-
We have tried every year to organize a kind of mini-workshop on project 3. In 2020 the contributions were (and some of these ended up in thesis work and/or publications, online only due to Covid-19!)
+A Support Vector Machine (SVM) is a very powerful and versatile +Machine Learning method, capable of performing linear or nonlinear +classification, regression, and even outlier detection. It is one of +the most popular models in Machine Learning, and anyone interested in +Machine Learning should have it in their toolbox. SVMs are +particularly well suited for classification of complex but small-sized or +medium-sized datasets. +
+ +The case with two well-separated classes only can be understood in an +intuitive way in terms of lines in a two-dimensional space separating +the two classes (see figure below). +
+ +The basic mathematics behind the SVM is however less familiar to most of us. +It relies on the definition of hyperplanes and the +definition of a margin which separates classes (in case of +classification problems) of variables. It is also used for regression +problems. +
+ +With SVMs we distinguish between hard margin and soft margins. The +latter introduces a so-called softening parameter to be discussed +below. We distinguish also between linear and non-linear +approaches. The latter are the most frequent ones since it is rather +unlikely that we can separate classes easily by say straight lines. +
-diff --git a/doc/pub/week46/html/._week46-bs003.html b/doc/pub/week46/html/._week46-bs003.html index 5b082001c..4635dbb96 100644 --- a/doc/pub/week46/html/._week46-bs003.html +++ b/doc/pub/week46/html/._week46-bs003.html @@ -8,8 +8,8 @@ doconce format html week46.do.txt --html_style=bootstrap --pygments_html_style=d - -
-
We would very much like to organize something similar this year as well. Feel free to come with suggesstions by Thursday November 17. We will then try to set up the various contributions for Friday November 18. -The presentations last normally 5-15 mins and span from loose ideas to more well-defined topics. Nothing pretentious, we wish to keep this as low-key as possible. +
The theory behind support vector machines (SVM hereafter) is based on +the mathematical description of so-called hyperplanes. Let us start +with a two-dimensional case. This will also allow us to introduce our +first SVM examples. These will be tailored to the case of two specific +classes, as displayed in the figure here based on the usage of the petal data.
+We assume here that our data set can be well separated into two +domains, where a straight line does the job in the separating the two +classes. Here the two classes are represented by either squares or +circles. +
+ + +from sklearn import datasets
+from sklearn.svm import SVC, LinearSVC
+from sklearn.linear_model import SGDClassifier
+from sklearn.preprocessing import StandardScaler
+import matplotlib
+import matplotlib.pyplot as plt
+plt.rcParams['axes.labelsize'] = 14
+plt.rcParams['xtick.labelsize'] = 12
+plt.rcParams['ytick.labelsize'] = 12
+
+
+iris = datasets.load_iris()
+X = iris["data"][:, (2, 3)] # petal length, petal width
+y = iris["target"]
+
+setosa_or_versicolor = (y == 0) | (y == 1)
+X = X[setosa_or_versicolor]
+y = y[setosa_or_versicolor]
+
+
+
+C = 5
+alpha = 1 / (C * len(X))
+
+lin_clf = LinearSVC(loss="hinge", C=C, random_state=42)
+svm_clf = SVC(kernel="linear", C=C)
+sgd_clf = SGDClassifier(loss="hinge", learning_rate="constant", eta0=0.001, alpha=alpha,
+ max_iter=100000, random_state=42)
+
+scaler = StandardScaler()
+X_scaled = scaler.fit_transform(X)
+
+lin_clf.fit(X_scaled, y)
+svm_clf.fit(X_scaled, y)
+sgd_clf.fit(X_scaled, y)
+
+print("LinearSVC: ", lin_clf.intercept_, lin_clf.coef_)
+print("SVC: ", svm_clf.intercept_, svm_clf.coef_)
+print("SGDClassifier(alpha={:.5f}):".format(sgd_clf.alpha), sgd_clf.intercept_, sgd_clf.coef_)
+
+# Compute the slope and bias of each decision boundary
+w1 = -lin_clf.coef_[0, 0]/lin_clf.coef_[0, 1]
+b1 = -lin_clf.intercept_[0]/lin_clf.coef_[0, 1]
+w2 = -svm_clf.coef_[0, 0]/svm_clf.coef_[0, 1]
+b2 = -svm_clf.intercept_[0]/svm_clf.coef_[0, 1]
+w3 = -sgd_clf.coef_[0, 0]/sgd_clf.coef_[0, 1]
+b3 = -sgd_clf.intercept_[0]/sgd_clf.coef_[0, 1]
+
+# Transform the decision boundary lines back to the original scale
+line1 = scaler.inverse_transform([[-10, -10 * w1 + b1], [10, 10 * w1 + b1]])
+line2 = scaler.inverse_transform([[-10, -10 * w2 + b2], [10, 10 * w2 + b2]])
+line3 = scaler.inverse_transform([[-10, -10 * w3 + b3], [10, 10 * w3 + b3]])
+
+# Plot all three decision boundaries
+plt.figure(figsize=(11, 4))
+plt.plot(line1[:, 0], line1[:, 1], "k:", label="LinearSVC")
+plt.plot(line2[:, 0], line2[:, 1], "b--", linewidth=2, label="SVC")
+plt.plot(line3[:, 0], line3[:, 1], "r-", label="SGDClassifier")
+plt.plot(X[:, 0][y==1], X[:, 1][y==1], "bs") # label="Iris-Versicolor"
+plt.plot(X[:, 0][y==0], X[:, 1][y==0], "yo") # label="Iris-Setosa"
+plt.xlabel("Petal length", fontsize=14)
+plt.ylabel("Petal width", fontsize=14)
+plt.legend(loc="upper center", fontsize=14)
+plt.axis([0, 5.5, 0, 2])
+
+plt.show()
+
+
-
A Support Vector Machine (SVM) is a very powerful and versatile -Machine Learning method, capable of performing linear or nonlinear -classification, regression, and even outlier detection. It is one of -the most popular models in Machine Learning, and anyone interested in -Machine Learning should have it in their toolbox. SVMs are -particularly well suited for classification of complex but small-sized or -medium-sized datasets. +
The aim of the SVM algorithm is to find a hyperplane in a +\( p \)-dimensional space, where \( p \) is the number of features that +distinctly classifies the data points.
-The case with two well-separated classes only can be understood in an -intuitive way in terms of lines in a two-dimensional space separating -the two classes (see figure below). +
In a \( p \)-dimensional space, a hyperplane is what we call an affine subspace of dimension of \( p-1 \). +As an example, in two dimension, a hyperplane is simply as straight line while in three dimensions it is +a two-dimensional subspace, or stated simply, a plane.
-The basic mathematics behind the SVM is however less familiar to most of us. -It relies on the definition of hyperplanes and the -definition of a margin which separates classes (in case of -classification problems) of variables. It is also used for regression -problems. +
In two dimensions, with the variables \( x_1 \) and \( x_2 \), the hyperplane is defined as
+$$ +b+w_1x_1+w_2x_2=0, +$$ + +where \( b \) is the intercept and \( w_1 \) and \( w_2 \) define the elements of a vector orthogonal to the line +\( b+w_1x_1+w_2x_2=0 \). +In two dimensions we define the vectors \( \boldsymbol{x} =[x1,x2] \) and \( \boldsymbol{w}=[w1,w2] \). +We can then rewrite the above equation as
-With SVMs we distinguish between hard margin and soft margins. The -latter introduces a so-called softening parameter to be discussed -below. We distinguish also between linear and non-linear -approaches. The latter are the most frequent ones since it is rather -unlikely that we can separate classes easily by say straight lines. -
+$$ +\boldsymbol{x}^T\boldsymbol{w}+b=0. +$$ +@@ -226,7 +302,7 @@ unlikely that we can separate classes easily by say straight lines.
-
The theory behind support vector machines (SVM hereafter) is based on -the mathematical description of so-called hyperplanes. Let us start -with a two-dimensional case. This will also allow us to introduce our -first SVM examples. These will be tailored to the case of two specific -classes, as displayed in the figure here based on the usage of the petal data. +
We limit ourselves to two classes of outputs \( y_i \) and assign these classes the values \( y_i = \pm 1 \). +In a \( p \)-dimensional space of say \( p \) features we have a hyperplane defines as +
+$$ +b+wx_1+w_2x_2+\dots +w_px_p=0. +$$ + +If we define a +matrix \( \boldsymbol{X}=\left[\boldsymbol{x}_1,\boldsymbol{x}_2,\dots, \boldsymbol{x}_p\right] \) +of dimension \( n\times p \), where \( n \) represents the observations for each feature and each vector \( x_i \) is a column vector of the matrix \( \boldsymbol{X} \), +
+$$ +\boldsymbol{x}_i = \begin{bmatrix} x_{i1} \\ x_{i2} \\ \dots \\ \dots \\ x_{ip} \end{bmatrix}. +$$ + +If the above condition is not met for a given vector \( \boldsymbol{x}_i \) we have
+$$ +b+w_1x_{i1}+w_2x_{i2}+\dots +w_px_{ip} >0, +$$ + +if our output \( y_i=1 \). +In this case we say that \( \boldsymbol{x}_i \) lies on one of the sides of the hyperplane and if +
+$$ +b+w_1x_{i1}+w_2x_{i2}+\dots +w_px_{ip} < 0, +$$ + +for the class of observations \( y_i=-1 \), +then \( \boldsymbol{x}_i \) lies on the other side.
-We assume here that our data set can be well separated into two -domains, where a straight line does the job in the separating the two -classes. Here the two classes are represented by either squares or -circles. -
- - -from sklearn import datasets
-from sklearn.svm import SVC, LinearSVC
-from sklearn.linear_model import SGDClassifier
-from sklearn.preprocessing import StandardScaler
-import matplotlib
-import matplotlib.pyplot as plt
-plt.rcParams['axes.labelsize'] = 14
-plt.rcParams['xtick.labelsize'] = 12
-plt.rcParams['ytick.labelsize'] = 12
-
-
-iris = datasets.load_iris()
-X = iris["data"][:, (2, 3)] # petal length, petal width
-y = iris["target"]
-
-setosa_or_versicolor = (y == 0) | (y == 1)
-X = X[setosa_or_versicolor]
-y = y[setosa_or_versicolor]
-
-
-
-C = 5
-alpha = 1 / (C * len(X))
-
-lin_clf = LinearSVC(loss="hinge", C=C, random_state=42)
-svm_clf = SVC(kernel="linear", C=C)
-sgd_clf = SGDClassifier(loss="hinge", learning_rate="constant", eta0=0.001, alpha=alpha,
- max_iter=100000, random_state=42)
-
-scaler = StandardScaler()
-X_scaled = scaler.fit_transform(X)
-
-lin_clf.fit(X_scaled, y)
-svm_clf.fit(X_scaled, y)
-sgd_clf.fit(X_scaled, y)
-
-print("LinearSVC: ", lin_clf.intercept_, lin_clf.coef_)
-print("SVC: ", svm_clf.intercept_, svm_clf.coef_)
-print("SGDClassifier(alpha={:.5f}):".format(sgd_clf.alpha), sgd_clf.intercept_, sgd_clf.coef_)
-
-# Compute the slope and bias of each decision boundary
-w1 = -lin_clf.coef_[0, 0]/lin_clf.coef_[0, 1]
-b1 = -lin_clf.intercept_[0]/lin_clf.coef_[0, 1]
-w2 = -svm_clf.coef_[0, 0]/svm_clf.coef_[0, 1]
-b2 = -svm_clf.intercept_[0]/svm_clf.coef_[0, 1]
-w3 = -sgd_clf.coef_[0, 0]/sgd_clf.coef_[0, 1]
-b3 = -sgd_clf.intercept_[0]/sgd_clf.coef_[0, 1]
-
-# Transform the decision boundary lines back to the original scale
-line1 = scaler.inverse_transform([[-10, -10 * w1 + b1], [10, 10 * w1 + b1]])
-line2 = scaler.inverse_transform([[-10, -10 * w2 + b2], [10, 10 * w2 + b2]])
-line3 = scaler.inverse_transform([[-10, -10 * w3 + b3], [10, 10 * w3 + b3]])
-
-# Plot all three decision boundaries
-plt.figure(figsize=(11, 4))
-plt.plot(line1[:, 0], line1[:, 1], "k:", label="LinearSVC")
-plt.plot(line2[:, 0], line2[:, 1], "b--", linewidth=2, label="SVC")
-plt.plot(line3[:, 0], line3[:, 1], "r-", label="SGDClassifier")
-plt.plot(X[:, 0][y==1], X[:, 1][y==1], "bs") # label="Iris-Versicolor"
-plt.plot(X[:, 0][y==0], X[:, 1][y==0], "yo") # label="Iris-Setosa"
-plt.xlabel("Petal length", fontsize=14)
-plt.ylabel("Petal width", fontsize=14)
-plt.legend(loc="upper center", fontsize=14)
-plt.axis([0, 5.5, 0, 2])
-
-plt.show()
-
-Equivalently, for the two classes of observations we have
+$$ +y_i\left(b+w_1x_{i1}+w_2x_{i2}+\dots +w_px_{ip}\right) > 0. +$$ +When we try to separate hyperplanes, if it exists, we can use it to construct a natural classifier: a test observation is assigned a given class depending on which side of the hyperplane it is located.
@@ -300,7 +315,7 @@ plt.show()
- -
The aim of the SVM algorithm is to find a hyperplane in a -\( p \)-dimensional space, where \( p \) is the number of features that -distinctly classifies the data points. +
Let us try to develop our intuition about SVMs by limiting ourselves to a two-dimensional +plane. To separate the two classes of data points, there are many +possible lines (hyperplanes if you prefer a more strict naming) +that could be chosen. Our objective is to find a +plane that has the maximum margin, i.e the maximum distance between +data points of both classes. Maximizing the margin distance provides +some reinforcement so that future data points can be classified with +more confidence.
-In a \( p \)-dimensional space, a hyperplane is what we call an affine subspace of dimension of \( p-1 \). -As an example, in two dimension, a hyperplane is simply as straight line while in three dimensions it is -a two-dimensional subspace, or stated simply, a plane. +
What a linear classifier attempts to accomplish is to split the +feature space into two half spaces by placing a hyperplane between the +data points. This hyperplane will be our decision boundary. All +points on one side of the plane will belong to class one and all points +on the other side of the plane will belong to the second class two.
-In two dimensions, with the variables \( x_1 \) and \( x_2 \), the hyperplane is defined as
-$$ -b+w_1x_1+w_2x_2=0, -$$ - -where \( b \) is the intercept and \( w_1 \) and \( w_2 \) define the elements of a vector orthogonal to the line -\( b+w_1x_1+w_2x_2=0 \). -In two dimensions we define the vectors \( \boldsymbol{x} =[x1,x2] \) and \( \boldsymbol{w}=[w1,w2] \). -We can then rewrite the above equation as +
Unfortunately there are many ways in which we can place a hyperplane +to divide the data. Below is an example of two candidate hyperplanes +for our data sample.
-$$ -\boldsymbol{x}^T\boldsymbol{w}+b=0. -$$ - -diff --git a/doc/pub/week46/html/._week46-bs007.html b/doc/pub/week46/html/._week46-bs007.html index 0aad22506..e4f65ba79 100644 --- a/doc/pub/week46/html/._week46-bs007.html +++ b/doc/pub/week46/html/._week46-bs007.html @@ -8,8 +8,8 @@ doconce format html week46.do.txt --html_style=bootstrap --pygments_html_style=d - -
-
We limit ourselves to two classes of outputs \( y_i \) and assign these classes the values \( y_i = \pm 1 \). -In a \( p \)-dimensional space of say \( p \) features we have a hyperplane defines as -
+Let us define the function
$$ -b+wx_1+w_2x_2+\dots +w_px_p=0. +f(x) = \boldsymbol{w}^T\boldsymbol{x}+b = 0, $$ -If we define a -matrix \( \boldsymbol{X}=\left[\boldsymbol{x}_1,\boldsymbol{x}_2,\dots, \boldsymbol{x}_p\right] \) -of dimension \( n\times p \), where \( n \) represents the observations for each feature and each vector \( x_i \) is a column vector of the matrix \( \boldsymbol{X} \), -
+as the function that determines the line \( L \) that separates two classes (our two features), see the figure here.
+ +Any point defined by \( \boldsymbol{x}_i \) and \( \boldsymbol{x}_2 \) on the line \( L \) will satisfy \( \boldsymbol{w}^T(\boldsymbol{x}_1-\boldsymbol{x}_2)=0 \).
+ +The signed distance \( \delta \) from any point defined by a vector \( \boldsymbol{x} \) and a point \( \boldsymbol{x}_0 \) on the line \( L \) is then
$$ -\boldsymbol{x}_i = \begin{bmatrix} x_{i1} \\ x_{i2} \\ \dots \\ \dots \\ x_{ip} \end{bmatrix}. +\delta = \frac{1}{\vert\vert \boldsymbol{w}\vert\vert}(\boldsymbol{w}^T\boldsymbol{x}+b). $$ -If the above condition is not met for a given vector \( \boldsymbol{x}_i \) we have
-$$ -b+w_1x_{i1}+w_2x_{i2}+\dots +w_px_{ip} >0, -$$ - -if our output \( y_i=1 \). -In this case we say that \( \boldsymbol{x}_i \) lies on one of the sides of the hyperplane and if -
-$$ -b+w_1x_{i1}+w_2x_{i2}+\dots +w_px_{ip} < 0, -$$ - -for the class of observations \( y_i=-1 \), -then \( \boldsymbol{x}_i \) lies on the other side. -
- -Equivalently, for the two classes of observations we have
-$$ -y_i\left(b+w_1x_{i1}+w_2x_{i2}+\dots +w_px_{ip}\right) > 0. -$$ - -When we try to separate hyperplanes, if it exists, we can use it to construct a natural classifier: a test observation is assigned a given class depending on which side of the hyperplane it is located.
@@ -239,7 +294,7 @@ $$
- -
Let us try to develop our intuition about SVMs by limiting ourselves to a two-dimensional -plane. To separate the two classes of data points, there are many -possible lines (hyperplanes if you prefer a more strict naming) -that could be chosen. Our objective is to find a -plane that has the maximum margin, i.e the maximum distance between -data points of both classes. Maximizing the margin distance provides -some reinforcement so that future data points can be classified with -more confidence. +
How do we find the parameter \( b \) and the vector \( \boldsymbol{w} \)? What we could +do is to define a cost function which now contains the set of all +misclassified points \( M \) and attempt to minimize this function
-What a linear classifier attempts to accomplish is to split the -feature space into two half spaces by placing a hyperplane between the -data points. This hyperplane will be our decision boundary. All -points on one side of the plane will belong to class one and all points -on the other side of the plane will belong to the second class two. -
+$$ +C(\boldsymbol{w},b) = -\sum_{i\in M} y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b). +$$ + +We could now for example define all values \( y_i =1 \) as misclassified in case we have \( \boldsymbol{w}^T\boldsymbol{x}_i+b < 0 \) and the opposite if we have \( y_i=-1 \). Taking the derivatives gives us
+$$ +\frac{\partial C}{\partial b} = -\sum_{i\in M} y_i, +$$ + +and
+$$ +\frac{\partial C}{\partial \boldsymbol{w}} = -\sum_{i\in M} y_ix_i. +$$ -Unfortunately there are many ways in which we can place a hyperplane -to divide the data. Below is an example of two candidate hyperplanes -for our data sample. -
@@ -224,7 +300,7 @@ for our data sample.
-
Let us define the function
+We can now use the Newton-Raphson method or different variants of the gradient descent family (from plain gradient descent to various stochastic gradient descent approaches) to solve the equations
$$ -f(x) = \boldsymbol{w}^T\boldsymbol{x}+b = 0, +b \leftarrow b +\eta \frac{\partial C}{\partial b}, $$ -as the function that determines the line \( L \) that separates two classes (our two features), see the figure here.
- -Any point defined by \( \boldsymbol{x}_i \) and \( \boldsymbol{x}_2 \) on the line \( L \) will satisfy \( \boldsymbol{w}^T(\boldsymbol{x}_1-\boldsymbol{x}_2)=0 \).
- -The signed distance \( \delta \) from any point defined by a vector \( \boldsymbol{x} \) and a point \( \boldsymbol{x}_0 \) on the line \( L \) is then
+and
$$ -\delta = \frac{1}{\vert\vert \boldsymbol{w}\vert\vert}(\boldsymbol{w}^T\boldsymbol{x}+b). +\boldsymbol{w} \leftarrow \boldsymbol{w} +\eta \frac{\partial C}{\partial \boldsymbol{w}}, $$ +where \( \eta \) is our by now well-known learning rate.
@@ -218,7 +293,7 @@ $$
-
How do we find the parameter \( b \) and the vector \( \boldsymbol{w} \)? What we could -do is to define a cost function which now contains the set of all -misclassified points \( M \) and attempt to minimize this function +
The equations we discussed above can be coded rather easily (the +framework is similar to what we developed for logistic +regression). We are going to set up a simple case with two classes only and we want to find a line which separates them the best possible way.
-$$ -C(\boldsymbol{w},b) = -\sum_{i\in M} y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b). -$$ - -We could now for example define all values \( y_i =1 \) as misclassified in case we have \( \boldsymbol{w}^T\boldsymbol{x}_i+b < 0 \) and the opposite if we have \( y_i=-1 \). Taking the derivatives gives us
-$$ -\frac{\partial C}{\partial b} = -\sum_{i\in M} y_i, -$$ - -and
-$$ -\frac{\partial C}{\partial \boldsymbol{w}} = -\sum_{i\in M} y_ix_i. -$$ + +
+
+@@ -224,7 +310,7 @@ $$
-
We can now use the Newton-Raphson method or different variants of the gradient descent family (from plain gradient descent to various stochastic gradient descent approaches) to solve the equations
-$$ -b \leftarrow b +\eta \frac{\partial C}{\partial b}, -$$ +There are however problems with this approach, although it looks +pretty straightforward to implement. When running the above code, we see that we can easily end up with many diffeent lines which separate the two classes. +
-and
-$$ -\boldsymbol{w} \leftarrow \boldsymbol{w} +\eta \frac{\partial C}{\partial \boldsymbol{w}}, -$$ - -where \( \eta \) is our by now well-known learning rate.
+For small +gaps between the entries, we may also end up needing many iterations +before the solutions converge and if the data cannot be separated +properly into two distinct classes, we may not experience a converge +at all. +
@@ -216,7 +293,7 @@ $$
-
The equations we discussed above can be coded rather easily (the -framework is similar to what we developed for logistic -regression). We are going to set up a simple case with two classes only and we want to find a line which separates them the best possible way. +
A better approach is rather to try to define a large margin between +the two classes (if they are well separated from the beginning).
- -
-
-Thus, we wish to find a margin \( M \) with \( \boldsymbol{w} \) normalized to +\( \vert\vert \boldsymbol{w}\vert\vert =1 \) subject to the condition +
+$$ +y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) \geq M \hspace{0.1cm}\forall i=1,2,\dots, p. +$$ + +All points are thus at a signed distance from the decision boundary defined by the line \( L \). The parameters \( b \) and \( w_1 \) and \( w_2 \) define this line.
+ +We seek thus the largest value \( M \) defined by
+$$ +\frac{1}{\vert \vert \boldsymbol{w}\vert\vert}y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) \geq M \hspace{0.1cm}\forall i=1,2,\dots, n, +$$ + +or just
+$$ +y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) \geq M\vert \vert \boldsymbol{w}\vert\vert \hspace{0.1cm}\forall i. +$$ + +If we scale the equation so that \( \vert \vert \boldsymbol{w}\vert\vert = 1/M \), we have to find the minimum of +\( \boldsymbol{w}^T\boldsymbol{w}=\vert \vert \boldsymbol{w}\vert\vert \) (the norm) subject to the condition +
+$$ +y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) \geq 1 \hspace{0.1cm}\forall i. +$$ + +We have thus defined our margin as the invers of the norm of +\( \boldsymbol{w} \). We want to minimize the norm in order to have a as large as +possible margin \( M \). Before we proceed, we need to remind ourselves +about Lagrangian multipliers. +
@@ -232,7 +319,7 @@ regression). We are going to set up a simple case with two classes only and we w
-
There are however problems with this approach, although it looks -pretty straightforward to implement. When running the above code, we see that we can easily end up with many diffeent lines which separate the two classes. +
Consider a function of three independent variables \( f(x,y,z) \) . For the function \( f \) to be an +extreme we have +
+$$ +df=0. +$$ + +A necessary and sufficient condition is
+$$ +\frac{\partial f}{\partial x} =\frac{\partial f}{\partial y}=\frac{\partial f}{\partial z}=0, +$$ + +due to
+$$ +df = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz. +$$ + +In many problems the variables \( x,y,z \) are often subject to constraints (such as those above for the margin) +so that they are no longer all independent. It is possible at least in principle to use each +constraint to eliminate one variable +and to proceed with a new and smaller set of independent varables.
-For small -gaps between the entries, we may also end up needing many iterations -before the solutions converge and if the data cannot be separated -properly into two distinct classes, we may not experience a converge -at all. +
The use of so-called Lagrangian multipliers is an alternative technique when the elimination +of variables is incovenient or undesirable. Assume that we have an equation of constraint on +the variables \( x,y,z \) +
+$$ +\phi(x,y,z) = 0, +$$ + +resulting in
+$$ +d\phi = \frac{\partial \phi}{\partial x}dx+\frac{\partial \phi}{\partial y}dy+\frac{\partial \phi}{\partial z}dz =0. +$$ + +Now we cannot set anymore
+$$ +\frac{\partial f}{\partial x} =\frac{\partial f}{\partial y}=\frac{\partial f}{\partial z}=0, +$$ + +if \( df=0 \) is wanted +because there are now only two independent variables! Assume \( x \) and \( y \) are the independent +variables. +Then \( dz \) is no longer arbitrary.
@@ -215,7 +329,7 @@ at all.
-
A better approach is rather to try to define a large margin between -the two classes (if they are well separated from the beginning). -
- -Thus, we wish to find a margin \( M \) with \( \boldsymbol{w} \) normalized to -\( \vert\vert \boldsymbol{w}\vert\vert =1 \) subject to the condition -
+However, we can add to
$$ -y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) \geq M \hspace{0.1cm}\forall i=1,2,\dots, p. +df = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz, $$ -All points are thus at a signed distance from the decision boundary defined by the line \( L \). The parameters \( b \) and \( w_1 \) and \( w_2 \) define this line.
- -We seek thus the largest value \( M \) defined by
+a multiplum of \( d\phi \), viz. \( \lambda d\phi \), resulting in
$$ -\frac{1}{\vert \vert \boldsymbol{w}\vert\vert}y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) \geq M \hspace{0.1cm}\forall i=1,2,\dots, n, +df+\lambda d\phi = (\frac{\partial f}{\partial z}+\lambda +\frac{\partial \phi}{\partial x})dx+(\frac{\partial f}{\partial y}+\lambda\frac{\partial \phi}{\partial y})dy+ +(\frac{\partial f}{\partial z}+\lambda\frac{\partial \phi}{\partial z})dz =0. $$ -or just
+Our multiplier is chosen so that
$$ -y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) \geq M\vert \vert \boldsymbol{w}\vert\vert \hspace{0.1cm}\forall i. +\frac{\partial f}{\partial z}+\lambda\frac{\partial \phi}{\partial z} =0. $$ -If we scale the equation so that \( \vert \vert \boldsymbol{w}\vert\vert = 1/M \), we have to find the minimum of -\( \boldsymbol{w}^T\boldsymbol{w}=\vert \vert \boldsymbol{w}\vert\vert \) (the norm) subject to the condition +
We need to remember that we took \( dx \) and \( dy \) to be arbitrary and thus we must have
+$$ +\frac{\partial f}{\partial x}+\lambda\frac{\partial \phi}{\partial x} =0, +$$ + +and
+$$ +\frac{\partial f}{\partial y}+\lambda\frac{\partial \phi}{\partial y} =0. +$$ + +When all these equations are satisfied, \( df=0 \). We have four unknowns, \( x,y,z \) and +\( \lambda \). Actually we want only \( x,y,z \), \( \lambda \) needs not to be determined, +it is therefore often called +Lagrange's undetermined multiplier. +If we have a set of constraints \( \phi_k \) we have the equations
$$ -y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) \geq 1 \hspace{0.1cm}\forall i. +\frac{\partial f}{\partial x_i}+\sum_k\lambda_k\frac{\partial \phi_k}{\partial x_i} =0. $$ -We have thus defined our margin as the invers of the norm of -\( \boldsymbol{w} \). We want to minimize the norm in order to have a as large as -possible margin \( M \). Before we proceed, we need to remind ourselves -about Lagrangian multipliers. -
@@ -241,7 +320,7 @@ about Lagrangian multipliers.
-
In order to solve the above problem, we define the following Lagrangian function to be minimized
+$$ +{\cal L}(\lambda,b,\boldsymbol{w})=\frac{1}{2}\boldsymbol{w}^T\boldsymbol{w}-\sum_{i=1}^n\lambda_i\left[y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)-1\right], +$$ -Consider a function of three independent variables \( f(x,y,z) \) . For the function \( f \) to be an -extreme we have +
where \( \lambda_i \) is a so-called Lagrange multiplier subject to the condition \( \lambda_i \geq 0 \).
+ +Taking the derivatives with respect to \( b \) and \( \boldsymbol{w} \) we obtain
+$$ +\frac{\partial {\cal L}}{\partial b} = -\sum_{i} \lambda_iy_i=0, +$$ + +and
+$$ +\frac{\partial {\cal L}}{\partial \boldsymbol{w}} = 0 = \boldsymbol{w}-\sum_{i} \lambda_iy_i\boldsymbol{x}_i. +$$ + +Inserting these constraints into the equation for \( {\cal L} \) we obtain
+$$ +{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j, +$$ + +subject to the constraints \( \lambda_i\geq 0 \) and \( \sum_i\lambda_iy_i=0 \). +We must in addition satisfy the Karush-Kuhn-Tucker (KKT) condition
$$ -df=0. +\lambda_i\left[y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) -1\right] \hspace{0.1cm}\forall i. $$ -A necessary and sufficient condition is
-$$ -\frac{\partial f}{\partial x} =\frac{\partial f}{\partial y}=\frac{\partial f}{\partial z}=0, -$$ - -due to
-$$ -df = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz. -$$ - -In many problems the variables \( x,y,z \) are often subject to constraints (such as those above for the margin) -so that they are no longer all independent. It is possible at least in principle to use each -constraint to eliminate one variable -and to proceed with a new and smaller set of independent varables. -
- -The use of so-called Lagrangian multipliers is an alternative technique when the elimination -of variables is incovenient or undesirable. Assume that we have an equation of constraint on -the variables \( x,y,z \) -
-$$ -\phi(x,y,z) = 0, -$$ - -resulting in
-$$ -d\phi = \frac{\partial \phi}{\partial x}dx+\frac{\partial \phi}{\partial y}dy+\frac{\partial \phi}{\partial z}dz =0. -$$ - -Now we cannot set anymore
-$$ -\frac{\partial f}{\partial x} =\frac{\partial f}{\partial y}=\frac{\partial f}{\partial z}=0, -$$ - -if \( df=0 \) is wanted -because there are now only two independent variables! Assume \( x \) and \( y \) are the independent -variables. -Then \( dz \) is no longer arbitrary. -
+When \( \lambda_i > 0 \), the vectors \( \boldsymbol{x}_i \) are called support vectors. They are the vectors closest to the line (or hyperplane) and define the margin \( M \).
@@ -251,7 +316,7 @@ Then \( dz \) is no longer arbitrary.
-
However, we can add to
+We can rewrite
$$ -df = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz, +{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j, $$ -a multiplum of \( d\phi \), viz. \( \lambda d\phi \), resulting in
+and its constraints in terms of a matrix-vector problem where we minimize w.r.t. \( \lambda \) the following problem
$$ -df+\lambda d\phi = (\frac{\partial f}{\partial z}+\lambda -\frac{\partial \phi}{\partial x})dx+(\frac{\partial f}{\partial y}+\lambda\frac{\partial \phi}{\partial y})dy+ -(\frac{\partial f}{\partial z}+\lambda\frac{\partial \phi}{\partial z})dz =0. +\frac{1}{2} \boldsymbol{\lambda}^T\begin{bmatrix} y_1y_1\boldsymbol{x}_1^T\boldsymbol{x}_1 & y_1y_2\boldsymbol{x}_1^T\boldsymbol{x}_2 & \dots & \dots & y_1y_n\boldsymbol{x}_1^T\boldsymbol{x}_n \\ +y_2y_1\boldsymbol{x}_2^T\boldsymbol{x}_1 & y_2y_2\boldsymbol{x}_2^T\boldsymbol{x}_2 & \dots & \dots & y_1y_n\boldsymbol{x}_2^T\boldsymbol{x}_n \\ +\dots & \dots & \dots & \dots & \dots \\ +\dots & \dots & \dots & \dots & \dots \\ +y_ny_1\boldsymbol{x}_n^T\boldsymbol{x}_1 & y_ny_2\boldsymbol{x}_n^T\boldsymbol{x}_2 & \dots & \dots & y_ny_n\boldsymbol{x}_n^T\boldsymbol{x}_n \\ +\end{bmatrix}\boldsymbol{\lambda}-\mathbb{1}\boldsymbol{\lambda}, $$ -Our multiplier is chosen so that
-$$ -\frac{\partial f}{\partial z}+\lambda\frac{\partial \phi}{\partial z} =0. -$$ - -We need to remember that we took \( dx \) and \( dy \) to be arbitrary and thus we must have
-$$ -\frac{\partial f}{\partial x}+\lambda\frac{\partial \phi}{\partial x} =0, -$$ - -and
-$$ -\frac{\partial f}{\partial y}+\lambda\frac{\partial \phi}{\partial y} =0. -$$ - -When all these equations are satisfied, \( df=0 \). We have four unknowns, \( x,y,z \) and -\( \lambda \). Actually we want only \( x,y,z \), \( \lambda \) needs not to be determined, -it is therefore often called -Lagrange's undetermined multiplier. -If we have a set of constraints \( \phi_k \) we have the equations +
subject to \( \boldsymbol{y}^T\boldsymbol{\lambda}=0 \). Here we defined the vectors \( \boldsymbol{\lambda} =[\lambda_1,\lambda_2,\dots,\lambda_n] \) and +\( \boldsymbol{y}=[y_1,y_2,\dots,y_n] \).
-$$ -\frac{\partial f}{\partial x_i}+\sum_k\lambda_k\frac{\partial \phi_k}{\partial x_i} =0. -$$ -@@ -242,7 +301,7 @@ $$
-
In order to solve the above problem, we define the following Lagrangian function to be minimized
-$$ -{\cal L}(\lambda,b,\boldsymbol{w})=\frac{1}{2}\boldsymbol{w}^T\boldsymbol{w}-\sum_{i=1}^n\lambda_i\left[y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)-1\right], -$$ +where \( \lambda_i \) is a so-called Lagrange multiplier subject to the condition \( \lambda_i \geq 0 \).
- -Taking the derivatives with respect to \( b \) and \( \boldsymbol{w} \) we obtain
-$$ -\frac{\partial {\cal L}}{\partial b} = -\sum_{i} \lambda_iy_i=0, -$$ - -and
-$$ -\frac{\partial {\cal L}}{\partial \boldsymbol{w}} = 0 = \boldsymbol{w}-\sum_{i} \lambda_iy_i\boldsymbol{x}_i. -$$ - -Inserting these constraints into the equation for \( {\cal L} \) we obtain
-$$ -{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j, -$$ - -subject to the constraints \( \lambda_i\geq 0 \) and \( \sum_i\lambda_iy_i=0 \). -We must in addition satisfy the Karush-Kuhn-Tucker (KKT) condition +
Solving the above problem, yields the values of \( \lambda_i \). +To find the coefficients of your hyperplane we need simply to compute
$$ -\lambda_i\left[y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) -1\right] \hspace{0.1cm}\forall i. +\boldsymbol{w}=\sum_{i} \lambda_iy_i\boldsymbol{x}_i. $$ -When \( \lambda_i > 0 \), the vectors \( \boldsymbol{x}_i \) are called support vectors. They are the vectors closest to the line (or hyperplane) and define the margin \( M \).
+With our vector \( \boldsymbol{w} \) we can in turn find the value of the intercept \( b \) (here in two dimensions) via
+$$ +y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)=1, +$$ + +resulting in
+$$ +b = \frac{1}{y_i}-\boldsymbol{w}^T\boldsymbol{x}_i, +$$ + +or if we write it out in terms of the support vectors only, with \( N_s \) being their number, we have
+$$ +b = \frac{1}{N_s}\sum_{j\in N_s}\left(y_j-\sum_{i=1}^n\lambda_iy_i\boldsymbol{x}_i^T\boldsymbol{x}_j\right). +$$ + +With our hyperplane coefficients we can use our classifier to assign any observation by simply using
+$$ +y_i = \mathrm{sign}(\boldsymbol{w}^T\boldsymbol{x}_i+b). +$$ + +Below we discuss how to find the optimal values of \( \lambda_i \). Before we proceed however, we discuss now the so-called soft classifier.
@@ -238,7 +311,7 @@ $$
-
We can rewrite
+Till now, the margin is strictly defined by the support vectors. This defines what is called a hard classifier, that is the margins are well defined.
+ +Suppose now that classes overlap in feature space, as shown in the +figure here. One way to deal with this problem before we define the +so-called kernel approach, is to allow a kind of slack in the sense +that we allow some points to be on the wrong side of the margin. +
+ +We introduce thus the so-called slack variables \( \boldsymbol{\xi} =[\xi_1,x_2,\dots,x_n] \) and +modify our previous equation +
$$ -{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j, +y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)=1, $$ -and its constraints in terms of a matrix-vector problem where we minimize w.r.t. \( \lambda \) the following problem
+to
$$ -\frac{1}{2} \boldsymbol{\lambda}^T\begin{bmatrix} y_1y_1\boldsymbol{x}_1^T\boldsymbol{x}_1 & y_1y_2\boldsymbol{x}_1^T\boldsymbol{x}_2 & \dots & \dots & y_1y_n\boldsymbol{x}_1^T\boldsymbol{x}_n \\ -y_2y_1\boldsymbol{x}_2^T\boldsymbol{x}_1 & y_2y_2\boldsymbol{x}_2^T\boldsymbol{x}_2 & \dots & \dots & y_1y_n\boldsymbol{x}_2^T\boldsymbol{x}_n \\ -\dots & \dots & \dots & \dots & \dots \\ -\dots & \dots & \dots & \dots & \dots \\ -y_ny_1\boldsymbol{x}_n^T\boldsymbol{x}_1 & y_ny_2\boldsymbol{x}_n^T\boldsymbol{x}_2 & \dots & \dots & y_ny_n\boldsymbol{x}_n^T\boldsymbol{x}_n \\ -\end{bmatrix}\boldsymbol{\lambda}-\mathbb{1}\boldsymbol{\lambda}, +y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)=1-\xi_i, $$ -subject to \( \boldsymbol{y}^T\boldsymbol{\lambda}=0 \). Here we defined the vectors \( \boldsymbol{\lambda} =[\lambda_1,\lambda_2,\dots,\lambda_n] \) and -\( \boldsymbol{y}=[y_1,y_2,\dots,y_n] \). +
with the requirement \( \xi_i\geq 0 \). The total violation is now \( \sum_i\xi \). +The value \( \xi_i \) in the constraint the last constraint corresponds to the amount by which the prediction +\( y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)=1 \) is on the wrong side of its margin. Hence by bounding the sum \( \sum_i \xi_i \), +we bound the total amount by which predictions fall on the wrong side of their margins. +
+ +Misclassifications occur when \( \xi_i > 1 \). Thus bounding the total sum by some value \( C \) bounds in turn the total number of +misclassifications.
@@ -223,7 +312,7 @@ $$
-
Solving the above problem, yields the values of \( \lambda_i \). -To find the coefficients of your hyperplane we need simply to compute +
This has in turn the consequences that we change our optmization problem to finding the minimum of
+$$ +{\cal L}=\frac{1}{2}\boldsymbol{w}^T\boldsymbol{w}-\sum_{i=1}^n\lambda_i\left[y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)-(1-\xi_)\right]+C\sum_{i=1}^n\xi_i-\sum_{i=1}^n\gamma_i\xi_i, +$$ + +subject to
+$$ +y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)=1-\xi_i \hspace{0.1cm}\forall i, +$$ + +with the requirement \( \xi_i\geq 0 \).
+ +Taking the derivatives with respect to \( b \) and \( \boldsymbol{w} \) we obtain
+$$ +\frac{\partial {\cal L}}{\partial b} = -\sum_{i} \lambda_iy_i=0, +$$ + +and
+$$ +\frac{\partial {\cal L}}{\partial \boldsymbol{w}} = 0 = \boldsymbol{w}-\sum_{i} \lambda_iy_i\boldsymbol{x}_i, +$$ + +and
+$$ +\lambda_i = C-\gamma_i \hspace{0.1cm}\forall i. +$$ + +Inserting these constraints into the equation for \( {\cal L} \) we obtain the same equation as before
+$$ +{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j, +$$ + +but now subject to the constraints \( \lambda_i\geq 0 \), \( \sum_i\lambda_iy_i=0 \) and \( 0\leq\lambda_i \leq C \). +We must in addition satisfy the Karush-Kuhn-Tucker condition which now reads
$$ -\boldsymbol{w}=\sum_{i} \lambda_iy_i\boldsymbol{x}_i. +\lambda_i\left[y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) -(1-\xi_)\right]=0 \hspace{0.1cm}\forall i, $$ -With our vector \( \boldsymbol{w} \) we can in turn find the value of the intercept \( b \) (here in two dimensions) via
$$ -y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)=1, +\gamma_i\xi_i = 0, $$ -resulting in
+and
$$ -b = \frac{1}{y_i}-\boldsymbol{w}^T\boldsymbol{x}_i, +y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) -(1-\xi_) \geq 0 \hspace{0.1cm}\forall i. $$ -or if we write it out in terms of the support vectors only, with \( N_s \) being their number, we have
-$$ -b = \frac{1}{N_s}\sum_{j\in N_s}\left(y_j-\sum_{i=1}^n\lambda_iy_i\boldsymbol{x}_i^T\boldsymbol{x}_j\right). -$$ - -With our hyperplane coefficients we can use our classifier to assign any observation by simply using
-$$ -y_i = \mathrm{sign}(\boldsymbol{w}^T\boldsymbol{x}_i+b). -$$ - -Below we discuss how to find the optimal values of \( \lambda_i \). Before we proceed however, we discuss now the so-called soft classifier.
@@ -233,7 +331,7 @@ $$
-
Till now, the margin is strictly defined by the support vectors. This defines what is called a hard classifier, that is the margins are well defined.
- -Suppose now that classes overlap in feature space, as shown in the -figure here. One way to deal with this problem before we define the -so-called kernel approach, is to allow a kind of slack in the sense -that we allow some points to be on the wrong side of the margin. +
The cases we have studied till now, were all characterized by two classes +with a close to linear separability. The classifiers we have described +so far find linear boundaries in our input feature space. It is +possible to make our procedure more flexible by exploring the feature +space using other basis expansions such as higher-order polynomials, +wavelets, splines etc.
-We introduce thus the so-called slack variables \( \boldsymbol{\xi} =[\xi_1,x_2,\dots,x_n] \) and -modify our previous equation -
-$$ -y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)=1, -$$ - -to
-$$ -y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)=1-\xi_i, -$$ - -with the requirement \( \xi_i\geq 0 \). The total violation is now \( \sum_i\xi \). -The value \( \xi_i \) in the constraint the last constraint corresponds to the amount by which the prediction -\( y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)=1 \) is on the wrong side of its margin. Hence by bounding the sum \( \sum_i \xi_i \), -we bound the total amount by which predictions fall on the wrong side of their margins. +
If our feature space is not easy to separate, as shown in the figure +here, we can achieve a better separation by introducing more complex +basis functions. The ideal would be, as shown in the next figure, to, via a specific transformation to +obtain a separation between the classes which is almost linear.
-Misclassifications occur when \( \xi_i > 1 \). Thus bounding the total sum by some value \( C \) bounds in turn the total number of -misclassifications. +
The change of basis, from \( x\rightarrow z=\phi(x) \) leads to the same type of equations to be solved, except that +we need to introduce for example a polynomial transformation to a two-dimensional training set.
+ + +import numpy as np
+import os
+
+np.random.seed(42)
+
+# To plot pretty figures
+import matplotlib
+import matplotlib.pyplot as plt
+plt.rcParams['axes.labelsize'] = 14
+plt.rcParams['xtick.labelsize'] = 12
+plt.rcParams['ytick.labelsize'] = 12
+
+
+from sklearn.svm import SVC
+from sklearn import datasets
+
+
+
+X1D = np.linspace(-4, 4, 9).reshape(-1, 1)
+X2D = np.c_[X1D, X1D**2]
+y = np.array([0, 0, 1, 1, 1, 1, 1, 0, 0])
+
+plt.figure(figsize=(11, 4))
+
+plt.subplot(121)
+plt.grid(True, which='both')
+plt.axhline(y=0, color='k')
+plt.plot(X1D[:, 0][y==0], np.zeros(4), "bs")
+plt.plot(X1D[:, 0][y==1], np.zeros(5), "g^")
+plt.gca().get_yaxis().set_ticks([])
+plt.xlabel(r"$x_1$", fontsize=20)
+plt.axis([-4.5, 4.5, -0.2, 0.2])
+
+plt.subplot(122)
+plt.grid(True, which='both')
+plt.axhline(y=0, color='k')
+plt.axvline(x=0, color='k')
+plt.plot(X2D[:, 0][y==0], X2D[:, 1][y==0], "bs")
+plt.plot(X2D[:, 0][y==1], X2D[:, 1][y==1], "g^")
+plt.xlabel(r"$x_1$", fontsize=20)
+plt.ylabel(r"$x_2$", fontsize=20, rotation=0)
+plt.gca().get_yaxis().set_ticks([0, 4, 8, 12, 16])
+plt.plot([-4.5, 4.5], [6.5, 6.5], "r--", linewidth=3)
+plt.axis([-4.5, 4.5, -1, 17])
+plt.subplots_adjust(right=1)
+plt.show()
+
+diff --git a/doc/pub/week46/html/._week46-bs021.html b/doc/pub/week46/html/._week46-bs021.html index e9adcc6a1..02476b33d 100644 --- a/doc/pub/week46/html/._week46-bs021.html +++ b/doc/pub/week46/html/._week46-bs021.html @@ -8,8 +8,8 @@ doconce format html week46.do.txt --html_style=bootstrap --pygments_html_style=d - -
-
This has in turn the consequences that we change our optmization problem to finding the minimum of
+Suppose we define a polynomial transformation of degree two only (we continue to live in a plane with \( x_i \) and \( y_i \) as variables)
$$ -{\cal L}=\frac{1}{2}\boldsymbol{w}^T\boldsymbol{w}-\sum_{i=1}^n\lambda_i\left[y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)-(1-\xi_)\right]+C\sum_{i=1}^n\xi_i-\sum_{i=1}^n\gamma_i\xi_i, +z = \phi(x_i) =\left(x_i^2, y_i^2, \sqrt{2}x_iy_i\right). $$ -subject to
+With our new basis, the equations we solved earlier are basically the same, that is we have now (without the slack option for simplicity)
$$ -y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)=1-\xi_i \hspace{0.1cm}\forall i, +{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{z}_i^T\boldsymbol{z}_j, $$ -with the requirement \( \xi_i\geq 0 \).
- -Taking the derivatives with respect to \( b \) and \( \boldsymbol{w} \) we obtain
+subject to the constraints \( \lambda_i\geq 0 \), \( \sum_i\lambda_iy_i=0 \), and for the support vectors
$$ -\frac{\partial {\cal L}}{\partial b} = -\sum_{i} \lambda_iy_i=0, +y_i(\boldsymbol{w}^T\boldsymbol{z}_i+b)= 1 \hspace{0.1cm}\forall i, $$ -and
-$$ -\frac{\partial {\cal L}}{\partial \boldsymbol{w}} = 0 = \boldsymbol{w}-\sum_{i} \lambda_iy_i\boldsymbol{x}_i, -$$ - -and
-$$ -\lambda_i = C-\gamma_i \hspace{0.1cm}\forall i. -$$ - -Inserting these constraints into the equation for \( {\cal L} \) we obtain the same equation as before
-$$ -{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j, -$$ - -but now subject to the constraints \( \lambda_i\geq 0 \), \( \sum_i\lambda_iy_i=0 \) and \( 0\leq\lambda_i \leq C \). -We must in addition satisfy the Karush-Kuhn-Tucker condition which now reads +
from which we also find \( b \). +To compute \( \boldsymbol{z}_i^T\boldsymbol{z}_j \) we define the kernel \( K(\boldsymbol{x}_i,\boldsymbol{x}_j) \) as
$$ -\lambda_i\left[y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) -(1-\xi_)\right]=0 \hspace{0.1cm}\forall i, +K(\boldsymbol{x}_i,\boldsymbol{x}_j)=\boldsymbol{z}_i^T\boldsymbol{z}_j= \phi(\boldsymbol{x}_i)^T\phi(\boldsymbol{x}_j). $$ +For the above example, the kernel reads
$$ -\gamma_i\xi_i = 0, +K(\boldsymbol{x}_i,\boldsymbol{x}_j)=[x_i^2, y_i^2, \sqrt{2}x_iy_i]^T\begin{bmatrix} x_j^2 \\ y_j^2 \\ \sqrt{2}x_jy_j \end{bmatrix}=x_i^2x_j^2+2x_ix_jy_iy_j+y_i^2y_j^2. $$ -and
-$$ -y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) -(1-\xi_) \geq 0 \hspace{0.1cm}\forall i. -$$ +We note that this is nothing but the dot product of the two original +vectors \( (\boldsymbol{x}_i^T\boldsymbol{x}_j)^2 \). Instead of thus computing the +product in the Lagrangian of \( \boldsymbol{z}_i^T\boldsymbol{z}_j \) we simply compute +the dot product \( (\boldsymbol{x}_i^T\boldsymbol{x}_j)^2 \). +
+This leads to the so-called +kernel trick and the result leads to the same as if we went through +the trouble of performing the transformation +\( \phi(\boldsymbol{x}_i)^T\phi(\boldsymbol{x}_j) \) during the SVM calculations. +
@@ -252,6 +320,8 @@ $$
-
Using our definition of the kernel We can rewrite again the Lagrangian
+$$ +{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{z}_j, +$$ -The cases we have studied till now, were all characterized by two classes -with a close to linear separability. The classifiers we have described -so far find linear boundaries in our input feature space. It is -possible to make our procedure more flexible by exploring the feature -space using other basis expansions such as higher-order polynomials, -wavelets, splines etc. +
subject to the constraints \( \lambda_i\geq 0 \), \( \sum_i\lambda_iy_i=0 \) in terms of a convex optimization problem
+$$ +\frac{1}{2} \boldsymbol{\lambda}^T\begin{bmatrix} y_1y_1K(\boldsymbol{x}_1,\boldsymbol{x}_1) & y_1y_2K(\boldsymbol{x}_1,\boldsymbol{x}_2) & \dots & \dots & y_1y_nK(\boldsymbol{x}_1,\boldsymbol{x}_n) \\ +y_2y_1K(\boldsymbol{x}_2,\boldsymbol{x}_1) & y_2y_2(\boldsymbol{x}_2,\boldsymbol{x}_2) & \dots & \dots & y_1y_nK(\boldsymbol{x}_2,\boldsymbol{x}_n) \\ +\dots & \dots & \dots & \dots & \dots \\ +\dots & \dots & \dots & \dots & \dots \\ +y_ny_1K(\boldsymbol{x}_n,\boldsymbol{x}_1) & y_ny_2K(\boldsymbol{x}_n\boldsymbol{x}_2) & \dots & \dots & y_ny_nK(\boldsymbol{x}_n,\boldsymbol{x}_n) \\ +\end{bmatrix}\boldsymbol{\lambda}-\mathbb{1}\boldsymbol{\lambda}, +$$ + +subject to \( \boldsymbol{y}^T\boldsymbol{\lambda}=0 \). Here we defined the vectors \( \boldsymbol{\lambda} =[\lambda_1,\lambda_2,\dots,\lambda_n] \) and +\( \boldsymbol{y}=[y_1,y_2,\dots,y_n] \). +If we add the slack constants this leads to the additional constraint \( 0\leq \lambda_i \leq C \).
-If our feature space is not easy to separate, as shown in the figure -here, we can achieve a better separation by introducing more complex -basis functions. The ideal would be, as shown in the next figure, to, via a specific transformation to -obtain a separation between the classes which is almost linear. +
We can rewrite this (see the solutions below) in terms of a convex optimization problem of the type
+$$ +\begin{align*} + &\mathrm{min}_{\lambda}\hspace{0.2cm} \frac{1}{2}\boldsymbol{\lambda}^T\boldsymbol{P}\boldsymbol{\lambda}+\boldsymbol{q}^T\boldsymbol{\lambda},\\ \nonumber + &\mathrm{subject\hspace{0.1cm}to} \hspace{0.2cm} \boldsymbol{G}\boldsymbol{\lambda} \preceq \boldsymbol{h} \hspace{0.2cm} \wedge \boldsymbol{A}\boldsymbol{\lambda}=f. +\end{align*} +$$ + +Below we discuss how to solve these equations. Here we note that the matrix \( \boldsymbol{P} \) has matrix elements \( p_{ij}=y_iy_jK(\boldsymbol{x}_i,\boldsymbol{x}_j) \). +Given a kernel \( K \) and the targets \( y_i \) this matrix is easy to set up. The constraint \( \boldsymbol{y}^T\boldsymbol{\lambda}=0 \) leads to \( f=0 \) and \( \boldsymbol{A}=\boldsymbol{y} \). How to set up the matrix \( \boldsymbol{G} \) is discussed later. Here note that the inequalities \( 0\leq \lambda_i \leq C \) can be split up into +\( 0\leq \lambda_i \) and \( \lambda_i \leq C \). These two inequalities define then the matrix \( \boldsymbol{G} \) and the vector \( \boldsymbol{h} \).
-The change of basis, from \( x\rightarrow z=\phi(x) \) leads to the same type of equations to be solved, except that -we need to introduce for example a polynomial transformation to a two-dimensional training set. -
- - - -import numpy as np
-import os
-
-np.random.seed(42)
-
-# To plot pretty figures
-import matplotlib
-import matplotlib.pyplot as plt
-plt.rcParams['axes.labelsize'] = 14
-plt.rcParams['xtick.labelsize'] = 12
-plt.rcParams['ytick.labelsize'] = 12
-
-
-from sklearn.svm import SVC
-from sklearn import datasets
-
-
-
-X1D = np.linspace(-4, 4, 9).reshape(-1, 1)
-X2D = np.c_[X1D, X1D**2]
-y = np.array([0, 0, 1, 1, 1, 1, 1, 0, 0])
-
-plt.figure(figsize=(11, 4))
-
-plt.subplot(121)
-plt.grid(True, which='both')
-plt.axhline(y=0, color='k')
-plt.plot(X1D[:, 0][y==0], np.zeros(4), "bs")
-plt.plot(X1D[:, 0][y==1], np.zeros(5), "g^")
-plt.gca().get_yaxis().set_ticks([])
-plt.xlabel(r"$x_1$", fontsize=20)
-plt.axis([-4.5, 4.5, -0.2, 0.2])
-
-plt.subplot(122)
-plt.grid(True, which='both')
-plt.axhline(y=0, color='k')
-plt.axvline(x=0, color='k')
-plt.plot(X2D[:, 0][y==0], X2D[:, 1][y==0], "bs")
-plt.plot(X2D[:, 0][y==1], X2D[:, 1][y==1], "g^")
-plt.xlabel(r"$x_1$", fontsize=20)
-plt.ylabel(r"$x_2$", fontsize=20, rotation=0)
-plt.gca().get_yaxis().set_ticks([0, 4, 8, 12, 16])
-plt.plot([-4.5, 4.5], [6.5, 6.5], "r--", linewidth=3)
-plt.axis([-4.5, 4.5, -1, 17])
-plt.subplots_adjust(right=1)
-plt.show()
-
-diff --git a/doc/pub/week46/html/._week46-bs023.html b/doc/pub/week46/html/._week46-bs023.html index 3e9e1b396..e5d4994c7 100644 --- a/doc/pub/week46/html/._week46-bs023.html +++ b/doc/pub/week46/html/._week46-bs023.html @@ -8,8 +8,8 @@ doconce format html week46.do.txt --html_style=bootstrap --pygments_html_style=d - -
-
Suppose we define a polynomial transformation of degree two only (we continue to live in a plane with \( x_i \) and \( y_i \) as variables)
-$$ -z = \phi(x_i) =\left(x_i^2, y_i^2, \sqrt{2}x_iy_i\right). -$$ +There are several popular kernels being used. These are
+and many other ones.
-With our new basis, the equations we solved earlier are basically the same, that is we have now (without the slack option for simplicity)
-$$ -{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{z}_i^T\boldsymbol{z}_j, -$$ - -subject to the constraints \( \lambda_i\geq 0 \), \( \sum_i\lambda_iy_i=0 \), and for the support vectors
-$$ -y_i(\boldsymbol{w}^T\boldsymbol{z}_i+b)= 1 \hspace{0.1cm}\forall i, -$$ - -from which we also find \( b \). -To compute \( \boldsymbol{z}_i^T\boldsymbol{z}_j \) we define the kernel \( K(\boldsymbol{x}_i,\boldsymbol{x}_j) \) as -
-$$ -K(\boldsymbol{x}_i,\boldsymbol{x}_j)=\boldsymbol{z}_i^T\boldsymbol{z}_j= \phi(\boldsymbol{x}_i)^T\phi(\boldsymbol{x}_j). -$$ - -For the above example, the kernel reads
-$$ -K(\boldsymbol{x}_i,\boldsymbol{x}_j)=[x_i^2, y_i^2, \sqrt{2}x_iy_i]^T\begin{bmatrix} x_j^2 \\ y_j^2 \\ \sqrt{2}x_jy_j \end{bmatrix}=x_i^2x_j^2+2x_ix_jy_iy_j+y_i^2y_j^2. -$$ - -We note that this is nothing but the dot product of the two original -vectors \( (\boldsymbol{x}_i^T\boldsymbol{x}_j)^2 \). Instead of thus computing the -product in the Lagrangian of \( \boldsymbol{z}_i^T\boldsymbol{z}_j \) we simply compute -the dot product \( (\boldsymbol{x}_i^T\boldsymbol{x}_j)^2 \). +
An important theorem for us is Mercer's +theorem. The +theorem states that if a kernel function \( K \) is symmetric, continuous +and leads to a positive semi-definite matrix \( \boldsymbol{P} \) then there +exists a function \( \phi \) that maps \( \boldsymbol{x}_i \) and \( \boldsymbol{x}_j \) into +another space (possibly with much higher dimensions) such that
-This leads to the so-called -kernel trick and the result leads to the same as if we went through -the trouble of performing the transformation -\( \phi(\boldsymbol{x}_i)^T\phi(\boldsymbol{x}_j) \) during the SVM calculations. +$$ +K(\boldsymbol{x}_i,\boldsymbol{x}_j)=\phi(\boldsymbol{x}_i)^T\phi(\boldsymbol{x}_j). +$$ + +
So you can use \( K \) as a kernel since you know \( \phi \) exists, even if +you don’t know what \( \phi \) is. +
+ +Note that some frequently used kernels (such as the Sigmoid kernel) +don’t respect all of Mercer’s conditions, yet they generally work well +in practice.
@@ -240,6 +309,10 @@ the trouble of performing the transformation
-
Using our definition of the kernel We can rewrite again the Lagrangian
-$$ -{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{z}_j, -$$ +subject to the constraints \( \lambda_i\geq 0 \), \( \sum_i\lambda_iy_i=0 \) in terms of a convex optimization problem
-$$ -\frac{1}{2} \boldsymbol{\lambda}^T\begin{bmatrix} y_1y_1K(\boldsymbol{x}_1,\boldsymbol{x}_1) & y_1y_2K(\boldsymbol{x}_1,\boldsymbol{x}_2) & \dots & \dots & y_1y_nK(\boldsymbol{x}_1,\boldsymbol{x}_n) \\ -y_2y_1K(\boldsymbol{x}_2,\boldsymbol{x}_1) & y_2y_2(\boldsymbol{x}_2,\boldsymbol{x}_2) & \dots & \dots & y_1y_nK(\boldsymbol{x}_2,\boldsymbol{x}_n) \\ -\dots & \dots & \dots & \dots & \dots \\ -\dots & \dots & \dots & \dots & \dots \\ -y_ny_1K(\boldsymbol{x}_n,\boldsymbol{x}_1) & y_ny_2K(\boldsymbol{x}_n\boldsymbol{x}_2) & \dots & \dots & y_ny_nK(\boldsymbol{x}_n,\boldsymbol{x}_n) \\ -\end{bmatrix}\boldsymbol{\lambda}-\mathbb{1}\boldsymbol{\lambda}, -$$ + +from __future__ import division, print_function, unicode_literals
-subject to \( \boldsymbol{y}^T\boldsymbol{\lambda}=0 \). Here we defined the vectors \( \boldsymbol{\lambda} =[\lambda_1,\lambda_2,\dots,\lambda_n] \) and
-\( \boldsymbol{y}=[y_1,y_2,\dots,y_n] \).
-If we add the slack constants this leads to the additional constraint \( 0\leq \lambda_i \leq C \).
-
+import numpy as np
+np.random.seed(42)
-We can rewrite this (see the solutions below) in terms of a convex optimization problem of the type
-$$
-\begin{align*}
- &\mathrm{min}_{\lambda}\hspace{0.2cm} \frac{1}{2}\boldsymbol{\lambda}^T\boldsymbol{P}\boldsymbol{\lambda}+\boldsymbol{q}^T\boldsymbol{\lambda},\\ \nonumber
- &\mathrm{subject\hspace{0.1cm}to} \hspace{0.2cm} \boldsymbol{G}\boldsymbol{\lambda} \preceq \boldsymbol{h} \hspace{0.2cm} \wedge \boldsymbol{A}\boldsymbol{\lambda}=f.
-\end{align*}
-$$
+import matplotlib
+import matplotlib.pyplot as plt
+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.pipeline import Pipeline
+from sklearn.preprocessing import StandardScaler
+from sklearn.svm import LinearSVC
+
+
+from sklearn.datasets import make_moons
+X, y = make_moons(n_samples=100, noise=0.15, random_state=42)
+
+def plot_dataset(X, y, axes):
+ plt.plot(X[:, 0][y==0], X[:, 1][y==0], "bs")
+ plt.plot(X[:, 0][y==1], X[:, 1][y==1], "g^")
+ plt.axis(axes)
+ plt.grid(True, which='both')
+ plt.xlabel(r"$x_1$", fontsize=20)
+ plt.ylabel(r"$x_2$", fontsize=20, rotation=0)
+
+plot_dataset(X, y, [-1.5, 2.5, -1, 1.5])
+plt.show()
+
+from sklearn.datasets import make_moons
+from sklearn.pipeline import Pipeline
+from sklearn.preprocessing import PolynomialFeatures
+
+polynomial_svm_clf = Pipeline([
+ ("poly_features", PolynomialFeatures(degree=3)),
+ ("scaler", StandardScaler()),
+ ("svm_clf", LinearSVC(C=10, loss="hinge", random_state=42))
+ ])
+
+polynomial_svm_clf.fit(X, y)
+
+def plot_predictions(clf, axes):
+ x0s = np.linspace(axes[0], axes[1], 100)
+ x1s = np.linspace(axes[2], axes[3], 100)
+ x0, x1 = np.meshgrid(x0s, x1s)
+ X = np.c_[x0.ravel(), x1.ravel()]
+ y_pred = clf.predict(X).reshape(x0.shape)
+ y_decision = clf.decision_function(X).reshape(x0.shape)
+ plt.contourf(x0, x1, y_pred, cmap=plt.cm.brg, alpha=0.2)
+ plt.contourf(x0, x1, y_decision, cmap=plt.cm.brg, alpha=0.1)
+
+plot_predictions(polynomial_svm_clf, [-1.5, 2.5, -1, 1.5])
+plot_dataset(X, y, [-1.5, 2.5, -1, 1.5])
+
+plt.show()
+
+
+from sklearn.svm import SVC
+
+poly_kernel_svm_clf = Pipeline([
+ ("scaler", StandardScaler()),
+ ("svm_clf", SVC(kernel="poly", degree=3, coef0=1, C=5))
+ ])
+poly_kernel_svm_clf.fit(X, y)
+
+poly100_kernel_svm_clf = Pipeline([
+ ("scaler", StandardScaler()),
+ ("svm_clf", SVC(kernel="poly", degree=10, coef0=100, C=5))
+ ])
+poly100_kernel_svm_clf.fit(X, y)
+
+plt.figure(figsize=(11, 4))
+
+plt.subplot(121)
+plot_predictions(poly_kernel_svm_clf, [-1.5, 2.5, -1, 1.5])
+plot_dataset(X, y, [-1.5, 2.5, -1, 1.5])
+plt.title(r"$d=3, r=1, C=5$", fontsize=18)
+
+plt.subplot(122)
+plot_predictions(poly100_kernel_svm_clf, [-1.5, 2.5, -1, 1.5])
+plot_dataset(X, y, [-1.5, 2.5, -1, 1.5])
+plt.title(r"$d=10, r=100, C=5$", fontsize=18)
+
+plt.show()
+
+def gaussian_rbf(x, landmark, gamma):
+ return np.exp(-gamma * np.linalg.norm(x - landmark, axis=1)**2)
+
+gamma = 0.3
+
+x1s = np.linspace(-4.5, 4.5, 200).reshape(-1, 1)
+x2s = gaussian_rbf(x1s, -2, gamma)
+x3s = gaussian_rbf(x1s, 1, gamma)
+
+XK = np.c_[gaussian_rbf(X1D, -2, gamma), gaussian_rbf(X1D, 1, gamma)]
+yk = np.array([0, 0, 1, 1, 1, 1, 1, 0, 0])
+
+plt.figure(figsize=(11, 4))
+
+plt.subplot(121)
+plt.grid(True, which='both')
+plt.axhline(y=0, color='k')
+plt.scatter(x=[-2, 1], y=[0, 0], s=150, alpha=0.5, c="red")
+plt.plot(X1D[:, 0][yk==0], np.zeros(4), "bs")
+plt.plot(X1D[:, 0][yk==1], np.zeros(5), "g^")
+plt.plot(x1s, x2s, "g--")
+plt.plot(x1s, x3s, "b:")
+plt.gca().get_yaxis().set_ticks([0, 0.25, 0.5, 0.75, 1])
+plt.xlabel(r"$x_1$", fontsize=20)
+plt.ylabel(r"Similarity", fontsize=14)
+plt.annotate(r'$\mathbf{x}$',
+ xy=(X1D[3, 0], 0),
+ xytext=(-0.5, 0.20),
+ ha="center",
+ arrowprops=dict(facecolor='black', shrink=0.1),
+ fontsize=18,
+ )
+plt.text(-2, 0.9, "$x_2$", ha="center", fontsize=20)
+plt.text(1, 0.9, "$x_3$", ha="center", fontsize=20)
+plt.axis([-4.5, 4.5, -0.1, 1.1])
+
+plt.subplot(122)
+plt.grid(True, which='both')
+plt.axhline(y=0, color='k')
+plt.axvline(x=0, color='k')
+plt.plot(XK[:, 0][yk==0], XK[:, 1][yk==0], "bs")
+plt.plot(XK[:, 0][yk==1], XK[:, 1][yk==1], "g^")
+plt.xlabel(r"$x_2$", fontsize=20)
+plt.ylabel(r"$x_3$ ", fontsize=20, rotation=0)
+plt.annotate(r'$\phi\left(\mathbf{x}\right)$',
+ xy=(XK[3, 0], XK[3, 1]),
+ xytext=(0.65, 0.50),
+ ha="center",
+ arrowprops=dict(facecolor='black', shrink=0.1),
+ fontsize=18,
+ )
+plt.plot([-0.1, 1.1], [0.57, -0.1], "r--", linewidth=3)
+plt.axis([-0.1, 1.1, -0.1, 1.1])
+
+plt.subplots_adjust(right=1)
+
+plt.show()
+
+
+x1_example = X1D[3, 0]
+for landmark in (-2, 1):
+ k = gaussian_rbf(np.array([[x1_example]]), np.array([[landmark]]), gamma)
+ print("Phi({}, {}) = {}".format(x1_example, landmark, k))
+
+rbf_kernel_svm_clf = Pipeline([
+ ("scaler", StandardScaler()),
+ ("svm_clf", SVC(kernel="rbf", gamma=5, C=0.001))
+ ])
+rbf_kernel_svm_clf.fit(X, y)
+
+
+from sklearn.svm import SVC
+
+gamma1, gamma2 = 0.1, 5
+C1, C2 = 0.001, 1000
+hyperparams = (gamma1, C1), (gamma1, C2), (gamma2, C1), (gamma2, C2)
+
+svm_clfs = []
+for gamma, C in hyperparams:
+ rbf_kernel_svm_clf = Pipeline([
+ ("scaler", StandardScaler()),
+ ("svm_clf", SVC(kernel="rbf", gamma=gamma, C=C))
+ ])
+ rbf_kernel_svm_clf.fit(X, y)
+ svm_clfs.append(rbf_kernel_svm_clf)
+
+plt.figure(figsize=(11, 7))
+
+for i, svm_clf in enumerate(svm_clfs):
+ plt.subplot(221 + i)
+ plot_predictions(svm_clf, [-1.5, 2.5, -1, 1.5])
+ plot_dataset(X, y, [-1.5, 2.5, -1, 1.5])
+ gamma, C = hyperparams[i]
+ plt.title(r"$\gamma = {}, C = {}$".format(gamma, C), fontsize=16)
+
+plt.show()
+
+Below we discuss how to solve these equations. Here we note that the matrix \( \boldsymbol{P} \) has matrix elements \( p_{ij}=y_iy_jK(\boldsymbol{x}_i,\boldsymbol{x}_j) \). -Given a kernel \( K \) and the targets \( y_i \) this matrix is easy to set up. The constraint \( \boldsymbol{y}^T\boldsymbol{\lambda}=0 \) leads to \( f=0 \) and \( \boldsymbol{A}=\boldsymbol{y} \). How to set up the matrix \( \boldsymbol{G} \) is discussed later. Here note that the inequalities \( 0\leq \lambda_i \leq C \) can be split up into -\( 0\leq \lambda_i \) and \( \lambda_i \leq C \). These two inequalities define then the matrix \( \boldsymbol{G} \) and the vector \( \boldsymbol{h} \). -
@@ -232,6 +488,11 @@ Given a kernel \( K \) and the targets \( y_i \) this matrix is easy to set up.
-
There are several popular kernels being used. These are
-and many other ones.
- -An important theorem for us is Mercer's -theorem. The -theorem states that if a kernel function \( K \) is symmetric, continuous -and leads to a positive semi-definite matrix \( \boldsymbol{P} \) then there -exists a function \( \phi \) that maps \( \boldsymbol{x}_i \) and \( \boldsymbol{x}_j \) into -another space (possibly with much higher dimensions) such that -
+A mathematical (quadratic) optimization problem, or just optimization problem, has the form
$$ -K(\boldsymbol{x}_i,\boldsymbol{x}_j)=\phi(\boldsymbol{x}_i)^T\phi(\boldsymbol{x}_j). +\begin{align*} + &\mathrm{min}_{\lambda}\hspace{0.2cm} \frac{1}{2}\boldsymbol{\lambda}^T\boldsymbol{P}\boldsymbol{\lambda}+\boldsymbol{q}^T\boldsymbol{\lambda},\\ \nonumber + &\mathrm{subject\hspace{0.1cm}to} \hspace{0.2cm} \boldsymbol{G}\boldsymbol{\lambda} \preceq \boldsymbol{h} \wedge \boldsymbol{A}\boldsymbol{\lambda}=f. +\end{align*} $$ -So you can use \( K \) as a kernel since you know \( \phi \) exists, even if -you don’t know what \( \phi \) is. +
subject to some constraints for say a selected set \( i=1,2,\dots, n \). +In our case we are optimizing with respect to the Lagrangian multipliers \( \lambda_i \), and the +vector \( \boldsymbol{\lambda}=[\lambda_1, \lambda_2,\dots, \lambda_n] \) is the optimization variable we are dealing with.
-Note that some frequently used kernels (such as the Sigmoid kernel) -don’t respect all of Mercer’s conditions, yet they generally work well -in practice. +
In our case we are particularly interested in a class of optimization problems called convex optmization problems. +In our discussion on gradient descent methods we discussed at length the definition of a convex function.
+Convex optimization problems play a central role in applied mathematics and we recommend strongly Boyd and Vandenberghe's text on the topics.
+diff --git a/doc/pub/week46/html/._week46-bs026.html b/doc/pub/week46/html/._week46-bs026.html index 69efb92e4..1e9543627 100644 --- a/doc/pub/week46/html/._week46-bs026.html +++ b/doc/pub/week46/html/._week46-bs026.html @@ -8,8 +8,8 @@ doconce format html week46.do.txt --html_style=bootstrap --pygments_html_style=d - -
-
If we use Python as programming language and wish to venture beyond +scikit-learn, tensorflow and similar software which makes our +lives so much easier, we need to dive into the wonderful world of +quadratic programming. We can, if we wish, solve the minimization +problem using say standard gradient methods or conjugate gradient +methods. However, these methods tend to exhibit a rather slow +converge. So, welcome to the promised land of quadratic programming. +
+ +The functions we need are contained in the quadratic programming library CVXOPT and we need to import it together with numpy as
+from __future__ import division, print_function, unicode_literals
-
-import numpy as np
-np.random.seed(42)
-
-import matplotlib
-import matplotlib.pyplot as plt
-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.pipeline import Pipeline
-from sklearn.preprocessing import StandardScaler
-from sklearn.svm import LinearSVC
-
-
-from sklearn.datasets import make_moons
-X, y = make_moons(n_samples=100, noise=0.15, random_state=42)
-
-def plot_dataset(X, y, axes):
- plt.plot(X[:, 0][y==0], X[:, 1][y==0], "bs")
- plt.plot(X[:, 0][y==1], X[:, 1][y==1], "g^")
- plt.axis(axes)
- plt.grid(True, which='both')
- plt.xlabel(r"$x_1$", fontsize=20)
- plt.ylabel(r"$x_2$", fontsize=20, rotation=0)
-
-plot_dataset(X, y, [-1.5, 2.5, -1, 1.5])
-plt.show()
-
-from sklearn.datasets import make_moons
-from sklearn.pipeline import Pipeline
-from sklearn.preprocessing import PolynomialFeatures
-
-polynomial_svm_clf = Pipeline([
- ("poly_features", PolynomialFeatures(degree=3)),
- ("scaler", StandardScaler()),
- ("svm_clf", LinearSVC(C=10, loss="hinge", random_state=42))
- ])
-
-polynomial_svm_clf.fit(X, y)
-
-def plot_predictions(clf, axes):
- x0s = np.linspace(axes[0], axes[1], 100)
- x1s = np.linspace(axes[2], axes[3], 100)
- x0, x1 = np.meshgrid(x0s, x1s)
- X = np.c_[x0.ravel(), x1.ravel()]
- y_pred = clf.predict(X).reshape(x0.shape)
- y_decision = clf.decision_function(X).reshape(x0.shape)
- plt.contourf(x0, x1, y_pred, cmap=plt.cm.brg, alpha=0.2)
- plt.contourf(x0, x1, y_decision, cmap=plt.cm.brg, alpha=0.1)
-
-plot_predictions(polynomial_svm_clf, [-1.5, 2.5, -1, 1.5])
-plot_dataset(X, y, [-1.5, 2.5, -1, 1.5])
-
-plt.show()
-
-
-from sklearn.svm import SVC
-
-poly_kernel_svm_clf = Pipeline([
- ("scaler", StandardScaler()),
- ("svm_clf", SVC(kernel="poly", degree=3, coef0=1, C=5))
- ])
-poly_kernel_svm_clf.fit(X, y)
-
-poly100_kernel_svm_clf = Pipeline([
- ("scaler", StandardScaler()),
- ("svm_clf", SVC(kernel="poly", degree=10, coef0=100, C=5))
- ])
-poly100_kernel_svm_clf.fit(X, y)
-
-plt.figure(figsize=(11, 4))
-
-plt.subplot(121)
-plot_predictions(poly_kernel_svm_clf, [-1.5, 2.5, -1, 1.5])
-plot_dataset(X, y, [-1.5, 2.5, -1, 1.5])
-plt.title(r"$d=3, r=1, C=5$", fontsize=18)
-
-plt.subplot(122)
-plot_predictions(poly100_kernel_svm_clf, [-1.5, 2.5, -1, 1.5])
-plot_dataset(X, y, [-1.5, 2.5, -1, 1.5])
-plt.title(r"$d=10, r=100, C=5$", fontsize=18)
-
-plt.show()
-
-def gaussian_rbf(x, landmark, gamma):
- return np.exp(-gamma * np.linalg.norm(x - landmark, axis=1)**2)
-
-gamma = 0.3
-
-x1s = np.linspace(-4.5, 4.5, 200).reshape(-1, 1)
-x2s = gaussian_rbf(x1s, -2, gamma)
-x3s = gaussian_rbf(x1s, 1, gamma)
-
-XK = np.c_[gaussian_rbf(X1D, -2, gamma), gaussian_rbf(X1D, 1, gamma)]
-yk = np.array([0, 0, 1, 1, 1, 1, 1, 0, 0])
-
-plt.figure(figsize=(11, 4))
-
-plt.subplot(121)
-plt.grid(True, which='both')
-plt.axhline(y=0, color='k')
-plt.scatter(x=[-2, 1], y=[0, 0], s=150, alpha=0.5, c="red")
-plt.plot(X1D[:, 0][yk==0], np.zeros(4), "bs")
-plt.plot(X1D[:, 0][yk==1], np.zeros(5), "g^")
-plt.plot(x1s, x2s, "g--")
-plt.plot(x1s, x3s, "b:")
-plt.gca().get_yaxis().set_ticks([0, 0.25, 0.5, 0.75, 1])
-plt.xlabel(r"$x_1$", fontsize=20)
-plt.ylabel(r"Similarity", fontsize=14)
-plt.annotate(r'$\mathbf{x}$',
- xy=(X1D[3, 0], 0),
- xytext=(-0.5, 0.20),
- ha="center",
- arrowprops=dict(facecolor='black', shrink=0.1),
- fontsize=18,
- )
-plt.text(-2, 0.9, "$x_2$", ha="center", fontsize=20)
-plt.text(1, 0.9, "$x_3$", ha="center", fontsize=20)
-plt.axis([-4.5, 4.5, -0.1, 1.1])
-
-plt.subplot(122)
-plt.grid(True, which='both')
-plt.axhline(y=0, color='k')
-plt.axvline(x=0, color='k')
-plt.plot(XK[:, 0][yk==0], XK[:, 1][yk==0], "bs")
-plt.plot(XK[:, 0][yk==1], XK[:, 1][yk==1], "g^")
-plt.xlabel(r"$x_2$", fontsize=20)
-plt.ylabel(r"$x_3$ ", fontsize=20, rotation=0)
-plt.annotate(r'$\phi\left(\mathbf{x}\right)$',
- xy=(XK[3, 0], XK[3, 1]),
- xytext=(0.65, 0.50),
- ha="center",
- arrowprops=dict(facecolor='black', shrink=0.1),
- fontsize=18,
- )
-plt.plot([-0.1, 1.1], [0.57, -0.1], "r--", linewidth=3)
-plt.axis([-0.1, 1.1, -0.1, 1.1])
-
-plt.subplots_adjust(right=1)
-
-plt.show()
-
-
-x1_example = X1D[3, 0]
-for landmark in (-2, 1):
- k = gaussian_rbf(np.array([[x1_example]]), np.array([[landmark]]), gamma)
- print("Phi({}, {}) = {}".format(x1_example, landmark, k))
-
-rbf_kernel_svm_clf = Pipeline([
- ("scaler", StandardScaler()),
- ("svm_clf", SVC(kernel="rbf", gamma=5, C=0.001))
- ])
-rbf_kernel_svm_clf.fit(X, y)
-
-
-from sklearn.svm import SVC
-
-gamma1, gamma2 = 0.1, 5
-C1, C2 = 0.001, 1000
-hyperparams = (gamma1, C1), (gamma1, C2), (gamma2, C1), (gamma2, C2)
-
-svm_clfs = []
-for gamma, C in hyperparams:
- rbf_kernel_svm_clf = Pipeline([
- ("scaler", StandardScaler()),
- ("svm_clf", SVC(kernel="rbf", gamma=gamma, C=C))
- ])
- rbf_kernel_svm_clf.fit(X, y)
- svm_clfs.append(rbf_kernel_svm_clf)
-
-plt.figure(figsize=(11, 7))
-
-for i, svm_clf in enumerate(svm_clfs):
- plt.subplot(221 + i)
- plot_predictions(svm_clf, [-1.5, 2.5, -1, 1.5])
- plot_dataset(X, y, [-1.5, 2.5, -1, 1.5])
- gamma, C = hyperparams[i]
- plt.title(r"$\gamma = {}, C = {}$".format(gamma, C), fontsize=16)
-
-plt.show()
+ import numpy
+import cvxopt
This will make our life much easier. You don't need t0 write your own optimizer.
@@ -408,6 +313,13 @@ plt.show()
-
A mathematical (quadratic) optimization problem, or just optimization problem, has the form
+We remind ourselves about the general problem we want to solve
$$ \begin{align*} - &\mathrm{min}_{\lambda}\hspace{0.2cm} \frac{1}{2}\boldsymbol{\lambda}^T\boldsymbol{P}\boldsymbol{\lambda}+\boldsymbol{q}^T\boldsymbol{\lambda},\\ \nonumber - &\mathrm{subject\hspace{0.1cm}to} \hspace{0.2cm} \boldsymbol{G}\boldsymbol{\lambda} \preceq \boldsymbol{h} \wedge \boldsymbol{A}\boldsymbol{\lambda}=f. + &\mathrm{min}_{x}\hspace{0.2cm} \frac{1}{2}\boldsymbol{x}^T\boldsymbol{P}\boldsymbol{x}+\boldsymbol{q}^T\boldsymbol{x},\\ \nonumber + &\mathrm{subject\hspace{0.1cm} to} \hspace{0.2cm} \boldsymbol{G}\boldsymbol{x} \preceq \boldsymbol{h} \wedge \boldsymbol{A}\boldsymbol{x}=f. \end{align*} $$ -subject to some constraints for say a selected set \( i=1,2,\dots, n \). -In our case we are optimizing with respect to the Lagrangian multipliers \( \lambda_i \), and the -vector \( \boldsymbol{\lambda}=[\lambda_1, \lambda_2,\dots, \lambda_n] \) is the optimization variable we are dealing with. +
Let us show how to perform the optmization using a simple case. Assume we want to optimize the following problem
+$$ +\begin{align*} + &\mathrm{min}_{x}\hspace{0.2cm} \frac{1}{2}x^2+5x+3y \\ \nonumber + &\mathrm{subject}\hspace{0.5cm} \mathrm{to} \\ \nonumber + &x, y \geq 0 \\ \nonumber + &x+3y \geq 15 \\ \nonumber + &2x+5y \leq 100 \\ \nonumber + &3x+4y \leq 80. \\ \nonumber +\end{align*} +$$ + +The minimization problem can be rewritten in terms of vectors and matrices as (with \( x \) and \( y \) being the unknowns)
+$$ +\frac{1}{2}\begin{bmatrix} x\\ y \end{bmatrix}^T \begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix}3\\ 4 \end{bmatrix}^T \begin{bmatrix}x \\ y \end{bmatrix}. +$$ + +Similarly, we can now set up the inequalities (we need to change \( \geq \) to \( \leq \) by multiplying with \( -1 \) on bot sides) as the following matrix-vector equation
+$$ +\begin{bmatrix} -1 & 0 \\ 0 & -1 \\ -1 & -3 \\ 2 & 5 \\ 3 & 4\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix} \preceq \begin{bmatrix}0 \\ 0\\ -15 \\ 100 \\ 80\end{bmatrix}. +$$ + +We have collapsed all the inequalities into a single matrix \( \boldsymbol{G} \). We see also that our matrix
+$$ +\boldsymbol{P} =\begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix} +$$ + +is clearly positive semi-definite (all eigenvalues larger or equal zero). +Finally, the vector \( \boldsymbol{h} \) is defined as +
+$$ +\boldsymbol{h} = \begin{bmatrix}0 \\ 0\\ -15 \\ 100 \\ 80\end{bmatrix}. +$$ + +Since we don't have any equalities the matrix \( \boldsymbol{A} \) is set to zero +The following code solves the equations for us
-In our case we are particularly interested in a class of optimization problems called convex optmization problems. -In our discussion on gradient descent methods we discussed at length the definition of a convex function. -
+ +# Import the necessary packages
+import numpy
+from cvxopt import matrix
+from cvxopt import solvers
+P = matrix(numpy.diag([1,0]), tc='d')
+q = matrix(numpy.array([3,4]), tc='d')
+G = matrix(numpy.array([[-1,0],[0,-1],[-1,-3],[2,5],[3,4]]), tc='d')
+h = matrix(numpy.array([0,0,-15,100,80]), tc='d')
+# Construct the QP, invoke solver
+sol = solvers.qp(P,q,G,h)
+# Extract optimal value and solution
+sol['x']
+sol['primal objective']
+
+Convex optimization problems play a central role in applied mathematics and we recommend strongly Boyd and Vandenberghe's text on the topics.
@@ -216,6 +356,14 @@ In our discussion on gradient descent methods we discussed at length the definit
-
If we use Python as programming language and wish to venture beyond -scikit-learn, tensorflow and similar software which makes our -lives so much easier, we need to dive into the wonderful world of -quadratic programming. We can, if we wish, solve the minimization -problem using say standard gradient methods or conjugate gradient -methods. However, these methods tend to exhibit a rather slow -converge. So, welcome to the promised land of quadratic programming. +
We are now ready to return to our setup of the optmization problem for a more realistic case. Introducing the slack parameter \( C \) we have
+$$ +\frac{1}{2} \boldsymbol{\lambda}^T\begin{bmatrix} y_1y_1K(\boldsymbol{x}_1,\boldsymbol{x}_1) & y_1y_2K(\boldsymbol{x}_1,\boldsymbol{x}_2) & \dots & \dots & y_1y_nK(\boldsymbol{x}_1,\boldsymbol{x}_n) \\ +y_2y_1K(\boldsymbol{x}_2,\boldsymbol{x}_1) & y_2y_2K(\boldsymbol{x}_2,\boldsymbol{x}_2) & \dots & \dots & y_1y_nK(\boldsymbol{x}_2,\boldsymbol{x}_n) \\ +\dots & \dots & \dots & \dots & \dots \\ +\dots & \dots & \dots & \dots & \dots \\ +y_ny_1K(\boldsymbol{x}_n,\boldsymbol{x}_1) & y_ny_2K(\boldsymbol{x}_n\boldsymbol{x}_2) & \dots & \dots & y_ny_nK(\boldsymbol{x}_n,\boldsymbol{x}_n) \\ +\end{bmatrix}\boldsymbol{\lambda}-\mathbb{I}\boldsymbol{\lambda}, +$$ + +subject to \( \boldsymbol{y}^T\boldsymbol{\lambda}=0 \). Here we defined the vectors \( \boldsymbol{\lambda} =[\lambda_1,\lambda_2,\dots,\lambda_n] \) and +\( \boldsymbol{y}=[y_1,y_2,\dots,y_n] \). +With the slack constants this leads to the additional constraint \( 0\leq \lambda_i \leq C \).
-The functions we need are contained in the quadratic programming package CVXOPT and we need to import it together with numpy as
- - - -import numpy
-import cvxopt
-
-This will make our life much easier. You don't need t write your own optimizer.
+Using the CVXOPT library, the matrix \( P \) would then be defined by the
@@ -233,6 +291,15 @@ converge. So, welcome to the promised land of quadratic programming.
-
We remind ourselves about the general problem we want to solve
-$$ -\begin{align*} - &\mathrm{min}_{x}\hspace{0.2cm} \frac{1}{2}\boldsymbol{x}^T\boldsymbol{P}\boldsymbol{x}+\boldsymbol{q}^T\boldsymbol{x},\\ \nonumber - &\mathrm{subject\hspace{0.1cm} to} \hspace{0.2cm} \boldsymbol{G}\boldsymbol{x} \preceq \boldsymbol{h} \wedge \boldsymbol{A}\boldsymbol{x}=f. -\end{align*} -$$ - -Let us show how to perform the optmization using a simple case. Assume we want to optimize the following problem
-$$ -\begin{align*} - &\mathrm{min}_{x}\hspace{0.2cm} \frac{1}{2}x^2+5x+3y \\ \nonumber - &\mathrm{subject to} \\ \nonumber - &x, y \geq 0 \\ \nonumber - &x+3y \geq 15 \\ \nonumber - &2x+5y \leq 100 \\ \nonumber - &3x+4y \leq 80. \\ \nonumber -\end{align*} -$$ - -The minimization problem can be rewritten in terms of vectors and matrices as (with \( x \) and \( y \) being the unknowns)
-$$ -\frac{1}{2}\begin{bmatrix} x\\ y \end{bmatrix}^T \begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix}3\\ 4 \end{bmatrix}^T \begin{bmatrix}x \\ y \end{bmatrix}. -$$ - -Similarly, we can now set up the inequalities (we need to change \( \geq \) to \( \leq \) by multiplying with \( -1 \) on bot sides) as the following matrix-vector equation
-$$ -\begin{bmatrix} -1 & 0 \\ 0 & -1 \\ -1 & -3 \\ 2 & 5 \\ 3 & 4\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix} \preceq \begin{bmatrix}0 \\ 0\\ -15 \\ 100 \\ 80\end{bmatrix}. -$$ - -We have collapsed all the inequalities into a single matrix \( \boldsymbol{G} \). We see also that our matrix
-$$ -\boldsymbol{P} =\begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix} -$$ - -is clearly positive semi-definite (all eigenvalues larger or equal zero). -Finally, the vector \( \boldsymbol{h} \) is defined as -
-$$ -\boldsymbol{h} = \begin{bmatrix}0 \\ 0\\ -15 \\ 100 \\ 80\end{bmatrix}. -$$ - -Since we don't have any equalities the matrix \( \boldsymbol{A} \) is set to zero -The following code solves the equations for us +
The principal component analysis deals with the problem of fitting a +low-dimensional affine subspace \( S \) of dimension \( d \) much smaller than +the total dimension \( D \) 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. +Historically, the PCA was first formulated in a statistical setting in order to estimate the principal component of a multivariate random variable.
- -# Import the necessary packages
-import numpy
-from cvxopt import matrix
-from cvxopt import solvers
-P = matrix(numpy.diag([1,0]), tc='d')
-q = matrix(numpy.array([3,4]), tc='d')
-G = matrix(numpy.array([[-1,0],[0,-1],[-1,-3],[2,5],[3,4]]), tc='d')
-h = matrix(numpy.array([0,0,-15,100,80]), tc='d')
-# Construct the QP, invoke solver
-sol = solvers.qp(P,q,G,h)
-# Extract optimal value and solution
-sol['x']
-sol['primal objective']
-
-We have a data set defined by a design/feature matrix \( \boldsymbol{X} \) (see below for its definition)
+A good read is for example Vidal, Ma and Sastry.
@@ -276,6 +290,16 @@ sol['primal objective']
-
We have tried every year to organize a kind of mini-workshop on project 3. In 2020 the contributions were (and some of these ended up in thesis work and/or publications, online only due to Covid-19!)
- --
We would very much like to organize something similar this year as well. Feel free to come with suggesstions by Thursday November 17. We will then try to set up the various contributions for Friday November 18. -The presentations last normally 5-15 mins and span from loose ideas to more well-defined topics. Nothing pretentious, we wish to keep this as low-key as possible. -
-The functions we need are contained in the quadratic programming package CVXOPT and we need to import it together with numpy as
+The functions we need are contained in the quadratic programming library CVXOPT and we need to import it together with numpy as
@@ -1515,7 +1483,7 @@ converge. So, welcome to the promised land of quadratic programming.This will make our life much easier. You don't need t write your own optimizer.
+This will make our life much easier. You don't need t0 write your own optimizer.
Using the CVXOPT library, the matrix \( P \) would then be defined by the
+The principal component analysis deals with the problem of fitting a +low-dimensional affine subspace \( S \) of dimension \( d \) much smaller than +the total dimension \( D \) 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. +Historically, the PCA was first formulated in a statistical setting in order to estimate the principal component of a multivariate random variable. +
+ +We have a data set defined by a design/feature matrix \( \boldsymbol{X} \) (see below for its definition)
++
A good read is for example Vidal, Ma and Sastry.
+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 \( \boldsymbol{C} \) is defined as +
+
+$$
+\boldsymbol{C}[\boldsymbol{x},\boldsymbol{y}] = \begin{bmatrix} \mathrm{cov}[\boldsymbol{x},\boldsymbol{x}] & \mathrm{cov}[\boldsymbol{x},\boldsymbol{y}] \\
+ \mathrm{cov}[\boldsymbol{y},\boldsymbol{x}] & \mathrm{cov}[\boldsymbol{y},\boldsymbol{y}] \\
+ \end{bmatrix},
+$$
+
+
+
where for example
+
+$$
+\mathrm{cov}[\boldsymbol{x},\boldsymbol{y}] =\frac{1}{n} \sum_{i=0}^{n-1}(x_i- \overline{x})(y_i- \overline{y}).
+$$
+
+
+
With this definition and recalling that the variance is defined as
+
+$$
+\mathrm{var}[\boldsymbol{x}]=\frac{1}{n} \sum_{i=0}^{n-1}(x_i- \overline{x})^2,
+$$
+
+
+
we can rewrite the covariance matrix as
+
+$$
+\boldsymbol{C}[\boldsymbol{x},\boldsymbol{y}] = \begin{bmatrix} \mathrm{var}[\boldsymbol{x}] & \mathrm{cov}[\boldsymbol{x},\boldsymbol{y}] \\
+ \mathrm{cov}[\boldsymbol{x},\boldsymbol{y}] & \mathrm{var}[\boldsymbol{y}] \\
+ \end{bmatrix}.
+$$
+
+
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 +
+ +
+$$
+\mathrm{corr}[\boldsymbol{x},\boldsymbol{y}]=\frac{\mathrm{cov}[\boldsymbol{x},\boldsymbol{y}]}{\sqrt{\mathrm{var}[\boldsymbol{x}] \mathrm{var}[\boldsymbol{y}]}}.
+$$
+
+
+
The correlation function is then given by values \( \mathrm{corr}[\boldsymbol{x},\boldsymbol{y}] +\in [-1,1] \). This avoids eventual problems with too large values. We +can then define the correlation matrix for the two vectors \( \boldsymbol{x} \) +and \( \boldsymbol{y} \) as +
+ +
+$$
+\boldsymbol{K}[\boldsymbol{x},\boldsymbol{y}] = \begin{bmatrix} 1 & \mathrm{corr}[\boldsymbol{x},\boldsymbol{y}] \\
+ \mathrm{corr}[\boldsymbol{y},\boldsymbol{x}] & 1 \\
+ \end{bmatrix},
+$$
+
+
+
In the above example this is the function we constructed using pandas.
+In our derivation of the various regression algorithms like Ordinary Least Squares or Ridge regression +we defined the design/feature matrix \( \boldsymbol{X} \) as +
+ +
+$$
+\boldsymbol{X}=\begin{bmatrix}
+x_{0,0} & x_{0,1} & x_{0,2}& \dots & \dots x_{0,p-1}\\
+x_{1,0} & x_{1,1} & x_{1,2}& \dots & \dots x_{1,p-1}\\
+x_{2,0} & x_{2,1} & x_{2,2}& \dots & \dots x_{2,p-1}\\
+\dots & \dots & \dots & \dots \dots & \dots \\
+x_{n-2,0} & x_{n-2,1} & x_{n-2,2}& \dots & \dots x_{n-2,p-1}\\
+x_{n-1,0} & x_{n-1,1} & x_{n-1,2}& \dots & \dots x_{n-1,p-1}\\
+\end{bmatrix},
+$$
+
+
+
with \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \), 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 +
+
+$$
+\boldsymbol{X}=\begin{bmatrix} \boldsymbol{x}_0 & \boldsymbol{x}_1 & \boldsymbol{x}_2 & \dots & \dots & \boldsymbol{x}_{p-1}\end{bmatrix},
+$$
+
+
+
with a given vector
+
+$$
+\boldsymbol{x}_i^T = \begin{bmatrix}x_{0,i} & x_{1,i} & x_{2,i}& \dots & \dots x_{n-1,i}\end{bmatrix}.
+$$
+
+
With these definitions, we can now rewrite our \( 2\times 2 \) +correlation/covariance matrix in terms of a moe general design/feature +matrix \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \). This leads to a \( p\times p \) +covariance matrix for the vectors \( \boldsymbol{x}_i \) with \( i=0,1,\dots,p-1 \) +
+ +
+$$
+\boldsymbol{C}[\boldsymbol{x}] = \begin{bmatrix}
+\mathrm{var}[\boldsymbol{x}_0] & \mathrm{cov}[\boldsymbol{x}_0,\boldsymbol{x}_1] & \mathrm{cov}[\boldsymbol{x}_0,\boldsymbol{x}_2] & \dots & \dots & \mathrm{cov}[\boldsymbol{x}_0,\boldsymbol{x}_{p-1}]\\
+\mathrm{cov}[\boldsymbol{x}_1,\boldsymbol{x}_0] & \mathrm{var}[\boldsymbol{x}_1] & \mathrm{cov}[\boldsymbol{x}_1,\boldsymbol{x}_2] & \dots & \dots & \mathrm{cov}[\boldsymbol{x}_1,\boldsymbol{x}_{p-1}]\\
+\mathrm{cov}[\boldsymbol{x}_2,\boldsymbol{x}_0] & \mathrm{cov}[\boldsymbol{x}_2,\boldsymbol{x}_1] & \mathrm{var}[\boldsymbol{x}_2] & \dots & \dots & \mathrm{cov}[\boldsymbol{x}_2,\boldsymbol{x}_{p-1}]\\
+\dots & \dots & \dots & \dots & \dots & \dots \\
+\dots & \dots & \dots & \dots & \dots & \dots \\
+\mathrm{cov}[\boldsymbol{x}_{p-1},\boldsymbol{x}_0] & \mathrm{cov}[\boldsymbol{x}_{p-1},\boldsymbol{x}_1] & \mathrm{cov}[\boldsymbol{x}_{p-1},\boldsymbol{x}_{2}] & \dots & \dots & \mathrm{var}[\boldsymbol{x}_{p-1}]\\
+\end{bmatrix},
+$$
+
+
and the correlation matrix
+
+$$
+\boldsymbol{K}[\boldsymbol{x}] = \begin{bmatrix}
+1 & \mathrm{corr}[\boldsymbol{x}_0,\boldsymbol{x}_1] & \mathrm{corr}[\boldsymbol{x}_0,\boldsymbol{x}_2] & \dots & \dots & \mathrm{corr}[\boldsymbol{x}_0,\boldsymbol{x}_{p-1}]\\
+\mathrm{corr}[\boldsymbol{x}_1,\boldsymbol{x}_0] & 1 & \mathrm{corr}[\boldsymbol{x}_1,\boldsymbol{x}_2] & \dots & \dots & \mathrm{corr}[\boldsymbol{x}_1,\boldsymbol{x}_{p-1}]\\
+\mathrm{corr}[\boldsymbol{x}_2,\boldsymbol{x}_0] & \mathrm{corr}[\boldsymbol{x}_2,\boldsymbol{x}_1] & 1 & \dots & \dots & \mathrm{corr}[\boldsymbol{x}_2,\boldsymbol{x}_{p-1}]\\
+\dots & \dots & \dots & \dots & \dots & \dots \\
+\dots & \dots & \dots & \dots & \dots & \dots \\
+\mathrm{corr}[\boldsymbol{x}_{p-1},\boldsymbol{x}_0] & \mathrm{corr}[\boldsymbol{x}_{p-1},\boldsymbol{x}_1] & \mathrm{corr}[\boldsymbol{x}_{p-1},\boldsymbol{x}_{2}] & \dots & \dots & 1\\
+\end{bmatrix},
+$$
+
+
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 \( \boldsymbol{W} \) +
+ +
+$$
+\boldsymbol{W}^T = \begin{bmatrix} x_0 & y_0 \\
+ x_1 & y_1 \\
+ x_2 & y_2\\
+ \dots & \dots \\
+ x_{n-2} & y_{n-2}\\
+ x_{n-1} & y_{n-1} &
+ \end{bmatrix},
+$$
+
+
+
which in turn is converted into into the \( 2\times 2 \) covariance matrix +\( \boldsymbol{C} \) via the Numpy function np.cov(). We note that we can also calculate +the mean value of each set of samples \( \boldsymbol{x} \) etc using the Numpy +function np.mean(x). We can also extract the eigenvalues of the +covariance matrix through the np.linalg.eig() function. +
+ + + +# 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)
+
+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). +
+ + + +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)
+
+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.
+We whow here how we can set up the correlation matrix using pandas, as done in this simple code
+ + +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)
+
+We expand this model to the Franke function discussed above.
+ + + +# 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)
+
+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 \( n \)). We would however not include the intercept +and wee can simply +drop these elements and construct a correlation +matrix without them by centering our matrix elements by subtracting the mean of each column. +
+We can rewrite the covariance matrix in a more compact form in terms of the design/feature matrix \( \boldsymbol{X} \) as
+
+$$
+\boldsymbol{C}[\boldsymbol{x}] = \frac{1}{n}\boldsymbol{X}^T\boldsymbol{X}= \mathbb{E}[\boldsymbol{X}^T\boldsymbol{X}].
+$$
+
+
+
To see this let us simply look at a design matrix \( \boldsymbol{X}\in {\mathbb{R}}^{2\times 2} \)
+
+$$
+\boldsymbol{X}=\begin{bmatrix}
+x_{00} & x_{01}\\
+x_{10} & x_{11}\\
+\end{bmatrix}=\begin{bmatrix}
+\boldsymbol{x}_{0} & \boldsymbol{x}_{1}\\
+\end{bmatrix}.
+$$
+
+
If we then compute the expectation value
+
+$$
+\mathbb{E}[\boldsymbol{X}^T\boldsymbol{X}] = \frac{1}{n}\boldsymbol{X}^T\boldsymbol{X}=\begin{bmatrix}
+x_{00}^2+x_{01}^2 & x_{00}x_{10}+x_{01}x_{11}\\
+x_{10}x_{00}+x_{11}x_{01} & x_{10}^2+x_{11}^2\\
+\end{bmatrix},
+$$
+
+
+
which is just
+
+$$
+\boldsymbol{C}[\boldsymbol{x}_0,\boldsymbol{x}_1] = \boldsymbol{C}[\boldsymbol{x}]=\begin{bmatrix} \mathrm{var}[\boldsymbol{x}_0] & \mathrm{cov}[\boldsymbol{x}_0,\boldsymbol{x}_1] \\
+ \mathrm{cov}[\boldsymbol{x}_1,\boldsymbol{x}_0] & \mathrm{var}[\boldsymbol{x}_1] \\
+ \end{bmatrix},
+$$
+
+
+
where we wrote
+$$\boldsymbol{C}[\boldsymbol{x}_0,\boldsymbol{x}_1] = \boldsymbol{C}[\boldsymbol{x}]$$
+
to indicate that this the covariance of the vectors \( \boldsymbol{x} \) of the design/feature matrix \( \boldsymbol{X} \).
It is easy to generalize this to a matrix \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \).
+We have that the covariance matrix (the correlation matrix involves a simple rescaling) is given as
+
+$$
+\boldsymbol{C}[\boldsymbol{x}] = \frac{1}{n}\boldsymbol{X}^T\boldsymbol{X}= \mathbb{E}[\boldsymbol{X}^T\boldsymbol{X}].
+$$
+
+
+
Let us now assume that we can perform a series of orthogonal transformations where we employ some orthogonal matrices \( \boldsymbol{S} \). +These matrices are defined as \( \boldsymbol{S}\in {\mathbb{R}}^{p\times p} \) and obey the orthogonality requirements \( \boldsymbol{S}\boldsymbol{S}^T=\boldsymbol{S}^T\boldsymbol{S}=\boldsymbol{I} \). The matrix can be written out in terms of the column vectors \( \boldsymbol{s}_i \) as \( \boldsymbol{S}=[\boldsymbol{s}_0,\boldsymbol{s}_1,\dots,\boldsymbol{s}_{p-1}] \) and \( \boldsymbol{s}_i \in {\mathbb{R}}^{p} \). +
+ +Assume also that there is a transformation \( \boldsymbol{S}^T\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{S}=\boldsymbol{C}[\boldsymbol{y}] \) such that the new matrix \( \boldsymbol{C}[\boldsymbol{y}] \) is diagonal with elements \( [\lambda_0,\lambda_1,\lambda_2,\dots,\lambda_{p-1}] \).
+ +That is we have
+
+$$
+\boldsymbol{C}[\boldsymbol{y}] = \mathbb{E}[\boldsymbol{S}^T\boldsymbol{X}^T\boldsymbol{X}T\boldsymbol{S}]=\boldsymbol{S}^T\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{S},
+$$
+
+
+
since the matrix \( \boldsymbol{S} \) is not a data dependent matrix. Multiplying with \( \boldsymbol{S} \) from the left we have
+
+$$
+\boldsymbol{S}\boldsymbol{C}[\boldsymbol{y}] = \boldsymbol{C}[\boldsymbol{x}]\boldsymbol{S},
+$$
+
+
+
and since \( \boldsymbol{C}[\boldsymbol{y}] \) is diagonal we have for a given eigenvalue \( i \) of the covariance matrix that
+ +
+$$
+\boldsymbol{S}_i\lambda_i = \boldsymbol{C}[\boldsymbol{x}]\boldsymbol{S}_i.
+$$
+
+
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. +
+ +Here is how we would proceed in setting up the algorithm for the PCA, see also discussion below here.
+We will use a simple example first with two-dimensional data +drawn from a multivariate normal distribution with the following mean and covariance matrix (we have fixed these quantities but will play around with them below): +
+
+$$
+\mu = (-1,2) \qquad \Sigma = \begin{bmatrix} 4 & 2 \\
+2 & 2
+\end{bmatrix}
+$$
+
+
+
Note that the mean refers to each column of data. +We will generate \( n = 10000 \) points \( X = \{ x_1, \ldots, x_N \} \) from +this distribution, and store them in the \( 1000 \times 2 \) matrix \( \boldsymbol{X} \). This is our design matrix where we have forced the covariance and mean values to take specific values. +
+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 \( n \). +
+ + +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)
+
+Now we are going to implement the PCA algorithm. We will break it down into various substeps.
+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
+
+$$
+\mu_n = \frac{1}{n} \sum_{i=1}^n x_i
+$$
+
+
+
and the mean-centered data \( \bar{X} = \{ \bar{x}_1, \ldots, \bar{x}_n \} \) takes the form
+
+$$
+\bar{x}_i = x_i - \mu_n.
+$$
+
+
+
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! +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. +
+ + +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)
+
+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. +
+Now we are going to use the mean centered data to compute the sample covariance of the data by using the following equation
+
+$$
+\begin{equation*}
+\Sigma_n = \frac{1}{n-1} \sum_{i=1}^n \bar{x}_i^T \bar{x}_i = \frac{1}{n-1} \sum_{i=1}^n (x_i - \mu_n)^T (x_i - \mu_n)
+\end{equation*}
+$$
+
+
+
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 +
+ + +print(df.cov())
+print(np.cov(X_centered.T))
+
+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. +
+ + +# 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()
+
+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? Try to change the covariance and the mean values. For example, try to make the variance of the first element much larger than that of the second diagonal element. Try also to shrink the covariance (the non-diagonal elements) and see how the data points are distributed. +
+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: +
+ ++
+$$
+\begin{equation*}
+x_i \approx \tilde{x}_i = \mu_n + \langle x_i, v_0 \rangle v_0
+\end{equation*}
+$$
+
+
+
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. +
+ + + +# 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])
+
+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?
+We assume now that we have a design matrix \( \boldsymbol{X} \) 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 \( [\boldsymbol{x}_0,\boldsymbol{x}_1,\dots, \boldsymbol{x}_{p-1}] \) each with dimension +\( \boldsymbol{x}\in {\mathbb{R}}^{n} \). +
+ +The PCA theorem states that minimizing the above reconstruction error +corresponds to setting \( \boldsymbol{W}=\boldsymbol{S} \), 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 +\( \boldsymbol{z}_i \) 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). +
+To show the PCA theorem let us start with the assumption that there is one vector \( \boldsymbol{s}_0 \) 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 \( \boldsymbol{w}_0 \) and \( \boldsymbol{z}_0 \) as
+ +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. +
+ +We could trivially maximize the variance of the projection (and +thereby minimize the error in the reconstruction function) by letting +the norm-2 of \( \boldsymbol{w}_0 \) go to infinity. However, this norm since we +want the matrix \( \boldsymbol{W} \) to be an orthogonal matrix, is constrained by +\( \vert\vert \boldsymbol{w}_0 \vert\vert_2^2=1 \). Imposing this condition via a +Lagrange multiplier we can then in turn maximize +
+ +
+$$
+J(\boldsymbol{w}_0)= \boldsymbol{w}_0^T\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{w}_0+\lambda_0(1-\boldsymbol{w}_0^T\boldsymbol{w}_0).
+$$
+
+
+
Taking the derivative with respect to \( \boldsymbol{w}_0 \) we obtain
+ +
+$$
+\frac{\partial J(\boldsymbol{w}_0)}{\partial \boldsymbol{w}_0}= 2\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{w}_0-2\lambda_0\boldsymbol{w}_0=0,
+$$
+
+
+
meaning that
+
+$$
+\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{w}_0=\lambda_0\boldsymbol{w}_0.
+$$
+
+
+
The direction that maximizes the variance (or minimizes the construction error) is an eigenvector of the covariance matrix! If we left multiply with \( \boldsymbol{w}_0^T \) we have the variance of the projected data is
+
+$$
+\boldsymbol{w}_0^T\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{w}_0=\lambda_0.
+$$
+
+
+
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 \( \boldsymbol{X} \). +
+ +The proof +for the other eigenvectors \( \boldsymbol{w}_1,\boldsymbol{w}_2,\dots \) can be +established by applying the above arguments and using the fact that +our basis of eigenvectors is orthogonal, see Murphy chapter +12.2. 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. +
+ +For more details, see for example Vidal, Ma and Sastry, chapter 2.
+For a detailed demonstration of the geometric interpretation, see Vidal, Ma and Sastry, section 2.1.2.
+ +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 +
+ + +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)
+
+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. +
+ + +W2 = V.T[:, :2]
+X2D = X_centered.dot(W2)
+
+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): +
+ + +#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)
+
+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 +
+ + +pca.components_.T[:, 0]
+
+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. +
+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. +
+ + +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)))
+
+We see that our training data after the PCA decomposition has a performance similar to the non-scaled data.
+ +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: +
+ + +pca = PCA()
+pca.fit(X)
+cumsum = np.cumsum(pca.explained_variance_ratio_)
+d = np.argmax(cumsum >= 0.95) + 1
+
+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: +
+ + +pca = PCA(n_components=0.95)
+X_reduced = pca.fit_transform(X)
+
+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). +
+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 \( n \). +
+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 +
+ + +from sklearn.decomposition import KernelPCA
+rbf_pca = KernelPCA(n_components = 2, kernel="rbf", gamma=0.04)
+X_reduced = rbf_pca.fit_transform(X)
+
+There are many other dimensionality reduction techniques, several of which are available in Scikit-Learn.
+ +Here are some of the most popular:
+We have tried every year to organize a kind of mini-workshop on project 3. In 2020 the contributions were (and some of these ended up in thesis work and/or publications, online only due to Covid-19!)
- -We would very much like to organize something similar this year as well. Feel free to come with suggesstions by Thursday November 17. We will then try to set up the various contributions for Friday November 18. -The presentations last normally 5-15 mins and span from loose ideas to more well-defined topics. Nothing pretentious, we wish to keep this as low-key as possible. -
-The functions we need are contained in the quadratic programming package CVXOPT and we need to import it together with numpy as
+The functions we need are contained in the quadratic programming library CVXOPT and we need to import it together with numpy as
@@ -1341,7 +1362,7 @@ converge. So, welcome to the promised land of quadratic programming.This will make our life much easier. You don't need t write your own optimizer.
+This will make our life much easier. You don't need t0 write your own optimizer.
Using the CVXOPT library, the matrix \( P \) would then be defined by the
+ +The principal component analysis deals with the problem of fitting a +low-dimensional affine subspace \( S \) of dimension \( d \) much smaller than +the total dimension \( D \) 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. +Historically, the PCA was first formulated in a statistical setting in order to estimate the principal component of a multivariate random variable. +
+ +We have a data set defined by a design/feature matrix \( \boldsymbol{X} \) (see below for its definition)
+A good read is for example Vidal, Ma and Sastry.
+ +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 \( \boldsymbol{C} \) is defined as +
+$$ +\boldsymbol{C}[\boldsymbol{x},\boldsymbol{y}] = \begin{bmatrix} \mathrm{cov}[\boldsymbol{x},\boldsymbol{x}] & \mathrm{cov}[\boldsymbol{x},\boldsymbol{y}] \\ + \mathrm{cov}[\boldsymbol{y},\boldsymbol{x}] & \mathrm{cov}[\boldsymbol{y},\boldsymbol{y}] \\ + \end{bmatrix}, +$$ + +where for example
+$$ +\mathrm{cov}[\boldsymbol{x},\boldsymbol{y}] =\frac{1}{n} \sum_{i=0}^{n-1}(x_i- \overline{x})(y_i- \overline{y}). +$$ + +With this definition and recalling that the variance is defined as
+$$ +\mathrm{var}[\boldsymbol{x}]=\frac{1}{n} \sum_{i=0}^{n-1}(x_i- \overline{x})^2, +$$ + +we can rewrite the covariance matrix as
+$$ +\boldsymbol{C}[\boldsymbol{x},\boldsymbol{y}] = \begin{bmatrix} \mathrm{var}[\boldsymbol{x}] & \mathrm{cov}[\boldsymbol{x},\boldsymbol{y}] \\ + \mathrm{cov}[\boldsymbol{x},\boldsymbol{y}] & \mathrm{var}[\boldsymbol{y}] \\ + \end{bmatrix}. +$$ + + +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 +
+ +$$ +\mathrm{corr}[\boldsymbol{x},\boldsymbol{y}]=\frac{\mathrm{cov}[\boldsymbol{x},\boldsymbol{y}]}{\sqrt{\mathrm{var}[\boldsymbol{x}] \mathrm{var}[\boldsymbol{y}]}}. +$$ + +The correlation function is then given by values \( \mathrm{corr}[\boldsymbol{x},\boldsymbol{y}] +\in [-1,1] \). This avoids eventual problems with too large values. We +can then define the correlation matrix for the two vectors \( \boldsymbol{x} \) +and \( \boldsymbol{y} \) as +
+ +$$ +\boldsymbol{K}[\boldsymbol{x},\boldsymbol{y}] = \begin{bmatrix} 1 & \mathrm{corr}[\boldsymbol{x},\boldsymbol{y}] \\ + \mathrm{corr}[\boldsymbol{y},\boldsymbol{x}] & 1 \\ + \end{bmatrix}, +$$ + +In the above example this is the function we constructed using pandas.
+ +In our derivation of the various regression algorithms like Ordinary Least Squares or Ridge regression +we defined the design/feature matrix \( \boldsymbol{X} \) as +
+ +$$ +\boldsymbol{X}=\begin{bmatrix} +x_{0,0} & x_{0,1} & x_{0,2}& \dots & \dots x_{0,p-1}\\ +x_{1,0} & x_{1,1} & x_{1,2}& \dots & \dots x_{1,p-1}\\ +x_{2,0} & x_{2,1} & x_{2,2}& \dots & \dots x_{2,p-1}\\ +\dots & \dots & \dots & \dots \dots & \dots \\ +x_{n-2,0} & x_{n-2,1} & x_{n-2,2}& \dots & \dots x_{n-2,p-1}\\ +x_{n-1,0} & x_{n-1,1} & x_{n-1,2}& \dots & \dots x_{n-1,p-1}\\ +\end{bmatrix}, +$$ + +with \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \), 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 +
+$$ +\boldsymbol{X}=\begin{bmatrix} \boldsymbol{x}_0 & \boldsymbol{x}_1 & \boldsymbol{x}_2 & \dots & \dots & \boldsymbol{x}_{p-1}\end{bmatrix}, +$$ + +with a given vector
+$$ +\boldsymbol{x}_i^T = \begin{bmatrix}x_{0,i} & x_{1,i} & x_{2,i}& \dots & \dots x_{n-1,i}\end{bmatrix}. +$$ + + +With these definitions, we can now rewrite our \( 2\times 2 \) +correlation/covariance matrix in terms of a moe general design/feature +matrix \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \). This leads to a \( p\times p \) +covariance matrix for the vectors \( \boldsymbol{x}_i \) with \( i=0,1,\dots,p-1 \) +
+ +$$ +\boldsymbol{C}[\boldsymbol{x}] = \begin{bmatrix} +\mathrm{var}[\boldsymbol{x}_0] & \mathrm{cov}[\boldsymbol{x}_0,\boldsymbol{x}_1] & \mathrm{cov}[\boldsymbol{x}_0,\boldsymbol{x}_2] & \dots & \dots & \mathrm{cov}[\boldsymbol{x}_0,\boldsymbol{x}_{p-1}]\\ +\mathrm{cov}[\boldsymbol{x}_1,\boldsymbol{x}_0] & \mathrm{var}[\boldsymbol{x}_1] & \mathrm{cov}[\boldsymbol{x}_1,\boldsymbol{x}_2] & \dots & \dots & \mathrm{cov}[\boldsymbol{x}_1,\boldsymbol{x}_{p-1}]\\ +\mathrm{cov}[\boldsymbol{x}_2,\boldsymbol{x}_0] & \mathrm{cov}[\boldsymbol{x}_2,\boldsymbol{x}_1] & \mathrm{var}[\boldsymbol{x}_2] & \dots & \dots & \mathrm{cov}[\boldsymbol{x}_2,\boldsymbol{x}_{p-1}]\\ +\dots & \dots & \dots & \dots & \dots & \dots \\ +\dots & \dots & \dots & \dots & \dots & \dots \\ +\mathrm{cov}[\boldsymbol{x}_{p-1},\boldsymbol{x}_0] & \mathrm{cov}[\boldsymbol{x}_{p-1},\boldsymbol{x}_1] & \mathrm{cov}[\boldsymbol{x}_{p-1},\boldsymbol{x}_{2}] & \dots & \dots & \mathrm{var}[\boldsymbol{x}_{p-1}]\\ +\end{bmatrix}, +$$ + + +and the correlation matrix
+$$ +\boldsymbol{K}[\boldsymbol{x}] = \begin{bmatrix} +1 & \mathrm{corr}[\boldsymbol{x}_0,\boldsymbol{x}_1] & \mathrm{corr}[\boldsymbol{x}_0,\boldsymbol{x}_2] & \dots & \dots & \mathrm{corr}[\boldsymbol{x}_0,\boldsymbol{x}_{p-1}]\\ +\mathrm{corr}[\boldsymbol{x}_1,\boldsymbol{x}_0] & 1 & \mathrm{corr}[\boldsymbol{x}_1,\boldsymbol{x}_2] & \dots & \dots & \mathrm{corr}[\boldsymbol{x}_1,\boldsymbol{x}_{p-1}]\\ +\mathrm{corr}[\boldsymbol{x}_2,\boldsymbol{x}_0] & \mathrm{corr}[\boldsymbol{x}_2,\boldsymbol{x}_1] & 1 & \dots & \dots & \mathrm{corr}[\boldsymbol{x}_2,\boldsymbol{x}_{p-1}]\\ +\dots & \dots & \dots & \dots & \dots & \dots \\ +\dots & \dots & \dots & \dots & \dots & \dots \\ +\mathrm{corr}[\boldsymbol{x}_{p-1},\boldsymbol{x}_0] & \mathrm{corr}[\boldsymbol{x}_{p-1},\boldsymbol{x}_1] & \mathrm{corr}[\boldsymbol{x}_{p-1},\boldsymbol{x}_{2}] & \dots & \dots & 1\\ +\end{bmatrix}, +$$ + + +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 \( \boldsymbol{W} \) +
+ +$$ +\boldsymbol{W}^T = \begin{bmatrix} x_0 & y_0 \\ + x_1 & y_1 \\ + x_2 & y_2\\ + \dots & \dots \\ + x_{n-2} & y_{n-2}\\ + x_{n-1} & y_{n-1} & + \end{bmatrix}, +$$ + +which in turn is converted into into the \( 2\times 2 \) covariance matrix +\( \boldsymbol{C} \) via the Numpy function np.cov(). We note that we can also calculate +the mean value of each set of samples \( \boldsymbol{x} \) etc using the Numpy +function np.mean(x). We can also extract the eigenvalues of the +covariance matrix through the np.linalg.eig() function. +
+ + + +# 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)
+
+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). +
+ + + +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)
+
+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.
+ +We whow here how we can set up the correlation matrix using pandas, as done in this simple code
+ + +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)
+
+We expand this model to the Franke function discussed above.
+ + + +# 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)
+
+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 \( n \)). We would however not include the intercept +and wee can simply +drop these elements and construct a correlation +matrix without them by centering our matrix elements by subtracting the mean of each column. +
+ +We can rewrite the covariance matrix in a more compact form in terms of the design/feature matrix \( \boldsymbol{X} \) as
+$$ +\boldsymbol{C}[\boldsymbol{x}] = \frac{1}{n}\boldsymbol{X}^T\boldsymbol{X}= \mathbb{E}[\boldsymbol{X}^T\boldsymbol{X}]. +$$ + +To see this let us simply look at a design matrix \( \boldsymbol{X}\in {\mathbb{R}}^{2\times 2} \)
+$$ +\boldsymbol{X}=\begin{bmatrix} +x_{00} & x_{01}\\ +x_{10} & x_{11}\\ +\end{bmatrix}=\begin{bmatrix} +\boldsymbol{x}_{0} & \boldsymbol{x}_{1}\\ +\end{bmatrix}. +$$ + + +If we then compute the expectation value
+$$ +\mathbb{E}[\boldsymbol{X}^T\boldsymbol{X}] = \frac{1}{n}\boldsymbol{X}^T\boldsymbol{X}=\begin{bmatrix} +x_{00}^2+x_{01}^2 & x_{00}x_{10}+x_{01}x_{11}\\ +x_{10}x_{00}+x_{11}x_{01} & x_{10}^2+x_{11}^2\\ +\end{bmatrix}, +$$ + +which is just
+$$ +\boldsymbol{C}[\boldsymbol{x}_0,\boldsymbol{x}_1] = \boldsymbol{C}[\boldsymbol{x}]=\begin{bmatrix} \mathrm{var}[\boldsymbol{x}_0] & \mathrm{cov}[\boldsymbol{x}_0,\boldsymbol{x}_1] \\ + \mathrm{cov}[\boldsymbol{x}_1,\boldsymbol{x}_0] & \mathrm{var}[\boldsymbol{x}_1] \\ + \end{bmatrix}, +$$ + +where we wrote $$\boldsymbol{C}[\boldsymbol{x}_0,\boldsymbol{x}_1] = \boldsymbol{C}[\boldsymbol{x}]$$ to indicate that this the covariance of the vectors \( \boldsymbol{x} \) of the design/feature matrix \( \boldsymbol{X} \).
+ +It is easy to generalize this to a matrix \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \).
+ +We have that the covariance matrix (the correlation matrix involves a simple rescaling) is given as
+$$ +\boldsymbol{C}[\boldsymbol{x}] = \frac{1}{n}\boldsymbol{X}^T\boldsymbol{X}= \mathbb{E}[\boldsymbol{X}^T\boldsymbol{X}]. +$$ + +Let us now assume that we can perform a series of orthogonal transformations where we employ some orthogonal matrices \( \boldsymbol{S} \). +These matrices are defined as \( \boldsymbol{S}\in {\mathbb{R}}^{p\times p} \) and obey the orthogonality requirements \( \boldsymbol{S}\boldsymbol{S}^T=\boldsymbol{S}^T\boldsymbol{S}=\boldsymbol{I} \). The matrix can be written out in terms of the column vectors \( \boldsymbol{s}_i \) as \( \boldsymbol{S}=[\boldsymbol{s}_0,\boldsymbol{s}_1,\dots,\boldsymbol{s}_{p-1}] \) and \( \boldsymbol{s}_i \in {\mathbb{R}}^{p} \). +
+ +Assume also that there is a transformation \( \boldsymbol{S}^T\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{S}=\boldsymbol{C}[\boldsymbol{y}] \) such that the new matrix \( \boldsymbol{C}[\boldsymbol{y}] \) is diagonal with elements \( [\lambda_0,\lambda_1,\lambda_2,\dots,\lambda_{p-1}] \).
+ +That is we have
+$$ +\boldsymbol{C}[\boldsymbol{y}] = \mathbb{E}[\boldsymbol{S}^T\boldsymbol{X}^T\boldsymbol{X}T\boldsymbol{S}]=\boldsymbol{S}^T\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{S}, +$$ + +since the matrix \( \boldsymbol{S} \) is not a data dependent matrix. Multiplying with \( \boldsymbol{S} \) from the left we have
+$$ +\boldsymbol{S}\boldsymbol{C}[\boldsymbol{y}] = \boldsymbol{C}[\boldsymbol{x}]\boldsymbol{S}, +$$ + +and since \( \boldsymbol{C}[\boldsymbol{y}] \) is diagonal we have for a given eigenvalue \( i \) of the covariance matrix that
+ +$$ +\boldsymbol{S}_i\lambda_i = \boldsymbol{C}[\boldsymbol{x}]\boldsymbol{S}_i. +$$ + + +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. +
+ +Here is how we would proceed in setting up the algorithm for the PCA, see also discussion below here.
+We will use a simple example first with two-dimensional data +drawn from a multivariate normal distribution with the following mean and covariance matrix (we have fixed these quantities but will play around with them below): +
+$$ +\mu = (-1,2) \qquad \Sigma = \begin{bmatrix} 4 & 2 \\ +2 & 2 +\end{bmatrix} +$$ + +Note that the mean refers to each column of data. +We will generate \( n = 10000 \) points \( X = \{ x_1, \ldots, x_N \} \) from +this distribution, and store them in the \( 1000 \times 2 \) matrix \( \boldsymbol{X} \). This is our design matrix where we have forced the covariance and mean values to take specific values. +
+ +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 \( n \). +
+ + +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)
+
+Now we are going to implement the PCA algorithm. We will break it down into various substeps.
+ +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
+$$ +\mu_n = \frac{1}{n} \sum_{i=1}^n x_i +$$ + +and the mean-centered data \( \bar{X} = \{ \bar{x}_1, \ldots, \bar{x}_n \} \) takes the form
+$$ +\bar{x}_i = x_i - \mu_n. +$$ + +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! +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. +
+ + +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)
+
+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. +
+ +Now we are going to use the mean centered data to compute the sample covariance of the data by using the following equation
+$$ +\begin{equation*} +\Sigma_n = \frac{1}{n-1} \sum_{i=1}^n \bar{x}_i^T \bar{x}_i = \frac{1}{n-1} \sum_{i=1}^n (x_i - \mu_n)^T (x_i - \mu_n) +\end{equation*} +$$ + +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 +
+ + +print(df.cov())
+print(np.cov(X_centered.T))
+
+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. +
+ + +# 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()
+
+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? Try to change the covariance and the mean values. For example, try to make the variance of the first element much larger than that of the second diagonal element. Try also to shrink the covariance (the non-diagonal elements) and see how the data points are distributed. +
+ +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: +
+ +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. +
+ + + +# 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])
+
+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?
+ +We assume now that we have a design matrix \( \boldsymbol{X} \) 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 \( [\boldsymbol{x}_0,\boldsymbol{x}_1,\dots, \boldsymbol{x}_{p-1}] \) each with dimension +\( \boldsymbol{x}\in {\mathbb{R}}^{n} \). +
+ +The PCA theorem states that minimizing the above reconstruction error +corresponds to setting \( \boldsymbol{W}=\boldsymbol{S} \), 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 +\( \boldsymbol{z}_i \) 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). +
+ +To show the PCA theorem let us start with the assumption that there is one vector \( \boldsymbol{s}_0 \) 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 \( \boldsymbol{w}_0 \) and \( \boldsymbol{z}_0 \) as
+ +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. +
+ +We could trivially maximize the variance of the projection (and +thereby minimize the error in the reconstruction function) by letting +the norm-2 of \( \boldsymbol{w}_0 \) go to infinity. However, this norm since we +want the matrix \( \boldsymbol{W} \) to be an orthogonal matrix, is constrained by +\( \vert\vert \boldsymbol{w}_0 \vert\vert_2^2=1 \). Imposing this condition via a +Lagrange multiplier we can then in turn maximize +
+ +$$ +J(\boldsymbol{w}_0)= \boldsymbol{w}_0^T\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{w}_0+\lambda_0(1-\boldsymbol{w}_0^T\boldsymbol{w}_0). +$$ + +Taking the derivative with respect to \( \boldsymbol{w}_0 \) we obtain
+ +$$ +\frac{\partial J(\boldsymbol{w}_0)}{\partial \boldsymbol{w}_0}= 2\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{w}_0-2\lambda_0\boldsymbol{w}_0=0, +$$ + +meaning that
+$$ +\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{w}_0=\lambda_0\boldsymbol{w}_0. +$$ + +The direction that maximizes the variance (or minimizes the construction error) is an eigenvector of the covariance matrix! If we left multiply with \( \boldsymbol{w}_0^T \) we have the variance of the projected data is
+$$ +\boldsymbol{w}_0^T\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{w}_0=\lambda_0. +$$ + +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 \( \boldsymbol{X} \). +
+ +The proof +for the other eigenvectors \( \boldsymbol{w}_1,\boldsymbol{w}_2,\dots \) can be +established by applying the above arguments and using the fact that +our basis of eigenvectors is orthogonal, see Murphy chapter +12.2. 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. +
+ +For more details, see for example Vidal, Ma and Sastry, chapter 2.
+ +For a detailed demonstration of the geometric interpretation, see Vidal, Ma and Sastry, section 2.1.2.
+ +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 +
+ + +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)
+
+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. +
+ + +W2 = V.T[:, :2]
+X2D = X_centered.dot(W2)
+
+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): +
+ + +#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)
+
+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 +
+ + +pca.components_.T[:, 0]
+
+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. +
+ +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. +
+ + +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)))
+
+We see that our training data after the PCA decomposition has a performance similar to the non-scaled data.
+ +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: +
+ + +pca = PCA()
+pca.fit(X)
+cumsum = np.cumsum(pca.explained_variance_ratio_)
+d = np.argmax(cumsum >= 0.95) + 1
+
+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: +
+ + +pca = PCA(n_components=0.95)
+X_reduced = pca.fit_transform(X)
+
+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). +
+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 \( n \). +
+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 +
+ + +from sklearn.decomposition import KernelPCA
+rbf_pca = KernelPCA(n_components = 2, kernel="rbf", gamma=0.04)
+X_reduced = rbf_pca.fit_transform(X)
+
+There are many other dimensionality reduction techniques, several of which are available in Scikit-Learn.
+ +Here are some of the most popular:
+