More on the PCA ===== - -Instead of arbitrarily choosing the number of dimensions to reduce down to, it is generally preferable to -choose the number of dimensions that add up to a sufficiently large portion of the variance (e.g., 95%). -Unless, of course, you are reducing dimensionality for data visualization — in that case you will -generally want to reduce the dimensionality down to 2 or 3. -The following code computes PCA without reducing dimensionality, then computes the minimum number -of dimensions required to preserve 95% of the training set’s variance: -!bc pycod -pca = PCA() -pca.fit(X) -cumsum = np.cumsum(pca.explained_variance_ratio_) -d = np.argmax(cumsum >= 0.95) + 1 -!ec -You could then set $n\_components=d$ and run PCA again. However, there is a much better option: instead -of specifying the number of principal components you want to preserve, you can set $n\_components$ to be -a float between 0.0 and 1.0, indicating the ratio of variance you wish to preserve: -!bc pycod -pca = PCA(n_components=0.95) -X_reduced = pca.fit_transform(X) -!ec - -!split -===== Incremental PCA ===== - -One problem with the preceding implementation of PCA is that it requires the whole training set to fit in -memory in order for the SVD algorithm to run. Fortunately, Incremental PCA (IPCA) algorithms have -been developed: you can split the training set into mini-batches and feed an IPCA algorithm one minibatch -at a time. This is useful for large training sets, and also to apply PCA online (i.e., on the fly, as new -instances arrive). - -!split -===== Randomized PCA ===== - -Scikit-Learn offers yet another option to perform PCA, called Randomized PCA. This is a stochastic -algorithm that quickly finds an approximation of the first d principal components. Its computational -complexity is $O(m \times d^2)+O(d^3)$, instead of $O(m \times n^2) + O(n^3)$, so it is dramatically faster than the -previous algorithms when $d$ is much smaller than $n$. - - - - -!split -===== Kernel PCA ===== -!bblock - -The kernel trick is a mathematical technique that implicitly maps instances into a -very high-dimensional space (called the feature space), enabling nonlinear classification and regression -with Support Vector Machines. Recall that a linear decision boundary in the high-dimensional feature -space corresponds to a complex nonlinear decision boundary in the original space. -It turns out that the same trick can be applied to PCA, making it possible to perform complex nonlinear -projections for dimensionality reduction. This is called Kernel PCA (kPCA). It is often good at -preserving clusters of instances after projection, or sometimes even unrolling datasets that lie close to a -twisted manifold. -For example, the following code uses Scikit-Learn’s KernelPCA class to perform kPCA with an -!bc pycod -from sklearn.decomposition import KernelPCA -rbf_pca = KernelPCA(n_components = 2, kernel="rbf", gamma=0.04) -X_reduced = rbf_pca.fit_transform(X) -!ec - -!eblock - - -!split -===== LLE ===== - -Locally Linear Embedding (LLE) is another very powerful nonlinear dimensionality reduction -(NLDR) technique. It is a Manifold Learning technique that does not rely on projections like the previous -algorithms. In a nutshell, LLE works by first measuring how each training instance linearly relates to its -closest neighbors (c.n.), and then looking for a low-dimensional representation of the training set where -these local relationships are best preserved (more details shortly). - - - -!split -===== Other techniques ===== - - -There are many other dimensionality reduction techniques, several of which are available in Scikit-Learn. - -Here are some of the most popular: -* _Multidimensional Scaling (MDS)_ reduces dimensionality while trying to preserve the distances between the instances. -* _Isomap_ creates a graph by connecting each instance to its nearest neighbors, then reduces dimensionality while trying to preserve the geodesic distances between the instances. -* _t-Distributed Stochastic Neighbor Embedding_ (t-SNE) reduces dimensionality while trying to keep similar instances close and dissimilar instances apart. It is mostly used for visualization, in particular to visualize clusters of instances in high-dimensional space (e.g., to visualize the MNIST images in 2D). -* Linear Discriminant Analysis (LDA) is actually a classification algorithm, but during training it learns the most discriminative axes between the classes, and these axes can then be used to define a hyperplane onto which to project the data. The benefit is that the projection will keep classes as far apart as possible, so LDA is a good technique to reduce dimensionality before running another classification algorithm such as a Support Vector Machine (SVM) classifier discussed in the SVM lectures. - - -!split -===== Differential equations ===== - -The Universal Approximation Theorem states that a neural network can -approximate any function at a single hidden layer along with one input -and output layer to any given precision. Having this in mind, we will -look closer at whether a neural network manages to solve for a -function in an equation. - - -!split -===== Description of the equation to solve for ===== -A differential equation is a equation where the solution is a function. -The equation describes how the derivatives of the function behaves in a given domain along with some conditions. - -Given a differential equation, it is desirable to know how to -reformulate it into an equation a neural network can solve. Having -decided on which activation functions each layer should use, along -with the number of hidden layers and neurons within each layer, the -changeable parameters of a neural network are the weights and biases -for each neuron in every layer in the net. If a differential equation -is reformulated into an equation where minimization of some parameters -must be done, a neural net could possibly solve this equation. - -A trial solution might be tricky to find in general. Due to the -Universal Approximation Theorem, one could hope that outcome of the -deep neural net might solve a given differential equation, even though -it is used in a simple trial solution. Let us try this idea on some -well-known ordinary differential equations and thereafter try to solve -for functions defined by two variables, giving partial differential -equations. - -!split -===== Ordinary Differential Equations ===== - -An ordinary differential equation (ODE) is an equation involving functions having one variable. - -In general, an ordinary differential equation looks like - -!bt -\begin{equation} \label{ode} -f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right) = 0 -\end{equation} -!et - -where $g(x)$ is the function to find, and $g^{(n)}(x)$ is the $n$-th derivative of $g(x)$. - -The $f\left(x, g(x), g'(x), g''(x), \, \dots \, , g^{(n)}(x)\right)$ is just a way to write that there is an expression involving $x$ and $g(x), \ g'(x), \ g''(x), \, \dots \, , \text{ and } g^{(n)}(x)$ on the left side of the equality sign in (ref{ode}). -The highest order of derivative, that is the value of $n$, determines to the order of the equation. -The equation is referred to as a $n$-th order ODE. -Along with (ref{ode}), some additional conditions of the function $g(x)$ are typically given -for the solution to be unique. - -!split -===== The trial solution ===== - -Let the trial solution $g_t(x)$ be - -!bt -\begin{equation} - g_t(x) = h_1(x) + h_2(x,N(x,P)) -\end{equation} -!et - -where $h_1(x)$ is a function that makes $g_t(x)$ satisfy a given set of conditions, $N(x,P)$ a neural network with weights and biases described by $P$ and $h_2(x, N(x,P))$ some expression involving the neural network. -The role of the function $h_2(x, N(x,P))$, is to ensure that the output from $N(x,P)$ is zero when $g_t(x)$ is evaluated at the values of $x$ where the given conditions must be satisfied. -The function $h_1(x)$ should alone make $g_t(x)$ satisfy the conditions. - -But what about the network $N(x,P)$? -As described previously, an optimization method could be used to minimize the parameters of a neural network, that being its weights and biases, through backward propagation. -For the minimization to be defined, we need to have a cost function at hand to minimize. - -It is given that $f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right)$ should be equal to zero in (ref{ode}). -We can choose to consider the mean squared error as the cost function for an input $x$. -Since we are looking at one input, the cost function is just $f$ squared. -The cost function $c\left(x, P \right)$ can therefore be expressed as - -!bt -c\left(x, P\right) = \big(f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right)\big)^2 -!et - -If $N$ inputs are given as a vector $\vec x$ with elements $x_i$ for $i = 1,\dots,N$, -the cost function becomes - -!bt -\begin{equation} \label{cost} - c\left(\vec x, P\right) = \frac{1}{N} \sum_{i=1}^N \big(f\left(x_i, \, g(x_i), \, g'(x_i), \, g''(x_i), \, \dots \, , \, g^{(n)}(x_i)\right)\big)^2 -\end{equation} -!et - -The neural net should then find some parameters $P$ that minimizes the cost function in -(ref{cost}) for a set of $N$ training samples $x_i$. - -!split -===== Minimizing the cost function using gradient descent and automatic differentiation ===== -To perform the minimization using gradient descent, the gradient of $c\left(\vec x, P\right)$ is needed. -It might happen so that finding an analytical expression of the gradient of $c(\vec x, P)$ from (ref{cost}) gets too messy, depending on which cost function one desires to use. - -Luckily, there exists libraries that makes the job for us through automatic differentiation. -Automatic differentiation is a method of finding the derivatives numerically with very high precision. - -In the forthcoming examples presenting possible usages of Autograd and TensorFlow, -it is shown how one could set up a neural network using gradient descent solving a differential -equation. - -!split -===== Example: Exponential decay and setting up the network using Autograd ===== -An exponential decay of a quantity $g(x)$ is described by the equation - -!bt -\begin{equation} \label{solve_expdec} - g'(x) = -\gamma g(x) -\end{equation} -!et - -with $g(0) = g_0$ for some chosen initial value $g_0$. - -The analytical solution of (ref{solve_expdec}) is - -!bt -\begin{equation} - g(x) = g_0 \exp\left(-\gamma x\right) -\end{equation} -!et - -Having an analytical solution at hand, it is possible to use it to compare how well a neural network finds a solution of (ref{solve_expdec}). - -In this example, a neural network will be implemented using Autograd in order to perform backpropagation. - -!split -===== The function to solve for ===== - -The program will use a neural network to solve - -!bt -\begin{equation} \label{solveode} -g'(x) = -\gamma g(x) -\end{equation} -!et - -where $g(0) = g_0$ with $\gamma$ and $g_0$ being some chosen values. - -In this example, $\gamma = 2$ and $g_0 = 10$. - -!split -===== The trial solution ===== -To begin with, a trial solution $g_t(t)$ must be chosen. A general trial solution for ordinary differential equations could be - -!bt -g_t(x, P) = h_1(x) + h_2(x, N(x, P)) -!et - -with $h_1(x)$ ensuring that $g_t(x)$ satisfies some conditions and $h_2(x,N(x, P))$ an expression involving $x$ and the output from the neural network $N(x,P)$ with $P $ being the collection of the weights and biases for each layer. For now, it is assumed that the network consists of one input layer, one hidden layer, and one output layer. - -In this network, there are no weights and bias at the input layer, so $P = \{ P_{\text{hidden}}, P_{\text{output}} \}$. -If there are $N_{\text{hidden} }$ neurons in the hidden layer, then $P_{\text{hidden}}$ is a $N_{\text{hidden} } \times (1 + N_{\text{input}})$ matrix, given that there are $N_{\text{input}}$ neurons in the input layer. - -The first column in $P_{\text{hidden} }$ represents the bias for each neuron in the hidden layer and the second column represents the weights for each neuron in the hidden layer from the input layer. -If there are $N_{\text{output} }$ neurons in the output layer, then $P_{\text{output}} $ is a $N_{\text{output} } \times (1 + N_{\text{hidden} })$ matrix. - -Its first column represents the bias of each neuron and the remaining columns represents the weights to each neuron. - -It is given that $g(0) = g_0$. The trial solution must fulfill this condition to be a proper solution of (ref{solveode}). A possible way to ensure that $g_t(0, P) = g_0$, is to let $F(N(x,P)) = x \cdot N(x,P)$ and $A(x) = g_0$. This gives the following trial solution: - -!bt -\begin{equation} \label{trial} -g_t(x, P) = g_0 + x \cdot N(x, P) -\end{equation} -!et - -!split -===== Reformulating the problem ===== -We wish that our neural network manages to minimize a given cost function. - -A reformulation of out equation, (ref{solveode}), must therefore be done, -such that it describes the problem a neural network can solve for. - -The neural network must find the set of weights and biases $P$ such that the trial solution in (ref{trial}) satisfies (ref{solveode}). - -The trial solution - -!bt -g_t(x, P) = g_0 + x \cdot N(x, P) -!et - -has been chosen such that it already solves the condition $g(0) = g_0$. What remains, is to find $P$ such that - -!bt -\begin{equation} \label{nnmin} -g_t'(x, P) = - \gamma g_t(x, P) -\end{equation} -!et - -is fulfilled as *best as possible*. - -The left hand side and right hand side of (ref{nnmin}) must be computed separately, and then the neural network must choose weights and biases, contained in $P$, such that the sides are equal as best as possible. -This means that the absolute or squared difference between the sides must be as close to zero, ideally equal to zero. -In this case, the difference squared shows to be an appropriate measurement of how erroneous the trial solution is with respect to $P$ of the neural network. - -This gives the following cost function our neural network must solve for: - -!bt -\min_{P}\Big\{ \big(g_t'(x, P) - ( -\gamma g_t(x, P) \big)^2 \Big\} -!et - -(the notation $\min_{P}\{ f(x, P) \}$ means that we desire to find $P$ that yields the minimum of $f(x, P)$) - -or, in terms of weights and biases for the hidden and output layer in our network: - -!bt -\min_{P_{\text{hidden} }, \ P_{\text{output} }}\Big\{ \big(g_t'(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) - ( -\gamma g_t(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) \big)^2 \Big\} -!et - -for an input value $x$. - -If the neural network evaluates $g_t(x, P)$ at more values for $x$, say $N$ values $x_i$ for $i = 1, \dots, N$, then the *total* error to minimize becomes - -!bt -\begin{equation} \label{min} -\min_{P}\Big\{\frac{1}{N} \sum_{i=1}^N \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 \Big\} -\end{equation} -!et - -Letting $\vec x$ be a vector with elements $x_i$ and $c(\vec x, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2$ denote the cost function, the minimization problem that our network must solve, becomes - -!bt -\min_{P} c(\vec x, P) -!et - -In terms of $P_{\text{hidden} }$ and $P_{\text{output} }$, this could also be expressed as - -$$ -\min_{P_{\text{hidden} }, \ P_{\text{output} }} c(\vec x, \{P_{\text{hidden} }, P_{\text{output} }\}) -$$ - -!split -===== A possible implementation of a neural network using Autograd ===== - -For simplicity, it is assumed that the input is an array $\vec x = (x_1, \dots, x_N)$ with $N$ elements. It is at these points the neural network should find $P$ such that it fulfills (ref{min}). - -First, the neural network must feed forward the inputs. -This means that $\vec x$ must be passed through an input layer, a hidden layer and a output layer. The input layer in this case, does not need to process the data any further. -The input layer will consist of $N_{\text{input} }$ neurons, passing its element to each neuron in the hidden layer. The number of neurons in the hidden layer will be $N_{\text{hidden} }$. - -For the $i$-th in the hidden layer with weight $w_i^{\text{hidden} }$ and bias $b_i^{\text{hidden} }$, the weighting from the $j$-th neuron at the input layer is: - -!bt -\begin{aligned} -z_{i,j}^{\text{hidden}} &= b_i^{\text{hidden}} + w_i^{\text{hidden}}x_j \\ -&= -\begin{pmatrix} -b_i^{\text{hidden}} & w_i^{\text{hidden}} -\end{pmatrix} -\begin{pmatrix} -1 \\ -x_j -\end{pmatrix} -\end{aligned} -!et - -The result after weighting the inputs at the $i$-th hidden neuron can be written as a vector: - -!bt -\begin{aligned} -\vec{z}_{i}^{\text{hidden}} &= \Big( b_i^{\text{hidden}} + w_i^{\text{hidden}}x_1 , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_2, \ \dots \, , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_N\Big) \\ -&= -\begin{pmatrix} - b_i^{\text{hidden}} & w_i^{\text{hidden}} -\end{pmatrix} -\begin{pmatrix} -1 & 1 & \dots & 1 \\ -x_1 & x_2 & \dots & x_N -\end{pmatrix} \\ -&= \vec{p}_{i, \text{hidden}}^T X -\end{aligned} -!et - -The vector $\vec{p}_{i, \text{hidden}}^T$ constitutes each row in $P_{\text{hidden} }$, which contains the weights for the neural network to minimize according to (ref{min}). - -After having found $\vec{z}_{i}^{\text{hidden}} $ for every $i$-th neuron within the hidden layer, the vector will be sent to an activation function $a_i(\vec{z})$. - -In this example, the sigmoid function has been chosen to be the activation function for each hidden neuron: - -!bt -f(z) = \frac{1}{1 + \exp{(-z)}} -!et - -It is possible to use other activations functions for the hidden layer also. - -The output $\vec{x}_i^{\text{hidden} }$from each $i$-th hidden neuron is: - -$$ -\vec{x}_i^{\text{hidden} } = f\big( \vec{z}_{i}^{\text{hidden}} \big) -$$ - -The outputs $\vec{x}_i^{\text{hidden} } $ are then sent to the output layer. - -The output layer consists of one neuron in this case, and combines the output from each of the neurons in the hidden layers. The output layer combines the results from the hidden layer using some weights $ w_i^{\text{output}}$ and biases $b_i^{\text{output}}$. In this case, it is assumes that the number of neurons in the output layer is one. - -The procedure of weighting the output neuron $j$ in the hidden layer to the $i$-th neuron in the output layer is similar as for the hidden layer described previously. - -!bt -\begin{aligned} -z_{1,j}^{\text{output}} & = -\begin{pmatrix} -b_1^{\text{output}} & \vec{w}_1^{\text{output}} -\end{pmatrix} -\begin{pmatrix} -1 \\ -\vec{x}_j^{\text{hidden}} -\end{pmatrix} -\end{aligned} -!et - -Expressing $z_{1,j}^{\text{output}}$ as a vector gives the following way of weighting the inputs from the hidden layer: - -!bt -\vec{z}_{1}^{\text{output}} = -\begin{pmatrix} -b_1^{\text{output}} & \vec{w}_1^{\text{output}} -\end{pmatrix} -\begin{pmatrix} -1 & 1 & \dots & 1 \\ -\vec{x}_1^{\text{hidden}} & \vec{x}_2^{\text{hidden}} & \dots & \vec{x}_N^{\text{hidden}} -\end{pmatrix} -!et - -In this case we seek a continuous range of values since we are approximating a function. This means that after computing $\vec{z}_{1}^{\text{output}}$ the neural network has finished its feed forward step, and $\vec{z}_{1}^{\text{output}}$ is the final output of the network. - -!split -===== Backpropagation using Autograd ===== -The next step is to decide how the parameters should be changed such that they minimize the cost function. - -The chosen cost function for this problem is - -!bt -c(\vec x, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 -!et - -In order to minimize the cost function, an optimization method must be chosen. - -Here, gradient descent with a constant step size has been chosen. - -!split -===== Gradient descent ===== -The idea of the gradient descent algorithm is to update parameters in direction where the cost function decreases goes to a minimum. - -In general, the update of some parameters $\vec \omega$ given a cost function defined by some weights $\vec \omega$, $c(\vec x, \vec \omega)$, goes as follows: - -!bt -\vec \omega_{\text{new} } = \vec \omega - \lambda \nabla_{\vec \omega} c(\vec x, \vec \omega) -!et - -for a number of iterations or until $ \big|\big| \vec \omega_{\text{new} } - \vec \omega \big|\big|$ becomes smaller than some given tolerance. - -The value of $\lambda$ decides how large steps the algorithm must take in the direction of $ \nabla_{\vec \omega} c(\vec x, \vec \omega)$. -The notation $\nabla_{\vec \omega}$ express the gradient with respect to the elements in $\vec \omega$. - -In our case, we have to minimize the cost function $c(\vec x, P)$ with respect to the two sets of weights and biases, that is for the hidden layer $P_{\text{hidden} }$ and for the output layer $P_{\text{output} }$ . - -This means that $P_{\text{hidden} }$ and $P_{\text{output} }$ is updated by - -!bt -\begin{aligned} -P_{\text{hidden},\text{new}} &= P_{\text{hidden}} - \lambda \nabla_{P_{\text{hidden}}} c(\vec x, P) \\ -P_{\text{output},\text{new}} &= P_{\text{output}} - \lambda \nabla_{P_{\text{output}}} c(\vec x, P) -\end{aligned} -!et - -In general, one could risk using a cost function having gradients that are cumbersome to derive analytically. -For our case, the cost functions are just the mean squared error. -One could employ an implementation of the back propagation for this case, but we will emphasis -on how one could use automatic differentiation in order to train the network. - -However, it might be useful to know how automatic differentiation can be used, e.g through Autograd, in order to test an implementation. - -!split -===== The network with one input, hidden, and output layer ===== - -!bc pycod -# Autograd will be used for later, so the numpy wrapper for Autograd must be imported -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -# Assuming one input, hidden, and output layer -def neural_network(params, x): - - # Find the weights (including and biases) for the hidden and output layer. - # Assume that params is a list of parameters for each layer. - # The biases are the first element for each array in params, - # and the weights are the remaning elements in each array in params. - - w_hidden = params[0] - w_output = params[1] - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - ## Hidden layer: - - # Add a row of ones to include bias - x_input = np.concatenate((np.ones((1,num_values)), x_input ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_input) - x_hidden = sigmoid(z_hidden) - - ## Output layer: - - # Include bias: - x_hidden = np.concatenate((np.ones((1,num_values)), x_hidden ), axis = 0) - - z_output = np.matmul(w_output, x_hidden) - x_output = z_output - - return x_output - -# The trial solution using the deep neural network: -def g_trial(x,params, g0 = 10): - return g0 + x*neural_network(params,x) - -# The right side of the ODE: -def g(x, g_trial, gamma = 2): - return -gamma*g_trial - -# The cost function: -def cost_function(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial(x,P) - - # Find the derivative w.r.t x of the neural network - d_net_out = elementwise_grad(neural_network,1)(P,x) - - # Find the derivative w.r.t x of the trial function - d_g_t = elementwise_grad(g_trial,0)(x,P) - - # The right side of the ODE - func = g(x, g_t) - - err_sqr = (d_g_t - func)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum / np.size(err_sqr) - -# Solve the exponential decay ODE using neural network with one input, hidden, and output layer -def solve_ode_neural_network(x, num_neurons_hidden, num_iter, lmb): - ## Set up initial weights and biases - - # For the hidden layer - p0 = npr.randn(num_neurons_hidden, 2 ) - - # For the output layer - p1 = npr.randn(1, num_neurons_hidden + 1 ) # +1 since bias is included - - P = [p0, p1] - - print('Initial cost: %g'%cost_function(P, x)) - - ## Start finding the optimal weights using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_grad = grad(cost_function,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of two arrays; - # one for the gradient w.r.t P_hidden and - # one for the gradient w.r.t P_output - cost_grad = cost_function_grad(P, x) - - P[0] = P[0] - lmb * cost_grad[0] - P[1] = P[1] - lmb * cost_grad[1] - - print('Final cost: %g'%cost_function(P, x)) - - return P - -def g_analytic(x, gamma = 2, g0 = 10): - return g0*np.exp(-gamma*x) - -# Solve the given problem -if __name__ == '__main__': - # Set seed such that the weight are initialized - # with same weights and biases for every run. - npr.seed(15) - - ## Decide the vales of arguments to the function to solve - N = 10 - x = np.linspace(0, 1, N) - - ## Set up the initial parameters - num_hidden_neurons = 10 - num_iter = 10000 - lmb = 0.001 - - # Use the network - P = solve_ode_neural_network(x, num_hidden_neurons, num_iter, lmb) - - # Print the deviation from the trial solution and true solution - res = g_trial(x,P) - res_analytical = g_analytic(x) - - print('Max absolute difference: %g'%np.max(np.abs(res - res_analytical))) - - # Plot the results - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(x, res_analytical) - plt.plot(x, res[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('x') - plt.ylabel('g(x)') - plt.show() -!ec - -!split -===== The network with one input layer, specified number of hidden layers, and one output layer output layer ===== - -It is also possible to extend the construction of our network into a more general one, allowing the network to contain more than one hidden layers. - -The number of neurons within each hidden layer are given as a list of integers in the program below. - -!bc pycod -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -# The neural network with one input layer and one output layer, -# but with number of hidden layers specified by the user. -def deep_neural_network(deep_params, x): - # N_hidden is the number of hidden layers - - N_hidden = np.size(deep_params) - 1 # -1 since params consists of - # parameters to all the hidden - # layers AND the output layer. - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - # Due to multiple hidden layers, define a variable referencing to the - # output of the previous layer: - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output - -# The trial solution using the deep neural network: -def g_trial_deep(x,params, g0 = 10): - return g0 + x*deep_neural_network(params, x) - -# The right side of the ODE: -def g(x, g_trial, gamma = 2): - return -gamma*g_trial - -# The same cost function as before, but calls deep_neural_network instead. -def cost_function_deep(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial_deep(x,P) - - # Find the derivative w.r.t x of the neural network - d_net_out = elementwise_grad(deep_neural_network,1)(P,x) - - # Find the derivative w.r.t x of the trial function - d_g_t = elementwise_grad(g_trial_deep,0)(x,P) - - # The right side of the ODE - func = g(x, g_t) - - err_sqr = (d_g_t - func)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum / np.size(err_sqr) - -# Solve the exponential decay ODE using neural network with one input and one output layer, -# but with specified number of hidden layers from the user. -def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): - # num_hidden_neurons is now a list of number of neurons within each hidden layer - - # The number of elements in the list num_hidden_neurons thus represents - # the number of hidden layers. - - # Find the number of hidden layers: - N_hidden = np.size(num_neurons) - - ## Set up initial weights and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 ) - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: %g'%cost_function_deep(P, x)) - - ## Start finding the optimal weights using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_deep_grad = grad(cost_function_deep,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases - # in the hidden layers and output layers evaluated at x. - cost_deep_grad = cost_function_deep_grad(P, x) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_deep_grad[l] - - print('Final cost: %g'%cost_function_deep(P, x)) - - return P - -def g_analytic(x, gamma = 2, g0 = 10): - return g0*np.exp(-gamma*x) - -# Solve the given problem -if __name__ == '__main__': - npr.seed(15) - - ## Decide the vales of arguments to the function to solve - N = 10 - x = np.linspace(0, 1, N) - - ## Set up the initial parameters - num_hidden_neurons = np.array([10,10]) - num_iter = 10000 - lmb = 0.001 - - P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb) - - res = g_trial_deep(x,P) - res_analytical = g_analytic(x) - - plt.figure(figsize=(10,10)) - - plt.title('Performance of a deep neural network solving an ODE compared to the analytical solution') - plt.plot(x, res_analytical) - plt.plot(x, res[0,:]) - plt.legend(['analytical','dnn']) - plt.ylabel('g(x)') - plt.show() -!ec - -!split -===== Example: Population growth, comparing Autograd, TensorFlow, and Euler's scheme ===== - -A logistic model of population growth assumes that a population converges toward an equilibrium. -The population growth can be modeled by - -!bt -\begin{equation} \label{log} - g'(t) = \alpha g(t)(A - g(t)) -\end{equation} -!et - -where $g(t)$ is the population density at time $t$, $\alpha > 0$ the growth rate and $A > 0$ is the maximum population number in the environment. -Also, at $t = 0$ the population has the size $g(0) = g_0$, where $g_0$ is some chosen constant. - -In this example, similar network as for the exponential decay using Autograd has been used to solve the equation. However, as the implementation might suffer from e.g numerical instability -and high execution time (this might be more apparent in the examples solving PDEs), -a network has been constructed using TensorFlow also. -For comparison, the forward Euler method has been implemented in order to see how the networks performs compared to a numerical scheme. - -!split -===== Setting up the problem ===== - -Here, we will model a population $g(t)$ in an environment having carrying capacity $A$. -The population follows the model - -!bt -\begin{equation} \label{solveode_population} -g'(t) = \alpha g(t)(A - g(t)) -\end{equation} -!et - -where $g(0) = g_0$. - -In this example, we let $\alpha = 2$, $A = 1$, and $g_0 = 1.2$. - -!split -===== The trial solution ===== -We will get a slightly different trial solution, as the boundary conditions are different -compared to the case for exponential decay. - -A possible trial solution satisfying the condition $g(0) = g_0$ could be - -$$ -h_1(t) = g_0 + t \cdot N(t,P) -$$ - -with $N(t,P)$ being the output from the neural network with weights and biases for each layer collected in the set $P$. - -The analytical solution is - -$$ -g(t) = \frac{Ag_0}{g_0 + (A - g_0)\exp(-\alpha A t)} -$$ - -!split -===== The program using Autograd ===== - -The network will be the similar as for the exponential decay example, but with some small modifications for our problem. - -!bc pycod -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -# Function to get the parameters. -# Done such that one can easily change the paramaters after one's liking. -def get_parameters(): - alpha = 2 - A = 1 - g0 = 1.2 - return alpha, A, g0 - -def deep_neural_network(P, x): - # N_hidden is the number of hidden layers - N_hidden = np.size(P) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - # Due to multiple hidden layers, define a variable referencing to the - # output of the previous layer: - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = P[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = P[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output - - -def cost_function_deep(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial_deep(x,P) - - # Find the derivative w.r.t x of the trial function - d_g_t = elementwise_grad(g_trial_deep,0)(x,P) - - # The right side of the ODE - func = f(x, g_t) - - err_sqr = (d_g_t - func)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum / np.size(err_sqr) - -# The right side of the ODE: -def f(x, g_trial): - alpha,A, g0 = get_parameters() - return alpha*g_trial*(A - g_trial) - -# The trial solution using the deep neural network: -def g_trial_deep(x, params): - alpha,A, g0 = get_parameters() - return g0 + x*deep_neural_network(params,x) - -# The analytical solution: -def g_analytic(t): - alpha,A, g0 = get_parameters() - return A*g0/(g0 + (A - g0)*np.exp(-alpha*A*t)) - -def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): - # num_hidden_neurons is now a list of number of neurons within each hidden layer - - # Find the number of hidden layers: - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 ) - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: %g'%cost_function_deep(P, x)) - - ## Start finding the optimal weigths using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_deep_grad = grad(cost_function_deep,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases - # in the hidden layers and output layers evaluated at x. - cost_deep_grad = cost_function_deep_grad(P, x) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_deep_grad[l] - - print('Final cost: %g'%cost_function_deep(P, x)) - - return P - -if __name__ == '__main__': - npr.seed(4155) - - ## Decide the vales of arguments to the function to solve - Nt = 10 - T = 1 - t = np.linspace(0,T, Nt) - - ## Set up the initial parameters - num_hidden_neurons = [100, 50, 25] - num_iter = 1000 - lmb = 1e-3 - - P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb) - - g_dnn_ag = g_trial_deep(t,P) - g_analytical = g_analytic(t) - - # Find the maximum absolute difference between the solutons: - diff_ag = np.max(np.abs(g_dnn_ag - g_analytical)) - print("The max absolute difference between the solutions is: %g"%diff_ag) - - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(t, g_analytical) - plt.plot(t, g_dnn_ag[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('t') - plt.ylabel('g(t)') - - plt.show() -!ec - -!split -===== Using forward Euler to solve the ODE ===== - -A straight-forward way of solving an ODE numerically, is to use Euler's method. - -Euler's method uses Taylor series to approximate the value at a function $f$ at a step $\Delta x$ from $x$: - -$$ -f(x + \Delta x) \approx f(x) + \Delta x f'(x) -$$ - -In our case, using Euler's method to approximate the value of $g$ at a step $\Delta t$ from $t$ yields - -!bt -\begin{aligned} - g(t + \Delta t) &\approx g(t) + \Delta t g'(t) \\ - &= g(t) + \Delta t \big(\alpha g(t)(A - g(t))\big) -\end{aligned} -!et -along with the condition that $g(0) = g_0$. - -Let $t_i = i \cdot \Delta t$ where $\Delta t = \frac{T}{N_t-1}$ where $T$ is the final time our solver must solve for and $N_t$ the number of values for $t \in [0, T]$ for $i = 0, \dots, N_t-1$. - -For $i \geq 1$, we have that -!bt -\begin{aligned} -t_i &= i\Delta t \\ -&= (i - 1)\Delta t + \Delta t \\ -&= t_{i-1} + \Delta t -\end{aligned} -!et - -Now, if $g_i = g(t_i)$ then - -!bt -\begin{equation} - \begin{aligned} - g_i &= g(t_i) \\ - &= g(t_{i-1} + \Delta t) \\ - &\approx g(t_{i-1}) + \Delta t \big(\alpha g(t_{i-1})(A - g(t_{i-1}))\big) \\ - &= g_{i-1} + \Delta t \big(\alpha g_{i-1}(A - g_{i-1})\big) - \end{aligned} -\end{equation} \label{odenum} -!et -for $i \geq 1$ and $g_0 = g(t_0) = g(0) = g_0$. - -Equation (ref{odenum}) could be implemented in the following way, -extending the program that uses the network using Autograd: - -!bc pycod -# Assume that all function definitions from the example program using Autograd -# are located here. - -if __name__ == '__main__': - npr.seed(4155) - - ## Decide the vales of arguments to the function to solve - Nt = 10 - T = 1 - t = np.linspace(0,T, Nt) - - ## Set up the initial parameters - num_hidden_neurons = [100,50,25] - num_iter = 1000 - lmb = 1e-3 - - P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb) - - g_dnn_ag = g_trial_deep(t,P) - g_analytical = g_analytic(t) - - # Find the maximum absolute difference between the solutons: - diff_ag = np.max(np.abs(g_dnn_ag - g_analytical)) - print("The max absolute difference between the solutions is: %g"%diff_ag) - - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(t, g_analytical) - plt.plot(t, g_dnn_ag[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('t') - plt.ylabel('g(t)') - - ## Find an approximation to the funtion using forward Euler - - alpha, A, g0 = get_parameters() - dt = T/(Nt - 1) - - # Perform forward Euler to solve the ODE - g_euler = np.zeros(Nt) - g_euler[0] = g0 - - for i in range(1,Nt): - g_euler[i] = g_euler[i-1] + dt*(alpha*g_euler[i-1]*(A - g_euler[i-1])) - - # Print the errors done by each method - diff1 = np.max(np.abs(g_euler - g_analytical)) - diff2 = np.max(np.abs(g_dnn_ag[0,:] - g_analytical)) - - print('Max absolute difference between Euler method and analytical: %g'%diff1) - print('Max absolute difference between deep neural network and analytical: %g'%diff2) - - # Plot results - plt.figure(figsize=(10,10)) - - plt.plot(t,g_euler) - plt.plot(t,g_analytical) - plt.plot(t,g_dnn_ag[0,:]) - - plt.legend(['euler','analytical','dnn']) - plt.xlabel('Time t') - plt.ylabel('g(t)') - - plt.show() -!ec - -Running the program gives - -!bc -Max absolute difference between Euler method and analytical: 0.011225 -Max absolute difference between deep neural network and analytical: 0.00424909 -!ec - -!split -===== Using TensorFlow to model logistic population growth ===== - -TensorFlow is a library widely used in the machine learning community. -A neural network can be set up in a flexible manner, where various optimization algorithms are implemented and different types of networks can be used, making it easier to experiment on solving differential equations using neural networks. - -!split -===== The general program flow in TensorFlow ===== - -Usually, a program in TensorFlow is divided into two parts; the *construction phase* and the *execution phase*. -In the construction phase, the computational graph that TensorFlow uses to perform its calculations are set up. -In the execution phase, TensorFlow evaluates any procedure that was defined in the construction phase. - -===== Program flow in TensorFlow - Construction phase ===== - -Here, the architecture for the neural network will be set up, along with the cost function and an optimizer class used during training of the network. -Note that TensorFlow uses a different convention for the weighting done in each neuron in each layer within the network than in the implementation using Autograd. -The matrix-vector multiplication between the input from the previous layer and the weighting at the neuron at current layer in the program using Autograd, is the transpose of the convention used in TensorFlow. But it will not affect that much our construction, as TensorFlow takes care of most of the computations. The only thing we have to be aware of, is how the dimensions are for our inputs. - -===== Program flow in TensorFlow - Execution phase ===== - -The computation graph has been defined, and is ready to be evaluated. -In order to get access to the graph, it has to be initialized and be runned within a Session. - -===== The full program modeling logistic population growth using TensorFlow ===== - -!bc pycod -import tensorflow as tf -import numpy as np -import matplotlib.pyplot as plt - -# Just to reset the graph such that it is possible to rerun this in a -# Jupyter cell without resetting the whole kernel. -tf.reset_default_graph() - -# Set a seed to ensure getting the same results from every run -tf.set_random_seed(4155) - -Nt = 10 -T = 1 -t = np.linspace(0,T, Nt) - -## The construction phase - -# Convert the values the trial solution is evaluated at to a tensor. -t_tf = tf.convert_to_tensor(t.reshape(-1,1),dtype=tf.float64) -zeros = tf.reshape(tf.convert_to_tensor(np.zeros(t.shape)),shape=(-1,1)) - -# Define the parameters of the equation -alpha = tf.constant(2.,dtype=tf.float64) -A = tf.constant(1.,dtype=tf.float64) -g0 = tf.constant(1.2,dtype=tf.float64) - -num_iter = 100000 - -# Define the number of neurons at each hidden layer -num_hidden_neurons = [100,50,25] -num_hidden_layers = np.size(num_hidden_neurons) - -# Construct the network. -# tf.name_scope is used to group each step in the construction, -# just for a more organized visualization in TensorBoard -with tf.name_scope('dnn'): - - # Input layer - previous_layer = t_tf - - # Hidden layers - for l in range(num_hidden_layers): - current_layer = tf.layers.dense(previous_layer, num_hidden_neurons[l], name='hidden%d'%(l+1), activation=tf.nn.sigmoid) - previous_layer = current_layer - - # Output layer - dnn_output = tf.layers.dense(previous_layer, 1, name='output') - -# Define the cost function -with tf.name_scope('cost'): - g_trial = g0 + t_tf*dnn_output - d_g_trial = tf.gradients(g_trial,t_tf) - - func = alpha*g_trial*(A - g_trial) - cost = tf.losses.mean_squared_error(zeros, d_g_trial[0] - func) - - -# Choose the method to minimize the cost function, along with a learning rate -learning_rate = 1e-2 -with tf.name_scope('train'): - optimizer = tf.train.GradientDescentOptimizer(learning_rate) - traning_op = optimizer.minimize(cost) - -# Set up a referance to the result from the neural network: -g_dnn_tf = None - -# Define a node that initializes all of the other nodes in the computational graph -# used by TensorFlow: -init = tf.global_variables_initializer() - -## Execution phase - -# Start a session where the graph defined from the construction phase can be evaluated at: -with tf.Session() as sess: - # Initialize the whole graph - init.run() - - # Evaluate the initial cost: - print('Initial cost: %g'%cost.eval()) - - # The training of the network: - for i in range(num_iter): - sess.run(traning_op) - - # If one desires to see how the cost function behaves for each iteration: - #if i % 1000 == 0: - # print(cost.eval()) - - # Training is done, and we have an approximate solution to the ODE - print('Final cost: %g'%cost.eval()) - - # Store the result - g_dnn_tf = g_trial.eval() - -# Compare with analytical solution -def get_parameters(): - alpha = 2 - A = 1 - g0 = 1.2 - return alpha, A, g0 - -def g_analytic(t): - alpha,A, g0 = get_parameters() - return A*g0/(g0 + (A - g0)*np.exp(-alpha*A*t)) - -g_analytical = g_analytic(t) -diff_tf = g_dnn_tf - g_analytical.reshape(-1,1) - -print('\nMax absolute difference between the analytical solution and solution from TensorFlow DNN: %g'%np.max(np.abs(diff_tf))) - -# Plot the result -plt.figure(figsize=(10,10)) - -plt.title('Numerical solutions of the ODE') - -plt.plot(t, g_dnn_tf) -plt.plot(t, g_analytical) - -plt.legend(['dnn, tensorflow', 'exact']) -plt.xlabel('Time t') -plt.ylabel('g(t)') - -plt.show() - -!ec - -!split -===== Example: Solving the one dimensional Poisson equation using Autograd and TensorFlow ===== - -The Poisson equation for $g(x)$ in one dimension is - -!bt -\begin{equation} \label{poisson} - -g''(x) = f(x) -\end{equation} -!et - -where $f(x)$ is a given function for $x \in (0,1)$. - -The conditions that $g(x)$ is chosen to fulfill, are -!bt -\begin{align*} - g(0) &= 0 \\ - g(1) &= 0 -\end{align*} -!et - -This equation can be solved numerically using programs where e.g Autograd and TensorFlow are used. -The results from the networks can then be compared to the analytical solution. -In addition, it could be interesting to see how a typical method for numerically solving second order ODEs compares to the neural networks. - -There exists many different optimization methods implemented in TensorFlow. -In the examples program using TensorFlow, it could also be of interest to see how -the choice of an optimization method affects our results. -In the "TensorFlow documentation about optimizers":"https://www.tensorflow.org/versions/r1.2/api_guides/python/train#Optimizers", a list over available optimization methods are shown. - -!split -===== The specific equation to solve for ===== - -Here, the function $g(x)$ to solve for follows the equation - -!bt --g''(x) = f(x),\qquad x \in (0,1) -!et - -where $f(x)$ is a given function, along with the chosen conditions - -!bt -\begin{aligned} -g(0) = g(1) = 0 -\end{aligned}\label{cond} -!et - -In this example, we consider the case when $f(x) = (3x + x^2)\exp(x)$. - -For this case, a possible trial solution satisfying the conditions could be - -!bt -g_t(x) = x \cdot (1-x) \cdot N(P,x) -!et - -The analytical solution for this problem is - -!bt -g(x) = x(1 - x)\exp(x) -!et - -!split -===== Solving the equation using Autograd ===== - -!bc pycod -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -def deep_neural_network(deep_params, x): - # N_hidden is the number of hidden layers - N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - # Due to multiple hidden layers, define a variable referencing to the - # output of the previous layer: - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output - -def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): - # num_hidden_neurons is now a list of number of neurons within each hidden layer - - # Find the number of hidden layers: - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 ) - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: %g'%cost_function_deep(P, x)) - - ## Start finding the optimal weigths using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_deep_grad = grad(cost_function_deep,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases - # in the hidden layers and output layers evaluated at x. - cost_deep_grad = cost_function_deep_grad(P, x) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_deep_grad[l] - - print('Final cost: %g'%cost_function_deep(P, x)) - - return P - -## Set up the cost function specified for this Poisson equation: - -# The right side of the ODE -def f(x): - return (3*x + x**2)*np.exp(x) - -def cost_function_deep(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial_deep(x,P) - - # Find the derivative w.r.t x of the trial function - d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P) - - right_side = f(x) - - err_sqr = (-d2_g_t - right_side)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum/np.size(err_sqr) - -# The trial solution: -def g_trial_deep(x,P): - return x*(1-x)*deep_neural_network(P,x) - -# The analytic solution; -def g_analytic(x): - return x*(1-x)*np.exp(x) - -if __name__ == '__main__': - npr.seed(4155) - - ## Decide the vales of arguments to the function to solve - Nx = 10 - x = np.linspace(0,1, Nx) - - ## Set up the initial parameters - num_hidden_neurons = [200,100] - num_iter = 1000 - lmb = 1e-3 - - P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb) - - g_dnn_ag = g_trial_deep(x,P) - g_analytical = g_analytic(x) - - # Find the maximum absolute difference between the solutons: - max_diff = np.max(np.abs(g_dnn_ag - g_analytical)) - print("The max absolute difference between the solutions is: %g"%max_diff) - - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(x, g_analytical) - plt.plot(x, g_dnn_ag[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('x') - plt.ylabel('g(x)') - plt.show() -!ec - -!split -===== Comparing with a numerical scheme ===== - -The Poisson equation is possible to solve using Taylor series to approximate the second derivative. - -Using Taylor series, the second derivative can be expressed as - -$$ -g''(x) = \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2} + E_{\Delta x}(x) -$$ - -where $\Delta x$ is a small step size and $E_{\Delta x}(x)$ being the error term. - -Looking away from the error terms gives an approximation to the second derivative: - -!bt -\begin{equation} \label{approx} -g''(x) \approx \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2} -\end{equation} -!et - -If $x_i = i \Delta x = x_{i-1} + \Delta x$ and $g_i = g(x_i)$ for $i = 1,\dots N_x - 2$ with $N_x$ being the number of values for $x$, (ref{approx}) becomes - -!bt -\begin{aligned} -g''(x_i) &\approx \frac{g(x_i + \Delta x) - 2g(x_i) + g(x_i -\Delta x)}{\Delta x^2} \\ -&= \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2} -\end{aligned} -!et - -Since we know from our problem that - -!bt -\begin{aligned} --g''(x) &= f(x) \\ -&= (3x + x^2)\exp(x) -\end{aligned} -!et - -along with the conditions $g(0) = g(1) = 0$, -the following scheme can be used to find an approximate solution for $g(x)$ numerically: - -!bt -\begin{equation} - \begin{aligned} - -\Big( \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2} \Big) &= f(x_i) \\ - -g_{i+1} + 2g_i - g_{i-1} &= \Delta x^2 f(x_i) - \end{aligned} -\end{equation} \label{odesys} -!et - -for $i = 1, \dots, N_x - 2$ where $g_0 = g_{N_x - 1} = 0$ and $f(x_i) = (3x_i + x_i^2)\exp(x_i)$, which is given for our specific problem. - -The equation can be rewritten into a matrix equation: - -!bt -\begin{aligned} -\begin{pmatrix} -2 & -1 & 0 & \dots & 0 \\ --1 & 2 & -1 & \dots & 0 \\ -\vdots & & \ddots & & \vdots \\ -0 & \dots & -1 & 2 & -1 \\ -0 & \dots & 0 & -1 & 2\\ -\end{pmatrix} -\begin{pmatrix} -g_1 \\ -g_2 \\ -\vdots \\ -g_{N_x - 3} \\ -g_{N_x - 2} -\end{pmatrix} -&= -\Delta x^2 -\begin{pmatrix} -f(x_1) \\ -f(x_2) \\ -\vdots \\ -f(x_{N_x - 3}) \\ -f(x_{N_x - 2}) -\end{pmatrix} \\ -A\vec{g} &= \vec{f} -\end{aligned} -!et - -which makes it possible to solve for the vector $\vec{g}$. - -We can then compare the result from this numerical scheme with the output from our network using Autograd: - -!bc pycod -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -def deep_neural_network(deep_params, x): - # N_hidden is the number of hidden layers - N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - # Due to multiple hidden layers, define a variable referencing to the - # output of the previous layer: - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output - -def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): - # num_hidden_neurons is now a list of number of neurons within each hidden layer - - # Find the number of hidden layers: - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 ) - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: %g'%cost_function_deep(P, x)) - - ## Start finding the optimal weigths using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_deep_grad = grad(cost_function_deep,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases - # in the hidden layers and output layers evaluated at x. - cost_deep_grad = cost_function_deep_grad(P, x) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_deep_grad[l] - - print('Final cost: %g'%cost_function_deep(P, x)) - - return P - -## Set up the cost function specified for this Poisson equation: - -# The right side of the ODE -def f(x): - return (3*x + x**2)*np.exp(x) - -def cost_function_deep(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial_deep(x,P) - - # Find the derivative w.r.t x of the trial function - d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P) - - right_side = f(x) - - err_sqr = (-d2_g_t - right_side)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum/np.size(err_sqr) - -# The trial solution: -def g_trial_deep(x,P): - return x*(1-x)*deep_neural_network(P,x) - -# The analytic solution; -def g_analytic(x): - return x*(1-x)*np.exp(x) - -if __name__ == '__main__': - npr.seed(4155) - - ## Decide the vales of arguments to the function to solve - Nx = 10 - x = np.linspace(0,1, Nx) - - ## Set up the initial parameters - num_hidden_neurons = [200,100] - num_iter = 1000 - lmb = 1e-3 - - P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb) - - g_dnn_ag = g_trial_deep(x,P) - g_analytical = g_analytic(x) - - # Find the maximum absolute difference between the solutons: - - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(x, g_analytical) - plt.plot(x, g_dnn_ag[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('x') - plt.ylabel('g(x)') - - ## Perform the computation using the numerical scheme - - dx = 1/(Nx - 1) - - # Set up the matrix A - A = np.zeros((Nx-2,Nx-2)) - - A[0,0] = 2 - A[0,1] = -1 - - for i in range(1,Nx-3): - A[i,i-1] = -1 - A[i,i] = 2 - A[i,i+1] = -1 - - A[Nx - 3, Nx - 4] = -1 - A[Nx - 3, Nx - 3] = 2 - - # Set up the vector f - f_vec = dx**2 * f(x[1:-1]) - - # Solve the equation - g_res = np.linalg.solve(A,f_vec) - - g_vec = np.zeros(Nx) - g_vec[1:-1] = g_res - - # Print the differences between each method - max_diff1 = np.max(np.abs(g_dnn_ag - g_analytical)) - max_diff2 = np.max(np.abs(g_vec - g_analytical)) - print("The max absolute difference between the analytical solution and DNN Autograd: %g"%max_diff1) - print("The max absolute difference between the analytical solution and numerical scheme: %g"%max_diff2) - - # Plot the results - plt.figure(figsize=(10,10)) - - plt.plot(x,g_vec) - plt.plot(x,g_analytical) - plt.plot(x,g_dnn_ag[0,:]) - - plt.legend(['numerical scheme','analytical','dnn']) - plt.show() - -!ec - -The program prints out: -!bc -The max absolute difference between the analytical solution and DNN Autograd: 0.000464088 -The max absolute difference between the analytical solution and numerical scheme: 0.00266858 -!ec - -!split -===== Using gradient descent in TensorFlow to solve Poisson equation ===== -The program follows the similar idea as for the logistic population model. - -What has changed, is what the cost function minimizes and the trial solution. - -!bc pycod -import tensorflow as tf -import numpy as np -import matplotlib.pyplot as plt -## Construction phase - -# Just to reset the graph such that it is possible to rerun this in a -# Jupyter cell without resetting the whole kernel. -tf.reset_default_graph() - -tf.set_random_seed(4155) - -# Convert the values the trial solution is evaluated at to a tensor. -Nx = 10 -x = np.linspace(0,1, Nx) -x_tf = tf.convert_to_tensor(x.reshape(-1,1),dtype=tf.float64) - - -num_iter = 10000 - -# Define the number of neurons at each hidden layer -num_hidden_neurons = [20,10] -num_hidden_layers = np.size(num_hidden_neurons) - -# Construct the network. -# tf.name_scope is used to group each step in the construction, -# just for a more organized visualization in TensorBoard -with tf.name_scope('dnn'): - - # Input layer - previous_layer = x_tf - - # Hidden layers - for l in range(num_hidden_layers): - current_layer = tf.layers.dense(previous_layer, num_hidden_neurons[l], name='hidden%d'%(l+1), activation=tf.nn.sigmoid) - previous_layer = current_layer - - # Output layer - dnn_output = tf.layers.dense(previous_layer, 1, name='output') - -# Define the cost function -with tf.name_scope('cost'): - g_trial = x_tf*(1-x_tf)*dnn_output - d_g_trial = tf.gradients(g_trial,x_tf) - d2_g_trial = tf.gradients(d_g_trial,x_tf) - - right_side = (3*x_tf + x_tf**2)*tf.exp(x_tf) - - err = tf.square( -d2_g_trial[0] - right_side) - cost = tf.reduce_sum(err, name = 'cost') - -# Choose the method to minimize the cost function, along with a learning rate -learning_rate = 1e-2 -with tf.name_scope('train'): - optimizer = tf.train.GradientDescentOptimizer(learning_rate) - traning_op = optimizer.minimize(cost) - -g_dnn_tf = None - -# Define a node that initializes all of the other nodes in the computational graph -# used by TensorFlow: -init = tf.global_variables_initializer() - - -## Execution phase - -# Start a session where the graph defined from the construction phase can be evaluated at: - -with tf.Session() as sess: - # Initialize the whole graph - init.run() - - # Evaluate the initial cost: - print('Initial cost: %g'%cost.eval()) - - # The traning of the network: - for i in range(num_iter): - sess.run(traning_op) - - # Training is done, and we have an approximate solution to the ODE - print('Final cost: %g'%cost.eval()) - - # Store the result - g_dnn_tf = g_trial.eval() - - writer = tf.summary.FileWriter("./output", sess.graph) - writer.close() - -# Evaluate the analytical function to compare with -def g_analytic(x): - return x*(1-x)*np.exp(x) - -g_analytical = g_analytic(x) - -diff_tf = g_dnn_tf - g_analytical.reshape(-1,1) - -print('\nMax absolute difference between the analytical solution and solution from TensorFlow DNN: %g'%np.max(np.abs(diff_tf))) - -# Plot the result -plt.figure(figsize=(10,10)) - -plt.title('Numerical solutions of the ODE') - -plt.plot(x, g_dnn_tf) -plt.plot(x, g_analytical) - -plt.legend(['dnn, tensorflow','exact']) -plt.xlabel('x') -plt.ylabel('g(x)') - -plt.show() - -!ec - -!split -===== Using a different optimization algorithm implemented in TensorFlow to solve Poisson equation ===== - -We can see that the results using GradientDescentOptimizer seems to converge towards the analytical solution. -But there exists many other methods for optimization also, see "the TensorFlow documentation on Optimizers":"https://www.tensorflow.org/versions/r1.2/api_guides/python/train#Optimizers". - -Adam is an optimization algorithm that changes its learning rates accordingly to the function it tries to minimize for every iteration. -The algorithm is described in "this paper":"https://arxiv.org/pdf/1412.6980.pdf". -How much an optimization algorithm has to say for the network to converge, could be interesting to experiment with. -Using the same TensorFlow program as before, the only change to do, is to replace the variable *optimizer*. - -In the program that uses TensorFlow to solve for the Poisson equation, change the line - -!bc -optimizer = tf.train.GradientDescentOptimizer(learning_rate) -!ec - -to - -!bc -optimizer = tf.train.AdamOptimizer(learning_rate) -!ec - - -The program using the Adam optimizer with a different initial learning rate yields indeed an interesting result: -!bc -Max absolute difference between the analytical solution and solution from TensorFlow DNN: 7.11243e-05 -!ec - -!split -===== Partial Differential Equations ===== -A partial differential equation (PDE) has a solution here the function is defined by multiple variables. -The equation may involve all kinds of combinations of which variables the function is differentiated with respect to. - -In general, a partial differential equation for a function $g(x_1,\dots,x_N)$ with $N$ variables may be expressed as - -!bt -\begin{equation} \label{PDE} - f\left(x_1, \, \dots \, , x_N, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1}, \dots , \frac{\partial g(x_1,\dots,x_N) }{\partial x_N}, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(x_1,\dots,x_N) }{\partial x_N^n} \right) = 0 -\end{equation} -!et - -where $f$ is an expression involving all kinds of possible mixed derivatives of $g(x_1,\dots,x_N)$ up to an order $n$. In order for the solution to be unique, some additional conditions must also be given. - -The problem our network must solve for, is similar to the ODE case. -We must have a trial solution $g_t$ at hand. - -For instance, the trial solution could be expressed as -!bt -\begin{align*} - g_t(x_1,\dots,x_N) = h_1(x_1,\dots,x_N) + h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P)) -\end{align*} -!et -where $h_1(x_1,\dots,x_N)$ is a function that ensures $g_t(x_1,\dots,x_N)$ satisfies some given conditions. -The neural network $N(x_1,\dots,x_N,P)$ has weights and biases described by $P$ and $h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P))$ is an expression using the output from the neural network in some way. - -The role of the function $h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P))$, is to ensure that the output of $N(x_1,\dots,x_N,P)$ is zero when $g_t(x_1,\dots,x_N)$ is evaluated at the values of $x_1,\dots,x_N$ where the given conditions must be satisfied. The function $h_1(x_1,\dots,x_N)$ should alone make $g_t(x_1,\dots,x_N)$ satisfy the conditions. - -The network tries then the minimize the cost function following the same ideas as described for the ODE case, but now with more than one variables to consider. -The concept still remains the same; find a set of parameters $P$ such that the expression $f$ in (ref{PDE}) is as close to zero as possible. - -As for the ODE case, the cost function is the mean squared error that the network must try to minimize. The cost function for the network to minimize is - -!bt -\begin{equation*} -c\left(x_1, \dots, x_N, P\right) = \left( f\left(x_1, \, \dots \, , x_N, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1}, \dots , \frac{\partial g(x_1,\dots,x_N) }{\partial x_N}, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(x_1,\dots,x_N) }{\partial x_N^n} \right) \right)^2 -\end{equation*} -!et - -If we let $\vec x = \big( x_1, \dots, x_N \big)$ be an array containing the values for $x_1, \dots, x_N$ respectively, the cost function can be reformulated into the following: -!bt -\begin{equation*} - c\left(\vec{x}, P\right) = f\left( \left( \vec{x}, \frac{\partial g(\vec x) }{\partial x_1}, \dots , \frac{\partial g(\vec x) }{\partial x_N}, \frac{\partial g(\vec x) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(\vec x) }{\partial x_N^n} \right) \right)^2 -\end{equation*} -!et - -If we also have $M$ different sets of values for $x_1, \dots, x_N$, that is $\vec{x}_i = \big(x_1^{(i)}, \dots, x_N^{(i)}\big)$ for $i = 1,\dots,M$ being the rows in matrix $X$, the cost function can be generalized into -!bt -\begin{equation*} -c\left(X, P \right) = \sum_{i=1}^M f\left( \left( \vec{x}_i, \frac{\partial g(\vec{x}_i) }{\partial x_1}, \dots , \frac{\partial g(\vec{x}_i) }{\partial x_N}, \frac{\partial g(\vec{x}_i) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(\vec{x}_i) }{\partial x_N^n} \right) \right)^2 -\end{equation*} -!et - -!split -===== Example: The diffusion equation ===== - -In one spatial dimension, the equation reads -!bt -\begin{equation*} - \frac{\partial g(x,t)}{\partial t} = \frac{\partial^2 g(x,t)}{\partial x^2} -\end{equation*} -!et - -where a possible choice of conditions are -!bt -\begin{align*} -g(0,t) &= 0 ,\qquad t \geq 0 \\ -g(1,t) &= 0, \qquad t \geq 0 \\ -g(x,0) &= u(x),\qquad x\in [0,1] -\end{align*} -!et -with $u(x)$ being some given function. - -!split -===== Defining the problem ===== - -For this case, we want to find $g(x,t)$ such that - -!bt -\begin{equation} - \frac{\partial g(x,t)}{\partial t} = \frac{\partial^2 g(x,t)}{\partial x^2} -\end{equation} \label{diffonedim} -!et - -and - -!bt -\begin{align*} -g(0,t) &= 0 ,\qquad t \geq 0 \\ -g(1,t) &= 0, \qquad t \geq 0 \\ -g(x,0) &= u(x),\qquad x\in [0,1] -\end{align*} -!et -with $u(x) = \sin(\pi x)$. - -First, let us set up the deep neural network. -The deep neural network will follow the same structure as discussed in the examples solving the ODEs. -First, we will look into how Autograd could be used in a network tailored to solve for bivariate functions. - - - -!split -===== Setting up the network using Autograd ===== - -The only change to do here, is to extend our network such that functions of multiple parameters are correctly handled. -In this case we have two variables in our function to solve for, that is time $t$ and position $x$. -The variables will be represented by a one-dimensional array in the program. -The program will evaluate the network at each possible pair $(x,t)$, given an array for the desired $x$-values and $t$-values to approximate the solution at. - -!bc pycod -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -def deep_neural_network(deep_params, x): - # x is now a point and a 1D numpy array; make it a column vector - num_coordinates = np.size(x,0) - x = x.reshape(num_coordinates,-1) - - num_points = np.size(x,1) - - # N_hidden is the number of hidden layers - N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assume that the input layer does nothing to the input x - x_input = x - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output[0][0] -!ec - -!split -===== Setting up the network using Autograd; The trial solution ===== -The cost function must then iterate through the given arrays containing values for $x$ and $t$, defines a point $(x,t)$ the deep neural network and the trial solution is evaluated at, and then finds the Jacobian of the trial solution. - -A possible trial solution for this PDE is - -$$ -g_t(x,t) = h_1(x,t) + x(1-x)tN(x,t,P) -$$ - -with $A(x,t)$ being a function ensuring that $g_t(x,t)$ satisfies our given conditions, and $N(x,t,P)$ being the output from the deep neural network using weights and biases for each layer from $P$. - -To fulfill the conditions, $A(x,t)$ could be: - -$$ -h_1(x,t) = (1-t)\Big(u(x) - \big((1-x)u(0) + x u(1)\big)\Big) = (1-t)u(x) = (1-t)\sin(\pi x) -$$ -since $(0) = u(1) = 0$ and $u(x) = \sin(\pi x)$. - -The Jacobian is used because the program must find the derivative of the trial solution with respect to $x$ and $t$. - -This gives the necessity of computing the Jacobian matrix, as we want to evaluate the gradient with respect to $x$ and $t$ (note that the Jacobian of a scalar-valued multivariate function is simply its gradient). - -In Autograd, the differentiation is by default done with respect to the first input argument of your Python function. Since the points is an array representing $x$ and $t$, the Jacobian is calculated using the values of $x$ and $t$. - -To find the second derivative with respect to $x$ and $t$, the Jacobian can be found for the second time. The result is a Hessian matrix, which is the matrix containing all the possible second order mixed derivatives of $g(x,t)$. - -!bc pycod -# Set up the trial function: -def u(x): - return np.sin(np.pi*x) - -def g_trial(point,P): - x,t = point - return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point) - -# The right side of the ODE: -def f(point): - return 0. - -# The cost function: -def cost_function(P, x, t): - cost_sum = 0 - - g_t_jacobian_func = jacobian(g_trial) - g_t_hessian_func = hessian(g_trial) - - for x_ in x: - for t_ in t: - point = np.array([x_,t_]) - - g_t = g_trial(point,P) - g_t_jacobian = g_t_jacobian_func(point,P) - g_t_hessian = g_t_hessian_func(point,P) - - g_t_dt = g_t_jacobian[1] - g_t_d2x = g_t_hessian[0][0] - - func = f(point) - - err_sqr = ( (g_t_dt - g_t_d2x) - func)**2 - cost_sum += err_sqr - - return cost_sum -!ec - -!split -===== Setting up the network using Autograd; The full program ===== -Having set up the network, along with the trial solution and cost function, we can now see how the deep neural network performs by comparing the results to the analytical solution. - -The analytical solution of our problem is - -$$ -g(x,t) = \exp(-\pi^2 t)\sin(\pi x) -$$ - -A possible way to implement a neural network solving the PDE, is given below. -Be aware, though, that it is fairly slow for the parameters used. -A better result is possible, but requires more iterations, and thus longer time to complete. - -Using only 20 neurons in one hidden layer, the program managed to make the trial solution have the maximum absolute error of 0.0075. The execution time, however, was approximately one day and 14 hours on a computer having Intel i7-7560U 2.4 GHz CPU. - -Indeed, the program below is not optimal in its implementation, but rather serves as an example on how to implement and use a neural network to solve a PDE. -Using TensorFlow in the next example sovling the wave equation, has a much better execution time. - -!bc pycod -import autograd.numpy as np -from autograd import jacobian,hessian,grad -import autograd.numpy.random as npr -from matplotlib import cm -from matplotlib import pyplot as plt -from mpl_toolkits.mplot3d import axes3d - -## Set up the network - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -def deep_neural_network(deep_params, x): - # x is now a point and a 1D numpy array; make it a column vector - num_coordinates = np.size(x,0) - x = x.reshape(num_coordinates,-1) - - num_points = np.size(x,1) - - # N_hidden is the number of hidden layers - N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assume that the input layer does nothing to the input x - x_input = x - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output[0][0] - -## Define the trial solution and cost function -def u(x): - return np.sin(np.pi*x) - -def g_trial(point,P): - x,t = point - return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point) - -# The right side of the ODE: -def f(point): - return 0. - -# The cost function: -def cost_function(P, x, t): - cost_sum = 0 - - g_t_jacobian_func = jacobian(g_trial) - g_t_hessian_func = hessian(g_trial) - - for x_ in x: - for t_ in t: - point = np.array([x_,t_]) - - g_t = g_trial(point,P) - g_t_jacobian = g_t_jacobian_func(point,P) - g_t_hessian = g_t_hessian_func(point,P) - - g_t_dt = g_t_jacobian[1] - g_t_d2x = g_t_hessian[0][0] - - func = f(point) - - err_sqr = ( (g_t_dt - g_t_d2x) - func)**2 - cost_sum += err_sqr - - return cost_sum /( np.size(x)*np.size(t) ) - -## For comparison, define the analytical solution -def g_analytic(point): - x,t = point - return np.exp(-np.pi**2*t)*np.sin(np.pi*x) - -## Set up a function for training the network to solve for the equation -def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb): - ## Set up initial weigths and biases - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: ',cost_function(P, x, t)) - - cost_function_grad = grad(cost_function,0) - - # Let the update be done num_iter times - for i in range(num_iter): - cost_grad = cost_function_grad(P, x , t) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_grad[l] - - print('Final cost: ',cost_function(P, x, t)) - - return P - -if __name__ == '__main__': - ### Use the neural network: - npr.seed(15) - - ## Decide the vales of arguments to the function to solve - Nx = 10; Nt = 10 - x = np.linspace(0, 1, Nx) - t = np.linspace(0,1,Nt) - - ## Set up the parameters for the network - num_hidden_neurons = [100, 25] - num_iter = 250 - lmb = 0.01 - - P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb) - - ## Store the results - g_dnn_ag = np.zeros((Nx, Nt)) - G_analytical = np.zeros((Nx, Nt)) - for i,x_ in enumerate(x): - for j, t_ in enumerate(t): - point = np.array([x_, t_]) - g_dnn_ag[i,j] = g_trial(point,P) - - G_analytical[i,j] = g_analytic(point) - - # Find the map difference between the analytical and the computed solution - diff_ag = np.abs(g_dnn_ag - G_analytical) - print('Max absolute difference between the analytical solution and the network: %g'%np.max(diff_ag)) - - ## Plot the solutions in two dimensions, that being in position and time - - T,X = np.meshgrid(t,x) - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons)) - s = ax.plot_surface(T,X,g_dnn_ag,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Analytical solution') - s = ax.plot_surface(T,X,G_analytical,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Difference') - s = ax.plot_surface(T,X,diff_ag,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - ## Take some slices of the 3D plots just to see the solutions at particular times - indx1 = 0 - indx2 = int(Nt/2) - indx3 = Nt-1 - - t1 = t[indx1] - t2 = t[indx2] - t3 = t[indx3] - - # Slice the results from the DNN - res1 = g_dnn_ag[:,indx1] - res2 = g_dnn_ag[:,indx2] - res3 = g_dnn_ag[:,indx3] - - # Slice the analytical results - res_analytical1 = G_analytical[:,indx1] - res_analytical2 = G_analytical[:,indx2] - res_analytical3 = G_analytical[:,indx3] - - # Plot the slices - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t1) - plt.plot(x, res1) - plt.plot(x,res_analytical1) - plt.legend(['dnn','analytical']) - - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t2) - plt.plot(x, res2) - plt.plot(x,res_analytical2) - plt.legend(['dnn','analytical']) - - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t3) - plt.plot(x, res3) - plt.plot(x,res_analytical3) - plt.legend(['dnn','analytical']) - - plt.show() -!ec - -!split -===== Example: Solving the wave equation using Autograd and TensorFlow ===== - -The wave equation is -!bt -\begin{equation*} - \frac{\partial^2 g(x,t)}{\partial t^2} = c^2\frac{\partial^2 g(x,t)}{\partial x^2} -\end{equation*} -!et - -with $c$ being the specified wave speed. - -Here, the chosen conditions are -!bt -\begin{align*} - g(0,t) &= 0 \\ - g(1,t) &= 0 \\ - g(x,0) &= u(x) \\ - \frac{\partial g(x,t)}{\partial t} \Big |_{t = 0} &= v(x) -\end{align*} -!et -where $\frac{\partial g(x,t)}{\partial t} \Big |_{t = 0}$ means the derivative of $g(x,t)$ with respect to $t$ is evaluated at $t = 0$, and $u(x)$ and $v(x)$ being given functions. - -!split -===== The problem to solve for ===== - -The wave equation to solve for, is - -!bt -\begin{equation} \label{wave} -\frac{\partial^2 g(x,t)}{\partial t^2} = c^2 \frac{\partial^2 g(x,t)}{\partial x^2} -\end{equation} -!et - -where $c$ is the given wave speed. -The chosen conditions for this equation are - -!bt -\begin{aligned} -g(0,t) &= 0, &t \geq 0 \\ -g(1,t) &= 0, &t \geq 0 \\ -g(x,0) &= u(x), &x\in[0,1] \\ -\frac{\partial g(x,t)}{\partial t}\Big |_{t = 0} &= v(x), &x \in [0,1] -\end{aligned} \label{condwave} -!et - -In this example, let $c = 1$ and $u(x) = \sin(\pi x)$ and $v(x) = -\pi\sin(\pi x)$. - - -!split -===== The trial solution ===== -Setting up the network is done in similar matter as for the example of solving the diffusion equation. -The only things we have to change, is the trial solution such that it satisfies the conditions from (ref{condwave}) and the cost function. - -The trial solution becomes slightly different since we have other conditions than in the example of solving the diffusion equation. Here, a possible trial solution $g_t(x,t)$ is - -$$ -g_t(x,t) = h_1(x,t) + x(1-x)t^2N(x,t,P) -$$ - -where - -$$ -h_1(x,t) = (1-t^2)u(x) + tv(x) -$$ - -Note that this trial solution satisfies the conditions only if $u(0) = v(0) = u(1) = v(1) = 0$, which is the case in this example. - -!split -===== The analytical solution ===== - -The analytical solution for our specific problem, is - -$$ -g(x,t) = \sin(\pi x)\cos(\pi t) - \sin(\pi x)\sin(\pi t) -$$ - -!split -===== Solving the wave equation - the full program using Autograd ===== - -!bc pycod -import autograd.numpy as np -from autograd import hessian,grad -import autograd.numpy.random as npr -from matplotlib import cm -from matplotlib import pyplot as plt -from mpl_toolkits.mplot3d import axes3d - -## Set up the trial function: -def u(x): - return np.sin(np.pi*x) - -def v(x): - return -np.pi*np.sin(np.pi*x) - -def h1(point): - x,t = point - return (1 - t**2)*u(x) + t*v(x) - -def g_trial(point,P): - x,t = point - return h1(point) + x*(1-x)*t**2*deep_neural_network(P,point) - -## Define the cost function -def cost_function(P, x, t): - cost_sum = 0 - - g_t_hessian_func = hessian(g_trial) - - for x_ in x: - for t_ in t: - point = np.array([x_,t_]) - - g_t_hessian = g_t_hessian_func(point,P) - - g_t_d2x = g_t_hessian[0][0] - g_t_d2t = g_t_hessian[1][1] - - err_sqr = ( (g_t_d2t - g_t_d2x) )**2 - cost_sum += err_sqr - - return cost_sum / (np.size(t) * np.size(x)) - -## The neural network -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -def deep_neural_network(deep_params, x): - # x is now a point and a 1D numpy array; make it a column vector - num_coordinates = np.size(x,0) - x = x.reshape(num_coordinates,-1) - - num_points = np.size(x,1) - - # N_hidden is the number of hidden layers - N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assume that the input layer does nothing to the input x - x_input = x - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output[0][0] - -## The analytical solution -def g_analytic(point): - x,t = point - return np.sin(np.pi*x)*np.cos(np.pi*t) - np.sin(np.pi*x)*np.sin(np.pi*t) - -def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb): - ## Set up initial weigths and biases - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: ',cost_function(P, x, t)) - - cost_function_grad = grad(cost_function,0) - - # Let the update be done num_iter times - for i in range(num_iter): - cost_grad = cost_function_grad(P, x , t) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_grad[l] - - - print('Final cost: ',cost_function(P, x, t)) - - return P - -if __name__ == '__main__': - ### Use the neural network: - npr.seed(15) - - ## Decide the vales of arguments to the function to solve - Nx = 10; Nt = 10 - x = np.linspace(0, 1, Nx) - t = np.linspace(0,1,Nt) - - ## Set up the parameters for the network - num_hidden_neurons = [50,20] - num_iter = 1000 - lmb = 0.01 - - P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb) - - ## Store the results - res = np.zeros((Nx, Nt)) - res_analytical = np.zeros((Nx, Nt)) - for i,x_ in enumerate(x): - for j, t_ in enumerate(t): - point = np.array([x_, t_]) - res[i,j] = g_trial(point,P) - - res_analytical[i,j] = g_analytic(point) - - diff = np.abs(res - res_analytical) - print("Max difference between analytical and solution from nn: %g"%np.max(diff)) - - ## Plot the solutions in two dimensions, that being in position and time - - T,X = np.meshgrid(t,x) - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons)) - s = ax.plot_surface(T,X,res,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Analytical solution') - s = ax.plot_surface(T,X,res_analytical,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Difference') - s = ax.plot_surface(T,X,diff,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - ## Take some slices of the 3D plots just to see the solutions at particular times - indx1 = 0 - indx2 = int(Nt/2) - indx3 = Nt-1 - - t1 = t[indx1] - t2 = t[indx2] - t3 = t[indx3] - - # Slice the results from the DNN - res1 = res[:,indx1] - res2 = res[:,indx2] - res3 = res[:,indx3] - - # Slice the analytical results - res_analytical1 = res_analytical[:,indx1] - res_analytical2 = res_analytical[:,indx2] - res_analytical3 = res_analytical[:,indx3] - - # Plot the slices - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t1) - plt.plot(x, res1) - plt.plot(x,res_analytical1) - plt.legend(['dnn','analytical']) - - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t2) - plt.plot(x, res2) - plt.plot(x,res_analytical2) - plt.legend(['dnn','analytical']) - - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t3) - plt.plot(x, res3) - plt.plot(x,res_analytical3) - plt.legend(['dnn','analytical']) - - plt.show() -!ec - -!split -===== Solving the wave equation - the full program using TensorFlow ===== -As the program using Autograd is fairly slow, one could hope that using TensorFlow -could make a naive implementation faster, and more numerically robust. - -In addition, having TensorFlow at hand, it could be easier to experiment with different -optimization algorithms, and other constructions of the network. - -The following program solves the given wave equation much faster, - -!bc pycod -import tensorflow as tf -import numpy as np -from matplotlib import cm -from matplotlib import pyplot as plt -from mpl_toolkits.mplot3d import axes3d - -Nx = 10 -x_np = np.linspace(0,1,Nx) - -Nt = 10 -t_np = np.linspace(0,1,Nt) - -X,T = np.meshgrid(x_np, t_np) - -x = X.ravel() -t = T.ravel() - -## The construction phase - -zeros = tf.reshape(tf.convert_to_tensor(np.zeros(x.shape)),shape=(-1,1)) -x = tf.reshape(tf.convert_to_tensor(x),shape=(-1,1)) -t = tf.reshape(tf.convert_to_tensor(t),shape=(-1,1)) - -points = tf.concat([x,t],1) - -num_iter = 100000 -num_hidden_neurons = [90] - -X = tf.convert_to_tensor(X) -T = tf.convert_to_tensor(T) - - -with tf.variable_scope('dnn'): - num_hidden_layers = np.size(num_hidden_neurons) - - previous_layer = points - - for l in range(num_hidden_layers): - current_layer = tf.layers.dense(previous_layer, num_hidden_neurons[l],activation=tf.nn.sigmoid) - previous_layer = current_layer - - dnn_output = tf.layers.dense(previous_layer, 1) - - -def u(x): - return tf.sin(np.pi*x) - -def v(x): - return -np.pi*tf.sin(np.pi*x) - -with tf.name_scope('loss'): - g_trial = (1 - t**2)*u(x) + t*v(x) + x*(1-x)*t**2*dnn_output - - g_trial_d2t = tf.gradients(tf.gradients(g_trial,t),t) - g_trial_d2x = tf.gradients(tf.gradients(g_trial,x),x) - - loss = tf.losses.mean_squared_error(zeros, g_trial_d2t[0] - g_trial_d2x[0]) - -learning_rate = 0.01 -with tf.name_scope('train'): - optimizer = tf.train.GradientDescentOptimizer(learning_rate) - traning_op = optimizer.minimize(loss) - -init = tf.global_variables_initializer() - -g_analytic = tf.sin(np.pi*x)*tf.cos(np.pi*t) - tf.sin(np.pi*x)*tf.sin(np.pi*t) -g_dnn = None - -## The execution phase -with tf.Session() as sess: - init.run() - for i in range(num_iter): - sess.run(traning_op) - - # If one desires to see how the cost function behaves during training - #if i % 100 == 0: - # print(loss.eval()) - - g_analytic = g_analytic.eval() - g_dnn = g_trial.eval() - - -## Compare with the analutical solution -diff = np.abs(g_analytic - g_dnn) -print('Max absolute difference between analytical solution and TensorFlow DNN = ',np.max(diff)) - -G_analytic = g_analytic.reshape((Nt,Nx)) -G_dnn = g_dnn.reshape((Nt,Nx)) - -diff = np.abs(G_analytic - G_dnn) - -# Plot the results - -X,T = np.meshgrid(x_np, t_np) - -fig = plt.figure(figsize=(10,10)) -ax = fig.gca(projection='3d') -ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons)) -s = ax.plot_surface(X,T,G_dnn,linewidth=0,antialiased=False,cmap=cm.viridis) -ax.set_xlabel('Time $t$') -ax.set_ylabel('Position $x$'); - -fig = plt.figure(figsize=(10,10)) -ax = fig.gca(projection='3d') -ax.set_title('Analytical solution') -s = ax.plot_surface(X,T,G_analytic,linewidth=0,antialiased=False,cmap=cm.viridis) -ax.set_xlabel('Time $t$') -ax.set_ylabel('Position $x$'); - -fig = plt.figure(figsize=(10,10)) -ax = fig.gca(projection='3d') -ax.set_title('Difference') -s = ax.plot_surface(X,T,diff,linewidth=0,antialiased=False,cmap=cm.viridis) -ax.set_xlabel('Time $t$') -ax.set_ylabel('Position $x$'); - -## Take some 3D slices - -indx1 = 0 -indx2 = int(Nt/2) -indx3 = Nt-1 - -t1 = t_np[indx1] -t2 = t_np[indx2] -t3 = t_np[indx3] - -# Slice the results from the DNN -res1 = G_dnn[indx1,:] -res2 = G_dnn[indx2,:] -res3 = G_dnn[indx3,:] - -# Slice the analytical results -res_analytical1 = G_analytic[indx1,:] -res_analytical2 = G_analytic[indx2,:] -res_analytical3 = G_analytic[indx3,:] - -# Plot the slices -plt.figure(figsize=(10,10)) -plt.title("Computed solutions at time = %g"%t1) -plt.plot(x_np, res1) -plt.plot(x_np,res_analytical1) -plt.legend(['dnn','analytical']) - -plt.figure(figsize=(10,10)) -plt.title("Computed solutions at time = %g"%t2) -plt.plot(x_np, res2) -plt.plot(x_np,res_analytical2) -plt.legend(['dnn','analytical']) - -plt.figure(figsize=(10,10)) -plt.title("Computed solutions at time = %g"%t3) -plt.plot(x_np, res3) -plt.plot(x_np,res_analytical3) -plt.legend(['dnn','analytical']) - -plt.show() -!ec - -The program manages to find a solution having max absolute difference to the analytical -at approximately 0.0059, by just using some minutes! -It was found, by some testing, that one hidden layer with 90 neurons actually performed well. - -!split -===== Resources ===== - -o "Artificial neural networks for solving ordinary and partial differential equations by I.E. Lagaris et al":"https://pdfs.semanticscholar.org/d061/df393e0e8fbfd0ea24976458b7d42419040d.pdf" -o "Neural networks for solving differential equations by A. Honchar":"https://becominghuman.ai/neural-networks-for-solving-differential-equations-fa230ac5e04c" -o "Solving differential equations using neural networks by M.M Chiaramonte and M. Kiener":"http://cs229.stanford.edu/proj2013/ChiaramonteKiener-SolvingDifferentialEquationsUsingNeuralNetworks.pdf" -o "Introduction to Partial Differential Equations by A. Tveitio, R. Winther":"https://www.springer.com/us/book/9783540225515" - diff --git a/doc/src/week43/programs/wave.py~ b/doc/src/week43/programs/wave.py~ deleted file mode 100644 index 2b6c2d667..000000000 --- a/doc/src/week43/programs/wave.py~ +++ /dev/null @@ -1,3759 +0,0 @@ -TITLE: Week 43: Solving Differential Equations with Deep Learning and Dimensionality Reduction methods -AUTHOR: Morten Hjorth-Jensen {copyright, 1999-present|CC BY-NC} at Department of Physics, University of Oslo & Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University -DATE: today - -!split - -* Thursday: Wrapping up Recurrent Neural Networks and solving differential equations. -* Friday: Principal Component Analysis and Dimensionality Reduction - -Reading suggestions for both days: "Aurelien Geron's chapters 8 - -!split -===== Recurrent Neural Networks ===== - - -!split -===== Solving ODEs with Deep Learning ===== - - - - - - - -!split -===== Basic ideas of the Principal Component Analysis (PCA) ===== - -The principal component analysis deals with the problem of fitting a -low-dimensional affine subspace $S$ of dimension $d$ much smaller than -the totaldimension $D$ of the problem at hand (our data -set). Mathematically it can be formulated as a statistical problem or -a geometric problem. In our discussion of the theorem for the -classical PCA, we will stay with a statistical approach. This is also -what set the scene historically which for the PCA. - -We have a data set defined by a design/feature matrix $\bm{X}$ (see below for its definition) -* Each data point is determined by $p$ extrinsic (measurement) variables -* We may want to ask the following question: Are there fewer intrinsic variables (say $d << p$) that still approximately describe the data? -* If so, these intrinsic variables may tell us something important and finding these intrinsic variables is what dimension reduction methods do. - - -!split -===== Introducing the Covariance and Correlation functions ===== - -Before we discuss the PCA theorem, we need to remind ourselves about -the definition of the covariance and the correlation function. These are quantities - -Suppose we have defined two vectors -$\hat{x}$ and $\hat{y}$ with $n$ elements each. The covariance matrix $\bm{C}$ is defined as -!bt -\[ -\bm{C}[\bm{x},\bm{y}] = \begin{bmatrix} \mathrm{cov}[\bm{x},\bm{x}] & \mathrm{cov}[\bm{x},\bm{y}] \\ - \mathrm{cov}[\bm{y},\bm{x}] & \mathrm{cov}[\bm{y},\bm{y}] \\ - \end{bmatrix}, -\] -!et -where for example -!bt -\[ -\mathrm{cov}[\bm{x},\bm{y}] =\frac{1}{n} \sum_{i=0}^{n-1}(x_i- \overline{x})(y_i- \overline{y}). -\] -!et -With this definition and recalling that the variance is defined as -!bt -\[ -\mathrm{var}[\bm{x}]=\frac{1}{n} \sum_{i=0}^{n-1}(x_i- \overline{x})^2, -\] -!et -we can rewrite the covariance matrix as -!bt -\[ -\bm{C}[\bm{x},\bm{y}] = \begin{bmatrix} \mathrm{var}[\bm{x}] & \mathrm{cov}[\bm{x},\bm{y}] \\ - \mathrm{cov}[\bm{x},\bm{y}] & \mathrm{var}[\bm{y}] \\ - \end{bmatrix}. -\] -!et - -The covariance takes values between zero and infinity and may thus -lead to problems with loss of numerical precision for particularly -large values. It is common to scale the covariance matrix by -introducing instead the correlation matrix defined via the so-called -correlation function - -!bt -\[ -\mathrm{corr}[\bm{x},\bm{y}]=\frac{\mathrm{cov}[\bm{x},\bm{y}]}{\sqrt{\mathrm{var}[\bm{x}] \mathrm{var}[\bm{y}]}}. -\] -!et - -The correlation function is then given by values $\mathrm{corr}[\bm{x},\bm{y}] -\in [-1,1]$. This avoids eventual problems with too large values. We -can then define the correlation matrix for the two vectors $\bm{x}$ -and $\bm{y}$ as - -!bt -\[ -\bm{K}[\bm{x},\bm{y}] = \begin{bmatrix} 1 & \mathrm{corr}[\bm{x},\bm{y}] \\ - \mathrm{corr}[\bm{y},\bm{x}] & 1 \\ - \end{bmatrix}, -\] -!et - -In the above example this is the function we constructed using _pandas_. - -!split -===== Correlation Function and Design/Feature Matrix ===== - -In our derivation of the various regression algorithms like _Ordinary Least Squares_ or _Ridge regression_ -we defined the design/feature matrix $\bm{X}$ as - -!bt -\[ -\bm{X}=\begin{bmatrix} -x_{0,0} & x_{0,1} & x_{0,2}& \dots & \dots x_{0,p-1}\\ -x_{1,0} & x_{1,1} & x_{1,2}& \dots & \dots x_{1,p-1}\\ -x_{2,0} & x_{2,1} & x_{2,2}& \dots & \dots x_{2,p-1}\\ -\dots & \dots & \dots & \dots \dots & \dots \\ -x_{n-2,0} & x_{n-2,1} & x_{n-2,2}& \dots & \dots x_{n-2,p-1}\\ -x_{n-1,0} & x_{n-1,1} & x_{n-1,2}& \dots & \dots x_{n-1,p-1}\\ -\end{bmatrix}, -\] -!et -with $\bm{X}\in {\mathbb{R}}^{n\times p}$, with the predictors/features $p$ refering to the column numbers and the -entries $n$ being the row elements. -We can rewrite the design/feature matrix in terms of its column vectors as -!bt -\[ -\bm{X}=\begin{bmatrix} \bm{x}_0 & \bm{x}_1 & \bm{x}_2 & \dots & \dots & \bm{x}_{p-1}\end{bmatrix}, -\] -!et -with a given vector -!bt -\[ -\bm{x}_i^T = \begin{bmatrix}x_{0,i} & x_{1,i} & x_{2,i}& \dots & \dots x_{n-1,i}\end{bmatrix}. -\] -!et - -With these definitions, we can now rewrite our $2\times 2$ -correaltion/covariance matrix in terms of a moe general design/feature -matrix $\bm{X}\in {\mathbb{R}}^{n\times p}$. This leads to a $p\times p$ -covariance matrix for the vectors $\bm{x}_i$ with $i=0,1,\dots,p-1$ - -!bt -\[ -\bm{C}[\bm{x}] = \begin{bmatrix} -\mathrm{var}[\bm{x}_0] & \mathrm{cov}[\bm{x}_0,\bm{x}_1] & \mathrm{cov}[\bm{x}_0,\bm{x}_2] & \dots & \dots & \mathrm{cov}[\bm{x}_0,\bm{x}_{p-1}]\\ -\mathrm{cov}[\bm{x}_1,\bm{x}_0] & \mathrm{var}[\bm{x}_1] & \mathrm{cov}[\bm{x}_1,\bm{x}_2] & \dots & \dots & \mathrm{cov}[\bm{x}_1,\bm{x}_{p-1}]\\ -\mathrm{cov}[\bm{x}_2,\bm{x}_0] & \mathrm{cov}[\bm{x}_2,\bm{x}_1] & \mathrm{var}[\bm{x}_2] & \dots & \dots & \mathrm{cov}[\bm{x}_2,\bm{x}_{p-1}]\\ -\dots & \dots & \dots & \dots & \dots & \dots \\ -\dots & \dots & \dots & \dots & \dots & \dots \\ -\mathrm{cov}[\bm{x}_{p-1},\bm{x}_0] & \mathrm{cov}[\bm{x}_{p-1},\bm{x}_1] & \mathrm{cov}[\bm{x}_{p-1},\bm{x}_{2}] & \dots & \dots & \mathrm{var}[\bm{x}_{p-1}]\\ -\end{bmatrix}, -\] -!et -and the correlation matrix -!bt -\[ -\bm{K}[\bm{x}] = \begin{bmatrix} -1 & \mathrm{corr}[\bm{x}_0,\bm{x}_1] & \mathrm{corr}[\bm{x}_0,\bm{x}_2] & \dots & \dots & \mathrm{corr}[\bm{x}_0,\bm{x}_{p-1}]\\ -\mathrm{corr}[\bm{x}_1,\bm{x}_0] & 1 & \mathrm{corr}[\bm{x}_1,\bm{x}_2] & \dots & \dots & \mathrm{corr}[\bm{x}_1,\bm{x}_{p-1}]\\ -\mathrm{corr}[\bm{x}_2,\bm{x}_0] & \mathrm{corr}[\bm{x}_2,\bm{x}_1] & 1 & \dots & \dots & \mathrm{corr}[\bm{x}_2,\bm{x}_{p-1}]\\ -\dots & \dots & \dots & \dots & \dots & \dots \\ -\dots & \dots & \dots & \dots & \dots & \dots \\ -\mathrm{corr}[\bm{x}_{p-1},\bm{x}_0] & \mathrm{corr}[\bm{x}_{p-1},\bm{x}_1] & \mathrm{corr}[\bm{x}_{p-1},\bm{x}_{2}] & \dots & \dots & 1\\ -\end{bmatrix}, -\] -!et - - -!split -===== Covariance Matrix Examples ===== - - -The Numpy function _np.cov_ calculates the covariance elements using -the factor $1/(n-1)$ instead of $1/n$ since it assumes we do not have -the exact mean values. The following simple function uses the -_np.vstack_ function which takes each vector of dimension $1\times n$ -and produces a $2\times n$ matrix $\bm{W}$ - - -!bt -\[ -\bm{W} = \begin{bmatrix} x_0 & y_0 \\ - x_1 & y_1 \\ - x_2 & y_2\\ - \dots & \dots \\ - x_{n-2} & y_{n-2}\\ - x_{n-1} & y_{n-1} & - \end{bmatrix}, -\] -!et - -which in turn is converted into into the $2\times 2$ covariance matrix -$\bm{C}$ via the Numpy function _np.cov()_. We note that we can also calculate -the mean value of each set of samples $\bm{x}$ etc using the Numpy -function _np.mean(x)_. We can also extract the eigenvalues of the -covariance matrix through the _np.linalg.eig()_ function. - -!bc pycod -# Importing various packages -import numpy as np -n = 100 -x = np.random.normal(size=n) -print(np.mean(x)) -y = 4+3*x+np.random.normal(size=n) -print(np.mean(y)) -W = np.vstack((x, y)) -C = np.cov(W) -print(C) -!ec - -!split -===== Correlation Matrix ===== - -The previous example can be converted into the correlation matrix by -simply scaling the matrix elements with the variances. We should also -subtract the mean values for each column. This leads to the following -code which sets up the correlations matrix for the previous example in -a more brute force way. Here we scale the mean values for each column of the design matrix, calculate the relevant mean values and variances and then finally set up the $2\times 2$ correlation matrix (since we have only two vectors). - -!bc pycod -import numpy as np -n = 100 -# define two vectors -x = np.random.random(size=n) -y = 4+3*x+np.random.normal(size=n) -#scaling the x and y vectors -x = x - np.mean(x) -y = y - np.mean(y) -variance_x = np.sum(x@x)/n -variance_y = np.sum(y@y)/n -print(variance_x) -print(variance_y) -cov_xy = np.sum(x@y)/n -cov_xx = np.sum(x@x)/n -cov_yy = np.sum(y@y)/n -C = np.zeros((2,2)) -C[0,0]= cov_xx/variance_x -C[1,1]= cov_yy/variance_y -C[0,1]= cov_xy/np.sqrt(variance_y*variance_x) -C[1,0]= C[0,1] -print(C) -!ec - -We see that the matrix elements along the diagonal are one as they -should be and that the matrix is symmetric. Furthermore, diagonalizing -this matrix we easily see that it is a positive definite matrix. - -The above procedure with _numpy_ can be made more compact if we use _pandas_. - -!split -===== Correlation Matrix with Pandas ===== - -We whow here how we can set up the correlation matrix using _pandas_, as done in this simple code -!bc pycod -import numpy as np -import pandas as pd -n = 10 -x = np.random.normal(size=n) -x = x - np.mean(x) -y = 4+3*x+np.random.normal(size=n) -y = y - np.mean(y) -X = (np.vstack((x, y))).T -print(X) -Xpd = pd.DataFrame(X) -print(Xpd) -correlation_matrix = Xpd.corr() -print(correlation_matrix) -!ec - - -We expand this model to the Franke function discussed above. - -!split -===== Correlation Matrix with Pandas and the Franke function ===== - -!bc pycod -# Common imports -import numpy as np -import pandas as pd - - -def FrankeFunction(x,y): - term1 = 0.75*np.exp(-(0.25*(9*x-2)**2) - 0.25*((9*y-2)**2)) - term2 = 0.75*np.exp(-((9*x+1)**2)/49.0 - 0.1*(9*y+1)) - term3 = 0.5*np.exp(-(9*x-7)**2/4.0 - 0.25*((9*y-3)**2)) - term4 = -0.2*np.exp(-(9*x-4)**2 - (9*y-7)**2) - return term1 + term2 + term3 + term4 - - -def create_X(x, y, n ): - if len(x.shape) > 1: - x = np.ravel(x) - y = np.ravel(y) - - N = len(x) - l = int((n+1)*(n+2)/2) # Number of elements in beta - X = np.ones((N,l)) - - for i in range(1,n+1): - q = int((i)*(i+1)/2) - for k in range(i+1): - X[:,q+k] = (x**(i-k))*(y**k) - - return X - - -# Making meshgrid of datapoints and compute Franke's function -n = 4 -N = 100 -x = np.sort(np.random.uniform(0, 1, N)) -y = np.sort(np.random.uniform(0, 1, N)) -z = FrankeFunction(x, y) -X = create_X(x, y, n=n) - -Xpd = pd.DataFrame(X) -# subtract the mean values and set up the covariance matrix -Xpd = Xpd - Xpd.mean() -covariance_matrix = Xpd.cov() -print(covariance_matrix) -!ec - -We note here that the covariance is zero for the first rows and -columns since all matrix elements in the design matrix were set to one -(we are fitting the function in terms of a polynomial of degree $n$). - -This means that the variance for these elements will be zero and will -cause problems when we set up the correlation matrix. We can simply -drop these elements and construct a correlation -matrix without these elements. - - -!split -===== Rewriting the Covariance and/or Correlation Matrix ===== - -We can rewrite the covariance matrix in a more compact form in terms of the design/feature matrix $\bm{X}$ as -!bt -\[ -\bm{C}[\bm{x}] = \frac{1}{n}\bm{X}\bm{X}^T= \mathbb{E}[\bm{X}\bm{X}^T]. -\] -!et - -To see this let us simply look at a design matrix $\bm{X}\in {\mathbb{R}}^{2\times 2}$ -!bt -\[ -\bm{X}=\begin{bmatrix} -x_{00} & x_{01}\\ -x_{10} & x_{11}\\ -\end{bmatrix}=\begin{bmatrix} -\bm{x}_{0} & \bm{x}_{1}\\ -\end{bmatrix}. -\] -!et - -If we then compute the expectation value -!bt -\[ -\mathbb{E}[\bm{X}\bm{X}^T] = \frac{1}{n}\bm{X}\bm{X}^T=\begin{bmatrix} -x_{00}^2+x_{01}^2 & x_{00}x_{10}+x_{01}x_{11}\\ -x_{10}x_{00}+x_{11}x_{01} & x_{10}^2+x_{11}^2\\ -\end{bmatrix}, -\] -!et -which is just -!bt -\[ -\bm{C}[\bm{x}_0,\bm{x}_1] = \bm{C}[\bm{x}]=\begin{bmatrix} \mathrm{var}[\bm{x}_0] & \mathrm{cov}[\bm{x}_0,\bm{x}_1] \\ - \mathrm{cov}[\bm{x}_1,\bm{x}_0] & \mathrm{var}[\bm{x}_1] \\ - \end{bmatrix}, -\] -!et -where we wrote $$\bm{C}[\bm{x}_0,\bm{x}_1] = \bm{C}[\bm{x}]$$ to indicate that this the covariance of the vectors $\bm{x}$ of the design/feature matrix $\bm{X}$. - -It is easy to generalize this to a matrix $\bm{X}\in {\mathbb{R}}^{n\times p}$. - - -!split -===== Towards the PCA theorem ===== - -We have that the covariance matrix (the correlation matrix involves a simple rescaling) is given as -!bt -\[ -\bm{C}[\bm{x}] = \frac{1}{n}\bm{X}\bm{X}^T= \mathbb{E}[\bm{X}\bm{X}^T]. -\] -!et -Let us now assume that we can perform a series of orthogonal transformations where we employ some orthogonal matrices $\bm{S}$. -These matrices are defined as $\bm{S}\in {\mathbb{R}}^{p\times p}$ and obey the orthogonality requirements $\bm{S}\bm{S}^T=\bm{S}^T\bm{S}=\bm{I}$. The matrix can be written out in terms of the column vectors $\bm{s}_i$ as $\bm{S}=[\bm{s}_0,\bm{s}_1,\dots,\bm{s}_{p-1}]$ and $\bm{s}_i \in {\mathbb{R}}^{p}$. - -Assume also that there is a transformation $\bm{S}\bm{C}[\bm{x}]\bm{S}^T=\bm{C}[\bm{y}]$ such that the new matrix $\bm{C}[\bm{y}]$ is diagonal with elements $[\lambda_0,\lambda_1,\lambda_2,\dots,\lambda_{p-1}]$. - -That is we have -!bt -\[ -\bm{C}[\bm{y}] = \mathbb{E}[\bm{S}\bm{X}\bm{X}^T\bm{S}^T]=\bm{S}\bm{C}[\bm{x}]\bm{S}^T, -\] -!et -since the matrix $\bm{S}$ is not a data dependent matrix. Multiplying with $\bm{S}^T$ from the left we have -!bt -\[ -\bm{S}^T\bm{C}[\bm{y}] = \bm{C}[\bm{x}]\bm{S}^T, -\] -!et -and since $\bm{C}[\bm{y}]$ is diagonal we have for a given eigenvalue $i$ of the covariance matrix that - -!bt -\[ -\bm{S}^T_i\lambda_i = \bm{C}[\bm{x}]\bm{S}^T_i. -\] -!et - -In the derivation of the PCA theorem we will assume that the eigenvalues are ordered in descending order, that is -$\lambda_0 > \lambda_1 > \dots > \lambda_{p-1}$. - - -The eigenvalues tell us then how much we need to stretch the -corresponding eigenvectors. Dimensions with large eigenvalues have -thus large variations (large variance) and define therefore useful -dimensions. The data points are more spread out in the direction of -these eigenvectors. Smaller eigenvalues mean on the other hand that -the corresponding eigenvectors are shrunk accordingly and the data -points are tightly bunched together and there is not much variation in -these specific directions. Hopefully then we could leave it out -dimensions where the eigenvalues are very small. If $p$ is very large, -we could then aim at reducing $p$ to $l << p$ and handle only $l$ -features/predictors. - -!split -===== The Algorithm before the Theorem ===== - -Here's how we would proceed in setting up the algorithm for the PCA, see also discussion below here. -* Set up the datapoints for the design/feature matrix $\bm{X}$ with $\bm{X}\in {\mathbb{R}}^{n\times p}$, with the predictors/features $p$ referring to the column numbers and the entries $n$ being the row elements. -!bt -\[ -\bm{X}=\begin{bmatrix} -x_{0,0} & x_{0,1} & x_{0,2}& \dots & \dots x_{0,p-1}\\ -x_{1,0} & x_{1,1} & x_{1,2}& \dots & \dots x_{1,p-1}\\ -x_{2,0} & x_{2,1} & x_{2,2}& \dots & \dots x_{2,p-1}\\ -\dots & \dots & \dots & \dots \dots & \dots \\ -x_{n-2,0} & x_{n-2,1} & x_{n-2,2}& \dots & \dots x_{n-2,p-1}\\ -x_{n-1,0} & x_{n-1,1} & x_{n-1,2}& \dots & \dots x_{n-1,p-1}\\ -\end{bmatrix}, -\] -!et -* Center the data by subtracting the mean value for each column. This leads to a new matrix $\bm{X}\rightarrow \overline{\bm{X}}$. -* Compute then the covariance/correlation matrix $\mathbb{E}[\overline{\bm{X}}\overline{\bm{X}}^T]$. -* Find the eigenpairs of $\bm{C}$ with eigenvalues $[\lambda_0,\lambda_1,\dots,\lambda_{p-1}]$ and eigenvectors $[\bm{s}_0,\bm{s}_1,\dots,\bm{s}_{p-1}]$. -* Order the eigenvalue (and the eigenvectors accordingly) in order of decreasing eigenvalues. -* Keep only those $l$ eigenvalues larger than a selected threshold value, discarding thus $p-l$ features since we expect small variations in the data here. - - -!split -===== Writing our own PCA code ===== - -We will use a simple example first with two-dimensional data -drawn from a multivariate normal distribution with the following mean and covariance matrix: -!bt -\[ -\mu = (-1,2) \qquad \Sigma = \begin{bmatrix} 4 & 2 \\ -2 & 2 -\end{bmatrix} -\] -!et -Note that the mean refers to each column of data. -We will generate $n = 1000$ points $X = \{ x_1, \ldots, x_N \}$ from -this distribution, and store them in the $1000 \times 2$ matrix $\bm{X}$. - -The following Python code aids in setting up the data and writing out the design matrix. -Note that the function _multivariate_ returns also the covariance discussed above and that it is defined by dividing by $n-1$ instead of $n$. -!bc pycod -import numpy as np -import pandas as pd -import matplotlib.pyplot as plt -from IPython.display import display -n = 10000 -mean = (-1, 2) -cov = [[4, 2], [2, 2]] -X = np.random.multivariate_normal(mean, cov, n) -!ec - -Now we are going to implement the PCA algorithm. We will break it down into various substeps. - -=== Compute the sample mean and center the data === - -The first step of PCA is to compute the sample mean of the data and use it to center the data. Recall that the sample mean is -!bt -\[ -\mu_n = \frac{1}{n} \sum_{i=1}^n x_i -\] -!et -and the mean-centered data $\bar{X} = \{ \bar{x}_1, \ldots, \bar{x}_n \}$ takes the form -!bt -\[ -\bar{x}_i = x_i - \mu_n. -\] -!et -When you are done with these steps, print out $\mu_n$ to verify it is -close to $\mu$ and plot your mean centered data to verify it is -centered at the origin! Compare your code with the functionality from _Scikit-Learn_ discussed above. -The following code elements perform these operations using _pandas_ or using our own functionality for doing so. The latter, using _numpy_ is rather simple through the _mean()_ function. -!bc pycod -df = pd.DataFrame(X) -# Pandas does the centering for us -df = df -df.mean() -# we center it ourselves -X_centered = X - X.mean(axis=0) -!ec - -Alternatively, we could use the functions we discussed -earlier for scaling the data set. That is, we could have used the -_StandardScaler_ function in _Scikit-Learn_, a function which ensures -that for each feature/predictor we study the mean value is zero and -the variance is one (every column in the design/feature matrix). You -would then not get the same results, since we divide by the -variance. The diagonal covariance matrix elements will then be one, -while the non-diagonal ones need to be divided by $2\sqrt{2}$ for our -specific case. - -=== Compute the sample covariance === - -Now we are going to use the mean centered data to compute the sample covariance of the data by using the following equation -!bt -\begin{equation*} -\Sigma_n = \frac{1}{n-1} \sum_{i=1}^n \bar{x}_i^T \bar{x}_i = \frac{1}{n-1} \sum_{i=1}^n (x_i - \mu_n)^T (x_i - \mu_n) -\end{equation*} -!et -where the data points $x_i \in \mathbb{R}^p$ (here in this example $p = 2$) are column vectors and $x^T$ is the transpose of $x$. -We can write our own code or simply use either the functionaly of _numpy_ or that of _pandas_, as follows -!bc pycod -print(df.cov()) -print(np.cov(X_centered.T)) -!ec -Note that the way we define the covariance matrix here has a factor $n-1$ instead of $n$. This is included in the _cov()_ function by _numpy_ and _pandas_. -Our own code here is not very elegant and asks for obvious improvements. It is tailored to this specific $2\times 2$ covariance matrix. -!bc pycod -# extract the relevant columns from the centered design matrix of dim n x 2 -x = X_centered[:,0] -y = X_centered[:,1] -Cov = np.zeros((2,2)) -Cov[0,1] = np.sum(x.T@y)/(n-1.0) -Cov[0,0] = np.sum(x.T@x)/(n-1.0) -Cov[1,1] = np.sum(y.T@y)/(n-1.0) -Cov[1,0]= Cov[0,1] -print("Centered covariance using own code") -print(Cov) -plt.plot(x, y, 'x') -plt.axis('equal') -plt.show() -!ec - -Depending on the number of points $n$, we will get results that are close to the covariance values defined above. -The plot shows how the data are clustered around a line with slope close to one. Is this expected? - -=== Diagonalize the sample covariance matrix to obtain the principal components === - -Now we are ready to solve for the principal components! To do so we -diagonalize the sample covariance matrix $\Sigma$. We can use the -function _np.linalg.eig_ to do so. It will return the eigenvalues and -eigenvectors of $\Sigma$. Once we have these we can perform the -following tasks: - -* We compute the percentage of the total variance captured by the first principal component -* We plot the mean centered data and lines along the first and second principal components -* Then we project the mean centered data onto the first and second principal components, and plot the projected data. -* Finally, we approximate the data as - -!bt -\begin{equation*} -x_i \approx \tilde{x}_i = \mu_n + \langle x_i, v_0 \rangle v_0 -\end{equation*} -!et -where $v_0$ is the first principal component. - -Collecting all these steps we can write our own PCA function and -compare this with the functionality included in _Scikit-Learn_. - -The code here outlines some of the elements we could include in the -analysis. Feel free to extend upon this in order to address the above -questions. - -!bc pycod -# diagonalize and obtain eigenvalues, not necessarily sorted -EigValues, EigVectors = np.linalg.eig(Cov) -# sort eigenvectors and eigenvalues -#permute = EigValues.argsort() -#EigValues = EigValues[permute] -#EigVectors = EigVectors[:,permute] -print("Eigenvalues of Covariance matrix") -for i in range(2): - print(EigValues[i]) -FirstEigvector = EigVectors[:,0] -SecondEigvector = EigVectors[:,1] -print("First eigenvector") -print(FirstEigvector) -print("Second eigenvector") -print(SecondEigvector) -#thereafter we do a PCA with Scikit-learn -from sklearn.decomposition import PCA -pca = PCA(n_components = 2) -X2Dsl = pca.fit_transform(X) -print("Eigenvector of largest eigenvalue") -print(pca.components_.T[:, 0]) - -!ec -This code does not contain all the above elements, but it shows how we can use _Scikit-Learn_ to extract the eigenvector which corresponds to the largest eigenvalue. Try to address the questions we pose before the above code. Try also to change the values of the covariance matrix by making one of the diagonal elements much larger than the other. What do you observe then? - -!split -===== Classical PCA Theorem ===== - -We assume now that we have a design matrix $\bm{X}$ which has been -centered as discussed above. For the sake of simplicity we skip the -overline symbol. The matrix is defined in terms of the various column -vectors $[\bm{x}_0,\bm{x}_1,\dots, \bm{x}_{p-1}]$ each with dimension -$\bm{x}\in {\mathbb{R}}^{n}$. - -We assume also that we have an orthogonal transformation $\bm{W}\in {\mathbb{R}}^{p\times p}$. We define the reconstruction error (which is similar to the mean squared error we have seen before) as -!bt -\[ -J(\bm{W},\bm{Z}) = \frac{1}{n}\sum_i (\bm{x}_i - \overline{\bm{x}}_i)^2, -\] -!et -with $\overline{\bm{x}}_i = \bm{W}\bm{z}_i$, where $\bm{z}_i$ is a row vector with dimension ${\mathbb{R}}^{n}$ of the matrix -$\bm{Z}\in{\mathbb{R}}^{p\times n}$. When doing PCA we want to reduce this dimensionality. - -The PCA theorem states that minimizing the above reconstruction error -corresponds to setting $\bm{W}=\bm{S}$, the orthogonal matrix which -diagonalizes the empirical covariance(correlation) matrix. The optimal -low-dimensional encoding of the data is then given by a set of vectors -$\bm{z}_i$ with at most $l$ vectors, with $l << p$, defined by the -orthogonal projection of the data onto the columns spanned by the -eigenvectors of the covariance(correlations matrix). - -The proof which follows will be updated by mid January 2020. - -!split -===== Proof of the PCA Theorem ===== - -To show the PCA theorem let us start with the assumption that there is one vector $\bm{w}_0$ which corresponds to a solution which minimized the reconstruction error $J$. This is an orthogonal vector. It means that we now approximate the reconstruction error in terms of $\bm{w}_0$ and $\bm{z}_0$ as -!bt -\[ -J(\bm{w}_0,\bm{z}_0)= \frac{1}{n}\sum_i (\bm{x}_i - z_{i0}\bm{w}_0)^2=\frac{1}{n}\sum_i (\bm{x}_i^T\bm{x}_i - 2z_{i0}\bm{w}_0^T\bm{x}_i+z_{i0}^2\bm{w}_0^T\bm{w}_0), -\] -!et -which we can rewrite due to the orthogonality of $\bm{w}_i$ as -!bt -\[ -J(\bm{w}_0,\bm{z}_0)=\frac{1}{n}\sum_i (\bm{x}_i^T\bm{x}_i - 2z_{i0}\bm{w}_0^T\bm{x}_i+z_{i0}^2). -\] -!et -Minimizing $J$ with respect to the unknown parameters $z_{0i}$ we obtain that -!bt -\[ -z_{i0}=\bm{w}_0^T\bm{x}_i, -\] -!et -where the vectors on the rhs are known. - - -!split -===== PCA Proof continued ===== - -We have now found the unknown parameters $z_{i0}$. These correspond to the projected coordinates and we can write -!bt -\[ -J(\bm{w}_0)= \frac{1}{p}\sum_i (\bm{x}_i^T\bm{x}_i - z_{i0}^2)=\mathrm{const}-\frac{1}{n}\sum_i z_{i0}^2. -\] -!et - -We can show that the variance of the projected coordinates defined by $\bm{w}_0^T\bm{x}_i$ are given by -!bt -\[ -\mathrm{var}[\bm{w}_0^T\bm{x}_i] = \frac{1}{n}\sum_i z_{i0}^2, -\] -!et -since the expectation value of -!bt -\[ -\mathbb{E}[\bm{w}_0^T\bm{x}_i] = \mathbb{E}[z_{i0}]= \bm{w}_0^T\mathbb{E}[\bm{x}_i]=0, -\] -!et -where we have used the fact that our data are centered. - -Recalling our definition of the covariance as -!bt -\[ -\bm{C}[\bm{x}] = \frac{1}{n}\bm{X}\bm{X}^T=\mathbb{E}[\bm{X}\bm{X}^T], -\] -!et -we have thus that -!bt -\[ -\mathrm{var}[\bm{w}_0^T\bm{x}_i] = \frac{1}{n}\sum_i z_{i0}^2=\bm{w}_0^T\bm{C}[\bm{x}]\bm{w}_0. -\] -!et - -We are almost there, we have obtained a relation between minimizing -the reconstruction error and the variance and the covariance -matrix. Minimizing the error is equivalent to maximizing the variance -of the projected data. - -!split -===== The final step ===== - -We could trivially maximize the variance of the projection (and -thereby minimize the error in the reconstruction function) by letting -the norm-2 of $\bm{w}_0$ go to infinity. However, this norm since we -want the matrix $\bm{W}$ to be an orthogonal matrix, is constrained by -$\vert\vert \bm{w}_0 \vert\vert_2^2=1$. Imposing this condition via a -Lagrange multiplier we can then in turn maximize - -!bt -\[ -J(\bm{w}_0)= \bm{w}_0^T\bm{C}[\bm{x}]\bm{w}_0+\lambda_0(1-\bm{w}_0^T\bm{w}_0). -\] -!et -Taking the derivative with respect to $\bm{w}_0$ we obtain - -!bt -\[ -\frac{\partial J(\bm{w}_0)}{\partial \bm{w}_0}= 2\bm{C}[\bm{x}]\bm{w}_0-2\lambda_0\bm{w}_0=0, -\] -!et -meaning that -!bt -\[ -\bm{C}[\bm{x}]\bm{w}_0=\lambda_0\bm{w}_0. -\] -!et -_The direction that maximizes the variance (or minimizes the construction error) is an eigenvector of the covariance matrix_! If we left multiply with $\bm{w}_0^T$ we have the variance of the projected data is -!bt -\[ -\bm{w}_0^T\bm{C}[\bm{x}]\bm{w}_0=\lambda_0. -\] -!et - -If we want to maximize the variance (minimize the construction error) -we simply pick the eigenvector of the covariance matrix with the -largest eigenvalue. This establishes the link between the minimization -of the reconstruction function $J$ in terms of an orthogonal matrix -and the maximization of the variance and thereby the covariance of our -observations encoded in the design/feature matrix $\bm{X}$. - -The proof -for the other eigenvectors $\bm{w}_1,\bm{w}_2,\dots$ can be -established by applying the above arguments and using the fact that -our basis of eigenvectors is orthogonal, see "Murphy chapter -12.2":"https://mitpress.mit.edu/books/machine-learning-1". The -discussion in chapter 12.2 of Murphy's text has also a nice link with -the Singular Value Decomposition theorem. For categorical data, see -chapter 12.4 and discussion therein. - -Additional part of the proof for the other eigenvectors will be added by mid January 2020. - -!split -===== Geometric Interpretation and link with Singular Value Decomposition ===== - -This material will be added by mid January 2020. - - -!split -===== Principal Component Analysis ===== - -Principal Component Analysis (PCA) is by far the most popular dimensionality reduction algorithm. -First it identifies the hyperplane that lies closest to the data, and then it projects the data onto it. - -The following Python code uses NumPy’s _svd()_ function to obtain all the principal components of the -training set, then extracts the first two principal components. First we center the data using either _pandas_ or our own code -!bc pycod -import numpy as np -import pandas as pd -from IPython.display import display -np.random.seed(100) -# setting up a 10 x 5 vanilla matrix -rows = 10 -cols = 5 -X = np.random.randn(rows,cols) -df = pd.DataFrame(X) -# Pandas does the centering for us -df = df -df.mean() -display(df) - -# we center it ourselves -X_centered = X - X.mean(axis=0) -# Then check the difference between pandas and our own set up -print(X_centered-df) -#Now we do an SVD -U, s, V = np.linalg.svd(X_centered) -c1 = V.T[:, 0] -c2 = V.T[:, 1] -W2 = V.T[:, :2] -X2D = X_centered.dot(W2) -print(X2D) -!ec - -PCA assumes that the dataset is centered around the origin. Scikit-Learn’s PCA classes take care of centering -the data for you. However, if you implement PCA yourself (as in the preceding example), or if you use other libraries, don’t -forget to center the data first. - -Once you have identified all the principal components, you can reduce the dimensionality of the dataset -down to $d$ dimensions by projecting it onto the hyperplane defined by the first $d$ principal components. -Selecting this hyperplane ensures that the projection will preserve as much variance as possible. -!bc pycod -W2 = V.T[:, :2] -X2D = X_centered.dot(W2) -!ec - -!split -===== PCA and scikit-learn ===== - -Scikit-Learn’s PCA class implements PCA using SVD decomposition just like we did before. The -following code applies PCA to reduce the dimensionality of the dataset down to two dimensions (note -that it automatically takes care of centering the data): -!bc pycod -#thereafter we do a PCA with Scikit-learn -from sklearn.decomposition import PCA -pca = PCA(n_components = 2) -X2D = pca.fit_transform(X) -print(X2D) -!ec -After fitting the PCA transformer to the dataset, you can access the principal components using the -components variable (note that it contains the PCs as horizontal vectors, so, for example, the first -principal component is equal to -!bc pycod -pca.components_.T[:, 0]. -!ec -Another very useful piece of information is the explained variance ratio of each principal component, -available via the $explained\_variance\_ratio$ variable. It indicates the proportion of the dataset’s -variance that lies along the axis of each principal component. - -!split -===== Back to the Cancer Data ===== -We can now repeat the above but applied to real data, in this case our breast cancer data. -Here we compute performance scores on the training data using logistic regression. -!bc pycod -import matplotlib.pyplot as plt -import numpy as np -from sklearn.model_selection import train_test_split -from sklearn.datasets import load_breast_cancer -from sklearn.linear_model import LogisticRegression -cancer = load_breast_cancer() - -X_train, X_test, y_train, y_test = train_test_split(cancer.data,cancer.target,random_state=0) - -logreg = LogisticRegression() -logreg.fit(X_train, y_train) -print("Train set accuracy from Logistic Regression: {:.2f}".format(logreg.score(X_train,y_train))) -# We scale the data -from sklearn.preprocessing import StandardScaler -scaler = StandardScaler() -scaler.fit(X_train) -X_train_scaled = scaler.transform(X_train) -X_test_scaled = scaler.transform(X_test) -# Then perform again a log reg fit -logreg.fit(X_train_scaled, y_train) -print("Train set accuracy scaled data: {:.2f}".format(logreg.score(X_train_scaled,y_train))) -#thereafter we do a PCA with Scikit-learn -from sklearn.decomposition import PCA -pca = PCA(n_components = 2) -X2D_train = pca.fit_transform(X_train_scaled) -# and finally compute the log reg fit and the score on the training data -logreg.fit(X2D_train,y_train) -print("Train set accuracy scaled and PCA data: {:.2f}".format(logreg.score(X2D_train,y_train))) - -!ec - -We see that our training data after the PCA decomposition has a performance similar to the non-scaled data. - -!split -===== More on the PCA ===== - -Instead of arbitrarily choosing the number of dimensions to reduce down to, it is generally preferable to -choose the number of dimensions that add up to a sufficiently large portion of the variance (e.g., 95%). -Unless, of course, you are reducing dimensionality for data visualization — in that case you will -generally want to reduce the dimensionality down to 2 or 3. -The following code computes PCA without reducing dimensionality, then computes the minimum number -of dimensions required to preserve 95% of the training set’s variance: -!bc pycod -pca = PCA() -pca.fit(X) -cumsum = np.cumsum(pca.explained_variance_ratio_) -d = np.argmax(cumsum >= 0.95) + 1 -!ec -You could then set $n\_components=d$ and run PCA again. However, there is a much better option: instead -of specifying the number of principal components you want to preserve, you can set $n\_components$ to be -a float between 0.0 and 1.0, indicating the ratio of variance you wish to preserve: -!bc pycod -pca = PCA(n_components=0.95) -X_reduced = pca.fit_transform(X) -!ec - -!split -===== Incremental PCA ===== - -One problem with the preceding implementation of PCA is that it requires the whole training set to fit in -memory in order for the SVD algorithm to run. Fortunately, Incremental PCA (IPCA) algorithms have -been developed: you can split the training set into mini-batches and feed an IPCA algorithm one minibatch -at a time. This is useful for large training sets, and also to apply PCA online (i.e., on the fly, as new -instances arrive). - -!split -===== Randomized PCA ===== - -Scikit-Learn offers yet another option to perform PCA, called Randomized PCA. This is a stochastic -algorithm that quickly finds an approximation of the first d principal components. Its computational -complexity is $O(m \times d^2)+O(d^3)$, instead of $O(m \times n^2) + O(n^3)$, so it is dramatically faster than the -previous algorithms when $d$ is much smaller than $n$. - - - - -!split -===== Kernel PCA ===== -!bblock - -The kernel trick is a mathematical technique that implicitly maps instances into a -very high-dimensional space (called the feature space), enabling nonlinear classification and regression -with Support Vector Machines. Recall that a linear decision boundary in the high-dimensional feature -space corresponds to a complex nonlinear decision boundary in the original space. -It turns out that the same trick can be applied to PCA, making it possible to perform complex nonlinear -projections for dimensionality reduction. This is called Kernel PCA (kPCA). It is often good at -preserving clusters of instances after projection, or sometimes even unrolling datasets that lie close to a -twisted manifold. -For example, the following code uses Scikit-Learn’s KernelPCA class to perform kPCA with an -!bc pycod -from sklearn.decomposition import KernelPCA -rbf_pca = KernelPCA(n_components = 2, kernel="rbf", gamma=0.04) -X_reduced = rbf_pca.fit_transform(X) -!ec - -!eblock - - -!split -===== LLE ===== - -Locally Linear Embedding (LLE) is another very powerful nonlinear dimensionality reduction -(NLDR) technique. It is a Manifold Learning technique that does not rely on projections like the previous -algorithms. In a nutshell, LLE works by first measuring how each training instance linearly relates to its -closest neighbors (c.n.), and then looking for a low-dimensional representation of the training set where -these local relationships are best preserved (more details shortly). - - - -!split -===== Other techniques ===== - - -There are many other dimensionality reduction techniques, several of which are available in Scikit-Learn. - -Here are some of the most popular: -* _Multidimensional Scaling (MDS)_ reduces dimensionality while trying to preserve the distances between the instances. -* _Isomap_ creates a graph by connecting each instance to its nearest neighbors, then reduces dimensionality while trying to preserve the geodesic distances between the instances. -* _t-Distributed Stochastic Neighbor Embedding_ (t-SNE) reduces dimensionality while trying to keep similar instances close and dissimilar instances apart. It is mostly used for visualization, in particular to visualize clusters of instances in high-dimensional space (e.g., to visualize the MNIST images in 2D). -* Linear Discriminant Analysis (LDA) is actually a classification algorithm, but during training it learns the most discriminative axes between the classes, and these axes can then be used to define a hyperplane onto which to project the data. The benefit is that the projection will keep classes as far apart as possible, so LDA is a good technique to reduce dimensionality before running another classification algorithm such as a Support Vector Machine (SVM) classifier discussed in the SVM lectures. - - -!split -===== Differential equations ===== - -The Universal Approximation Theorem states that a neural network can -approximate any function at a single hidden layer along with one input -and output layer to any given precision. Having this in mind, we will -look closer at whether a neural network manages to solve for a -function in an equation. - - -!split -===== Description of the equation to solve for ===== -A differential equation is a equation where the solution is a function. -The equation describes how the derivatives of the function behaves in a given domain along with some conditions. - -Given a differential equation, it is desirable to know how to -reformulate it into an equation a neural network can solve. Having -decided on which activation functions each layer should use, along -with the number of hidden layers and neurons within each layer, the -changeable parameters of a neural network are the weights and biases -for each neuron in every layer in the net. If a differential equation -is reformulated into an equation where minimization of some parameters -must be done, a neural net could possibly solve this equation. - -A trial solution might be tricky to find in general. Due to the -Universal Approximation Theorem, one could hope that outcome of the -deep neural net might solve a given differential equation, even though -it is used in a simple trial solution. Let us try this idea on some -well-known ordinary differential equations and thereafter try to solve -for functions defined by two variables, giving partial differential -equations. - -!split -===== Ordinary Differential Equations ===== - -An ordinary differential equation (ODE) is an equation involving functions having one variable. - -In general, an ordinary differential equation looks like - -!bt -\begin{equation} \label{ode} -f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right) = 0 -\end{equation} -!et - -where $g(x)$ is the function to find, and $g^{(n)}(x)$ is the $n$-th derivative of $g(x)$. - -The $f\left(x, g(x), g'(x), g''(x), \, \dots \, , g^{(n)}(x)\right)$ is just a way to write that there is an expression involving $x$ and $g(x), \ g'(x), \ g''(x), \, \dots \, , \text{ and } g^{(n)}(x)$ on the left side of the equality sign in (ref{ode}). -The highest order of derivative, that is the value of $n$, determines to the order of the equation. -The equation is referred to as a $n$-th order ODE. -Along with (ref{ode}), some additional conditions of the function $g(x)$ are typically given -for the solution to be unique. - -!split -===== The trial solution ===== - -Let the trial solution $g_t(x)$ be - -!bt -\begin{equation} - g_t(x) = h_1(x) + h_2(x,N(x,P)) -\end{equation} -!et - -where $h_1(x)$ is a function that makes $g_t(x)$ satisfy a given set of conditions, $N(x,P)$ a neural network with weights and biases described by $P$ and $h_2(x, N(x,P))$ some expression involving the neural network. -The role of the function $h_2(x, N(x,P))$, is to ensure that the output from $N(x,P)$ is zero when $g_t(x)$ is evaluated at the values of $x$ where the given conditions must be satisfied. -The function $h_1(x)$ should alone make $g_t(x)$ satisfy the conditions. - -But what about the network $N(x,P)$? -As described previously, an optimization method could be used to minimize the parameters of a neural network, that being its weights and biases, through backward propagation. -For the minimization to be defined, we need to have a cost function at hand to minimize. - -It is given that $f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right)$ should be equal to zero in (ref{ode}). -We can choose to consider the mean squared error as the cost function for an input $x$. -Since we are looking at one input, the cost function is just $f$ squared. -The cost function $c\left(x, P \right)$ can therefore be expressed as - -!bt -c\left(x, P\right) = \big(f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right)\big)^2 -!et - -If $N$ inputs are given as a vector $\vec x$ with elements $x_i$ for $i = 1,\dots,N$, -the cost function becomes - -!bt -\begin{equation} \label{cost} - c\left(\vec x, P\right) = \frac{1}{N} \sum_{i=1}^N \big(f\left(x_i, \, g(x_i), \, g'(x_i), \, g''(x_i), \, \dots \, , \, g^{(n)}(x_i)\right)\big)^2 -\end{equation} -!et - -The neural net should then find some parameters $P$ that minimizes the cost function in -(ref{cost}) for a set of $N$ training samples $x_i$. - -!split -===== Minimizing the cost function using gradient descent and automatic differentiation ===== -To perform the minimization using gradient descent, the gradient of $c\left(\vec x, P\right)$ is needed. -It might happen so that finding an analytical expression of the gradient of $c(\vec x, P)$ from (ref{cost}) gets too messy, depending on which cost function one desires to use. - -Luckily, there exists libraries that makes the job for us through automatic differentiation. -Automatic differentiation is a method of finding the derivatives numerically with very high precision. - -In the forthcoming examples presenting possible usages of Autograd and TensorFlow, -it is shown how one could set up a neural network using gradient descent solving a differential -equation. - -!split -===== Example: Exponential decay and setting up the network using Autograd ===== -An exponential decay of a quantity $g(x)$ is described by the equation - -!bt -\begin{equation} \label{solve_expdec} - g'(x) = -\gamma g(x) -\end{equation} -!et - -with $g(0) = g_0$ for some chosen initial value $g_0$. - -The analytical solution of (ref{solve_expdec}) is - -!bt -\begin{equation} - g(x) = g_0 \exp\left(-\gamma x\right) -\end{equation} -!et - -Having an analytical solution at hand, it is possible to use it to compare how well a neural network finds a solution of (ref{solve_expdec}). - -In this example, a neural network will be implemented using Autograd in order to perform backpropagation. - -!split -===== The function to solve for ===== - -The program will use a neural network to solve - -!bt -\begin{equation} \label{solveode} -g'(x) = -\gamma g(x) -\end{equation} -!et - -where $g(0) = g_0$ with $\gamma$ and $g_0$ being some chosen values. - -In this example, $\gamma = 2$ and $g_0 = 10$. - -!split -===== The trial solution ===== -To begin with, a trial solution $g_t(t)$ must be chosen. A general trial solution for ordinary differential equations could be - -!bt -g_t(x, P) = h_1(x) + h_2(x, N(x, P)) -!et - -with $h_1(x)$ ensuring that $g_t(x)$ satisfies some conditions and $h_2(x,N(x, P))$ an expression involving $x$ and the output from the neural network $N(x,P)$ with $P $ being the collection of the weights and biases for each layer. For now, it is assumed that the network consists of one input layer, one hidden layer, and one output layer. - -In this network, there are no weights and bias at the input layer, so $P = \{ P_{\text{hidden}}, P_{\text{output}} \}$. -If there are $N_{\text{hidden} }$ neurons in the hidden layer, then $P_{\text{hidden}}$ is a $N_{\text{hidden} } \times (1 + N_{\text{input}})$ matrix, given that there are $N_{\text{input}}$ neurons in the input layer. - -The first column in $P_{\text{hidden} }$ represents the bias for each neuron in the hidden layer and the second column represents the weights for each neuron in the hidden layer from the input layer. -If there are $N_{\text{output} }$ neurons in the output layer, then $P_{\text{output}} $ is a $N_{\text{output} } \times (1 + N_{\text{hidden} })$ matrix. - -Its first column represents the bias of each neuron and the remaining columns represents the weights to each neuron. - -It is given that $g(0) = g_0$. The trial solution must fulfill this condition to be a proper solution of (ref{solveode}). A possible way to ensure that $g_t(0, P) = g_0$, is to let $F(N(x,P)) = x \cdot N(x,P)$ and $A(x) = g_0$. This gives the following trial solution: - -!bt -\begin{equation} \label{trial} -g_t(x, P) = g_0 + x \cdot N(x, P) -\end{equation} -!et - -!split -===== Reformulating the problem ===== -We wish that our neural network manages to minimize a given cost function. - -A reformulation of out equation, (ref{solveode}), must therefore be done, -such that it describes the problem a neural network can solve for. - -The neural network must find the set of weights and biases $P$ such that the trial solution in (ref{trial}) satisfies (ref{solveode}). - -The trial solution - -!bt -g_t(x, P) = g_0 + x \cdot N(x, P) -!et - -has been chosen such that it already solves the condition $g(0) = g_0$. What remains, is to find $P$ such that - -!bt -\begin{equation} \label{nnmin} -g_t'(x, P) = - \gamma g_t(x, P) -\end{equation} -!et - -is fulfilled as *best as possible*. - -The left hand side and right hand side of (ref{nnmin}) must be computed separately, and then the neural network must choose weights and biases, contained in $P$, such that the sides are equal as best as possible. -This means that the absolute or squared difference between the sides must be as close to zero, ideally equal to zero. -In this case, the difference squared shows to be an appropriate measurement of how erroneous the trial solution is with respect to $P$ of the neural network. - -This gives the following cost function our neural network must solve for: - -!bt -\min_{P}\Big\{ \big(g_t'(x, P) - ( -\gamma g_t(x, P) \big)^2 \Big\} -!et - -(the notation $\min_{P}\{ f(x, P) \}$ means that we desire to find $P$ that yields the minimum of $f(x, P)$) - -or, in terms of weights and biases for the hidden and output layer in our network: - -!bt -\min_{P_{\text{hidden} }, \ P_{\text{output} }}\Big\{ \big(g_t'(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) - ( -\gamma g_t(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) \big)^2 \Big\} -!et - -for an input value $x$. - -If the neural network evaluates $g_t(x, P)$ at more values for $x$, say $N$ values $x_i$ for $i = 1, \dots, N$, then the *total* error to minimize becomes - -!bt -\begin{equation} \label{min} -\min_{P}\Big\{\frac{1}{N} \sum_{i=1}^N \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 \Big\} -\end{equation} -!et - -Letting $\vec x$ be a vector with elements $x_i$ and $c(\vec x, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2$ denote the cost function, the minimization problem that our network must solve, becomes - -!bt -\min_{P} c(\vec x, P) -!et - -In terms of $P_{\text{hidden} }$ and $P_{\text{output} }$, this could also be expressed as - -$$ -\min_{P_{\text{hidden} }, \ P_{\text{output} }} c(\vec x, \{P_{\text{hidden} }, P_{\text{output} }\}) -$$ - -!split -===== A possible implementation of a neural network using Autograd ===== - -For simplicity, it is assumed that the input is an array $\vec x = (x_1, \dots, x_N)$ with $N$ elements. It is at these points the neural network should find $P$ such that it fulfills (ref{min}). - -First, the neural network must feed forward the inputs. -This means that $\vec x$ must be passed through an input layer, a hidden layer and a output layer. The input layer in this case, does not need to process the data any further. -The input layer will consist of $N_{\text{input} }$ neurons, passing its element to each neuron in the hidden layer. The number of neurons in the hidden layer will be $N_{\text{hidden} }$. - -For the $i$-th in the hidden layer with weight $w_i^{\text{hidden} }$ and bias $b_i^{\text{hidden} }$, the weighting from the $j$-th neuron at the input layer is: - -!bt -\begin{aligned} -z_{i,j}^{\text{hidden}} &= b_i^{\text{hidden}} + w_i^{\text{hidden}}x_j \\ -&= -\begin{pmatrix} -b_i^{\text{hidden}} & w_i^{\text{hidden}} -\end{pmatrix} -\begin{pmatrix} -1 \\ -x_j -\end{pmatrix} -\end{aligned} -!et - -The result after weighting the inputs at the $i$-th hidden neuron can be written as a vector: - -!bt -\begin{aligned} -\vec{z}_{i}^{\text{hidden}} &= \Big( b_i^{\text{hidden}} + w_i^{\text{hidden}}x_1 , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_2, \ \dots \, , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_N\Big) \\ -&= -\begin{pmatrix} - b_i^{\text{hidden}} & w_i^{\text{hidden}} -\end{pmatrix} -\begin{pmatrix} -1 & 1 & \dots & 1 \\ -x_1 & x_2 & \dots & x_N -\end{pmatrix} \\ -&= \vec{p}_{i, \text{hidden}}^T X -\end{aligned} -!et - -The vector $\vec{p}_{i, \text{hidden}}^T$ constitutes each row in $P_{\text{hidden} }$, which contains the weights for the neural network to minimize according to (ref{min}). - -After having found $\vec{z}_{i}^{\text{hidden}} $ for every $i$-th neuron within the hidden layer, the vector will be sent to an activation function $a_i(\vec{z})$. - -In this example, the sigmoid function has been chosen to be the activation function for each hidden neuron: - -!bt -f(z) = \frac{1}{1 + \exp{(-z)}} -!et - -It is possible to use other activations functions for the hidden layer also. - -The output $\vec{x}_i^{\text{hidden} }$from each $i$-th hidden neuron is: - -$$ -\vec{x}_i^{\text{hidden} } = f\big( \vec{z}_{i}^{\text{hidden}} \big) -$$ - -The outputs $\vec{x}_i^{\text{hidden} } $ are then sent to the output layer. - -The output layer consists of one neuron in this case, and combines the output from each of the neurons in the hidden layers. The output layer combines the results from the hidden layer using some weights $ w_i^{\text{output}}$ and biases $b_i^{\text{output}}$. In this case, it is assumes that the number of neurons in the output layer is one. - -The procedure of weighting the output neuron $j$ in the hidden layer to the $i$-th neuron in the output layer is similar as for the hidden layer described previously. - -!bt -\begin{aligned} -z_{1,j}^{\text{output}} & = -\begin{pmatrix} -b_1^{\text{output}} & \vec{w}_1^{\text{output}} -\end{pmatrix} -\begin{pmatrix} -1 \\ -\vec{x}_j^{\text{hidden}} -\end{pmatrix} -\end{aligned} -!et - -Expressing $z_{1,j}^{\text{output}}$ as a vector gives the following way of weighting the inputs from the hidden layer: - -!bt -\vec{z}_{1}^{\text{output}} = -\begin{pmatrix} -b_1^{\text{output}} & \vec{w}_1^{\text{output}} -\end{pmatrix} -\begin{pmatrix} -1 & 1 & \dots & 1 \\ -\vec{x}_1^{\text{hidden}} & \vec{x}_2^{\text{hidden}} & \dots & \vec{x}_N^{\text{hidden}} -\end{pmatrix} -!et - -In this case we seek a continuous range of values since we are approximating a function. This means that after computing $\vec{z}_{1}^{\text{output}}$ the neural network has finished its feed forward step, and $\vec{z}_{1}^{\text{output}}$ is the final output of the network. - -!split -===== Backpropagation using Autograd ===== -The next step is to decide how the parameters should be changed such that they minimize the cost function. - -The chosen cost function for this problem is - -!bt -c(\vec x, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 -!et - -In order to minimize the cost function, an optimization method must be chosen. - -Here, gradient descent with a constant step size has been chosen. - -!split -===== Gradient descent ===== -The idea of the gradient descent algorithm is to update parameters in direction where the cost function decreases goes to a minimum. - -In general, the update of some parameters $\vec \omega$ given a cost function defined by some weights $\vec \omega$, $c(\vec x, \vec \omega)$, goes as follows: - -!bt -\vec \omega_{\text{new} } = \vec \omega - \lambda \nabla_{\vec \omega} c(\vec x, \vec \omega) -!et - -for a number of iterations or until $ \big|\big| \vec \omega_{\text{new} } - \vec \omega \big|\big|$ becomes smaller than some given tolerance. - -The value of $\lambda$ decides how large steps the algorithm must take in the direction of $ \nabla_{\vec \omega} c(\vec x, \vec \omega)$. -The notation $\nabla_{\vec \omega}$ express the gradient with respect to the elements in $\vec \omega$. - -In our case, we have to minimize the cost function $c(\vec x, P)$ with respect to the two sets of weights and biases, that is for the hidden layer $P_{\text{hidden} }$ and for the output layer $P_{\text{output} }$ . - -This means that $P_{\text{hidden} }$ and $P_{\text{output} }$ is updated by - -!bt -\begin{aligned} -P_{\text{hidden},\text{new}} &= P_{\text{hidden}} - \lambda \nabla_{P_{\text{hidden}}} c(\vec x, P) \\ -P_{\text{output},\text{new}} &= P_{\text{output}} - \lambda \nabla_{P_{\text{output}}} c(\vec x, P) -\end{aligned} -!et - -In general, one could risk using a cost function having gradients that are cumbersome to derive analytically. -For our case, the cost functions are just the mean squared error. -One could employ an implementation of the back propagation for this case, but we will emphasis -on how one could use automatic differentiation in order to train the network. - -However, it might be useful to know how automatic differentiation can be used, e.g through Autograd, in order to test an implementation. - -!split -===== The network with one input, hidden, and output layer ===== - -!bc pycod -# Autograd will be used for later, so the numpy wrapper for Autograd must be imported -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -# Assuming one input, hidden, and output layer -def neural_network(params, x): - - # Find the weights (including and biases) for the hidden and output layer. - # Assume that params is a list of parameters for each layer. - # The biases are the first element for each array in params, - # and the weights are the remaning elements in each array in params. - - w_hidden = params[0] - w_output = params[1] - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - ## Hidden layer: - - # Add a row of ones to include bias - x_input = np.concatenate((np.ones((1,num_values)), x_input ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_input) - x_hidden = sigmoid(z_hidden) - - ## Output layer: - - # Include bias: - x_hidden = np.concatenate((np.ones((1,num_values)), x_hidden ), axis = 0) - - z_output = np.matmul(w_output, x_hidden) - x_output = z_output - - return x_output - -# The trial solution using the deep neural network: -def g_trial(x,params, g0 = 10): - return g0 + x*neural_network(params,x) - -# The right side of the ODE: -def g(x, g_trial, gamma = 2): - return -gamma*g_trial - -# The cost function: -def cost_function(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial(x,P) - - # Find the derivative w.r.t x of the neural network - d_net_out = elementwise_grad(neural_network,1)(P,x) - - # Find the derivative w.r.t x of the trial function - d_g_t = elementwise_grad(g_trial,0)(x,P) - - # The right side of the ODE - func = g(x, g_t) - - err_sqr = (d_g_t - func)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum / np.size(err_sqr) - -# Solve the exponential decay ODE using neural network with one input, hidden, and output layer -def solve_ode_neural_network(x, num_neurons_hidden, num_iter, lmb): - ## Set up initial weights and biases - - # For the hidden layer - p0 = npr.randn(num_neurons_hidden, 2 ) - - # For the output layer - p1 = npr.randn(1, num_neurons_hidden + 1 ) # +1 since bias is included - - P = [p0, p1] - - print('Initial cost: %g'%cost_function(P, x)) - - ## Start finding the optimal weights using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_grad = grad(cost_function,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of two arrays; - # one for the gradient w.r.t P_hidden and - # one for the gradient w.r.t P_output - cost_grad = cost_function_grad(P, x) - - P[0] = P[0] - lmb * cost_grad[0] - P[1] = P[1] - lmb * cost_grad[1] - - print('Final cost: %g'%cost_function(P, x)) - - return P - -def g_analytic(x, gamma = 2, g0 = 10): - return g0*np.exp(-gamma*x) - -# Solve the given problem -if __name__ == '__main__': - # Set seed such that the weight are initialized - # with same weights and biases for every run. - npr.seed(15) - - ## Decide the vales of arguments to the function to solve - N = 10 - x = np.linspace(0, 1, N) - - ## Set up the initial parameters - num_hidden_neurons = 10 - num_iter = 10000 - lmb = 0.001 - - # Use the network - P = solve_ode_neural_network(x, num_hidden_neurons, num_iter, lmb) - - # Print the deviation from the trial solution and true solution - res = g_trial(x,P) - res_analytical = g_analytic(x) - - print('Max absolute difference: %g'%np.max(np.abs(res - res_analytical))) - - # Plot the results - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(x, res_analytical) - plt.plot(x, res[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('x') - plt.ylabel('g(x)') - plt.show() -!ec - -!split -===== The network with one input layer, specified number of hidden layers, and one output layer output layer ===== - -It is also possible to extend the construction of our network into a more general one, allowing the network to contain more than one hidden layers. - -The number of neurons within each hidden layer are given as a list of integers in the program below. - -!bc pycod -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -# The neural network with one input layer and one output layer, -# but with number of hidden layers specified by the user. -def deep_neural_network(deep_params, x): - # N_hidden is the number of hidden layers - - N_hidden = np.size(deep_params) - 1 # -1 since params consists of - # parameters to all the hidden - # layers AND the output layer. - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - # Due to multiple hidden layers, define a variable referencing to the - # output of the previous layer: - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output - -# The trial solution using the deep neural network: -def g_trial_deep(x,params, g0 = 10): - return g0 + x*deep_neural_network(params, x) - -# The right side of the ODE: -def g(x, g_trial, gamma = 2): - return -gamma*g_trial - -# The same cost function as before, but calls deep_neural_network instead. -def cost_function_deep(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial_deep(x,P) - - # Find the derivative w.r.t x of the neural network - d_net_out = elementwise_grad(deep_neural_network,1)(P,x) - - # Find the derivative w.r.t x of the trial function - d_g_t = elementwise_grad(g_trial_deep,0)(x,P) - - # The right side of the ODE - func = g(x, g_t) - - err_sqr = (d_g_t - func)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum / np.size(err_sqr) - -# Solve the exponential decay ODE using neural network with one input and one output layer, -# but with specified number of hidden layers from the user. -def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): - # num_hidden_neurons is now a list of number of neurons within each hidden layer - - # The number of elements in the list num_hidden_neurons thus represents - # the number of hidden layers. - - # Find the number of hidden layers: - N_hidden = np.size(num_neurons) - - ## Set up initial weights and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 ) - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: %g'%cost_function_deep(P, x)) - - ## Start finding the optimal weights using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_deep_grad = grad(cost_function_deep,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases - # in the hidden layers and output layers evaluated at x. - cost_deep_grad = cost_function_deep_grad(P, x) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_deep_grad[l] - - print('Final cost: %g'%cost_function_deep(P, x)) - - return P - -def g_analytic(x, gamma = 2, g0 = 10): - return g0*np.exp(-gamma*x) - -# Solve the given problem -if __name__ == '__main__': - npr.seed(15) - - ## Decide the vales of arguments to the function to solve - N = 10 - x = np.linspace(0, 1, N) - - ## Set up the initial parameters - num_hidden_neurons = np.array([10,10]) - num_iter = 10000 - lmb = 0.001 - - P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb) - - res = g_trial_deep(x,P) - res_analytical = g_analytic(x) - - plt.figure(figsize=(10,10)) - - plt.title('Performance of a deep neural network solving an ODE compared to the analytical solution') - plt.plot(x, res_analytical) - plt.plot(x, res[0,:]) - plt.legend(['analytical','dnn']) - plt.ylabel('g(x)') - plt.show() -!ec - -!split -===== Example: Population growth, comparing Autograd, TensorFlow, and Euler's scheme ===== - -A logistic model of population growth assumes that a population converges toward an equilibrium. -The population growth can be modeled by - -!bt -\begin{equation} \label{log} - g'(t) = \alpha g(t)(A - g(t)) -\end{equation} -!et - -where $g(t)$ is the population density at time $t$, $\alpha > 0$ the growth rate and $A > 0$ is the maximum population number in the environment. -Also, at $t = 0$ the population has the size $g(0) = g_0$, where $g_0$ is some chosen constant. - -In this example, similar network as for the exponential decay using Autograd has been used to solve the equation. However, as the implementation might suffer from e.g numerical instability -and high execution time (this might be more apparent in the examples solving PDEs), -a network has been constructed using TensorFlow also. -For comparison, the forward Euler method has been implemented in order to see how the networks performs compared to a numerical scheme. - -!split -===== Setting up the problem ===== - -Here, we will model a population $g(t)$ in an environment having carrying capacity $A$. -The population follows the model - -!bt -\begin{equation} \label{solveode_population} -g'(t) = \alpha g(t)(A - g(t)) -\end{equation} -!et - -where $g(0) = g_0$. - -In this example, we let $\alpha = 2$, $A = 1$, and $g_0 = 1.2$. - -!split -===== The trial solution ===== -We will get a slightly different trial solution, as the boundary conditions are different -compared to the case for exponential decay. - -A possible trial solution satisfying the condition $g(0) = g_0$ could be - -$$ -h_1(t) = g_0 + t \cdot N(t,P) -$$ - -with $N(t,P)$ being the output from the neural network with weights and biases for each layer collected in the set $P$. - -The analytical solution is - -$$ -g(t) = \frac{Ag_0}{g_0 + (A - g_0)\exp(-\alpha A t)} -$$ - -!split -===== The program using Autograd ===== - -The network will be the similar as for the exponential decay example, but with some small modifications for our problem. - -!bc pycod -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -# Function to get the parameters. -# Done such that one can easily change the paramaters after one's liking. -def get_parameters(): - alpha = 2 - A = 1 - g0 = 1.2 - return alpha, A, g0 - -def deep_neural_network(P, x): - # N_hidden is the number of hidden layers - N_hidden = np.size(P) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - # Due to multiple hidden layers, define a variable referencing to the - # output of the previous layer: - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = P[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = P[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output - - -def cost_function_deep(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial_deep(x,P) - - # Find the derivative w.r.t x of the trial function - d_g_t = elementwise_grad(g_trial_deep,0)(x,P) - - # The right side of the ODE - func = f(x, g_t) - - err_sqr = (d_g_t - func)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum / np.size(err_sqr) - -# The right side of the ODE: -def f(x, g_trial): - alpha,A, g0 = get_parameters() - return alpha*g_trial*(A - g_trial) - -# The trial solution using the deep neural network: -def g_trial_deep(x, params): - alpha,A, g0 = get_parameters() - return g0 + x*deep_neural_network(params,x) - -# The analytical solution: -def g_analytic(t): - alpha,A, g0 = get_parameters() - return A*g0/(g0 + (A - g0)*np.exp(-alpha*A*t)) - -def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): - # num_hidden_neurons is now a list of number of neurons within each hidden layer - - # Find the number of hidden layers: - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 ) - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: %g'%cost_function_deep(P, x)) - - ## Start finding the optimal weigths using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_deep_grad = grad(cost_function_deep,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases - # in the hidden layers and output layers evaluated at x. - cost_deep_grad = cost_function_deep_grad(P, x) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_deep_grad[l] - - print('Final cost: %g'%cost_function_deep(P, x)) - - return P - -if __name__ == '__main__': - npr.seed(4155) - - ## Decide the vales of arguments to the function to solve - Nt = 10 - T = 1 - t = np.linspace(0,T, Nt) - - ## Set up the initial parameters - num_hidden_neurons = [100, 50, 25] - num_iter = 1000 - lmb = 1e-3 - - P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb) - - g_dnn_ag = g_trial_deep(t,P) - g_analytical = g_analytic(t) - - # Find the maximum absolute difference between the solutons: - diff_ag = np.max(np.abs(g_dnn_ag - g_analytical)) - print("The max absolute difference between the solutions is: %g"%diff_ag) - - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(t, g_analytical) - plt.plot(t, g_dnn_ag[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('t') - plt.ylabel('g(t)') - - plt.show() -!ec - -!split -===== Using forward Euler to solve the ODE ===== - -A straight-forward way of solving an ODE numerically, is to use Euler's method. - -Euler's method uses Taylor series to approximate the value at a function $f$ at a step $\Delta x$ from $x$: - -$$ -f(x + \Delta x) \approx f(x) + \Delta x f'(x) -$$ - -In our case, using Euler's method to approximate the value of $g$ at a step $\Delta t$ from $t$ yields - -!bt -\begin{aligned} - g(t + \Delta t) &\approx g(t) + \Delta t g'(t) \\ - &= g(t) + \Delta t \big(\alpha g(t)(A - g(t))\big) -\end{aligned} -!et -along with the condition that $g(0) = g_0$. - -Let $t_i = i \cdot \Delta t$ where $\Delta t = \frac{T}{N_t-1}$ where $T$ is the final time our solver must solve for and $N_t$ the number of values for $t \in [0, T]$ for $i = 0, \dots, N_t-1$. - -For $i \geq 1$, we have that -!bt -\begin{aligned} -t_i &= i\Delta t \\ -&= (i - 1)\Delta t + \Delta t \\ -&= t_{i-1} + \Delta t -\end{aligned} -!et - -Now, if $g_i = g(t_i)$ then - -!bt -\begin{equation} - \begin{aligned} - g_i &= g(t_i) \\ - &= g(t_{i-1} + \Delta t) \\ - &\approx g(t_{i-1}) + \Delta t \big(\alpha g(t_{i-1})(A - g(t_{i-1}))\big) \\ - &= g_{i-1} + \Delta t \big(\alpha g_{i-1}(A - g_{i-1})\big) - \end{aligned} -\end{equation} \label{odenum} -!et -for $i \geq 1$ and $g_0 = g(t_0) = g(0) = g_0$. - -Equation (ref{odenum}) could be implemented in the following way, -extending the program that uses the network using Autograd: - -!bc pycod -# Assume that all function definitions from the example program using Autograd -# are located here. - -if __name__ == '__main__': - npr.seed(4155) - - ## Decide the vales of arguments to the function to solve - Nt = 10 - T = 1 - t = np.linspace(0,T, Nt) - - ## Set up the initial parameters - num_hidden_neurons = [100,50,25] - num_iter = 1000 - lmb = 1e-3 - - P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb) - - g_dnn_ag = g_trial_deep(t,P) - g_analytical = g_analytic(t) - - # Find the maximum absolute difference between the solutons: - diff_ag = np.max(np.abs(g_dnn_ag - g_analytical)) - print("The max absolute difference between the solutions is: %g"%diff_ag) - - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(t, g_analytical) - plt.plot(t, g_dnn_ag[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('t') - plt.ylabel('g(t)') - - ## Find an approximation to the funtion using forward Euler - - alpha, A, g0 = get_parameters() - dt = T/(Nt - 1) - - # Perform forward Euler to solve the ODE - g_euler = np.zeros(Nt) - g_euler[0] = g0 - - for i in range(1,Nt): - g_euler[i] = g_euler[i-1] + dt*(alpha*g_euler[i-1]*(A - g_euler[i-1])) - - # Print the errors done by each method - diff1 = np.max(np.abs(g_euler - g_analytical)) - diff2 = np.max(np.abs(g_dnn_ag[0,:] - g_analytical)) - - print('Max absolute difference between Euler method and analytical: %g'%diff1) - print('Max absolute difference between deep neural network and analytical: %g'%diff2) - - # Plot results - plt.figure(figsize=(10,10)) - - plt.plot(t,g_euler) - plt.plot(t,g_analytical) - plt.plot(t,g_dnn_ag[0,:]) - - plt.legend(['euler','analytical','dnn']) - plt.xlabel('Time t') - plt.ylabel('g(t)') - - plt.show() -!ec - -Running the program gives - -!bc -Max absolute difference between Euler method and analytical: 0.011225 -Max absolute difference between deep neural network and analytical: 0.00424909 -!ec - -!split -===== Using TensorFlow to model logistic population growth ===== - -TensorFlow is a library widely used in the machine learning community. -A neural network can be set up in a flexible manner, where various optimization algorithms are implemented and different types of networks can be used, making it easier to experiment on solving differential equations using neural networks. - -!split -===== The general program flow in TensorFlow ===== - -Usually, a program in TensorFlow is divided into two parts; the *construction phase* and the *execution phase*. -In the construction phase, the computational graph that TensorFlow uses to perform its calculations are set up. -In the execution phase, TensorFlow evaluates any procedure that was defined in the construction phase. - -===== Program flow in TensorFlow - Construction phase ===== - -Here, the architecture for the neural network will be set up, along with the cost function and an optimizer class used during training of the network. -Note that TensorFlow uses a different convention for the weighting done in each neuron in each layer within the network than in the implementation using Autograd. -The matrix-vector multiplication between the input from the previous layer and the weighting at the neuron at current layer in the program using Autograd, is the transpose of the convention used in TensorFlow. But it will not affect that much our construction, as TensorFlow takes care of most of the computations. The only thing we have to be aware of, is how the dimensions are for our inputs. - -===== Program flow in TensorFlow - Execution phase ===== - -The computation graph has been defined, and is ready to be evaluated. -In order to get access to the graph, it has to be initialized and be runned within a Session. - -===== The full program modeling logistic population growth using TensorFlow ===== - -!bc pycod -import tensorflow as tf -import numpy as np -import matplotlib.pyplot as plt - -# Just to reset the graph such that it is possible to rerun this in a -# Jupyter cell without resetting the whole kernel. -tf.reset_default_graph() - -# Set a seed to ensure getting the same results from every run -tf.set_random_seed(4155) - -Nt = 10 -T = 1 -t = np.linspace(0,T, Nt) - -## The construction phase - -# Convert the values the trial solution is evaluated at to a tensor. -t_tf = tf.convert_to_tensor(t.reshape(-1,1),dtype=tf.float64) -zeros = tf.reshape(tf.convert_to_tensor(np.zeros(t.shape)),shape=(-1,1)) - -# Define the parameters of the equation -alpha = tf.constant(2.,dtype=tf.float64) -A = tf.constant(1.,dtype=tf.float64) -g0 = tf.constant(1.2,dtype=tf.float64) - -num_iter = 100000 - -# Define the number of neurons at each hidden layer -num_hidden_neurons = [100,50,25] -num_hidden_layers = np.size(num_hidden_neurons) - -# Construct the network. -# tf.name_scope is used to group each step in the construction, -# just for a more organized visualization in TensorBoard -with tf.name_scope('dnn'): - - # Input layer - previous_layer = t_tf - - # Hidden layers - for l in range(num_hidden_layers): - current_layer = tf.layers.dense(previous_layer, num_hidden_neurons[l], name='hidden%d'%(l+1), activation=tf.nn.sigmoid) - previous_layer = current_layer - - # Output layer - dnn_output = tf.layers.dense(previous_layer, 1, name='output') - -# Define the cost function -with tf.name_scope('cost'): - g_trial = g0 + t_tf*dnn_output - d_g_trial = tf.gradients(g_trial,t_tf) - - func = alpha*g_trial*(A - g_trial) - cost = tf.losses.mean_squared_error(zeros, d_g_trial[0] - func) - - -# Choose the method to minimize the cost function, along with a learning rate -learning_rate = 1e-2 -with tf.name_scope('train'): - optimizer = tf.train.GradientDescentOptimizer(learning_rate) - traning_op = optimizer.minimize(cost) - -# Set up a referance to the result from the neural network: -g_dnn_tf = None - -# Define a node that initializes all of the other nodes in the computational graph -# used by TensorFlow: -init = tf.global_variables_initializer() - -## Execution phase - -# Start a session where the graph defined from the construction phase can be evaluated at: -with tf.Session() as sess: - # Initialize the whole graph - init.run() - - # Evaluate the initial cost: - print('Initial cost: %g'%cost.eval()) - - # The training of the network: - for i in range(num_iter): - sess.run(traning_op) - - # If one desires to see how the cost function behaves for each iteration: - #if i % 1000 == 0: - # print(cost.eval()) - - # Training is done, and we have an approximate solution to the ODE - print('Final cost: %g'%cost.eval()) - - # Store the result - g_dnn_tf = g_trial.eval() - -# Compare with analytical solution -def get_parameters(): - alpha = 2 - A = 1 - g0 = 1.2 - return alpha, A, g0 - -def g_analytic(t): - alpha,A, g0 = get_parameters() - return A*g0/(g0 + (A - g0)*np.exp(-alpha*A*t)) - -g_analytical = g_analytic(t) -diff_tf = g_dnn_tf - g_analytical.reshape(-1,1) - -print('\nMax absolute difference between the analytical solution and solution from TensorFlow DNN: %g'%np.max(np.abs(diff_tf))) - -# Plot the result -plt.figure(figsize=(10,10)) - -plt.title('Numerical solutions of the ODE') - -plt.plot(t, g_dnn_tf) -plt.plot(t, g_analytical) - -plt.legend(['dnn, tensorflow', 'exact']) -plt.xlabel('Time t') -plt.ylabel('g(t)') - -plt.show() - -!ec - -!split -===== Example: Solving the one dimensional Poisson equation using Autograd and TensorFlow ===== - -The Poisson equation for $g(x)$ in one dimension is - -!bt -\begin{equation} \label{poisson} - -g''(x) = f(x) -\end{equation} -!et - -where $f(x)$ is a given function for $x \in (0,1)$. - -The conditions that $g(x)$ is chosen to fulfill, are -!bt -\begin{align*} - g(0) &= 0 \\ - g(1) &= 0 -\end{align*} -!et - -This equation can be solved numerically using programs where e.g Autograd and TensorFlow are used. -The results from the networks can then be compared to the analytical solution. -In addition, it could be interesting to see how a typical method for numerically solving second order ODEs compares to the neural networks. - -There exists many different optimization methods implemented in TensorFlow. -In the examples program using TensorFlow, it could also be of interest to see how -the choice of an optimization method affects our results. -In the "TensorFlow documentation about optimizers":"https://www.tensorflow.org/versions/r1.2/api_guides/python/train#Optimizers", a list over available optimization methods are shown. - -!split -===== The specific equation to solve for ===== - -Here, the function $g(x)$ to solve for follows the equation - -!bt --g''(x) = f(x),\qquad x \in (0,1) -!et - -where $f(x)$ is a given function, along with the chosen conditions - -!bt -\begin{aligned} -g(0) = g(1) = 0 -\end{aligned}\label{cond} -!et - -In this example, we consider the case when $f(x) = (3x + x^2)\exp(x)$. - -For this case, a possible trial solution satisfying the conditions could be - -!bt -g_t(x) = x \cdot (1-x) \cdot N(P,x) -!et - -The analytical solution for this problem is - -!bt -g(x) = x(1 - x)\exp(x) -!et - -!split -===== Solving the equation using Autograd ===== - -!bc pycod -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -def deep_neural_network(deep_params, x): - # N_hidden is the number of hidden layers - N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - # Due to multiple hidden layers, define a variable referencing to the - # output of the previous layer: - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output - -def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): - # num_hidden_neurons is now a list of number of neurons within each hidden layer - - # Find the number of hidden layers: - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 ) - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: %g'%cost_function_deep(P, x)) - - ## Start finding the optimal weigths using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_deep_grad = grad(cost_function_deep,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases - # in the hidden layers and output layers evaluated at x. - cost_deep_grad = cost_function_deep_grad(P, x) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_deep_grad[l] - - print('Final cost: %g'%cost_function_deep(P, x)) - - return P - -## Set up the cost function specified for this Poisson equation: - -# The right side of the ODE -def f(x): - return (3*x + x**2)*np.exp(x) - -def cost_function_deep(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial_deep(x,P) - - # Find the derivative w.r.t x of the trial function - d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P) - - right_side = f(x) - - err_sqr = (-d2_g_t - right_side)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum/np.size(err_sqr) - -# The trial solution: -def g_trial_deep(x,P): - return x*(1-x)*deep_neural_network(P,x) - -# The analytic solution; -def g_analytic(x): - return x*(1-x)*np.exp(x) - -if __name__ == '__main__': - npr.seed(4155) - - ## Decide the vales of arguments to the function to solve - Nx = 10 - x = np.linspace(0,1, Nx) - - ## Set up the initial parameters - num_hidden_neurons = [200,100] - num_iter = 1000 - lmb = 1e-3 - - P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb) - - g_dnn_ag = g_trial_deep(x,P) - g_analytical = g_analytic(x) - - # Find the maximum absolute difference between the solutons: - max_diff = np.max(np.abs(g_dnn_ag - g_analytical)) - print("The max absolute difference between the solutions is: %g"%max_diff) - - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(x, g_analytical) - plt.plot(x, g_dnn_ag[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('x') - plt.ylabel('g(x)') - plt.show() -!ec - -!split -===== Comparing with a numerical scheme ===== - -The Poisson equation is possible to solve using Taylor series to approximate the second derivative. - -Using Taylor series, the second derivative can be expressed as - -$$ -g''(x) = \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2} + E_{\Delta x}(x) -$$ - -where $\Delta x$ is a small step size and $E_{\Delta x}(x)$ being the error term. - -Looking away from the error terms gives an approximation to the second derivative: - -!bt -\begin{equation} \label{approx} -g''(x) \approx \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2} -\end{equation} -!et - -If $x_i = i \Delta x = x_{i-1} + \Delta x$ and $g_i = g(x_i)$ for $i = 1,\dots N_x - 2$ with $N_x$ being the number of values for $x$, (ref{approx}) becomes - -!bt -\begin{aligned} -g''(x_i) &\approx \frac{g(x_i + \Delta x) - 2g(x_i) + g(x_i -\Delta x)}{\Delta x^2} \\ -&= \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2} -\end{aligned} -!et - -Since we know from our problem that - -!bt -\begin{aligned} --g''(x) &= f(x) \\ -&= (3x + x^2)\exp(x) -\end{aligned} -!et - -along with the conditions $g(0) = g(1) = 0$, -the following scheme can be used to find an approximate solution for $g(x)$ numerically: - -!bt -\begin{equation} - \begin{aligned} - -\Big( \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2} \Big) &= f(x_i) \\ - -g_{i+1} + 2g_i - g_{i-1} &= \Delta x^2 f(x_i) - \end{aligned} -\end{equation} \label{odesys} -!et - -for $i = 1, \dots, N_x - 2$ where $g_0 = g_{N_x - 1} = 0$ and $f(x_i) = (3x_i + x_i^2)\exp(x_i)$, which is given for our specific problem. - -The equation can be rewritten into a matrix equation: - -!bt -\begin{aligned} -\begin{pmatrix} -2 & -1 & 0 & \dots & 0 \\ --1 & 2 & -1 & \dots & 0 \\ -\vdots & & \ddots & & \vdots \\ -0 & \dots & -1 & 2 & -1 \\ -0 & \dots & 0 & -1 & 2\\ -\end{pmatrix} -\begin{pmatrix} -g_1 \\ -g_2 \\ -\vdots \\ -g_{N_x - 3} \\ -g_{N_x - 2} -\end{pmatrix} -&= -\Delta x^2 -\begin{pmatrix} -f(x_1) \\ -f(x_2) \\ -\vdots \\ -f(x_{N_x - 3}) \\ -f(x_{N_x - 2}) -\end{pmatrix} \\ -A\vec{g} &= \vec{f} -\end{aligned} -!et - -which makes it possible to solve for the vector $\vec{g}$. - -We can then compare the result from this numerical scheme with the output from our network using Autograd: - -!bc pycod -import autograd.numpy as np -from autograd import grad, elementwise_grad -import autograd.numpy.random as npr -from matplotlib import pyplot as plt - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -def deep_neural_network(deep_params, x): - # N_hidden is the number of hidden layers - N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assumes input x being an one-dimensional array - num_values = np.size(x) - x = x.reshape(-1, num_values) - - # Assume that the input layer does nothing to the input x - x_input = x - - # Due to multiple hidden layers, define a variable referencing to the - # output of the previous layer: - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output - -def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb): - # num_hidden_neurons is now a list of number of neurons within each hidden layer - - # Find the number of hidden layers: - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 ) - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: %g'%cost_function_deep(P, x)) - - ## Start finding the optimal weigths using gradient descent - - # Find the Python function that represents the gradient of the cost function - # w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer - cost_function_deep_grad = grad(cost_function_deep,0) - - # Let the update be done num_iter times - for i in range(num_iter): - # Evaluate the gradient at the current weights and biases in P. - # The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases - # in the hidden layers and output layers evaluated at x. - cost_deep_grad = cost_function_deep_grad(P, x) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_deep_grad[l] - - print('Final cost: %g'%cost_function_deep(P, x)) - - return P - -## Set up the cost function specified for this Poisson equation: - -# The right side of the ODE -def f(x): - return (3*x + x**2)*np.exp(x) - -def cost_function_deep(P, x): - - # Evaluate the trial function with the current parameters P - g_t = g_trial_deep(x,P) - - # Find the derivative w.r.t x of the trial function - d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P) - - right_side = f(x) - - err_sqr = (-d2_g_t - right_side)**2 - cost_sum = np.sum(err_sqr) - - return cost_sum/np.size(err_sqr) - -# The trial solution: -def g_trial_deep(x,P): - return x*(1-x)*deep_neural_network(P,x) - -# The analytic solution; -def g_analytic(x): - return x*(1-x)*np.exp(x) - -if __name__ == '__main__': - npr.seed(4155) - - ## Decide the vales of arguments to the function to solve - Nx = 10 - x = np.linspace(0,1, Nx) - - ## Set up the initial parameters - num_hidden_neurons = [200,100] - num_iter = 1000 - lmb = 1e-3 - - P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb) - - g_dnn_ag = g_trial_deep(x,P) - g_analytical = g_analytic(x) - - # Find the maximum absolute difference between the solutons: - - plt.figure(figsize=(10,10)) - - plt.title('Performance of neural network solving an ODE compared to the analytical solution') - plt.plot(x, g_analytical) - plt.plot(x, g_dnn_ag[0,:]) - plt.legend(['analytical','nn']) - plt.xlabel('x') - plt.ylabel('g(x)') - - ## Perform the computation using the numerical scheme - - dx = 1/(Nx - 1) - - # Set up the matrix A - A = np.zeros((Nx-2,Nx-2)) - - A[0,0] = 2 - A[0,1] = -1 - - for i in range(1,Nx-3): - A[i,i-1] = -1 - A[i,i] = 2 - A[i,i+1] = -1 - - A[Nx - 3, Nx - 4] = -1 - A[Nx - 3, Nx - 3] = 2 - - # Set up the vector f - f_vec = dx**2 * f(x[1:-1]) - - # Solve the equation - g_res = np.linalg.solve(A,f_vec) - - g_vec = np.zeros(Nx) - g_vec[1:-1] = g_res - - # Print the differences between each method - max_diff1 = np.max(np.abs(g_dnn_ag - g_analytical)) - max_diff2 = np.max(np.abs(g_vec - g_analytical)) - print("The max absolute difference between the analytical solution and DNN Autograd: %g"%max_diff1) - print("The max absolute difference between the analytical solution and numerical scheme: %g"%max_diff2) - - # Plot the results - plt.figure(figsize=(10,10)) - - plt.plot(x,g_vec) - plt.plot(x,g_analytical) - plt.plot(x,g_dnn_ag[0,:]) - - plt.legend(['numerical scheme','analytical','dnn']) - plt.show() - -!ec - -The program prints out: -!bc -The max absolute difference between the analytical solution and DNN Autograd: 0.000464088 -The max absolute difference between the analytical solution and numerical scheme: 0.00266858 -!ec - -!split -===== Using gradient descent in TensorFlow to solve Poisson equation ===== -The program follows the similar idea as for the logistic population model. - -What has changed, is what the cost function minimizes and the trial solution. - -!bc pycod -import tensorflow as tf -import numpy as np -import matplotlib.pyplot as plt -## Construction phase - -# Just to reset the graph such that it is possible to rerun this in a -# Jupyter cell without resetting the whole kernel. -tf.reset_default_graph() - -tf.set_random_seed(4155) - -# Convert the values the trial solution is evaluated at to a tensor. -Nx = 10 -x = np.linspace(0,1, Nx) -x_tf = tf.convert_to_tensor(x.reshape(-1,1),dtype=tf.float64) - - -num_iter = 10000 - -# Define the number of neurons at each hidden layer -num_hidden_neurons = [20,10] -num_hidden_layers = np.size(num_hidden_neurons) - -# Construct the network. -# tf.name_scope is used to group each step in the construction, -# just for a more organized visualization in TensorBoard -with tf.name_scope('dnn'): - - # Input layer - previous_layer = x_tf - - # Hidden layers - for l in range(num_hidden_layers): - current_layer = tf.layers.dense(previous_layer, num_hidden_neurons[l], name='hidden%d'%(l+1), activation=tf.nn.sigmoid) - previous_layer = current_layer - - # Output layer - dnn_output = tf.layers.dense(previous_layer, 1, name='output') - -# Define the cost function -with tf.name_scope('cost'): - g_trial = x_tf*(1-x_tf)*dnn_output - d_g_trial = tf.gradients(g_trial,x_tf) - d2_g_trial = tf.gradients(d_g_trial,x_tf) - - right_side = (3*x_tf + x_tf**2)*tf.exp(x_tf) - - err = tf.square( -d2_g_trial[0] - right_side) - cost = tf.reduce_sum(err, name = 'cost') - -# Choose the method to minimize the cost function, along with a learning rate -learning_rate = 1e-2 -with tf.name_scope('train'): - optimizer = tf.train.GradientDescentOptimizer(learning_rate) - traning_op = optimizer.minimize(cost) - -g_dnn_tf = None - -# Define a node that initializes all of the other nodes in the computational graph -# used by TensorFlow: -init = tf.global_variables_initializer() - - -## Execution phase - -# Start a session where the graph defined from the construction phase can be evaluated at: - -with tf.Session() as sess: - # Initialize the whole graph - init.run() - - # Evaluate the initial cost: - print('Initial cost: %g'%cost.eval()) - - # The traning of the network: - for i in range(num_iter): - sess.run(traning_op) - - # Training is done, and we have an approximate solution to the ODE - print('Final cost: %g'%cost.eval()) - - # Store the result - g_dnn_tf = g_trial.eval() - - writer = tf.summary.FileWriter("./output", sess.graph) - writer.close() - -# Evaluate the analytical function to compare with -def g_analytic(x): - return x*(1-x)*np.exp(x) - -g_analytical = g_analytic(x) - -diff_tf = g_dnn_tf - g_analytical.reshape(-1,1) - -print('\nMax absolute difference between the analytical solution and solution from TensorFlow DNN: %g'%np.max(np.abs(diff_tf))) - -# Plot the result -plt.figure(figsize=(10,10)) - -plt.title('Numerical solutions of the ODE') - -plt.plot(x, g_dnn_tf) -plt.plot(x, g_analytical) - -plt.legend(['dnn, tensorflow','exact']) -plt.xlabel('x') -plt.ylabel('g(x)') - -plt.show() - -!ec - -!split -===== Using a different optimization algorithm implemented in TensorFlow to solve Poisson equation ===== - -We can see that the results using GradientDescentOptimizer seems to converge towards the analytical solution. -But there exists many other methods for optimization also, see "the TensorFlow documentation on Optimizers":"https://www.tensorflow.org/versions/r1.2/api_guides/python/train#Optimizers". - -Adam is an optimization algorithm that changes its learning rates accordingly to the function it tries to minimize for every iteration. -The algorithm is described in "this paper":"https://arxiv.org/pdf/1412.6980.pdf". -How much an optimization algorithm has to say for the network to converge, could be interesting to experiment with. -Using the same TensorFlow program as before, the only change to do, is to replace the variable *optimizer*. - -In the program that uses TensorFlow to solve for the Poisson equation, change the line - -!bc -optimizer = tf.train.GradientDescentOptimizer(learning_rate) -!ec - -to - -!bc -optimizer = tf.train.AdamOptimizer(learning_rate) -!ec - - -The program using the Adam optimizer with a different initial learning rate yields indeed an interesting result: -!bc -Max absolute difference between the analytical solution and solution from TensorFlow DNN: 7.11243e-05 -!ec - -!split -===== Partial Differential Equations ===== -A partial differential equation (PDE) has a solution here the function is defined by multiple variables. -The equation may involve all kinds of combinations of which variables the function is differentiated with respect to. - -In general, a partial differential equation for a function $g(x_1,\dots,x_N)$ with $N$ variables may be expressed as - -!bt -\begin{equation} \label{PDE} - f\left(x_1, \, \dots \, , x_N, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1}, \dots , \frac{\partial g(x_1,\dots,x_N) }{\partial x_N}, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(x_1,\dots,x_N) }{\partial x_N^n} \right) = 0 -\end{equation} -!et - -where $f$ is an expression involving all kinds of possible mixed derivatives of $g(x_1,\dots,x_N)$ up to an order $n$. In order for the solution to be unique, some additional conditions must also be given. - -The problem our network must solve for, is similar to the ODE case. -We must have a trial solution $g_t$ at hand. - -For instance, the trial solution could be expressed as -!bt -\begin{align*} - g_t(x_1,\dots,x_N) = h_1(x_1,\dots,x_N) + h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P)) -\end{align*} -!et -where $h_1(x_1,\dots,x_N)$ is a function that ensures $g_t(x_1,\dots,x_N)$ satisfies some given conditions. -The neural network $N(x_1,\dots,x_N,P)$ has weights and biases described by $P$ and $h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P))$ is an expression using the output from the neural network in some way. - -The role of the function $h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P))$, is to ensure that the output of $N(x_1,\dots,x_N,P)$ is zero when $g_t(x_1,\dots,x_N)$ is evaluated at the values of $x_1,\dots,x_N$ where the given conditions must be satisfied. The function $h_1(x_1,\dots,x_N)$ should alone make $g_t(x_1,\dots,x_N)$ satisfy the conditions. - -The network tries then the minimize the cost function following the same ideas as described for the ODE case, but now with more than one variables to consider. -The concept still remains the same; find a set of parameters $P$ such that the expression $f$ in (ref{PDE}) is as close to zero as possible. - -As for the ODE case, the cost function is the mean squared error that the network must try to minimize. The cost function for the network to minimize is - -!bt -\begin{equation*} -c\left(x_1, \dots, x_N, P\right) = \left( f\left(x_1, \, \dots \, , x_N, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1}, \dots , \frac{\partial g(x_1,\dots,x_N) }{\partial x_N}, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(x_1,\dots,x_N) }{\partial x_N^n} \right) \right)^2 -\end{equation*} -!et - -If we let $\vec x = \big( x_1, \dots, x_N \big)$ be an array containing the values for $x_1, \dots, x_N$ respectively, the cost function can be reformulated into the following: -!bt -\begin{equation*} - c\left(\vec{x}, P\right) = f\left( \left( \vec{x}, \frac{\partial g(\vec x) }{\partial x_1}, \dots , \frac{\partial g(\vec x) }{\partial x_N}, \frac{\partial g(\vec x) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(\vec x) }{\partial x_N^n} \right) \right)^2 -\end{equation*} -!et - -If we also have $M$ different sets of values for $x_1, \dots, x_N$, that is $\vec{x}_i = \big(x_1^{(i)}, \dots, x_N^{(i)}\big)$ for $i = 1,\dots,M$ being the rows in matrix $X$, the cost function can be generalized into -!bt -\begin{equation*} -c\left(X, P \right) = \sum_{i=1}^M f\left( \left( \vec{x}_i, \frac{\partial g(\vec{x}_i) }{\partial x_1}, \dots , \frac{\partial g(\vec{x}_i) }{\partial x_N}, \frac{\partial g(\vec{x}_i) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(\vec{x}_i) }{\partial x_N^n} \right) \right)^2 -\end{equation*} -!et - -!split -===== Example: The diffusion equation ===== - -In one spatial dimension, the equation reads -!bt -\begin{equation*} - \frac{\partial g(x,t)}{\partial t} = \frac{\partial^2 g(x,t)}{\partial x^2} -\end{equation*} -!et - -where a possible choice of conditions are -!bt -\begin{align*} -g(0,t) &= 0 ,\qquad t \geq 0 \\ -g(1,t) &= 0, \qquad t \geq 0 \\ -g(x,0) &= u(x),\qquad x\in [0,1] -\end{align*} -!et -with $u(x)$ being some given function. - -!split -===== Defining the problem ===== - -For this case, we want to find $g(x,t)$ such that - -!bt -\begin{equation} - \frac{\partial g(x,t)}{\partial t} = \frac{\partial^2 g(x,t)}{\partial x^2} -\end{equation} \label{diffonedim} -!et - -and - -!bt -\begin{align*} -g(0,t) &= 0 ,\qquad t \geq 0 \\ -g(1,t) &= 0, \qquad t \geq 0 \\ -g(x,0) &= u(x),\qquad x\in [0,1] -\end{align*} -!et -with $u(x) = \sin(\pi x)$. - -First, let us set up the deep neural network. -The deep neural network will follow the same structure as discussed in the examples solving the ODEs. -First, we will look into how Autograd could be used in a network tailored to solve for bivariate functions. - - - -!split -===== Setting up the network using Autograd ===== - -The only change to do here, is to extend our network such that functions of multiple parameters are correctly handled. -In this case we have two variables in our function to solve for, that is time $t$ and position $x$. -The variables will be represented by a one-dimensional array in the program. -The program will evaluate the network at each possible pair $(x,t)$, given an array for the desired $x$-values and $t$-values to approximate the solution at. - -!bc pycod -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -def deep_neural_network(deep_params, x): - # x is now a point and a 1D numpy array; make it a column vector - num_coordinates = np.size(x,0) - x = x.reshape(num_coordinates,-1) - - num_points = np.size(x,1) - - # N_hidden is the number of hidden layers - N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assume that the input layer does nothing to the input x - x_input = x - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output[0][0] -!ec - -!split -===== Setting up the network using Autograd; The trial solution ===== -The cost function must then iterate through the given arrays containing values for $x$ and $t$, defines a point $(x,t)$ the deep neural network and the trial solution is evaluated at, and then finds the Jacobian of the trial solution. - -A possible trial solution for this PDE is - -$$ -g_t(x,t) = h_1(x,t) + x(1-x)tN(x,t,P) -$$ - -with $A(x,t)$ being a function ensuring that $g_t(x,t)$ satisfies our given conditions, and $N(x,t,P)$ being the output from the deep neural network using weights and biases for each layer from $P$. - -To fulfill the conditions, $A(x,t)$ could be: - -$$ -h_1(x,t) = (1-t)\Big(u(x) - \big((1-x)u(0) + x u(1)\big)\Big) = (1-t)u(x) = (1-t)\sin(\pi x) -$$ -since $(0) = u(1) = 0$ and $u(x) = \sin(\pi x)$. - -The Jacobian is used because the program must find the derivative of the trial solution with respect to $x$ and $t$. - -This gives the necessity of computing the Jacobian matrix, as we want to evaluate the gradient with respect to $x$ and $t$ (note that the Jacobian of a scalar-valued multivariate function is simply its gradient). - -In Autograd, the differentiation is by default done with respect to the first input argument of your Python function. Since the points is an array representing $x$ and $t$, the Jacobian is calculated using the values of $x$ and $t$. - -To find the second derivative with respect to $x$ and $t$, the Jacobian can be found for the second time. The result is a Hessian matrix, which is the matrix containing all the possible second order mixed derivatives of $g(x,t)$. - -!bc pycod -# Set up the trial function: -def u(x): - return np.sin(np.pi*x) - -def g_trial(point,P): - x,t = point - return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point) - -# The right side of the ODE: -def f(point): - return 0. - -# The cost function: -def cost_function(P, x, t): - cost_sum = 0 - - g_t_jacobian_func = jacobian(g_trial) - g_t_hessian_func = hessian(g_trial) - - for x_ in x: - for t_ in t: - point = np.array([x_,t_]) - - g_t = g_trial(point,P) - g_t_jacobian = g_t_jacobian_func(point,P) - g_t_hessian = g_t_hessian_func(point,P) - - g_t_dt = g_t_jacobian[1] - g_t_d2x = g_t_hessian[0][0] - - func = f(point) - - err_sqr = ( (g_t_dt - g_t_d2x) - func)**2 - cost_sum += err_sqr - - return cost_sum -!ec - -!split -===== Setting up the network using Autograd; The full program ===== -Having set up the network, along with the trial solution and cost function, we can now see how the deep neural network performs by comparing the results to the analytical solution. - -The analytical solution of our problem is - -$$ -g(x,t) = \exp(-\pi^2 t)\sin(\pi x) -$$ - -A possible way to implement a neural network solving the PDE, is given below. -Be aware, though, that it is fairly slow for the parameters used. -A better result is possible, but requires more iterations, and thus longer time to complete. - -Using only 20 neurons in one hidden layer, the program managed to make the trial solution have the maximum absolute error of 0.0075. The execution time, however, was approximately one day and 14 hours on a computer having Intel i7-7560U 2.4 GHz CPU. - -Indeed, the program below is not optimal in its implementation, but rather serves as an example on how to implement and use a neural network to solve a PDE. -Using TensorFlow in the next example sovling the wave equation, has a much better execution time. - -!bc pycod -import autograd.numpy as np -from autograd import jacobian,hessian,grad -import autograd.numpy.random as npr -from matplotlib import cm -from matplotlib import pyplot as plt -from mpl_toolkits.mplot3d import axes3d - -## Set up the network - -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -def deep_neural_network(deep_params, x): - # x is now a point and a 1D numpy array; make it a column vector - num_coordinates = np.size(x,0) - x = x.reshape(num_coordinates,-1) - - num_points = np.size(x,1) - - # N_hidden is the number of hidden layers - N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assume that the input layer does nothing to the input x - x_input = x - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output[0][0] - -## Define the trial solution and cost function -def u(x): - return np.sin(np.pi*x) - -def g_trial(point,P): - x,t = point - return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point) - -# The right side of the ODE: -def f(point): - return 0. - -# The cost function: -def cost_function(P, x, t): - cost_sum = 0 - - g_t_jacobian_func = jacobian(g_trial) - g_t_hessian_func = hessian(g_trial) - - for x_ in x: - for t_ in t: - point = np.array([x_,t_]) - - g_t = g_trial(point,P) - g_t_jacobian = g_t_jacobian_func(point,P) - g_t_hessian = g_t_hessian_func(point,P) - - g_t_dt = g_t_jacobian[1] - g_t_d2x = g_t_hessian[0][0] - - func = f(point) - - err_sqr = ( (g_t_dt - g_t_d2x) - func)**2 - cost_sum += err_sqr - - return cost_sum /( np.size(x)*np.size(t) ) - -## For comparison, define the analytical solution -def g_analytic(point): - x,t = point - return np.exp(-np.pi**2*t)*np.sin(np.pi*x) - -## Set up a function for training the network to solve for the equation -def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb): - ## Set up initial weigths and biases - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: ',cost_function(P, x, t)) - - cost_function_grad = grad(cost_function,0) - - # Let the update be done num_iter times - for i in range(num_iter): - cost_grad = cost_function_grad(P, x , t) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_grad[l] - - print('Final cost: ',cost_function(P, x, t)) - - return P - -if __name__ == '__main__': - ### Use the neural network: - npr.seed(15) - - ## Decide the vales of arguments to the function to solve - Nx = 10; Nt = 10 - x = np.linspace(0, 1, Nx) - t = np.linspace(0,1,Nt) - - ## Set up the parameters for the network - num_hidden_neurons = [100, 25] - num_iter = 250 - lmb = 0.01 - - P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb) - - ## Store the results - g_dnn_ag = np.zeros((Nx, Nt)) - G_analytical = np.zeros((Nx, Nt)) - for i,x_ in enumerate(x): - for j, t_ in enumerate(t): - point = np.array([x_, t_]) - g_dnn_ag[i,j] = g_trial(point,P) - - G_analytical[i,j] = g_analytic(point) - - # Find the map difference between the analytical and the computed solution - diff_ag = np.abs(g_dnn_ag - G_analytical) - print('Max absolute difference between the analytical solution and the network: %g'%np.max(diff_ag)) - - ## Plot the solutions in two dimensions, that being in position and time - - T,X = np.meshgrid(t,x) - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons)) - s = ax.plot_surface(T,X,g_dnn_ag,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Analytical solution') - s = ax.plot_surface(T,X,G_analytical,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Difference') - s = ax.plot_surface(T,X,diff_ag,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - ## Take some slices of the 3D plots just to see the solutions at particular times - indx1 = 0 - indx2 = int(Nt/2) - indx3 = Nt-1 - - t1 = t[indx1] - t2 = t[indx2] - t3 = t[indx3] - - # Slice the results from the DNN - res1 = g_dnn_ag[:,indx1] - res2 = g_dnn_ag[:,indx2] - res3 = g_dnn_ag[:,indx3] - - # Slice the analytical results - res_analytical1 = G_analytical[:,indx1] - res_analytical2 = G_analytical[:,indx2] - res_analytical3 = G_analytical[:,indx3] - - # Plot the slices - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t1) - plt.plot(x, res1) - plt.plot(x,res_analytical1) - plt.legend(['dnn','analytical']) - - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t2) - plt.plot(x, res2) - plt.plot(x,res_analytical2) - plt.legend(['dnn','analytical']) - - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t3) - plt.plot(x, res3) - plt.plot(x,res_analytical3) - plt.legend(['dnn','analytical']) - - plt.show() -!ec - -!split -===== Example: Solving the wave equation using Autograd and TensorFlow ===== - -The wave equation is -!bt -\begin{equation*} - \frac{\partial^2 g(x,t)}{\partial t^2} = c^2\frac{\partial^2 g(x,t)}{\partial x^2} -\end{equation*} -!et - -with $c$ being the specified wave speed. - -Here, the chosen conditions are -!bt -\begin{align*} - g(0,t) &= 0 \\ - g(1,t) &= 0 \\ - g(x,0) &= u(x) \\ - \frac{\partial g(x,t)}{\partial t} \Big |_{t = 0} &= v(x) -\end{align*} -!et -where $\frac{\partial g(x,t)}{\partial t} \Big |_{t = 0}$ means the derivative of $g(x,t)$ with respect to $t$ is evaluated at $t = 0$, and $u(x)$ and $v(x)$ being given functions. - -!split -===== The problem to solve for ===== - -The wave equation to solve for, is - -!bt -\begin{equation} \label{wave} -\frac{\partial^2 g(x,t)}{\partial t^2} = c^2 \frac{\partial^2 g(x,t)}{\partial x^2} -\end{equation} -!et - -where $c$ is the given wave speed. -The chosen conditions for this equation are - -!bt -\begin{aligned} -g(0,t) &= 0, &t \geq 0 \\ -g(1,t) &= 0, &t \geq 0 \\ -g(x,0) &= u(x), &x\in[0,1] \\ -\frac{\partial g(x,t)}{\partial t}\Big |_{t = 0} &= v(x), &x \in [0,1] -\end{aligned} \label{condwave} -!et - -In this example, let $c = 1$ and $u(x) = \sin(\pi x)$ and $v(x) = -\pi\sin(\pi x)$. - - -!split -===== The trial solution ===== -Setting up the network is done in similar matter as for the example of solving the diffusion equation. -The only things we have to change, is the trial solution such that it satisfies the conditions from (ref{condwave}) and the cost function. - -The trial solution becomes slightly different since we have other conditions than in the example of solving the diffusion equation. Here, a possible trial solution $g_t(x,t)$ is - -$$ -g_t(x,t) = h_1(x,t) + x(1-x)t^2N(x,t,P) -$$ - -where - -$$ -h_1(x,t) = (1-t^2)u(x) + tv(x) -$$ - -Note that this trial solution satisfies the conditions only if $u(0) = v(0) = u(1) = v(1) = 0$, which is the case in this example. - -!split -===== The analytical solution ===== - -The analytical solution for our specific problem, is - -$$ -g(x,t) = \sin(\pi x)\cos(\pi t) - \sin(\pi x)\sin(\pi t) -$$ - -!split -===== Solving the wave equation - the full program using Autograd ===== - -!bc pycod -import autograd.numpy as np -from autograd import hessian,grad -import autograd.numpy.random as npr -from matplotlib import cm -from matplotlib import pyplot as plt -from mpl_toolkits.mplot3d import axes3d - -## Set up the trial function: -def u(x): - return np.sin(np.pi*x) - -def v(x): - return -np.pi*np.sin(np.pi*x) - -def h1(point): - x,t = point - return (1 - t**2)*u(x) + t*v(x) - -def g_trial(point,P): - x,t = point - return h1(point) + x*(1-x)*t**2*deep_neural_network(P,point) - -## Define the cost function -def cost_function(P, x, t): - cost_sum = 0 - - g_t_hessian_func = hessian(g_trial) - - for x_ in x: - for t_ in t: - point = np.array([x_,t_]) - - g_t_hessian = g_t_hessian_func(point,P) - - g_t_d2x = g_t_hessian[0][0] - g_t_d2t = g_t_hessian[1][1] - - err_sqr = ( (g_t_d2t - g_t_d2x) )**2 - cost_sum += err_sqr - - return cost_sum / (np.size(t) * np.size(x)) - -## The neural network -def sigmoid(z): - return 1/(1 + np.exp(-z)) - -def deep_neural_network(deep_params, x): - # x is now a point and a 1D numpy array; make it a column vector - num_coordinates = np.size(x,0) - x = x.reshape(num_coordinates,-1) - - num_points = np.size(x,1) - - # N_hidden is the number of hidden layers - N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer - - # Assume that the input layer does nothing to the input x - x_input = x - x_prev = x_input - - ## Hidden layers: - - for l in range(N_hidden): - # From the list of parameters P; find the correct weigths and bias for this layer - w_hidden = deep_params[l] - - # Add a row of ones to include bias - x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0) - - z_hidden = np.matmul(w_hidden, x_prev) - x_hidden = sigmoid(z_hidden) - - # Update x_prev such that next layer can use the output from this layer - x_prev = x_hidden - - ## Output layer: - - # Get the weights and bias for this layer - w_output = deep_params[-1] - - # Include bias: - x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0) - - z_output = np.matmul(w_output, x_prev) - x_output = z_output - - return x_output[0][0] - -## The analytical solution -def g_analytic(point): - x,t = point - return np.sin(np.pi*x)*np.cos(np.pi*t) - np.sin(np.pi*x)*np.sin(np.pi*t) - -def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb): - ## Set up initial weigths and biases - N_hidden = np.size(num_neurons) - - ## Set up initial weigths and biases - - # Initialize the list of parameters: - P = [None]*(N_hidden + 1) # + 1 to include the output layer - - P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias - for l in range(1,N_hidden): - P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias - - # For the output layer - P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included - - print('Initial cost: ',cost_function(P, x, t)) - - cost_function_grad = grad(cost_function,0) - - # Let the update be done num_iter times - for i in range(num_iter): - cost_grad = cost_function_grad(P, x , t) - - for l in range(N_hidden+1): - P[l] = P[l] - lmb * cost_grad[l] - - - print('Final cost: ',cost_function(P, x, t)) - - return P - -if __name__ == '__main__': - ### Use the neural network: - npr.seed(15) - - ## Decide the vales of arguments to the function to solve - Nx = 10; Nt = 10 - x = np.linspace(0, 1, Nx) - t = np.linspace(0,1,Nt) - - ## Set up the parameters for the network - num_hidden_neurons = [50,20] - num_iter = 1000 - lmb = 0.01 - - P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb) - - ## Store the results - res = np.zeros((Nx, Nt)) - res_analytical = np.zeros((Nx, Nt)) - for i,x_ in enumerate(x): - for j, t_ in enumerate(t): - point = np.array([x_, t_]) - res[i,j] = g_trial(point,P) - - res_analytical[i,j] = g_analytic(point) - - diff = np.abs(res - res_analytical) - print("Max difference between analytical and solution from nn: %g"%np.max(diff)) - - ## Plot the solutions in two dimensions, that being in position and time - - T,X = np.meshgrid(t,x) - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons)) - s = ax.plot_surface(T,X,res,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Analytical solution') - s = ax.plot_surface(T,X,res_analytical,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - - fig = plt.figure(figsize=(10,10)) - ax = fig.gca(projection='3d') - ax.set_title('Difference') - s = ax.plot_surface(T,X,diff,linewidth=0,antialiased=False,cmap=cm.viridis) - ax.set_xlabel('Time $t$') - ax.set_ylabel('Position $x$'); - - ## Take some slices of the 3D plots just to see the solutions at particular times - indx1 = 0 - indx2 = int(Nt/2) - indx3 = Nt-1 - - t1 = t[indx1] - t2 = t[indx2] - t3 = t[indx3] - - # Slice the results from the DNN - res1 = res[:,indx1] - res2 = res[:,indx2] - res3 = res[:,indx3] - - # Slice the analytical results - res_analytical1 = res_analytical[:,indx1] - res_analytical2 = res_analytical[:,indx2] - res_analytical3 = res_analytical[:,indx3] - - # Plot the slices - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t1) - plt.plot(x, res1) - plt.plot(x,res_analytical1) - plt.legend(['dnn','analytical']) - - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t2) - plt.plot(x, res2) - plt.plot(x,res_analytical2) - plt.legend(['dnn','analytical']) - - plt.figure(figsize=(10,10)) - plt.title("Computed solutions at time = %g"%t3) - plt.plot(x, res3) - plt.plot(x,res_analytical3) - plt.legend(['dnn','analytical']) - - plt.show() -!ec - -!split -===== Solving the wave equation - the full program using TensorFlow ===== -As the program using Autograd is fairly slow, one could hope that using TensorFlow -could make a naive implementation faster, and more numerically robust. - -In addition, having TensorFlow at hand, it could be easier to experiment with different -optimization algorithms, and other constructions of the network. - -The following program solves the given wave equation much faster, - -!bc pycod -import tensorflow as tf -import numpy as np -from matplotlib import cm -from matplotlib import pyplot as plt -from mpl_toolkits.mplot3d import axes3d - -Nx = 10 -x_np = np.linspace(0,1,Nx) - -Nt = 10 -t_np = np.linspace(0,1,Nt) - -X,T = np.meshgrid(x_np, t_np) - -x = X.ravel() -t = T.ravel() - -## The construction phase - -zeros = tf.reshape(tf.convert_to_tensor(np.zeros(x.shape)),shape=(-1,1)) -x = tf.reshape(tf.convert_to_tensor(x),shape=(-1,1)) -t = tf.reshape(tf.convert_to_tensor(t),shape=(-1,1)) - -points = tf.concat([x,t],1) - -num_iter = 100000 -num_hidden_neurons = [90] - -X = tf.convert_to_tensor(X) -T = tf.convert_to_tensor(T) - - -with tf.variable_scope('dnn'): - num_hidden_layers = np.size(num_hidden_neurons) - - previous_layer = points - - for l in range(num_hidden_layers): - current_layer = tf.layers.dense(previous_layer, num_hidden_neurons[l],activation=tf.nn.sigmoid) - previous_layer = current_layer - - dnn_output = tf.layers.dense(previous_layer, 1) - - -def u(x): - return tf.sin(np.pi*x) - -def v(x): - return -np.pi*tf.sin(np.pi*x) - -with tf.name_scope('loss'): - g_trial = (1 - t**2)*u(x) + t*v(x) + x*(1-x)*t**2*dnn_output - - g_trial_d2t = tf.gradients(tf.gradients(g_trial,t),t) - g_trial_d2x = tf.gradients(tf.gradients(g_trial,x),x) - - loss = tf.losses.mean_squared_error(zeros, g_trial_d2t[0] - g_trial_d2x[0]) - -learning_rate = 0.01 -with tf.name_scope('train'): - optimizer = tf.train.GradientDescentOptimizer(learning_rate) - traning_op = optimizer.minimize(loss) - -init = tf.global_variables_initializer() - -g_analytic = tf.sin(np.pi*x)*tf.cos(np.pi*t) - tf.sin(np.pi*x)*tf.sin(np.pi*t) -g_dnn = None - -## The execution phase -with tf.Session() as sess: - init.run() - for i in range(num_iter): - sess.run(traning_op) - - # If one desires to see how the cost function behaves during training - #if i % 100 == 0: - # print(loss.eval()) - - g_analytic = g_analytic.eval() - g_dnn = g_trial.eval() - - -## Compare with the analutical solution -diff = np.abs(g_analytic - g_dnn) -print('Max absolute difference between analytical solution and TensorFlow DNN = ',np.max(diff)) - -G_analytic = g_analytic.reshape((Nt,Nx)) -G_dnn = g_dnn.reshape((Nt,Nx)) - -diff = np.abs(G_analytic - G_dnn) - -# Plot the results - -X,T = np.meshgrid(x_np, t_np) - -fig = plt.figure(figsize=(10,10)) -ax = fig.gca(projection='3d') -ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons)) -s = ax.plot_surface(X,T,G_dnn,linewidth=0,antialiased=False,cmap=cm.viridis) -ax.set_xlabel('Time $t$') -ax.set_ylabel('Position $x$'); - -fig = plt.figure(figsize=(10,10)) -ax = fig.gca(projection='3d') -ax.set_title('Analytical solution') -s = ax.plot_surface(X,T,G_analytic,linewidth=0,antialiased=False,cmap=cm.viridis) -ax.set_xlabel('Time $t$') -ax.set_ylabel('Position $x$'); - -fig = plt.figure(figsize=(10,10)) -ax = fig.gca(projection='3d') -ax.set_title('Difference') -s = ax.plot_surface(X,T,diff,linewidth=0,antialiased=False,cmap=cm.viridis) -ax.set_xlabel('Time $t$') -ax.set_ylabel('Position $x$'); - -## Take some 3D slices - -indx1 = 0 -indx2 = int(Nt/2) -indx3 = Nt-1 - -t1 = t_np[indx1] -t2 = t_np[indx2] -t3 = t_np[indx3] - -# Slice the results from the DNN -res1 = G_dnn[indx1,:] -res2 = G_dnn[indx2,:] -res3 = G_dnn[indx3,:] - -# Slice the analytical results -res_analytical1 = G_analytic[indx1,:] -res_analytical2 = G_analytic[indx2,:] -res_analytical3 = G_analytic[indx3,:] - -# Plot the slices -plt.figure(figsize=(10,10)) -plt.title("Computed solutions at time = %g"%t1) -plt.plot(x_np, res1) -plt.plot(x_np,res_analytical1) -plt.legend(['dnn','analytical']) - -plt.figure(figsize=(10,10)) -plt.title("Computed solutions at time = %g"%t2) -plt.plot(x_np, res2) -plt.plot(x_np,res_analytical2) -plt.legend(['dnn','analytical']) - -plt.figure(figsize=(10,10)) -plt.title("Computed solutions at time = %g"%t3) -plt.plot(x_np, res3) -plt.plot(x_np,res_analytical3) -plt.legend(['dnn','analytical']) - -plt.show() -!ec - -The program manages to find a solution having max absolute difference to the analytical -at approximately 0.0059, by just using some minutes! -It was found, by some testing, that one hidden layer with 90 neurons actually performed well. - -!split -===== Resources ===== - -o "Artificial neural networks for solving ordinary and partial differential equations by I.E. Lagaris et al":"https://pdfs.semanticscholar.org/d061/df393e0e8fbfd0ea24976458b7d42419040d.pdf" -o "Neural networks for solving differential equations by A. Honchar":"https://becominghuman.ai/neural-networks-for-solving-differential-equations-fa230ac5e04c" -o "Solving differential equations using neural networks by M.M Chiaramonte and M. Kiener":"http://cs229.stanford.edu/proj2013/ChiaramonteKiener-SolvingDifferentialEquationsUsingNeuralNetworks.pdf" -o "Introduction to Partial Differential Equations by A. Tveitio, R. Winther":"https://www.springer.com/us/book/9783540225515" - diff --git a/doc/src/week44/week44.do.txt b/doc/src/week44/week44.do.txt index 6393c6123..ee1c95611 100644 --- a/doc/src/week44/week44.do.txt +++ b/doc/src/week44/week44.do.txt @@ -32,6 +32,123 @@ methods into the real-time experimental data processing loop to accelerate scientific discovery. +!split +===== A short Discussion of Project 2 ===== + +For neural networks and regression, should I use a design matrix with information about a polynomial fit or not? +Discuss pros and cons. The example here shows some of these issues. + +!bc pycod +""" +Code to test Ridge and NNs using Scikit-Learn only +""" + +import numpy as np +import pandas as pd +import matplotlib.pyplot as plt +from sklearn.model_selection import train_test_split +from sklearn import linear_model +from sklearn.neural_network import MLPRegressor +from sklearn.metrics import accuracy_score +import seaborn as sns + + +def MSE(y_data,y_model): + n = np.size(y_model) + return np.sum((y_data-y_model)**2)/n +# A seed just to ensure that the random numbers are the same for every run. +# Useful for eventual debugging. +np.random.seed(315) + +n = 100 +x = np.random.rand(n) +y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2) + +Maxpolydegree = 5 +X = np.zeros((n,Maxpolydegree-1)) + +for degree in range(1,Maxpolydegree): #No intercept column + X[:,degree-1] = x**(degree) + +# We split the data in test and training data +X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2) + +# Decide which values of lambda to use + +nlambdas = 10 +lmbd_vals = np.logspace(-4, 0, nlambdas) +MSERidgePredict = np.zeros(nlambdas) +for i in range(nlambdas): + lmb = lmbd_vals[i] + RegRidge = linear_model.Ridge(lmb) + RegRidge.fit(X_train,y_train) + ypredictRidge = RegRidge.predict(X_test) + MSERidgePredict[i] = MSE(y_test,ypredictRidge) + +plt.figure() +plt.plot(np.log10(lmbd_vals), MSERidgePredict, 'g--', label = 'MSE SL Ridge Test') +plt.xlabel('log10(lambda)') +plt.ylabel('MSE') +plt.legend() +plt.show() + +# Neural Network part + +n_hidden_neurons = 50 +epochs = 100 +# store models for later use +eta_vals = np.logspace(-4, 0, 10) +# store the models for later use +DNN_scikit = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object) +test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals))) +sns.set() +for i, eta in enumerate(eta_vals): + for j, lmbd in enumerate(lmbd_vals): + dnn = MLPRegressor(hidden_layer_sizes=(n_hidden_neurons), activation='logistic', + alpha=lmbd, learning_rate_init=eta, max_iter=epochs) + dnn.fit(X_train, y_train) + ypredictMLP = dnn.predict(X_test) + test_accuracy[i][j] = MSE(ypredictMLP, y_test) + +fig, ax = plt.subplots(figsize = (10, 10)) +sns.heatmap(test_accuracy, annot=True, ax=ax, cmap="viridis") +ax.set_title("Training Accuracy") +ax.set_ylabel("$\eta$") +ax.set_xlabel("$\lambda$") +plt.show() + +# Now we redefine our design matrix to include only the x-values and try out our NN + +X = np.zeros((n,1)) +X[:,0] = x + +# We split the data in test and training data again +X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2) +# Repeat the NN calculation +DNN_scikit = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object) +test_accuracy = np.zeros((len(eta_vals), len(lmbd_vals))) +sns.set() +for i, eta in enumerate(eta_vals): + for j, lmbd in enumerate(lmbd_vals): + dnn = MLPRegressor(hidden_layer_sizes=(n_hidden_neurons), activation='logistic', + alpha=lmbd, learning_rate_init=eta, max_iter=epochs) + dnn.fit(X_train, y_train) + ypredictMLP = dnn.predict(X_test) + test_accuracy[i][j] = MSE(ypredictMLP, y_test) + +fig, ax = plt.subplots(figsize = (10, 10)) +sns.heatmap(test_accuracy, annot=True, ax=ax, cmap="viridis") +ax.set_title("Training Accuracy") +ax.set_ylabel("$\eta$") +ax.set_xlabel("$\lambda$") +plt.show() + +!ec + + + + + !split ===== Thursday, Principal Component Analysis =====