diff --git a/doc/pub/week34/html/._week34-bs005.html b/doc/pub/week34/html/._week34-bs005.html index 4755be328..6446743b6 100644 --- a/doc/pub/week34/html/._week34-bs005.html +++ b/doc/pub/week34/html/._week34-bs005.html @@ -120,6 +120,7 @@ doconce format html week34.do.txt --html_style=bootstrap --pygments_html_style=d ('Numpy and arrays', 2, None, 'numpy-and-arrays'), ('Matrices in Python', 2, None, 'matrices-in-python'), ('Meet the Pandas', 2, None, 'meet-the-pandas'), + ('Pandas AI', 2, None, 'pandas-ai'), ('Simple linear regression model using _scikit-learn_', 3, None, @@ -189,10 +190,6 @@ doconce format html week34.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'interpretations-and-optimizing-our-parameters'), - ('Some useful matrix and vector expressions', - 2, - None, - 'some-useful-matrix-and-vector-expressions'), ('Interpretations and optimizing our parameters', 2, None, @@ -303,26 +300,26 @@ MathJax.Hub.Config({
-
We start with perhaps our simplest possible example, using Scikit-Learn to perform linear regression analysis on a data set produced by us.
- -What follows is a simple Python code where we have defined a function -\( y \) in terms of the variable \( x \). Both are defined as vectors with \( 100 \) entries. -The numbers in the vector \( \boldsymbol{x} \) are given -by random numbers generated with a uniform distribution with entries -\( x_i \in [0,1] \) (more about probability distribution functions -later). These values are then used to define a function \( y(x) \) -(tabulated again as a vector) with a linear dependence on \( x \) plus a -random noise added via the normal distribution. -
- -The Numpy functions are imported used the import numpy as np -statement and the random number generator for the uniform distribution -is called using the function np.random.rand(), where we specificy -that we want \( 100 \) random variables. Using Numpy we define -automatically an array with the specified number of elements, \( 100 \) in -our case. With the Numpy function randn() we can compute random -numbers with the normal distribution (mean value \( \mu \) equal to zero and -variance \( \sigma^2 \) set to one) and produce the values of \( y \) assuming a linear -dependence as function of \( x \) -
- -$$ -y = 2x+N(0,1), -$$ - -where \( N(0,1) \) represents random numbers generated by the normal -distribution. From Scikit-Learn we import then the -LinearRegression functionality and make a prediction \( \tilde{y} = -\alpha + \beta x \) using the function fit(x,y). We call the set of -data \( (\boldsymbol{x},\boldsymbol{y}) \) for our training data. The Python package -scikit-learn has also a functionality which extracts the above -fitting parameters \( \alpha \) and \( \beta \) (see below). Later we will -distinguish between training data and test data. -
- -For plotting we use the Python package -matplotlib which produces publication -quality figures. Feel free to explore the extensive -gallery of examples. In -this example we plot our original values of \( x \) and \( y \) as well as the -prediction ypredict (\( \tilde{y} \)), which attempts at fitting our -data with a straight line. -
- -The Python code follows here.
- - -# Importing various packages
-import numpy as np
-import matplotlib.pyplot as plt
-from sklearn.linear_model import LinearRegression
-
-x = np.random.rand(100,1)
-y = 2*x+np.random.randn(100,1)
-linreg = LinearRegression()
-linreg.fit(x,y)
-xnew = np.array([[0],[1]])
-ypredict = linreg.predict(xnew)
-
-plt.plot(xnew, ypredict, "r-")
-plt.plot(x, y ,'ro')
-plt.axis([0,1.0,0, 5.0])
-plt.xlabel(r'$x$')
-plt.ylabel(r'$y$')
-plt.title(r'Simple Linear Regression')
-plt.show()
-
-This example serves several aims. It allows us to demonstrate several -aspects of data analysis and later machine learning algorithms. The -immediate visualization shows that our linear fit is not -impressive. It goes through the data points, but there are many -outliers which are not reproduced by our linear regression. We could -now play around with this small program and change for example the -factor in front of \( x \) and the normal distribution. Try to change the -function \( y \) to -
- -$$ -y = 10x+0.01 \times N(0,1), -$$ - -where \( x \) is defined as before. Does the fit look better? Indeed, by -reducing the role of the noise given by the normal distribution we see immediately that -our linear prediction seemingly reproduces better the training -set. However, this testing 'by the eye' is obviouly not satisfactory in the -long run. Here we have only defined the training data and our model, and -have not discussed a more rigorous approach to the cost function. -
- -We need more rigorous criteria in defining whether we have succeeded or -not in modeling our training data. You will be surprised to see that -many scientists seldomly venture beyond this 'by the eye' approach. A -standard approach for the cost function is the so-called \( \chi^2 \) -function (a variant of the mean-squared error (MSE)) -
- -$$ \chi^2 = \frac{1}{n} -\sum_{i=0}^{n-1}\frac{(y_i-\tilde{y}_i)^2}{\sigma_i^2}, -$$ - -where \( \sigma_i^2 \) is the variance (to be defined later) of the entry -\( y_i \). We may not know the explicit value of \( \sigma_i^2 \), it serves -however the aim of scaling the equations and make the cost function -dimensionless. -
- -Minimizing the cost function is a central aspect of -our discussions to come. Finding its minima as function of the model -parameters (\( \alpha \) and \( \beta \) in our case) will be a recurring -theme in these series of lectures. Essentially all machine learning -algorithms we will discuss center around the minimization of the -chosen cost function. This depends in turn on our specific -model for describing the data, a typical situation in supervised -learning. Automatizing the search for the minima of the cost function is a -central ingredient in all algorithms. Typical methods which are -employed are various variants of gradient methods. These will be -discussed in more detail later. Again, you'll be surprised to hear that -many practitioners minimize the above function ''by the eye', popularly dubbed as -'chi by the eye'. That is, change a parameter and see (visually and numerically) that -the \( \chi^2 \) function becomes smaller. -
- -There are many ways to define the cost function. A simpler approach is to look at the relative difference between the training data and the predicted data, that is we define -the relative error (why would we prefer the MSE instead of the relative error?) as -
- -$$ -\epsilon_{\mathrm{relative}}= \frac{\vert \boldsymbol{y} -\boldsymbol{\tilde{y}}\vert}{\vert \boldsymbol{y}\vert}. -$$ - -The squared cost function results in an arithmetic mean-unbiased -estimator, and the absolute-value cost function results in a -median-unbiased estimator (in the one-dimensional case, and a -geometric median-unbiased estimator for the multi-dimensional -case). The squared cost function has the disadvantage that it has the tendency -to be dominated by outliers. -
- -We can modify easily the above Python code and plot the relative error instead
- - -import numpy as np
-import matplotlib.pyplot as plt
-from sklearn.linear_model import LinearRegression
-
-x = np.random.rand(100,1)
-y = 5*x+0.01*np.random.randn(100,1)
-linreg = LinearRegression()
-linreg.fit(x,y)
-ypredict = linreg.predict(x)
-
-plt.plot(x, np.abs(ypredict-y)/abs(y), "ro")
-plt.axis([0,1.0,0.0, 0.5])
-plt.xlabel(r'$x$')
-plt.ylabel(r'$\epsilon_{\mathrm{relative}}$')
-plt.title(r'Relative error')
-plt.show()
-
-Depending on the parameter in front of the normal distribution, we may -have a small or larger relative error. Try to play around with -different training data sets and study (graphically) the value of the -relative error. -
- -As mentioned above, Scikit-Learn has an impressive functionality. -We can for example extract the values of \( \alpha \) and \( \beta \) and -their error estimates, or the variance and standard deviation and many -other properties from the statistical data analysis. -
- -Here we show an -example of the functionality of Scikit-Learn. -
- - -import numpy as np
-import matplotlib.pyplot as plt
-from sklearn.linear_model import LinearRegression
-from sklearn.metrics import mean_squared_error, r2_score, mean_squared_log_error, mean_absolute_error
-
-x = np.random.rand(100,1)
-y = 2.0+ 5*x+0.5*np.random.randn(100,1)
-linreg = LinearRegression()
-linreg.fit(x,y)
-ypredict = linreg.predict(x)
-print('The intercept alpha: \n', linreg.intercept_)
-print('Coefficient beta : \n', linreg.coef_)
-# The mean squared error
-print("Mean squared error: %.2f" % mean_squared_error(y, ypredict))
-# Explained variance score: 1 is perfect prediction
-print('Variance score: %.2f' % r2_score(y, ypredict))
-# Mean squared log error
-print('Mean squared log error: %.2f' % mean_squared_log_error(y, ypredict) )
-# Mean absolute error
-print('Mean absolute error: %.2f' % mean_absolute_error(y, ypredict))
-plt.plot(x, ypredict, "r-")
-plt.plot(x, y ,'ro')
-plt.axis([0.0,1.0,1.5, 7.0])
-plt.xlabel(r'$x$')
-plt.ylabel(r'$y$')
-plt.title(r'Linear Regression fit ')
-plt.show()
-
-The function coef gives us the parameter \( \beta \) of our fit while intercept yields -\( \alpha \). Depending on the constant in front of the normal distribution, we get values near or far from \( \alpha =2 \) and \( \beta =5 \). Try to play around with different parameters in front of the normal distribution. The function meansquarederror gives us the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error or loss defined as -
-$$ MSE(\boldsymbol{y},\boldsymbol{\tilde{y}}) = \frac{1}{n} -\sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2, -$$ - -The smaller the value, the better the fit. Ideally we would like to -have an MSE equal zero. The attentive reader has probably recognized -this function as being similar to the \( \chi^2 \) function defined above. -
- -The r2score function computes \( R^2 \), the coefficient of -determination. It provides a measure of how well future samples are -likely to be predicted by the model. Best possible score is 1.0 and it -can be negative (because the model can be arbitrarily worse). A -constant model that always predicts the expected value of \( \boldsymbol{y} \), -disregarding the input features, would get a \( R^2 \) score of \( 0.0 \). -
- -If \( \tilde{\boldsymbol{y}}_i \) is the predicted value of the \( i-th \) sample and \( y_i \) is the corresponding true value, then the score \( R^2 \) is defined as
-$$ -R^2(\boldsymbol{y}, \tilde{\boldsymbol{y}}) = 1 - \frac{\sum_{i=0}^{n - 1} (y_i - \tilde{y}_i)^2}{\sum_{i=0}^{n - 1} (y_i - \bar{y})^2}, -$$ - -where we have defined the mean value of \( \boldsymbol{y} \) as
-$$ -\bar{y} = \frac{1}{n} \sum_{i=0}^{n - 1} y_i. -$$ - -Another quantity taht we will meet again in our discussions of regression analysis is - the mean absolute error (MAE), a risk metric corresponding to the expected value of the absolute error loss or what we call the \( l1 \)-norm loss. In our discussion above we presented the relative error. -The MAE is defined as follows -
-$$ -\text{MAE}(\boldsymbol{y}, \boldsymbol{\tilde{y}}) = \frac{1}{n} \sum_{i=0}^{n-1} \left| y_i - \tilde{y}_i \right|. -$$ - -We present the -squared logarithmic (quadratic) error -
-$$ -\text{MSLE}(\boldsymbol{y}, \boldsymbol{\tilde{y}}) = \frac{1}{n} \sum_{i=0}^{n - 1} (\log_e (1 + y_i) - \log_e (1 + \tilde{y}_i) )^2, -$$ - -where \( \log_e (x) \) stands for the natural logarithm of \( x \). This error -estimate is best to use when targets having exponential growth, such -as population counts, average sales of a commodity over a span of -years etc. -
- -Finally, another cost function is the Huber cost function used in robust regression.
- -The rationale behind this possible cost function is its reduced -sensitivity to outliers in the data set. In our discussions on -dimensionality reduction and normalization of data we will meet other -ways of dealing with outliers. -
- -The Huber cost function is defined as
-$$ -H_{\delta}(\boldsymbol{a})=\left\{\begin{array}{cc}\frac{1}{2} \boldsymbol{a}^{2}& \text{for }|\boldsymbol{a}|\leq \delta\\ \delta (|\boldsymbol{a}|-\frac{1}{2}\delta ),&\text{otherwise}.\end{array}\right. -$$ - -Here \( \boldsymbol{a}=\boldsymbol{y} - \boldsymbol{\tilde{y}} \).
- -We will discuss in more detail these and other functions in the -various lectures and lab sessions. -
-Let us now dive into nuclear physics and remind ourselves briefly about some basic features about binding -energies. A basic quantity which can be measured for the ground -states of nuclei is the atomic mass \( M(N, Z) \) of the neutral atom with -atomic mass number \( A \) and charge \( Z \). The number of neutrons is \( N \). There are indeed several sophisticated experiments worldwide which allow us to measure this quantity to high precision (parts per million even). -
- -Atomic masses are usually tabulated in terms of the mass excess defined by
-$$ -\Delta M(N, Z) = M(N, Z) - uA, -$$ - -where \( u \) is the Atomic Mass Unit
-$$ -u = M(^{12}\mathrm{C})/12 = 931.4940954(57) \hspace{0.1cm} \mathrm{MeV}/c^2. -$$ - -The nucleon masses are
-$$ -m_p = 1.00727646693(9)u, -$$ - -and
-$$ -m_n = 939.56536(8)\hspace{0.1cm} \mathrm{MeV}/c^2 = 1.0086649156(6)u. -$$ - -In the 2016 mass evaluation of by W.J.Huang, G.Audi, M.Wang, F.G.Kondev, S.Naimi and X.Xu -there are data on masses and decays of 3437 nuclei. -
- -The nuclear binding energy is defined as the energy required to break -up a given nucleus into its constituent parts of \( N \) neutrons and \( Z \) -protons. In terms of the atomic masses \( M(N, Z) \) the binding energy is -defined by -
- -$$ -BE(N, Z) = ZM_H c^2 + Nm_n c^2 - M(N, Z)c^2 , -$$ - -where \( M_H \) is the mass of the hydrogen atom and \( m_n \) is the mass of the neutron. -In terms of the mass excess the binding energy is given by -
-$$ -BE(N, Z) = Z\Delta_H c^2 + N\Delta_n c^2 -\Delta(N, Z)c^2 , -$$ - -where \( \Delta_H c^2 = 7.2890 \) MeV and \( \Delta_n c^2 = 8.0713 \) MeV.
- -A popular and physically intuitive model which can be used to parametrize -the experimental binding energies as function of \( A \), is the so-called -liquid drop model. The ansatz is based on the following expression -
- -$$ -BE(N,Z) = a_1A-a_2A^{2/3}-a_3\frac{Z^2}{A^{1/3}}-a_4\frac{(N-Z)^2}{A}, -$$ - -where \( A \) stands for the number of nucleons and the $a_i$s are parameters which are determined by a fit -to the experimental data. -
- -To arrive at the above expression we have assumed that we can make the following assumptions:
- -We could also add a so-called pairing term, which is a correction term that -arises from the tendency of proton pairs and neutron pairs to -occur. An even number of particles is more stable than an odd number. -
-Let us start with reading and organizing our data. -We start with the compilation of masses and binding energies from 2016. -After having downloaded this file to our own computer, we are now ready to read the file and start structuring our data. -
- -We start with preparing folders for storing our calculations and the data file over masses and binding energies. We import also various modules that we will find useful in order to present various Machine Learning methods. Here we focus mainly on the functionality of scikit-learn.
- - -# Common imports
-import numpy as np
-import pandas as pd
-import matplotlib.pyplot as plt
-import sklearn.linear_model as skl
-from sklearn.model_selection import train_test_split
-from sklearn.metrics import mean_squared_error, r2_score, mean_absolute_error
-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("MassEval2016.dat"),'r')
-
-Before we proceed, we define also a function for making our plots. You can obviously avoid this and simply set up various matplotlib commands every time you need them. You may however find it convenient to collect all such commands in one function and simply call this function.
- - -from pylab import plt, mpl
-plt.style.use('seaborn')
-mpl.rcParams['font.family'] = 'serif'
-
-def MakePlot(x,y, styles, labels, axlabels):
- plt.figure(figsize=(10,6))
- for i in range(len(x)):
- plt.plot(x[i], y[i], styles[i], label = labels[i])
- plt.xlabel(axlabels[0])
- plt.ylabel(axlabels[1])
- plt.legend(loc=0)
-
-Our next step is to read the data on experimental binding energies and -reorganize them as functions of the mass number \( A \), the number of -protons \( Z \) and neutrons \( N \) using pandas. Before we do this it is -always useful (unless you have a binary file or other types of compressed -data) to actually open the file and simply take a look at it! -
- -In particular, the program that outputs the final nuclear masses is written in Fortran with a specific format. It means that we need to figure out the format and which columns contain the data we are interested in. Pandas comes with a function that reads formatted output. After having admired the file, we are now ready to start massaging it with pandas. The file begins with some basic format information.
- - -"""
-This is taken from the data file of the mass 2016 evaluation.
-All files are 3436 lines long with 124 character per line.
- Headers are 39 lines long.
- col 1 : Fortran character control: 1 = page feed 0 = line feed
- format : a1,i3,i5,i5,i5,1x,a3,a4,1x,f13.5,f11.5,f11.3,f9.3,1x,a2,f11.3,f9.3,1x,i3,1x,f12.5,f11.5
- These formats are reflected in the pandas widths variable below, see the statement
- widths=(1,3,5,5,5,1,3,4,1,13,11,11,9,1,2,11,9,1,3,1,12,11,1),
- Pandas has also a variable header, with length 39 in this case.
-"""
-
-The data we are interested in are in columns 2, 3, 4 and 11, giving us -the number of neutrons, protons, mass numbers and binding energies, -respectively. We add also for the sake of completeness the element name. The data are in fixed-width formatted lines and we will -covert them into the pandas DataFrame structure. -
- - - -# Read the experimental data with Pandas
-Masses = pd.read_fwf(infile, usecols=(2,3,4,6,11),
- names=('N', 'Z', 'A', 'Element', 'Ebinding'),
- widths=(1,3,5,5,5,1,3,4,1,13,11,11,9,1,2,11,9,1,3,1,12,11,1),
- header=39,
- index_col=False)
-
-# Extrapolated values are indicated by '#' in place of the decimal place, so
-# the Ebinding column won't be numeric. Coerce to float and drop these entries.
-Masses['Ebinding'] = pd.to_numeric(Masses['Ebinding'], errors='coerce')
-Masses = Masses.dropna()
-# Convert from keV to MeV.
-Masses['Ebinding'] /= 1000
-
-# Group the DataFrame by nucleon number, A.
-Masses = Masses.groupby('A')
-# Find the rows of the grouped DataFrame with the maximum binding energy.
-Masses = Masses.apply(lambda t: t[t.Ebinding==t.Ebinding.max()])
-
-We have now read in the data, grouped them according to the variables we are interested in. -We see how easy it is to reorganize the data using pandas. If we -were to do these operations in C/C++ or Fortran, we would have had to -write various functions/subroutines which perform the above -reorganizations for us. Having reorganized the data, we can now start -to make some simple fits using both the functionalities in numpy and -Scikit-Learn afterwards. -
- -Now we define five variables which contain -the number of nucleons \( A \), the number of protons \( Z \) and the number of neutrons \( N \), the element name and finally the energies themselves. -
- - -A = Masses['A']
-Z = Masses['Z']
-N = Masses['N']
-Element = Masses['Element']
-Energies = Masses['Ebinding']
-print(Masses)
-
-The next step, and we will define this mathematically later, is to set up the so-called design matrix. We will throughout call this matrix \( \boldsymbol{X} \). -It has dimensionality \( p\times n \), where \( n \) is the number of data points and \( p \) are the so-called predictors. In our case here they are given by the number of polynomials in \( A \) we wish to include in the fit. -
- - -# Now we set up the design matrix X
-X = np.zeros((len(A),5))
-X[:,0] = 1
-X[:,1] = A
-X[:,2] = A**(2.0/3.0)
-X[:,3] = A**(-1.0/3.0)
-X[:,4] = A**(-1.0)
-
-With scikitlearn we are now ready to use linear regression and fit our data.
- - -clf = skl.LinearRegression().fit(X, Energies)
-fity = clf.predict(X)
-
-Pretty simple! -Now we can print measures of how our fit is doing, the coefficients from the fits and plot the final fit together with our data. -
- - -# The mean squared error
-print("Mean squared error: %.2f" % mean_squared_error(Energies, fity))
-# Explained variance score: 1 is perfect prediction
-print('Variance score: %.2f' % r2_score(Energies, fity))
-# Mean absolute error
-print('Mean absolute error: %.2f' % mean_absolute_error(Energies, fity))
-print(clf.coef_, clf.intercept_)
-
-Masses['Eapprox'] = fity
-# Generate a plot comparing the experimental with the fitted values values.
-fig, ax = plt.subplots()
-ax.set_xlabel(r'$A = N + Z$')
-ax.set_ylabel(r'$E_\mathrm{bind}\,/\mathrm{MeV}$')
-ax.plot(Masses['A'], Masses['Ebinding'], alpha=0.7, lw=2,
- label='Ame2016')
-ax.plot(Masses['A'], Masses['Eapprox'], alpha=0.7, lw=2, c='m',
- label='Fit')
-ax.legend()
-save_fig("Masses2016")
-plt.show()
-
-The seaborn package allows us to visualize data in an efficient way. Note that we use scikit-learn's multi-layer perceptron (or feed forward neural network) -functionality. -
- - -from sklearn.neural_network import MLPRegressor
-from sklearn.metrics import accuracy_score
-import seaborn as sns
-
-
-X_train = X
-Y_train = Energies
-n_hidden_neurons = 50
-epochs = 100
-# store models for later use
-eta_vals = np.logspace(-3, 0, 4)
-lmbd_vals = np.logspace(-3, 0, 4)
-# store the models for later use
-DNN_scikit = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)
-train_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='relu', solver='adam',
- alpha=lmbd, learning_rate_init=eta, max_iter=epochs)
- dnn.fit(X_train, Y_train)
- DNN_scikit[i][j] = dnn
- train_accuracy[i][j] = dnn.score(X_train, Y_train)
- fity = dnn.predict(X_train)
- MSE = mean_squared_error(Y_train, fity)
- print("Mean squared error: %.2f" % mean_squared_error(Y_train, fity))
- train_accuracy[i][j] = MSE
-fig, ax = plt.subplots(figsize = (10, 10))
-sns.heatmap(train_accuracy, annot=True, ax=ax, cmap="viridis")
-ax.set_title("Training Accuracy")
-ax.set_ylabel("$\eta$")
-ax.set_xlabel("$\lambda$")
-plt.show()
-print(train_accuracy)
-
-The aim behind these introductory words was to present to you various -Python libraries and their functionalities, in particular libraries like -numpy, pandas, xarray and matplotlib and other that make our life much easier -in handling various data sets and visualizing data. -
- -Furthermore, -Scikit-Learn allows us with few lines of code to implement popular -Machine Learning algorithms for supervised learning. Later we will meet Tensorflow, a powerful library for deep learning. -Now it is time to dive more into the details of various methods. We will start with linear regression and try to take a deeper look at what it entails. -
+Try out Pandas AI
diff --git a/doc/pub/week34/html/._week34-bs038.html b/doc/pub/week34/html/._week34-bs038.html index b0b3faebd..94d81a4b4 100644 --- a/doc/pub/week34/html/._week34-bs038.html +++ b/doc/pub/week34/html/._week34-bs038.html @@ -120,6 +120,7 @@ doconce format html week34.do.txt --html_style=bootstrap --pygments_html_style=d ('Numpy and arrays', 2, None, 'numpy-and-arrays'), ('Matrices in Python', 2, None, 'matrices-in-python'), ('Meet the Pandas', 2, None, 'meet-the-pandas'), + ('Pandas AI', 2, None, 'pandas-ai'), ('Simple linear regression model using _scikit-learn_', 3, None, @@ -189,10 +190,6 @@ doconce format html week34.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'interpretations-and-optimizing-our-parameters'), - ('Some useful matrix and vector expressions', - 2, - None, - 'some-useful-matrix-and-vector-expressions'), ('Interpretations and optimizing our parameters', 2, None, @@ -303,26 +300,26 @@ MathJax.Hub.Config({
-
We start with perhaps our simplest possible example, using Scikit-Learn to perform linear regression analysis on a data set produced by us.
+ +What follows is a simple Python code where we have defined a function +\( y \) in terms of the variable \( x \). Both are defined as vectors with \( 100 \) entries. +The numbers in the vector \( \boldsymbol{x} \) are given +by random numbers generated with a uniform distribution with entries +\( x_i \in [0,1] \) (more about probability distribution functions +later). These values are then used to define a function \( y(x) \) +(tabulated again as a vector) with a linear dependence on \( x \) plus a +random noise added via the normal distribution. +
+ +The Numpy functions are imported used the import numpy as np +statement and the random number generator for the uniform distribution +is called using the function np.random.rand(), where we specificy +that we want \( 100 \) random variables. Using Numpy we define +automatically an array with the specified number of elements, \( 100 \) in +our case. With the Numpy function randn() we can compute random +numbers with the normal distribution (mean value \( \mu \) equal to zero and +variance \( \sigma^2 \) set to one) and produce the values of \( y \) assuming a linear +dependence as function of \( x \) +
+ +$$ +y = 2x+N(0,1), +$$ + +where \( N(0,1) \) represents random numbers generated by the normal +distribution. From Scikit-Learn we import then the +LinearRegression functionality and make a prediction \( \tilde{y} = +\alpha + \beta x \) using the function fit(x,y). We call the set of +data \( (\boldsymbol{x},\boldsymbol{y}) \) for our training data. The Python package +scikit-learn has also a functionality which extracts the above +fitting parameters \( \alpha \) and \( \beta \) (see below). Later we will +distinguish between training data and test data. +
+ +For plotting we use the Python package +matplotlib which produces publication +quality figures. Feel free to explore the extensive +gallery of examples. In +this example we plot our original values of \( x \) and \( y \) as well as the +prediction ypredict (\( \tilde{y} \)), which attempts at fitting our +data with a straight line. +
+ +The Python code follows here.
+ + +# Importing various packages
+import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.linear_model import LinearRegression
+
+x = np.random.rand(100,1)
+y = 2*x+np.random.randn(100,1)
+linreg = LinearRegression()
+linreg.fit(x,y)
+xnew = np.array([[0],[1]])
+ypredict = linreg.predict(xnew)
+
+plt.plot(xnew, ypredict, "r-")
+plt.plot(x, y ,'ro')
+plt.axis([0,1.0,0, 5.0])
+plt.xlabel(r'$x$')
+plt.ylabel(r'$y$')
+plt.title(r'Simple Linear Regression')
+plt.show()
+
+This example serves several aims. It allows us to demonstrate several +aspects of data analysis and later machine learning algorithms. The +immediate visualization shows that our linear fit is not +impressive. It goes through the data points, but there are many +outliers which are not reproduced by our linear regression. We could +now play around with this small program and change for example the +factor in front of \( x \) and the normal distribution. Try to change the +function \( y \) to +
+ +$$ +y = 10x+0.01 \times N(0,1), +$$ + +where \( x \) is defined as before. Does the fit look better? Indeed, by +reducing the role of the noise given by the normal distribution we see immediately that +our linear prediction seemingly reproduces better the training +set. However, this testing 'by the eye' is obviouly not satisfactory in the +long run. Here we have only defined the training data and our model, and +have not discussed a more rigorous approach to the cost function. +
+ +We need more rigorous criteria in defining whether we have succeeded or +not in modeling our training data. You will be surprised to see that +many scientists seldomly venture beyond this 'by the eye' approach. A +standard approach for the cost function is the so-called \( \chi^2 \) +function (a variant of the mean-squared error (MSE)) +
+ +$$ \chi^2 = \frac{1}{n} +\sum_{i=0}^{n-1}\frac{(y_i-\tilde{y}_i)^2}{\sigma_i^2}, +$$ + +where \( \sigma_i^2 \) is the variance (to be defined later) of the entry +\( y_i \). We may not know the explicit value of \( \sigma_i^2 \), it serves +however the aim of scaling the equations and make the cost function +dimensionless. +
+ +Minimizing the cost function is a central aspect of +our discussions to come. Finding its minima as function of the model +parameters (\( \alpha \) and \( \beta \) in our case) will be a recurring +theme in these series of lectures. Essentially all machine learning +algorithms we will discuss center around the minimization of the +chosen cost function. This depends in turn on our specific +model for describing the data, a typical situation in supervised +learning. Automatizing the search for the minima of the cost function is a +central ingredient in all algorithms. Typical methods which are +employed are various variants of gradient methods. These will be +discussed in more detail later. Again, you'll be surprised to hear that +many practitioners minimize the above function ''by the eye', popularly dubbed as +'chi by the eye'. That is, change a parameter and see (visually and numerically) that +the \( \chi^2 \) function becomes smaller. +
+ +There are many ways to define the cost function. A simpler approach is to look at the relative difference between the training data and the predicted data, that is we define +the relative error (why would we prefer the MSE instead of the relative error?) as +
+ +$$ +\epsilon_{\mathrm{relative}}= \frac{\vert \boldsymbol{y} -\boldsymbol{\tilde{y}}\vert}{\vert \boldsymbol{y}\vert}. +$$ + +The squared cost function results in an arithmetic mean-unbiased +estimator, and the absolute-value cost function results in a +median-unbiased estimator (in the one-dimensional case, and a +geometric median-unbiased estimator for the multi-dimensional +case). The squared cost function has the disadvantage that it has the tendency +to be dominated by outliers. +
+ +We can modify easily the above Python code and plot the relative error instead
+ + +import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.linear_model import LinearRegression
+
+x = np.random.rand(100,1)
+y = 5*x+0.01*np.random.randn(100,1)
+linreg = LinearRegression()
+linreg.fit(x,y)
+ypredict = linreg.predict(x)
+
+plt.plot(x, np.abs(ypredict-y)/abs(y), "ro")
+plt.axis([0,1.0,0.0, 0.5])
+plt.xlabel(r'$x$')
+plt.ylabel(r'$\epsilon_{\mathrm{relative}}$')
+plt.title(r'Relative error')
+plt.show()
+
+Depending on the parameter in front of the normal distribution, we may +have a small or larger relative error. Try to play around with +different training data sets and study (graphically) the value of the +relative error. +
+ +As mentioned above, Scikit-Learn has an impressive functionality. +We can for example extract the values of \( \alpha \) and \( \beta \) and +their error estimates, or the variance and standard deviation and many +other properties from the statistical data analysis. +
+ +Here we show an +example of the functionality of Scikit-Learn. +
+ + +import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.linear_model import LinearRegression
+from sklearn.metrics import mean_squared_error, r2_score, mean_squared_log_error, mean_absolute_error
+
+x = np.random.rand(100,1)
+y = 2.0+ 5*x+0.5*np.random.randn(100,1)
+linreg = LinearRegression()
+linreg.fit(x,y)
+ypredict = linreg.predict(x)
+print('The intercept alpha: \n', linreg.intercept_)
+print('Coefficient beta : \n', linreg.coef_)
+# The mean squared error
+print("Mean squared error: %.2f" % mean_squared_error(y, ypredict))
+# Explained variance score: 1 is perfect prediction
+print('Variance score: %.2f' % r2_score(y, ypredict))
+# Mean squared log error
+print('Mean squared log error: %.2f' % mean_squared_log_error(y, ypredict) )
+# Mean absolute error
+print('Mean absolute error: %.2f' % mean_absolute_error(y, ypredict))
+plt.plot(x, ypredict, "r-")
+plt.plot(x, y ,'ro')
+plt.axis([0.0,1.0,1.5, 7.0])
+plt.xlabel(r'$x$')
+plt.ylabel(r'$y$')
+plt.title(r'Linear Regression fit ')
+plt.show()
+
+The function coef gives us the parameter \( \beta \) of our fit while intercept yields +\( \alpha \). Depending on the constant in front of the normal distribution, we get values near or far from \( \alpha =2 \) and \( \beta =5 \). Try to play around with different parameters in front of the normal distribution. The function meansquarederror gives us the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error or loss defined as +
+$$ MSE(\boldsymbol{y},\boldsymbol{\tilde{y}}) = \frac{1}{n} +\sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2, +$$ + +The smaller the value, the better the fit. Ideally we would like to +have an MSE equal zero. The attentive reader has probably recognized +this function as being similar to the \( \chi^2 \) function defined above. +
+ +The r2score function computes \( R^2 \), the coefficient of +determination. It provides a measure of how well future samples are +likely to be predicted by the model. Best possible score is 1.0 and it +can be negative (because the model can be arbitrarily worse). A +constant model that always predicts the expected value of \( \boldsymbol{y} \), +disregarding the input features, would get a \( R^2 \) score of \( 0.0 \). +
+ +If \( \tilde{\boldsymbol{y}}_i \) is the predicted value of the \( i-th \) sample and \( y_i \) is the corresponding true value, then the score \( R^2 \) is defined as
+$$ +R^2(\boldsymbol{y}, \tilde{\boldsymbol{y}}) = 1 - \frac{\sum_{i=0}^{n - 1} (y_i - \tilde{y}_i)^2}{\sum_{i=0}^{n - 1} (y_i - \bar{y})^2}, +$$ + +where we have defined the mean value of \( \boldsymbol{y} \) as
+$$ +\bar{y} = \frac{1}{n} \sum_{i=0}^{n - 1} y_i. +$$ + +Another quantity taht we will meet again in our discussions of regression analysis is + the mean absolute error (MAE), a risk metric corresponding to the expected value of the absolute error loss or what we call the \( l1 \)-norm loss. In our discussion above we presented the relative error. +The MAE is defined as follows +
+$$ +\text{MAE}(\boldsymbol{y}, \boldsymbol{\tilde{y}}) = \frac{1}{n} \sum_{i=0}^{n-1} \left| y_i - \tilde{y}_i \right|. +$$ + +We present the +squared logarithmic (quadratic) error +
+$$ +\text{MSLE}(\boldsymbol{y}, \boldsymbol{\tilde{y}}) = \frac{1}{n} \sum_{i=0}^{n - 1} (\log_e (1 + y_i) - \log_e (1 + \tilde{y}_i) )^2, +$$ + +where \( \log_e (x) \) stands for the natural logarithm of \( x \). This error +estimate is best to use when targets having exponential growth, such +as population counts, average sales of a commodity over a span of +years etc. +
+ +Finally, another cost function is the Huber cost function used in robust regression.
+ +The rationale behind this possible cost function is its reduced +sensitivity to outliers in the data set. In our discussions on +dimensionality reduction and normalization of data we will meet other +ways of dealing with outliers. +
+ +The Huber cost function is defined as
+$$ +H_{\delta}(\boldsymbol{a})=\left\{\begin{array}{cc}\frac{1}{2} \boldsymbol{a}^{2}& \text{for }|\boldsymbol{a}|\leq \delta\\ \delta (|\boldsymbol{a}|-\frac{1}{2}\delta ),&\text{otherwise}.\end{array}\right. +$$ + +Here \( \boldsymbol{a}=\boldsymbol{y} - \boldsymbol{\tilde{y}} \).
+ +We will discuss in more detail these and other functions in the +various lectures and lab sessions. +
+Let us now dive into nuclear physics and remind ourselves briefly about some basic features about binding +energies. A basic quantity which can be measured for the ground +states of nuclei is the atomic mass \( M(N, Z) \) of the neutral atom with +atomic mass number \( A \) and charge \( Z \). The number of neutrons is \( N \). There are indeed several sophisticated experiments worldwide which allow us to measure this quantity to high precision (parts per million even). +
+ +Atomic masses are usually tabulated in terms of the mass excess defined by
+$$ +\Delta M(N, Z) = M(N, Z) - uA, +$$ + +where \( u \) is the Atomic Mass Unit
+$$ +u = M(^{12}\mathrm{C})/12 = 931.4940954(57) \hspace{0.1cm} \mathrm{MeV}/c^2. +$$ + +The nucleon masses are
+$$ +m_p = 1.00727646693(9)u, +$$ + +and
+$$ +m_n = 939.56536(8)\hspace{0.1cm} \mathrm{MeV}/c^2 = 1.0086649156(6)u. +$$ + +In the 2016 mass evaluation of by W.J.Huang, G.Audi, M.Wang, F.G.Kondev, S.Naimi and X.Xu +there are data on masses and decays of 3437 nuclei. +
+ +The nuclear binding energy is defined as the energy required to break +up a given nucleus into its constituent parts of \( N \) neutrons and \( Z \) +protons. In terms of the atomic masses \( M(N, Z) \) the binding energy is +defined by +
+ +$$ +BE(N, Z) = ZM_H c^2 + Nm_n c^2 - M(N, Z)c^2 , +$$ + +where \( M_H \) is the mass of the hydrogen atom and \( m_n \) is the mass of the neutron. +In terms of the mass excess the binding energy is given by +
+$$ +BE(N, Z) = Z\Delta_H c^2 + N\Delta_n c^2 -\Delta(N, Z)c^2 , +$$ + +where \( \Delta_H c^2 = 7.2890 \) MeV and \( \Delta_n c^2 = 8.0713 \) MeV.
+ +A popular and physically intuitive model which can be used to parametrize +the experimental binding energies as function of \( A \), is the so-called +liquid drop model. The ansatz is based on the following expression +
+ +$$ +BE(N,Z) = a_1A-a_2A^{2/3}-a_3\frac{Z^2}{A^{1/3}}-a_4\frac{(N-Z)^2}{A}, +$$ + +where \( A \) stands for the number of nucleons and the $a_i$s are parameters which are determined by a fit +to the experimental data. +
+ +To arrive at the above expression we have assumed that we can make the following assumptions:
-Fitting a continuous function with linear parameterization in terms of the parameters \( \boldsymbol{\beta} \).
For more discussions of Ridge and Lasso regression, Wessel van Wieringen's article is highly recommended. -Similarly, Mehta et al's article is also recommended. +
We could also add a so-called pairing term, which is a correction term that +arises from the tendency of proton pairs and neutron pairs to +occur. An even number of particles is more stable than an odd number. +
+Let us start with reading and organizing our data. +We start with the compilation of masses and binding energies from 2016. +After having downloaded this file to our own computer, we are now ready to read the file and start structuring our data. +
+ +We start with preparing folders for storing our calculations and the data file over masses and binding energies. We import also various modules that we will find useful in order to present various Machine Learning methods. Here we focus mainly on the functionality of scikit-learn.
+ + +# Common imports
+import numpy as np
+import pandas as pd
+import matplotlib.pyplot as plt
+import sklearn.linear_model as skl
+from sklearn.model_selection import train_test_split
+from sklearn.metrics import mean_squared_error, r2_score, mean_absolute_error
+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("MassEval2016.dat"),'r')
+
+Before we proceed, we define also a function for making our plots. You can obviously avoid this and simply set up various matplotlib commands every time you need them. You may however find it convenient to collect all such commands in one function and simply call this function.
+ + +from pylab import plt, mpl
+plt.style.use('seaborn')
+mpl.rcParams['font.family'] = 'serif'
+
+def MakePlot(x,y, styles, labels, axlabels):
+ plt.figure(figsize=(10,6))
+ for i in range(len(x)):
+ plt.plot(x[i], y[i], styles[i], label = labels[i])
+ plt.xlabel(axlabels[0])
+ plt.ylabel(axlabels[1])
+ plt.legend(loc=0)
+
+Our next step is to read the data on experimental binding energies and +reorganize them as functions of the mass number \( A \), the number of +protons \( Z \) and neutrons \( N \) using pandas. Before we do this it is +always useful (unless you have a binary file or other types of compressed +data) to actually open the file and simply take a look at it! +
+ +In particular, the program that outputs the final nuclear masses is written in Fortran with a specific format. It means that we need to figure out the format and which columns contain the data we are interested in. Pandas comes with a function that reads formatted output. After having admired the file, we are now ready to start massaging it with pandas. The file begins with some basic format information.
+ + +"""
+This is taken from the data file of the mass 2016 evaluation.
+All files are 3436 lines long with 124 character per line.
+ Headers are 39 lines long.
+ col 1 : Fortran character control: 1 = page feed 0 = line feed
+ format : a1,i3,i5,i5,i5,1x,a3,a4,1x,f13.5,f11.5,f11.3,f9.3,1x,a2,f11.3,f9.3,1x,i3,1x,f12.5,f11.5
+ These formats are reflected in the pandas widths variable below, see the statement
+ widths=(1,3,5,5,5,1,3,4,1,13,11,11,9,1,2,11,9,1,3,1,12,11,1),
+ Pandas has also a variable header, with length 39 in this case.
+"""
+
+The data we are interested in are in columns 2, 3, 4 and 11, giving us +the number of neutrons, protons, mass numbers and binding energies, +respectively. We add also for the sake of completeness the element name. The data are in fixed-width formatted lines and we will +covert them into the pandas DataFrame structure. +
+ + + +# Read the experimental data with Pandas
+Masses = pd.read_fwf(infile, usecols=(2,3,4,6,11),
+ names=('N', 'Z', 'A', 'Element', 'Ebinding'),
+ widths=(1,3,5,5,5,1,3,4,1,13,11,11,9,1,2,11,9,1,3,1,12,11,1),
+ header=39,
+ index_col=False)
+
+# Extrapolated values are indicated by '#' in place of the decimal place, so
+# the Ebinding column won't be numeric. Coerce to float and drop these entries.
+Masses['Ebinding'] = pd.to_numeric(Masses['Ebinding'], errors='coerce')
+Masses = Masses.dropna()
+# Convert from keV to MeV.
+Masses['Ebinding'] /= 1000
+
+# Group the DataFrame by nucleon number, A.
+Masses = Masses.groupby('A')
+# Find the rows of the grouped DataFrame with the maximum binding energy.
+Masses = Masses.apply(lambda t: t[t.Ebinding==t.Ebinding.max()])
+
+We have now read in the data, grouped them according to the variables we are interested in. +We see how easy it is to reorganize the data using pandas. If we +were to do these operations in C/C++ or Fortran, we would have had to +write various functions/subroutines which perform the above +reorganizations for us. Having reorganized the data, we can now start +to make some simple fits using both the functionalities in numpy and +Scikit-Learn afterwards. +
+ +Now we define five variables which contain +the number of nucleons \( A \), the number of protons \( Z \) and the number of neutrons \( N \), the element name and finally the energies themselves. +
+ + +A = Masses['A']
+Z = Masses['Z']
+N = Masses['N']
+Element = Masses['Element']
+Energies = Masses['Ebinding']
+print(Masses)
+
+The next step, and we will define this mathematically later, is to set up the so-called design matrix. We will throughout call this matrix \( \boldsymbol{X} \). +It has dimensionality \( p\times n \), where \( n \) is the number of data points and \( p \) are the so-called predictors. In our case here they are given by the number of polynomials in \( A \) we wish to include in the fit. +
+ + +# Now we set up the design matrix X
+X = np.zeros((len(A),5))
+X[:,0] = 1
+X[:,1] = A
+X[:,2] = A**(2.0/3.0)
+X[:,3] = A**(-1.0/3.0)
+X[:,4] = A**(-1.0)
+
+With scikitlearn we are now ready to use linear regression and fit our data.
+ + +clf = skl.LinearRegression().fit(X, Energies)
+fity = clf.predict(X)
+
+Pretty simple! +Now we can print measures of how our fit is doing, the coefficients from the fits and plot the final fit together with our data. +
+ + +# The mean squared error
+print("Mean squared error: %.2f" % mean_squared_error(Energies, fity))
+# Explained variance score: 1 is perfect prediction
+print('Variance score: %.2f' % r2_score(Energies, fity))
+# Mean absolute error
+print('Mean absolute error: %.2f' % mean_absolute_error(Energies, fity))
+print(clf.coef_, clf.intercept_)
+
+Masses['Eapprox'] = fity
+# Generate a plot comparing the experimental with the fitted values values.
+fig, ax = plt.subplots()
+ax.set_xlabel(r'$A = N + Z$')
+ax.set_ylabel(r'$E_\mathrm{bind}\,/\mathrm{MeV}$')
+ax.plot(Masses['A'], Masses['Ebinding'], alpha=0.7, lw=2,
+ label='Ame2016')
+ax.plot(Masses['A'], Masses['Eapprox'], alpha=0.7, lw=2, c='m',
+ label='Fit')
+ax.legend()
+save_fig("Masses2016")
+plt.show()
+
+The seaborn package allows us to visualize data in an efficient way. Note that we use scikit-learn's multi-layer perceptron (or feed forward neural network) +functionality. +
+ + +from sklearn.neural_network import MLPRegressor
+from sklearn.metrics import accuracy_score
+import seaborn as sns
+
+
+X_train = X
+Y_train = Energies
+n_hidden_neurons = 50
+epochs = 100
+# store models for later use
+eta_vals = np.logspace(-3, 0, 4)
+lmbd_vals = np.logspace(-3, 0, 4)
+# store the models for later use
+DNN_scikit = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)
+train_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='relu', solver='adam',
+ alpha=lmbd, learning_rate_init=eta, max_iter=epochs)
+ dnn.fit(X_train, Y_train)
+ DNN_scikit[i][j] = dnn
+ train_accuracy[i][j] = dnn.score(X_train, Y_train)
+ fity = dnn.predict(X_train)
+ MSE = mean_squared_error(Y_train, fity)
+ print("Mean squared error: %.2f" % mean_squared_error(Y_train, fity))
+ train_accuracy[i][j] = MSE
+fig, ax = plt.subplots(figsize = (10, 10))
+sns.heatmap(train_accuracy, annot=True, ax=ax, cmap="viridis")
+ax.set_title("Training Accuracy")
+ax.set_ylabel("$\eta$")
+ax.set_xlabel("$\lambda$")
+plt.show()
+print(train_accuracy)
+
+The aim behind these introductory words was to present to you various +Python libraries and their functionalities, in particular libraries like +numpy, pandas, xarray and matplotlib and other that make our life much easier +in handling various data sets and visualizing data. +
+ +Furthermore, +Scikit-Learn allows us with few lines of code to implement popular +Machine Learning algorithms for supervised learning. Later we will meet Tensorflow, a powerful library for deep learning. +Now it is time to dive more into the details of various methods. We will start with linear regression and try to take a deeper look at what it entails.
diff --git a/doc/pub/week34/html/._week34-bs039.html b/doc/pub/week34/html/._week34-bs039.html index 3c0172398..f17e95e55 100644 --- a/doc/pub/week34/html/._week34-bs039.html +++ b/doc/pub/week34/html/._week34-bs039.html @@ -120,6 +120,7 @@ doconce format html week34.do.txt --html_style=bootstrap --pygments_html_style=d ('Numpy and arrays', 2, None, 'numpy-and-arrays'), ('Matrices in Python', 2, None, 'matrices-in-python'), ('Meet the Pandas', 2, None, 'meet-the-pandas'), + ('Pandas AI', 2, None, 'pandas-ai'), ('Simple linear regression model using _scikit-learn_', 3, None, @@ -189,10 +190,6 @@ doconce format html week34.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'interpretations-and-optimizing-our-parameters'), - ('Some useful matrix and vector expressions', - 2, - None, - 'some-useful-matrix-and-vector-expressions'), ('Interpretations and optimizing our parameters', 2, None, @@ -303,26 +300,26 @@ MathJax.Hub.Config({
-
Regression modeling deals with the description of the sampling distribution of a given random variable \( y \) and how it varies as function of another variable or a set of such variables \( \boldsymbol{x} =[x_0, x_1,\dots, x_{n-1}]^T \). -The first variable is called the dependent, the outcome or the response variable while the set of variables \( \boldsymbol{x} \) is called the independent variable, or the predictor variable or the explanatory variable, or simply just the inputs. -
- -A regression model aims at finding a likelihood function \( p(\boldsymbol{y}\vert \boldsymbol{x}) \) or in the more traditional sense a function \( \boldsymbol{y}(\boldsymbol{x}) \), that is the conditional distribution for \( \boldsymbol{y} \) with a given \( \boldsymbol{x} \). The estimation of \( p(\boldsymbol{y}\vert \boldsymbol{x}) \) is made using a data set with
+Fitting a continuous function with linear parameterization in terms of the parameters \( \boldsymbol{\beta} \).
The goal of the regression analysis is to extract/exploit relationship between \( \boldsymbol{y} \) and \( \boldsymbol{x} \) in order to infer specific dependencies, approximations to the likelihood functions, functional relationships and to make predictions, making fits and many other things.
-For more discussions of Ridge and Lasso regression, Wessel van Wieringen's article is highly recommended. +Similarly, Mehta et al's article is also recommended. +
diff --git a/doc/pub/week34/html/._week34-bs040.html b/doc/pub/week34/html/._week34-bs040.html index 407afaac4..50e2bd647 100644 --- a/doc/pub/week34/html/._week34-bs040.html +++ b/doc/pub/week34/html/._week34-bs040.html @@ -120,6 +120,7 @@ doconce format html week34.do.txt --html_style=bootstrap --pygments_html_style=d ('Numpy and arrays', 2, None, 'numpy-and-arrays'), ('Matrices in Python', 2, None, 'matrices-in-python'), ('Meet the Pandas', 2, None, 'meet-the-pandas'), + ('Pandas AI', 2, None, 'pandas-ai'), ('Simple linear regression model using _scikit-learn_', 3, None, @@ -189,10 +190,6 @@ doconce format html week34.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'interpretations-and-optimizing-our-parameters'), - ('Some useful matrix and vector expressions', - 2, - None, - 'some-useful-matrix-and-vector-expressions'), ('Interpretations and optimizing our parameters', 2, None, @@ -303,26 +300,26 @@ MathJax.Hub.Config({
-
Consider an experiment in which \( p \) characteristics/features of \( n \) samples are -measured. The data from this experiment, for various explanatory variables \( p \) are normally represented by a matrix -\( \mathbf{X} \). +
Regression modeling deals with the description of the sampling distribution of a given random variable \( y \) and how it varies as function of another variable or a set of such variables \( \boldsymbol{x} =[x_0, x_1,\dots, x_{n-1}]^T \). +The first variable is called the dependent, the outcome or the response variable while the set of variables \( \boldsymbol{x} \) is called the independent variable, or the predictor variable or the explanatory variable, or simply just the inputs.
-The matrix \( \mathbf{X} \) is called the design -matrix. Additional information of the samples is available in the -form of \( \boldsymbol{y} \) (also as above). The variable \( \boldsymbol{y} \) is -generally referred to as the response variable. The aim of -regression analysis is to explain \( \boldsymbol{y} \) in terms of -\( \boldsymbol{X} \) through a functional relationship like \( y_i = -f(\mathbf{X}_{i,\ast}) \). When no prior knowledge on the form of -\( f(\cdot) \) is available, it is common to assume a linear relationship -between \( \boldsymbol{X} \) and \( \boldsymbol{y} \). This assumption gives rise to -the linear regression model where \( \boldsymbol{\beta} = [\beta_0, \ldots, -\beta_{p-1}]^{T} \) are the regression parameters. -
- -Linear regression gives us a set of analytical equations for the parameters \( \beta_j \).
+A regression model aims at finding a likelihood function \( p(\boldsymbol{y}\vert \boldsymbol{x}) \) or in the more traditional sense a function \( \boldsymbol{y}(\boldsymbol{x}) \), that is the conditional distribution for \( \boldsymbol{y} \) with a given \( \boldsymbol{x} \). The estimation of \( p(\boldsymbol{y}\vert \boldsymbol{x}) \) is made using a data set with
+The goal of the regression analysis is to extract/exploit relationship between \( \boldsymbol{y} \) and \( \boldsymbol{x} \) in order to infer specific dependencies, approximations to the likelihood functions, functional relationships and to make predictions, making fits and many other things.
-
In order to understand the relation among the predictors (or features or properties) \( p \), the set of data \( n \) and the target (outcome, output etc) \( \boldsymbol{y} \), -consider the model we discussed for describing nuclear binding energies. + +
Consider an experiment in which \( p \) characteristics/features of \( n \) samples are +measured. The data from this experiment, for various explanatory variables \( p \) are normally represented by a matrix +\( \mathbf{X} \).
-There we assumed that we could parametrize the data using a polynomial approximation based on the liquid drop model. -Assuming -
-$$ -BE(A) = a_0+a_1A+a_2A^{2/3}+a_3A^{-1/3}+a_4A^{-1}, -$$ - -we have five predictors, that is the intercept, the \( A \) dependent term, the \( A^{2/3} \) term and the \( A^{-1/3} \) and \( A^{-1} \) terms. -This gives \( p=0,1,2,3,4 \). Furthermore we have \( n \) entries for each predictor. It means that our design matrix is a -\( p\times n \) matrix \( \boldsymbol{X} \). +
The matrix \( \mathbf{X} \) is called the design +matrix. Additional information of the samples is available in the +form of \( \boldsymbol{y} \) (also as above). The variable \( \boldsymbol{y} \) is +generally referred to as the response variable. The aim of +regression analysis is to explain \( \boldsymbol{y} \) in terms of +\( \boldsymbol{X} \) through a functional relationship like \( y_i = +f(\mathbf{X}_{i,\ast}) \). When no prior knowledge on the form of +\( f(\cdot) \) is available, it is common to assume a linear relationship +between \( \boldsymbol{X} \) and \( \boldsymbol{y} \). This assumption gives rise to +the linear regression model where \( \boldsymbol{\beta} = [\beta_0, \ldots, +\beta_{p-1}]^{T} \) are the regression parameters.
-Here the predictors are based on a model we have made. A popular data set which is widely encountered in ML applications is the -so-called credit card default data from Taiwan. The data set contains data on \( n=30000 \) credit card holders with predictors like gender, marital status, age, profession, education, etc. In total there are \( 24 \) such predictors or attributes leading to a design matrix of dimensionality \( 24 \times 30000 \). This is however a classification problem and we will come back to it when we discuss Logistic Regression. -
+Linear regression gives us a set of analytical equations for the parameters \( \beta_j \).
-
Before we proceed let us study a case where we aim at fitting a set of data \( \boldsymbol{y}=[y_0,y_1,\dots,y_{n-1}] \). We could think of these data as a result of an experiment or a complicated numerical experiment. These data are functions of a series of variables \( \boldsymbol{x}=[x_0,x_1,\dots,x_{n-1}] \), that is \( y_i = y(x_i) \) with \( i=0,1,2,\dots,n-1 \). The variables \( x_i \) could represent physical quantities like time, temperature, position etc. We assume that \( y(x) \) is a smooth function.
+In order to understand the relation among the predictors (or features or properties) \( p \), the set of data \( n \) and the target (outcome, output etc) \( \boldsymbol{y} \), +consider the model we discussed for describing nuclear binding energies. +
-Since obtaining these data points may not be trivial, we want to use these data to fit a function which can allow us to make predictions for values of \( y \) which are not in the present set. The perhaps simplest approach is to assume we can parametrize our function in terms of a polynomial of degree \( n-1 \) with \( n \) points, that is
+There we assumed that we could parametrize the data using a polynomial approximation based on the liquid drop model. +Assuming +
$$ -y=y(x) \rightarrow y(x_i)=\tilde{y}_i+\epsilon_i=\sum_{j=0}^{n-1} \beta_j x_i^j+\epsilon_i, +BE(A) = a_0+a_1A+a_2A^{2/3}+a_3A^{-1/3}+a_4A^{-1}, $$ -where \( \epsilon_i \) is the error in our approximation.
+we have five predictors, that is the intercept, the \( A \) dependent term, the \( A^{2/3} \) term and the \( A^{-1/3} \) and \( A^{-1} \) terms. +This gives \( p=0,1,2,3,4 \). Furthermore we have \( n \) entries for each predictor. It means that our design matrix is a +\( p\times n \) matrix \( \boldsymbol{X} \). +
+ +Here the predictors are based on a model we have made. A popular data set which is widely encountered in ML applications is the +so-called credit card default data from Taiwan. The data set contains data on \( n=30000 \) credit card holders with predictors like gender, marital status, age, profession, education, etc. In total there are \( 24 \) such predictors or attributes leading to a design matrix of dimensionality \( 24 \times 30000 \). This is however a classification problem and we will come back to it when we discuss Logistic Regression. +
-
For every set of values \( y_i,x_i \) we have thus the corresponding set of equations
+Before we proceed let us study a case where we aim at fitting a set of data \( \boldsymbol{y}=[y_0,y_1,\dots,y_{n-1}] \). We could think of these data as a result of an experiment or a complicated numerical experiment. These data are functions of a series of variables \( \boldsymbol{x}=[x_0,x_1,\dots,x_{n-1}] \), that is \( y_i = y(x_i) \) with \( i=0,1,2,\dots,n-1 \). The variables \( x_i \) could represent physical quantities like time, temperature, position etc. We assume that \( y(x) \) is a smooth function.
+ +Since obtaining these data points may not be trivial, we want to use these data to fit a function which can allow us to make predictions for values of \( y \) which are not in the present set. The perhaps simplest approach is to assume we can parametrize our function in terms of a polynomial of degree \( n-1 \) with \( n \) points, that is
$$ -\begin{align*} -y_0&=\beta_0+\beta_1x_0^1+\beta_2x_0^2+\dots+\beta_{n-1}x_0^{n-1}+\epsilon_0\\ -y_1&=\beta_0+\beta_1x_1^1+\beta_2x_1^2+\dots+\beta_{n-1}x_1^{n-1}+\epsilon_1\\ -y_2&=\beta_0+\beta_1x_2^1+\beta_2x_2^2+\dots+\beta_{n-1}x_2^{n-1}+\epsilon_2\\ -\dots & \dots \\ -y_{n-1}&=\beta_0+\beta_1x_{n-1}^1+\beta_2x_{n-1}^2+\dots+\beta_{n-1}x_{n-1}^{n-1}+\epsilon_{n-1}.\\ -\end{align*} +y=y(x) \rightarrow y(x_i)=\tilde{y}_i+\epsilon_i=\sum_{j=0}^{n-1} \beta_j x_i^j+\epsilon_i, $$ + +where \( \epsilon_i \) is the error in our approximation.
-
Defining the vectors
+For every set of values \( y_i,x_i \) we have thus the corresponding set of equations
$$ -\boldsymbol{y} = [y_0,y_1, y_2,\dots, y_{n-1}]^T, +\begin{align*} +y_0&=\beta_0+\beta_1x_0^1+\beta_2x_0^2+\dots+\beta_{n-1}x_0^{n-1}+\epsilon_0\\ +y_1&=\beta_0+\beta_1x_1^1+\beta_2x_1^2+\dots+\beta_{n-1}x_1^{n-1}+\epsilon_1\\ +y_2&=\beta_0+\beta_1x_2^1+\beta_2x_2^2+\dots+\beta_{n-1}x_2^{n-1}+\epsilon_2\\ +\dots & \dots \\ +y_{n-1}&=\beta_0+\beta_1x_{n-1}^1+\beta_2x_{n-1}^2+\dots+\beta_{n-1}x_{n-1}^{n-1}+\epsilon_{n-1}.\\ +\end{align*} $$ - -and
-$$ -\boldsymbol{\beta} = [\beta_0,\beta_1, \beta_2,\dots, \beta_{n-1}]^T, -$$ - -and
-$$ -\boldsymbol{\epsilon} = [\epsilon_0,\epsilon_1, \epsilon_2,\dots, \epsilon_{n-1}]^T, -$$ - -and the design matrix
-$$ -\boldsymbol{X}= -\begin{bmatrix} -1& x_{0}^1 &x_{0}^2& \dots & \dots &x_{0}^{n-1}\\ -1& x_{1}^1 &x_{1}^2& \dots & \dots &x_{1}^{n-1}\\ -1& x_{2}^1 &x_{2}^2& \dots & \dots &x_{2}^{n-1}\\ -\dots& \dots &\dots& \dots & \dots &\dots\\ -1& x_{n-1}^1 &x_{n-1}^2& \dots & \dots &x_{n-1}^{n-1}\\ -\end{bmatrix} -$$ - -we can rewrite our equations as
-$$ -\boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta}+\boldsymbol{\epsilon}. -$$ - -The above design matrix is called a Vandermonde matrix.
-
We are obviously not limited to the above polynomial expansions. We -could replace the various powers of \( x \) with elements of Fourier -series or instead of \( x_i^j \) we could have \( \cos{(j x_i)} \) or \( \sin{(j -x_i)} \), or time series or other orthogonal functions. For every set -of values \( y_i,x_i \) we can then generalize the equations to -
- +Defining the vectors
$$ -\begin{align*} -y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ -y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ -y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_2\\ -\dots & \dots \\ -y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_i\\ -\dots & \dots \\ -y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_{n-1}x_{n-1,n-1}+\epsilon_{n-1}.\\ -\end{align*} +\boldsymbol{y} = [y_0,y_1, y_2,\dots, y_{n-1}]^T, $$ +and
+$$ +\boldsymbol{\beta} = [\beta_0,\beta_1, \beta_2,\dots, \beta_{n-1}]^T, +$$ -Note that we have \( p=n \) here. The matrix is symmetric. This is generally not the case! +and
+$$ +\boldsymbol{\epsilon} = [\epsilon_0,\epsilon_1, \epsilon_2,\dots, \epsilon_{n-1}]^T, +$$ + +and the design matrix
+$$ +\boldsymbol{X}= +\begin{bmatrix} +1& x_{0}^1 &x_{0}^2& \dots & \dots &x_{0}^{n-1}\\ +1& x_{1}^1 &x_{1}^2& \dots & \dots &x_{1}^{n-1}\\ +1& x_{2}^1 &x_{2}^2& \dots & \dots &x_{2}^{n-1}\\ +\dots& \dots &\dots& \dots & \dots &\dots\\ +1& x_{n-1}^1 &x_{n-1}^2& \dots & \dots &x_{n-1}^{n-1}\\ +\end{bmatrix} +$$ + +we can rewrite our equations as
+$$ +\boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta}+\boldsymbol{\epsilon}. +$$ + +The above design matrix is called a Vandermonde matrix.
We redefine in turn the matrix \( \boldsymbol{X} \) as
+ +We are obviously not limited to the above polynomial expansions. We +could replace the various powers of \( x \) with elements of Fourier +series or instead of \( x_i^j \) we could have \( \cos{(j x_i)} \) or \( \sin{(j +x_i)} \), or time series or other orthogonal functions. For every set +of values \( y_i,x_i \) we can then generalize the equations to +
+ $$ -\boldsymbol{X}= -\begin{bmatrix} -x_{00}& x_{01} &x_{02}& \dots & \dots &x_{0,n-1}\\ -x_{10}& x_{11} &x_{12}& \dots & \dots &x_{1,n-1}\\ -x_{20}& x_{21} &x_{22}& \dots & \dots &x_{2,n-1}\\ -\dots& \dots &\dots& \dots & \dots &\dots\\ -x_{n-1,0}& x_{n-1,1} &x_{n-1,2}& \dots & \dots &x_{n-1,n-1}\\ -\end{bmatrix} +\begin{align*} +y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ +y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ +y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_2\\ +\dots & \dots \\ +y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_i\\ +\dots & \dots \\ +y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_{n-1}x_{n-1,n-1}+\epsilon_{n-1}.\\ +\end{align*} $$ -and without loss of generality we rewrite again our equations as
-$$ -\boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta}+\boldsymbol{\epsilon}. -$$ -The left-hand side of this equation is kwown. Our error vector \( \boldsymbol{\epsilon} \) and the parameter vector \( \boldsymbol{\beta} \) are our unknow quantities. How can we obtain the optimal set of \( \beta_i \) values?
+Note that we have \( p=n \) here. The matrix is symmetric. This is generally not the case!
-
We have defined the matrix \( \boldsymbol{X} \) via the equations
+We redefine in turn the matrix \( \boldsymbol{X} \) as
$$ -\begin{align*} -y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ -y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ -y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_1\\ -\dots & \dots \\ -y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_1\\ -\dots & \dots \\ -y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_{n-1}x_{n-1,n-1}+\epsilon_{n-1}.\\ -\end{align*} +\boldsymbol{X}= +\begin{bmatrix} +x_{00}& x_{01} &x_{02}& \dots & \dots &x_{0,n-1}\\ +x_{10}& x_{11} &x_{12}& \dots & \dots &x_{1,n-1}\\ +x_{20}& x_{21} &x_{22}& \dots & \dots &x_{2,n-1}\\ +\dots& \dots &\dots& \dots & \dots &\dots\\ +x_{n-1,0}& x_{n-1,1} &x_{n-1,2}& \dots & \dots &x_{n-1,n-1}\\ +\end{bmatrix} $$ -As we noted above, we stayed with a system with the design matrix - \( \boldsymbol{X}\in {\mathbb{R}}^{n\times n} \), that is we have \( p=n \). For reasons to come later (algorithmic arguments) we will hereafter define -our matrix as \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \), with the predictors refering to the column numbers and the entries \( n \) being the row elements. -
+and without loss of generality we rewrite again our equations as
+$$ +\boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta}+\boldsymbol{\epsilon}. +$$ + +The left-hand side of this equation is kwown. Our error vector \( \boldsymbol{\epsilon} \) and the parameter vector \( \boldsymbol{\beta} \) are our unknow quantities. How can we obtain the optimal set of \( \beta_i \) values?
-
In our introductory notes we looked at the so-called liquid drop model. Let us remind ourselves about what we did by looking at the code.
- -We restate the parts of the code we are most interested in.
- - -# Common imports
-import numpy as np
-import pandas as pd
-import matplotlib.pyplot as plt
-from IPython.display import display
-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("MassEval2016.dat"),'r')
-
-
-# Read the experimental data with Pandas
-Masses = pd.read_fwf(infile, usecols=(2,3,4,6,11),
- names=('N', 'Z', 'A', 'Element', 'Ebinding'),
- widths=(1,3,5,5,5,1,3,4,1,13,11,11,9,1,2,11,9,1,3,1,12,11,1),
- header=39,
- index_col=False)
-
-# Extrapolated values are indicated by '#' in place of the decimal place, so
-# the Ebinding column won't be numeric. Coerce to float and drop these entries.
-Masses['Ebinding'] = pd.to_numeric(Masses['Ebinding'], errors='coerce')
-Masses = Masses.dropna()
-# Convert from keV to MeV.
-Masses['Ebinding'] /= 1000
-
-# Group the DataFrame by nucleon number, A.
-Masses = Masses.groupby('A')
-# Find the rows of the grouped DataFrame with the maximum binding energy.
-Masses = Masses.apply(lambda t: t[t.Ebinding==t.Ebinding.max()])
-A = Masses['A']
-Z = Masses['Z']
-N = Masses['N']
-Element = Masses['Element']
-Energies = Masses['Ebinding']
-
-# Now we set up the design matrix X
-X = np.zeros((len(A),5))
-X[:,0] = 1
-X[:,1] = A
-X[:,2] = A**(2.0/3.0)
-X[:,3] = A**(-1.0/3.0)
-X[:,4] = A**(-1.0)
-# Then nice printout using pandas
-DesignMatrix = pd.DataFrame(X)
-DesignMatrix.index = A
-DesignMatrix.columns = ['1', 'A', 'A^(2/3)', 'A^(-1/3)', '1/A']
-display(DesignMatrix)
-
-With \( \boldsymbol{\beta}\in {\mathbb{R}}^{p\times 1} \), it means that we will hereafter write our equations for the approximation as
+We have defined the matrix \( \boldsymbol{X} \) via the equations
$$ -\boldsymbol{\tilde{y}}= \boldsymbol{X}\boldsymbol{\beta}, +\begin{align*} +y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ +y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ +y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_1\\ +\dots & \dots \\ +y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_1\\ +\dots & \dots \\ +y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_{n-1}x_{n-1,n-1}+\epsilon_{n-1}.\\ +\end{align*} $$ -throughout these lectures.
+As we noted above, we stayed with a system with the design matrix + \( \boldsymbol{X}\in {\mathbb{R}}^{n\times n} \), that is we have \( p=n \). For reasons to come later (algorithmic arguments) we will hereafter define +our matrix as \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \), with the predictors refering to the column numbers and the entries \( n \) being the row elements. +
+diff --git a/doc/pub/week34/html/._week34-bs049.html b/doc/pub/week34/html/._week34-bs049.html index e46e17785..930dce2d6 100644 --- a/doc/pub/week34/html/._week34-bs049.html +++ b/doc/pub/week34/html/._week34-bs049.html @@ -120,6 +120,7 @@ doconce format html week34.do.txt --html_style=bootstrap --pygments_html_style=d ('Numpy and arrays', 2, None, 'numpy-and-arrays'), ('Matrices in Python', 2, None, 'matrices-in-python'), ('Meet the Pandas', 2, None, 'meet-the-pandas'), + ('Pandas AI', 2, None, 'pandas-ai'), ('Simple linear regression model using _scikit-learn_', 3, None, @@ -189,10 +190,6 @@ doconce format html week34.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'interpretations-and-optimizing-our-parameters'), - ('Some useful matrix and vector expressions', - 2, - None, - 'some-useful-matrix-and-vector-expressions'), ('Interpretations and optimizing our parameters', 2, None, @@ -303,26 +300,26 @@ MathJax.Hub.Config({
-
With the above we use the design matrix to define the approximation \( \boldsymbol{\tilde{y}} \) via the unknown quantity \( \boldsymbol{\beta} \) as
+In our introductory notes we looked at the so-called liquid drop model. Let us remind ourselves about what we did by looking at the code.
+ +We restate the parts of the code we are most interested in.
+ + +# Common imports
+import numpy as np
+import pandas as pd
+import matplotlib.pyplot as plt
+from IPython.display import display
+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("MassEval2016.dat"),'r')
+
+
+# Read the experimental data with Pandas
+Masses = pd.read_fwf(infile, usecols=(2,3,4,6,11),
+ names=('N', 'Z', 'A', 'Element', 'Ebinding'),
+ widths=(1,3,5,5,5,1,3,4,1,13,11,11,9,1,2,11,9,1,3,1,12,11,1),
+ header=39,
+ index_col=False)
+
+# Extrapolated values are indicated by '#' in place of the decimal place, so
+# the Ebinding column won't be numeric. Coerce to float and drop these entries.
+Masses['Ebinding'] = pd.to_numeric(Masses['Ebinding'], errors='coerce')
+Masses = Masses.dropna()
+# Convert from keV to MeV.
+Masses['Ebinding'] /= 1000
+
+# Group the DataFrame by nucleon number, A.
+Masses = Masses.groupby('A')
+# Find the rows of the grouped DataFrame with the maximum binding energy.
+Masses = Masses.apply(lambda t: t[t.Ebinding==t.Ebinding.max()])
+A = Masses['A']
+Z = Masses['Z']
+N = Masses['N']
+Element = Masses['Element']
+Energies = Masses['Ebinding']
+
+# Now we set up the design matrix X
+X = np.zeros((len(A),5))
+X[:,0] = 1
+X[:,1] = A
+X[:,2] = A**(2.0/3.0)
+X[:,3] = A**(-1.0/3.0)
+X[:,4] = A**(-1.0)
+# Then nice printout using pandas
+DesignMatrix = pd.DataFrame(X)
+DesignMatrix.index = A
+DesignMatrix.columns = ['1', 'A', 'A^(2/3)', 'A^(-1/3)', '1/A']
+display(DesignMatrix)
+
+With \( \boldsymbol{\beta}\in {\mathbb{R}}^{p\times 1} \), it means that we will hereafter write our equations for the approximation as
$$ \boldsymbol{\tilde{y}}= \boldsymbol{X}\boldsymbol{\beta}, $$ -and in order to find the optimal parameters \( \beta_i \) instead of solving the above linear algebra problem, we define a function which gives a measure of the spread between the values \( y_i \) (which represent hopefully the exact values) and the parameterized values \( \tilde{y}_i \), namely
-$$ -C(\boldsymbol{\beta})=\frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)^T\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)\right\}, -$$ - -or using the matrix \( \boldsymbol{X} \) and in a more compact matrix-vector notation as
-$$ -C(\boldsymbol{\beta})=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}. -$$ - -This function is one possible way to define the so-called cost function.
- -It is also common to define -the function \( C \) as -
- -$$ -C(\boldsymbol{\beta})=\frac{1}{2n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2, -$$ - -since when taking the first derivative with respect to the unknown parameters \( \beta \), the factor of \( 2 \) cancels out.
-throughout these lectures.
diff --git a/doc/pub/week34/html/._week34-bs050.html b/doc/pub/week34/html/._week34-bs050.html index 84c05b464..93121d895 100644 --- a/doc/pub/week34/html/._week34-bs050.html +++ b/doc/pub/week34/html/._week34-bs050.html @@ -120,6 +120,7 @@ doconce format html week34.do.txt --html_style=bootstrap --pygments_html_style=d ('Numpy and arrays', 2, None, 'numpy-and-arrays'), ('Matrices in Python', 2, None, 'matrices-in-python'), ('Meet the Pandas', 2, None, 'meet-the-pandas'), + ('Pandas AI', 2, None, 'pandas-ai'), ('Simple linear regression model using _scikit-learn_', 3, None, @@ -189,10 +190,6 @@ doconce format html week34.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'interpretations-and-optimizing-our-parameters'), - ('Some useful matrix and vector expressions', - 2, - None, - 'some-useful-matrix-and-vector-expressions'), ('Interpretations and optimizing our parameters', 2, None, @@ -303,26 +300,26 @@ MathJax.Hub.Config({
-
The function
+With the above we use the design matrix to define the approximation \( \boldsymbol{\tilde{y}} \) via the unknown quantity \( \boldsymbol{\beta} \) as
$$ -C(\boldsymbol{\beta})=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}, +\boldsymbol{\tilde{y}}= \boldsymbol{X}\boldsymbol{\beta}, $$ -can be linked to the variance of the quantity \( y_i \) if we interpret the latter as the mean value. -When linking (see the discussion below) with the maximum likelihood approach below, we will indeed interpret \( y_i \) as a mean value -
+and in order to find the optimal parameters \( \beta_i \) instead of solving the above linear algebra problem, we define a function which gives a measure of the spread between the values \( y_i \) (which represent hopefully the exact values) and the parameterized values \( \tilde{y}_i \), namely
$$ -y_{i}=\langle y_i \rangle = \beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}+\epsilon_i, +C(\boldsymbol{\beta})=\frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)^T\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)\right\}, $$ -where \( \langle y_i \rangle \) is the mean value. Keep in mind also that -till now we have treated \( y_i \) as the exact value. Normally, the -response (dependent or outcome) variable \( y_i \) the outcome of a -numerical experiment or another type of experiment and is thus only an -approximation to the true value. It is then always accompanied by an -error estimate, often limited to a statistical error estimate given by -the standard deviation discussed earlier. In the discussion here we -will treat \( y_i \) as our exact value for the response variable. +
or using the matrix \( \boldsymbol{X} \) and in a more compact matrix-vector notation as
+$$ +C(\boldsymbol{\beta})=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}. +$$ + +This function is one possible way to define the so-called cost function.
+ +It is also common to define +the function \( C \) as
-In order to find the parameters \( \beta_i \) we will then minimize the spread of \( C(\boldsymbol{\beta}) \), that is we are going to solve the problem
$$ -{\displaystyle \min_{\boldsymbol{\beta}\in -{\mathbb{R}}^{p}}}\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}. +C(\boldsymbol{\beta})=\frac{1}{2n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2, $$ -In practical terms it means we will require
-$$ -\frac{\partial C(\boldsymbol{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)^2\right]=0, -$$ - -which results in
-$$ -\frac{\partial C(\boldsymbol{\beta})}{\partial \beta_j} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)\right]=0, -$$ - -or in a matrix-vector form as
-$$ -\frac{\partial C(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right). -$$ +since when taking the first derivative with respect to the unknown parameters \( \beta \), the factor of \( 2 \) cancels out.
We can rewrite
+ +The function
$$ -\frac{\partial C(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right), +C(\boldsymbol{\beta})=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}, $$ -as
-$$ -\boldsymbol{X}^T\boldsymbol{y} = \boldsymbol{X}^T\boldsymbol{X}\boldsymbol{\beta}, -$$ - -and if the matrix \( \boldsymbol{X}^T\boldsymbol{X} \) is invertible we have the solution
-$$ -\boldsymbol{\beta} =\left(\boldsymbol{X}^T\boldsymbol{X}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}. -$$ - -We note also that since our design matrix is defined as \( \boldsymbol{X}\in -{\mathbb{R}}^{n\times p} \), the product \( \boldsymbol{X}^T\boldsymbol{X} \in -{\mathbb{R}}^{p\times p} \). In the above case we have that \( p \ll n \), -in our case \( p=5 \) meaning that we end up with inverting a small -\( 5\times 5 \) matrix. This is a rather common situation, in many cases we end up with low-dimensional -matrices to invert. The methods discussed here and for many other -supervised learning algorithms like classification with logistic -regression or support vector machines, exhibit dimensionalities which -allow for the usage of direct linear algebra methods such as LU decomposition or Singular Value Decomposition (SVD) for finding the inverse of the matrix -\( \boldsymbol{X}^T\boldsymbol{X} \). +
can be linked to the variance of the quantity \( y_i \) if we interpret the latter as the mean value. +When linking (see the discussion below) with the maximum likelihood approach below, we will indeed interpret \( y_i \) as a mean value
-where \( \langle y_i \rangle \) is the mean value. Keep in mind also that +till now we have treated \( y_i \) as the exact value. Normally, the +response (dependent or outcome) variable \( y_i \) the outcome of a +numerical experiment or another type of experiment and is thus only an +approximation to the true value. It is then always accompanied by an +error estimate, often limited to a statistical error estimate given by +the standard deviation discussed earlier. In the discussion here we +will treat \( y_i \) as our exact value for the response variable. +
-Small question: Do you think the example we have at hand here (the nuclear binding energies) can lead to problems in inverting the matrix \( \boldsymbol{X}^T\boldsymbol{X} \)? What kind of problems can we expect?
+In order to find the parameters \( \beta_i \) we will then minimize the spread of \( C(\boldsymbol{\beta}) \), that is we are going to solve the problem
+$$ +{\displaystyle \min_{\boldsymbol{\beta}\in +{\mathbb{R}}^{p}}}\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}. +$$ + +In practical terms it means we will require
+$$ +\frac{\partial C(\boldsymbol{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)^2\right]=0, +$$ + +which results in
+$$ +\frac{\partial C(\boldsymbol{\beta})}{\partial \beta_j} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)\right]=0, +$$ + +or in a matrix-vector form as
+$$ +\frac{\partial C(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right). +$$
-
We can rewrite
+$$ +\frac{\partial C(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right), +$$ -See the handwritten notes at https://github.com/CompPhysics/MachineLearning/blob/master/doc/HandWrittenNotes/2022/NotesExercise5Week452022.pdf
+as
+$$ +\boldsymbol{X}^T\boldsymbol{y} = \boldsymbol{X}^T\boldsymbol{X}\boldsymbol{\beta}, +$$ + +and if the matrix \( \boldsymbol{X}^T\boldsymbol{X} \) is invertible we have the solution
+$$ +\boldsymbol{\beta} =\left(\boldsymbol{X}^T\boldsymbol{X}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}. +$$ + +We note also that since our design matrix is defined as \( \boldsymbol{X}\in +{\mathbb{R}}^{n\times p} \), the product \( \boldsymbol{X}^T\boldsymbol{X} \in +{\mathbb{R}}^{p\times p} \). In the above case we have that \( p \ll n \), +in our case \( p=5 \) meaning that we end up with inverting a small +\( 5\times 5 \) matrix. This is a rather common situation, in many cases we end up with low-dimensional +matrices to invert. The methods discussed here and for many other +supervised learning algorithms like classification with logistic +regression or support vector machines, exhibit dimensionalities which +allow for the usage of direct linear algebra methods such as LU decomposition or Singular Value Decomposition (SVD) for finding the inverse of the matrix +\( \boldsymbol{X}^T\boldsymbol{X} \). +
+Small question: Do you think the example we have at hand here (the nuclear binding energies) can lead to problems in inverting the matrix \( \boldsymbol{X}^T\boldsymbol{X} \)? What kind of problems can we expect?
+These notes will be discussed during one of the lectures.
diff --git a/doc/pub/week34/html/._week34-bs053.html b/doc/pub/week34/html/._week34-bs053.html index 180b75d03..c2773af3f 100644 --- a/doc/pub/week34/html/._week34-bs053.html +++ b/doc/pub/week34/html/._week34-bs053.html @@ -120,6 +120,7 @@ doconce format html week34.do.txt --html_style=bootstrap --pygments_html_style=d ('Numpy and arrays', 2, None, 'numpy-and-arrays'), ('Matrices in Python', 2, None, 'matrices-in-python'), ('Meet the Pandas', 2, None, 'meet-the-pandas'), + ('Pandas AI', 2, None, 'pandas-ai'), ('Simple linear regression model using _scikit-learn_', 3, None, @@ -189,10 +190,6 @@ doconce format html week34.do.txt --html_style=bootstrap --pygments_html_style=d 2, None, 'interpretations-and-optimizing-our-parameters'), - ('Some useful matrix and vector expressions', - 2, - None, - 'some-useful-matrix-and-vector-expressions'), ('Interpretations and optimizing our parameters', 2, None, @@ -303,26 +300,26 @@ MathJax.Hub.Config({
Try out Pandas AI
+ +See the handwritten notes at https://github.com/CompPhysics/MachineLearning/blob/master/doc/HandWrittenNotes/2022/NotesExercise5Week452022.pdf
- -These notes will be discussed during one of the lectures.
-Try out Pandas AI
+See the handwritten notes at https://github.com/CompPhysics/MachineLearning/blob/master/doc/HandWrittenNotes/2022/NotesExercise5Week452022.pdf
- -These notes will be discussed during one of the lectures.
-Try out Pandas AI
+See the handwritten notes at https://github.com/CompPhysics/MachineLearning/blob/master/doc/HandWrittenNotes/2022/NotesExercise5Week452022.pdf
- -These notes will be discussed during one of the lectures.
-