+ + +
+ +
+ +
+ + +
Reading recommendations:
++ +
+ +
+ + +
Last year we had a very interesting workshop with many presentations. These were (program 2020)
+ +We wish to organize something similar this coming Friday. The presentation last typically 5-10 minutes (some 3-5 slides) with time for questions afterwards. +Feel free to suggest topics. +
+ +The program will be available asap. It depends on input from you!
+ ++ +
+ +
+ + +
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. +
+ ++ +
+ +
+ + +
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()
+
++ +
+ +
+ + +
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. +
+ +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. +
+ +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 +
+ +$$ +\boldsymbol{x}^T\boldsymbol{w}+b=0. +$$ + + ++ +
+ +
+ + +
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. +
+ +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.
+ ++ +
+ +
+ + +
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. +
+ +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. +
+ +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. +
+ ++ +
+ +
+ + +
Let us define the function
+$$ +f(x) = \boldsymbol{w}^T\boldsymbol{x}+b = 0, +$$ + +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
+$$ +\delta = \frac{1}{\vert\vert \boldsymbol{w}\vert\vert}(\boldsymbol{w}^T\boldsymbol{x}+b). +$$ + + ++ +
+ +
+ + +
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 +
+ +$$ +C(\boldsymbol{w},b) = -\sum_{i\in M} y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b). +$$ + +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
+$$ +\frac{\partial C}{\partial \boldsymbol{w}} = -\sum_{i\in M} y_ix_i. +$$ + + ++ +
+ +
+ + +
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
+$$ +b \leftarrow b +\eta \frac{\partial C}{\partial b}, +$$ + +and
+$$ +\boldsymbol{w} \leftarrow \boldsymbol{w} +\eta \frac{\partial C}{\partial \boldsymbol{w}}, +$$ + +where \( \eta \) is our by now well-known learning rate.
+ ++ +
+ +
+ + +
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. +
+ + +
+
++ +
+ +
+ + +
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. +
+ +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. +
+ ++ +
+ +
+ + +
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. +
+ ++ +
+ +
+ + +
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. +
+ +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. +
+ ++ +
+ +
+ + +
However, we can add to
+$$ +df = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz, +$$ + +a multiplum of \( d\phi \), viz. \( \lambda d\phi \), resulting in
+$$ +df+\lambda d\phi = (\frac{\partial f}{\partial z}+\lambda +\frac{\partial \phi}{\partial x})dx+(\frac{\partial f}{\partial y}+\lambda\frac{\partial \phi}{\partial y})dy+ +(\frac{\partial f}{\partial z}+\lambda\frac{\partial \phi}{\partial z})dz =0. +$$ + +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 +
+$$ +\frac{\partial f}{\partial x_i}+\sum_k\lambda_k\frac{\partial \phi_k}{\partial x_i} =0. +$$ + + ++ +
+ +
+ + +
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 +
+$$ +\lambda_i\left[y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) -1\right] \hspace{0.1cm}\forall i. +$$ + +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 \( M \).
+ ++ +
+ +
+ + +
We can rewrite
+$$ +{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j, +$$ + +and its constraints in terms of a matrix-vector problem where we minimize w.r.t. \( \lambda \) the following problem
+$$ +\frac{1}{2} \boldsymbol{\lambda}^T\begin{bmatrix} y_1y_1\boldsymbol{x}_1^T\boldsymbol{x}_1 & y_1y_2\boldsymbol{x}_1^T\boldsymbol{x}_2 & \dots & \dots & y_1y_n\boldsymbol{x}_1^T\boldsymbol{x}_n \\ +y_2y_1\boldsymbol{x}_2^T\boldsymbol{x}_1 & y_2y_2\boldsymbol{x}_2^T\boldsymbol{x}_2 & \dots & \dots & y_1y_n\boldsymbol{x}_2^T\boldsymbol{x}_n \\ +\dots & \dots & \dots & \dots & \dots \\ +\dots & \dots & \dots & \dots & \dots \\ +y_ny_1\boldsymbol{x}_n^T\boldsymbol{x}_1 & y_ny_2\boldsymbol{x}_n^T\boldsymbol{x}_2 & \dots & \dots & y_ny_n\boldsymbol{x}_n^T\boldsymbol{x}_n \\ +\end{bmatrix}\boldsymbol{\lambda}-\mathbb{1}\boldsymbol{\lambda}, +$$ + +subject to \( \boldsymbol{y}^T\boldsymbol{\lambda}=0 \). Here we defined the vectors \( \boldsymbol{\lambda} =[\lambda_1,\lambda_2,\dots,\lambda_n] \) and +\( \boldsymbol{y}=[y_1,y_2,\dots,y_n] \). +
+ ++ +
+ +
+ + +
Solving the above problem, yields the values of \( \lambda_i \). +To find the coefficients of your hyperplane we need simply to compute +
+$$ +\boldsymbol{w}=\sum_{i} \lambda_iy_i\boldsymbol{x}_i. +$$ + +With our vector \( \boldsymbol{w} \) we can in turn find the value of the intercept \( b \) (here in two dimensions) via
+$$ +y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)=1, +$$ + +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.
+ ++ +
+ +
+ + +
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 +
+$$ +y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)=1, +$$ + +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. +
+ +Misclassifications occur when \( \xi_i > 1 \). Thus bounding the total sum by some value \( C \) bounds in turn the total number of +misclassifications. +
+ ++ +
+ +
+ + +
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 +
+$$ +\lambda_i\left[y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) -(1-\xi_)\right]=0 \hspace{0.1cm}\forall i, +$$ + +$$ +\gamma_i\xi_i = 0, +$$ + +and
+$$ +y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) -(1-\xi_) \geq 0 \hspace{0.1cm}\forall i. +$$ + + ++ +
+ +
+ + +
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. +
+ +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. +
+ +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()
+
++ +
+ +
+ + +
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)
+$$ +z = \phi(x_i) =\left(x_i^2, y_i^2, \sqrt{2}x_iy_i\right). +$$ + +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 \( (\boldsymbol{x}_i^T\boldsymbol{x}_j)^2 \). +
+ +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. +
+ ++ +
+ +
+ + +
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
+$$ +\frac{1}{2} \boldsymbol{\lambda}^T\begin{bmatrix} y_1y_1K(\boldsymbol{x}_1,\boldsymbol{x}_1) & y_1y_2K(\boldsymbol{x}_1,\boldsymbol{x}_2) & \dots & \dots & y_1y_nK(\boldsymbol{x}_1,\boldsymbol{x}_n) \\ +y_2y_1K(\boldsymbol{x}_2,\boldsymbol{x}_1) & y_2y_2(\boldsymbol{x}_2,\boldsymbol{x}_2) & \dots & \dots & y_1y_nK(\boldsymbol{x}_2,\boldsymbol{x}_n) \\ +\dots & \dots & \dots & \dots & \dots \\ +\dots & \dots & \dots & \dots & \dots \\ +y_ny_1K(\boldsymbol{x}_n,\boldsymbol{x}_1) & y_ny_2K(\boldsymbol{x}_n\boldsymbol{x}_2) & \dots & \dots & y_ny_nK(\boldsymbol{x}_n,\boldsymbol{x}_n) \\ +\end{bmatrix}\boldsymbol{\lambda}-\mathbb{1}\boldsymbol{\lambda}, +$$ + +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 \( 0\leq \lambda_i \leq C \). +
+ +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 \( \boldsymbol{h} \). +
+ ++ +
+ +
+ + +
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 +
+ +$$ +K(\boldsymbol{x}_i,\boldsymbol{x}_j)=\phi(\boldsymbol{x}_i)^T\phi(\boldsymbol{x}_j). +$$ + +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. +
+ ++ +
+ +
+ + +
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()
+
++ +
+ +
+ + +
A mathematical (quadratic) optimization problem, or just optimization problem, has the form
+$$ +\begin{align*} + &\mathrm{min}_{\lambda}\hspace{0.2cm} \frac{1}{2}\boldsymbol{\lambda}^T\boldsymbol{P}\boldsymbol{\lambda}+\boldsymbol{q}^T\boldsymbol{\lambda},\\ \nonumber + &\mathrm{subject\hspace{0.1cm}to} \hspace{0.2cm} \boldsymbol{G}\boldsymbol{\lambda} \preceq \boldsymbol{h} \wedge \boldsymbol{A}\boldsymbol{\lambda}=f. +\end{align*} +$$ + +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. +
+ +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.
+ ++ +
+ +
+ + +
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 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.
+ ++ +
+ +
+ + +
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 +
+ + +# 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 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 \( 0\leq \lambda_i \leq C \). +
+ +code will be added + ++ +
+ +
- - - -
-
+
-
@@ -202,29 +212,22 @@ MathJax.Hub.Config({
-
+
-
-
-
-See overview video on Support Vector Machines. See also this video. +
Reading recommendations:
++
+Last year we had a very interesting workshop with many presentations. These were (program 2020)
--The first lecture on Thursday is devoted to a summary from last week, with additional examples. This material is included in the lectures from week 45, see also Hastie et al. chapter 10.1-10.10. +
We wish to organize something similar this coming Friday. The presentation last typically 5-10 minutes (some 3-5 slides) with time for questions afterwards. +Feel free to suggest topics. +
--The second letter will be given by John M. Aiken, who recently defended his thesis on machine learning, and in particular using boosting methods, to data from the social sciences and science education. +
The program will be available asap. It depends on input from you!
-
+
-We start with a new topic and second-last topic covered in this course, support vector machines (SVM). - -
-
-
-
-A Support Vector Machine (SVM) is a very powerful and versatile +
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. +medium-sized datasets. +
--The case with two well-separated classes only can be understood in an +
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. +
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 +
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. +
-
+
-The theory behind support vector machines (SVM hereafter) is based on +
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 +
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 import datasets
from sklearn.svm import SVC, LinearSVC
from sklearn.linear_model import SGDClassifier
from sklearn.preprocessing import StandardScaler
@@ -267,87 +335,96 @@ plt.legend(loc="upper center", fon
plt.axis([0, 5.5, 0, 2])
plt.show()
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+-The aim of the SVM algorithm is to find a hyperplane in a +
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. +
--In a \( p \)-dimensional space, a hyperplane is what we call an affine subspace of dimension of \( p-1 \). +
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. +a two-dimensional subspace, or stated simply, a plane. +
--In two dimensions, with the variables \( x_1 \) and \( x_2 \), the hyperplane is defined as +
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 +
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 +
$$ \boldsymbol{x}^T\boldsymbol{w}+b=0. $$ -
+
+
-We limit ourselves to two classes of outputs \( y_i \) and assign these classes the values \( y_i = \pm 1 \). +
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 +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 +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 \( y_i=1 \). +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. +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 +
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. +
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.
-+
-Let us try to develop our intuition about SVMs by limiting ourselves to a two-dimensional +
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 @@ -355,302 +432,298 @@ 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. +
--What a linear classifier attempts to accomplish is to split the +
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. +
--Unfortunately there are many ways in which we can place a hyperplane +
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. +
-
+
-Let us define the function +
Let us define the function
$$ f(x) = \boldsymbol{w}^T\boldsymbol{x}+b = 0, $$ -as the function that determines the line \( L \) that separates two classes (our two features), see the figure here. +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 \( \boldsymbol{w}^T(\boldsymbol{x}_1-\boldsymbol{x}_2)=0 \). +
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 +
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
$$ \delta = \frac{1}{\vert\vert \boldsymbol{w}\vert\vert}(\boldsymbol{w}^T\boldsymbol{x}+b). $$ -
+
+
-How do we find the parameter \( b \) and the vector \( \boldsymbol{w} \)? What we could +
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 +
$$ C(\boldsymbol{w},b) = -\sum_{i\in M} y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b). $$ --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 +
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 +and
$$ \frac{\partial C}{\partial \boldsymbol{w}} = -\sum_{i\in M} y_ix_i. $$ -
+
+
-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 +
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
$$ b \leftarrow b +\eta \frac{\partial C}{\partial b}, $$ -and +and
$$ \boldsymbol{w} \leftarrow \boldsymbol{w} +\eta \frac{\partial C}{\partial \boldsymbol{w}}, $$ -where \( \eta \) is our by now well-known learning rate. +where \( \eta \) is our by now well-known learning rate.
-
+
-The equations we discussed above can be coded rather easily (the +
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. -
+
-
-+
+
+-There are however problems with this approach, although it looks +
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. +
--For small +
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. +
-
+
-A better approach is rather to try to define a large margin between +
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 +
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. +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 +
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 +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 +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 +
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. +
-
+
-Consider a function of three independent variables \( f(x,y,z) \) . For the function \( f \) to be an +
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 +A necessary and sufficient condition is
$$ \frac{\partial f}{\partial x} =\frac{\partial f}{\partial y}=\frac{\partial f}{\partial z}=0, $$ -due to +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) +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 +
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 +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 +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 +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. +
-
+
-However, we can add to +
However, we can add to
$$ df = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz, $$ -a multiplum of \( d\phi \), viz. \( \lambda d\phi \), resulting in +a multiplum of \( d\phi \), viz. \( \lambda d\phi \), resulting in
$$ df+\lambda d\phi = (\frac{\partial f}{\partial z}+\lambda \frac{\partial \phi}{\partial x})dx+(\frac{\partial f}{\partial y}+\lambda\frac{\partial \phi}{\partial y})dy+ (\frac{\partial f}{\partial z}+\lambda\frac{\partial \phi}{\partial z})dz =0. $$ -Our multiplier is chosen so that +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 +
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 +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 +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 +
$$ \frac{\partial f}{\partial x_i}+\sum_k\lambda_k\frac{\partial \phi_k}{\partial x_i} =0. $$ -
-
-
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 \( \lambda_i \geq 0 \). +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 +
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 +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 +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 \( \sum_i\lambda_iy_i=0 \). +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 +
$$ \lambda_i\left[y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) -1\right] \hspace{0.1cm}\forall i. $$ -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 \( M \).
-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 \( M \). - -
+
-We can rewrite +
We can rewrite
$$ {\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j, $$ -and its constraints in terms of a matrix-vector problem where we minimize w.r.t. \( \lambda \) the following problem +and its constraints in terms of a matrix-vector problem where we minimize w.r.t. \( \lambda \) the following problem
$$ \frac{1}{2} \boldsymbol{\lambda}^T\begin{bmatrix} y_1y_1\boldsymbol{x}_1^T\boldsymbol{x}_1 & y_1y_2\boldsymbol{x}_1^T\boldsymbol{x}_2 & \dots & \dots & y_1y_n\boldsymbol{x}_1^T\boldsymbol{x}_n \\ y_2y_1\boldsymbol{x}_2^T\boldsymbol{x}_1 & y_2y_2\boldsymbol{x}_2^T\boldsymbol{x}_2 & \dots & \dots & y_1y_n\boldsymbol{x}_2^T\boldsymbol{x}_n \\ @@ -660,119 +733,113 @@ y_ny_1\boldsymbol{x}_n^T\boldsymbol{x}_1 & y_ny_2\boldsymbol{x}_n^T\boldsymbol{x \end{bmatrix}\boldsymbol{\lambda}-\mathbb{1}\boldsymbol{\lambda}, $$ -subject to \( \boldsymbol{y}^T\boldsymbol{\lambda}=0 \). Here we defined the vectors \( \boldsymbol{\lambda} =[\lambda_1,\lambda_2,\dots,\lambda_n] \) and -\( \boldsymbol{y}=[y_1,y_2,\dots,y_n] \). +subject to \( \boldsymbol{y}^T\boldsymbol{\lambda}=0 \). Here we defined the vectors \( \boldsymbol{\lambda} =[\lambda_1,\lambda_2,\dots,\lambda_n] \) and +\( \boldsymbol{y}=[y_1,y_2,\dots,y_n] \). +
-
+
-Solving the above problem, yields the values of \( \lambda_i \). +
Solving the above problem, yields the values of \( \lambda_i \). To find the coefficients of your hyperplane we need simply to compute +
$$ \boldsymbol{w}=\sum_{i} \lambda_iy_i\boldsymbol{x}_i. $$ -With our vector \( \boldsymbol{w} \) we can in turn find the value of the intercept \( b \) (here in two dimensions) via +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 +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 +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 +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. +Below we discuss how to find the optimal values of \( \lambda_i \). Before we proceed however, we discuss now the so-called soft classifier.
-
+
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.
--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 +
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 +
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 +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 \( \sum_i\xi \). +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 occur when \( \xi_i > 1 \). Thus bounding the total sum by some value \( C \) bounds in turn the total number of misclassifications. +
-
+
-This has in turn the consequences that we change our optmization problem to finding the minimum of +
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 +subject to
$$ y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)=1-\xi_i \hspace{0.1cm}\forall i, $$ -with the requirement \( \xi_i\geq 0 \). +with the requirement \( \xi_i\geq 0 \).
--Taking the derivatives with respect to \( b \) and \( \boldsymbol{w} \) we obtain +
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 +and
$$ \frac{\partial {\cal L}}{\partial \boldsymbol{w}} = 0 = \boldsymbol{w}-\sum_{i} \lambda_iy_i\boldsymbol{x}_i, $$ -and +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 +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 \( 0\leq\lambda_i \leq C \). +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 +
$$ \lambda_i\left[y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) -(1-\xi_)\right]=0 \hspace{0.1cm}\forall i, $$ @@ -781,38 +848,41 @@ $$ \gamma_i\xi_i = 0, $$ -and +and
$$ y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) -(1-\xi_) \geq 0 \hspace{0.1cm}\forall i. $$ -
+
+
-The cases we have studied till now, were all characterized by two classes +
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. +
--If our feature space is not easy to separate, as shown in the figure +
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. +obtain a separation between the classes which is almost linear. +
--The change of basis, from \( x\rightarrow z=\phi(x) \) leads to the same type of equations to be solved, except that +
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 numpy as np
import os
np.random.seed(42)
@@ -858,62 +928,72 @@ plt.plot([-4.5, 4.5, 4.5, -1, 17])
plt.subplots_adjust(right=1)
plt.show()
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+-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) +
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)
$$ z = \phi(x_i) =\left(x_i^2, y_i^2, \sqrt{2}x_iy_i\right). $$ --With our new basis, the equations we solved earlier are basically the same, that is we have now (without the slack option for simplicity) +
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 +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 \( b \). +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 +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 +
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 +
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. +
-
-
-
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 +subject to the constraints \( \lambda_i\geq 0 \), \( \sum_i\lambda_iy_i=0 \) in terms of a convex optimization problem
$$ \frac{1}{2} \boldsymbol{\lambda}^T\begin{bmatrix} y_1y_1K(\boldsymbol{x}_1,\boldsymbol{x}_1) & y_1y_2K(\boldsymbol{x}_1,\boldsymbol{x}_2) & \dots & \dots & y_1y_nK(\boldsymbol{x}_1,\boldsymbol{x}_n) \\ y_2y_1K(\boldsymbol{x}_2,\boldsymbol{x}_1) & y_2y_2(\boldsymbol{x}_2,\boldsymbol{x}_2) & \dots & \dots & y_1y_nK(\boldsymbol{x}_2,\boldsymbol{x}_n) \\ @@ -923,12 +1003,12 @@ y_ny_1K(\boldsymbol{x}_n,\boldsymbol{x}_1) & y_ny_2K(\boldsymbol{x}_n\boldsymbol \end{bmatrix}\boldsymbol{\lambda}-\mathbb{1}\boldsymbol{\lambda}, $$ -subject to \( \boldsymbol{y}^T\boldsymbol{\lambda}=0 \). Here we defined the vectors \( \boldsymbol{\lambda} =[\lambda_1,\lambda_2,\dots,\lambda_n] \) and +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 \( 0\leq \lambda_i \leq C \). +
--We can rewrite this (see the solutions below) in terms of a convex optimization problem of the type +
We can rewrite this (see the solutions below) in terms of a convex optimization problem of the type
$$ \begin{align*} &\mathrm{min}_{\lambda}\hspace{0.2cm} \frac{1}{2}\boldsymbol{\lambda}^T\boldsymbol{P}\boldsymbol{\lambda}+\boldsymbol{q}^T\boldsymbol{\lambda},\\ \nonumber @@ -936,56 +1016,54 @@ $$ \end{align*} $$ -Below we discuss how to solve these equations. Here we note that the matrix \( \boldsymbol{P} \) has matrix elements \( p_{ij}=y_iy_jK(\boldsymbol{x}_i,\boldsymbol{x}_j) \). +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 \( \boldsymbol{h} \). +
-
+
-There are several popular kernels being used. These are - +
There are several popular kernels being used. These are
and many other ones.
-and many other ones. - -
-An important theorem for us is Mercer's
+ 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
+
-So you can use \( K \) as a kernel since you know \( \phi \) exists, even if
-you don’t know what \( \phi \) is.
+ 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)
+ 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.
+
+
+
+
-A mathematical (quadratic) optimization problem, or just optimization problem, has the form
+ A mathematical (quadratic) optimization problem, or just optimization problem, has the form 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.
+
-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.
+ 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.
+ Convex optimization problems play a central role in applied mathematics and we recommend strongly Boyd and Vandenberghe's text on the topics.
-If we use Python as programming language and wish to venture beyond
+ 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 package CVXOPT and we need to import it together with numpy as
+ The functions we need are contained in the quadratic programming package CVXOPT and we need to import it together with numpy as
-
-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 t write your own optimizer.
-We remind ourselves about the general problem we want to solve
+ We remind ourselves about the general problem we want to solve
-Let us show how to perform the optmization using a simple case. Assume we want to optimize the following problem
+ Let us show how to perform the optmization using a simple case. Assume we want to optimize the following problem The minimization problem can be rewritten in terms of vectors and matrices as (with \( x \) and \( y \) being the unknowns) 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 We have collapsed all the inequalities into a single matrix \( \boldsymbol{G} \). We see also that our matrix is clearly positive semi-definite (all eigenvalues larger or equal zero).
Finally, the vector \( \boldsymbol{h} \) is defined as
+
-Since we don't have any equalities the matrix \( \boldsymbol{A} \) is set to zero
+ Since we don't have any equalities the matrix \( \boldsymbol{A} \) is set to zero
The following code solves the equations for us
-
+
+
+
-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
+ 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 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 \( 0\leq \lambda_i \leq C \).
+
code will be added
-
-
-
-
-
+
-
-See overview video on Support Vector Machines. See also this video.
+ Reading recommendations:
+ Last year we had a very interesting workshop with many presentations. These were (program 2020)
-The first lecture on Thursday is devoted to a summary from last week, with additional examples. This material is included in the lectures from week 45, see also Hastie et al. chapter 10.1-10.10.
+ We wish to organize something similar this coming Friday. The presentation last typically 5-10 minutes (some 3-5 slides) with time for questions afterwards.
+Feel free to suggest topics.
+
-The second letter will be given by John M. Aiken, who recently defended his thesis on machine learning, and in particular using boosting methods, to data from the social sciences and science education.
+ The program will be available asap. It depends on input from you!
-We start with a new topic and second-last topic covered in this course, support vector machines (SVM).
-
-
-
-A Support Vector Machine (SVM) is a very powerful and versatile
+ 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.
+medium-sized datasets.
+
-The case with two well-separated classes only can be understood in an
+ 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.
+ 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
+ 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.
+
-The theory behind support vector machines (SVM hereafter) is based on
+ 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
+ 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.
-
+
+
+
-The aim of the SVM algorithm is to find a hyperplane in a
+ 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.
+
-In a \( p \)-dimensional space, a hyperplane is what we call an affine subspace of dimension of \( p-1 \).
+ 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.
+a two-dimensional subspace, or stated simply, a plane.
+
-In two dimensions, with the variables \( x_1 \) and \( x_2 \), the hyperplane is defined as
+ In two dimensions, with the variables \( x_1 \) and \( x_2 \), the hyperplane is defined as
-where \( b \) is the intercept and \( w_1 \) and \( w_2 \) define the elements of a vector orthogonal to the line
+ 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
+
+
-We limit ourselves to two classes of outputs \( y_i \) and assign these classes the values \( y_i = \pm 1 \).
+ 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
+ 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 \( \boldsymbol{X} \),
+ If the above condition is not met for a given vector \( \boldsymbol{x}_i \) we have 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
+ 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
+ Equivalently, for the two classes of observations we have
-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.
+ 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.
+
-Let us try to develop our intuition about SVMs by limiting ourselves to a two-dimensional
+ 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
@@ -360,302 +509,298 @@ 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.
+
-What a linear classifier attempts to accomplish is to split the
+ 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.
+
-Unfortunately there are many ways in which we can place a hyperplane
+ 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.
+
-Let us define the function
+ Let us define the function 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 \( \boldsymbol{w}^T(\boldsymbol{x}_1-\boldsymbol{x}_2)=0 \).
+ 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
+ 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
+
-How do we find the parameter \( b \) and the vector \( \boldsymbol{w} \)? What we could
+ 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
+
-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
+ 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 and
+
-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
+ 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 and where \( \eta \) is our by now well-known learning rate.
-The equations we discussed above can be coded rather easily (the
+ 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.
-
+
+
-There are however problems with this approach, although it looks
+ 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.
+
-For small
+ 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.
+
-A better approach is rather to try to define a large margin between
+ 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
+ Thus, we wish to find a margin \( M \) with \( \boldsymbol{w} \) normalized to
\( \vert\vert \boldsymbol{w}\vert\vert =1 \) subject to the condition
+ 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
+ We seek thus the largest value \( M \) defined by or just 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 have thus defined our margin as the invers of the norm of
+ 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.
+
-Consider a function of three independent variables \( f(x,y,z) \) . For the function \( f \) to be an
+ Consider a function of three independent variables \( f(x,y,z) \) . For the function \( f \) to be an
extreme we have
+ A necessary and sufficient condition is due to 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
+ 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 \( x,y,z \)
+ resulting in Now we cannot set anymore 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.
+
-However, we can add to
+ However, we can add to a multiplum of \( d\phi \), viz. \( \lambda d\phi \), resulting in Our multiplier is chosen so that
-We need to remember that we took \( dx \) and \( dy \) to be arbitrary and thus we must have
+ We need to remember that we took \( dx \) and \( dy \) to be arbitrary and thus we must have and 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
+
- In order to solve the above problem, we define the following Lagrangian function to be minimized 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
+ Taking the derivatives with respect to \( b \) and \( \boldsymbol{w} \) we obtain and Inserting these constraints into the equation for \( {\cal L} \) we obtain 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
+ 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 \( M \).
-We can rewrite
+ We can rewrite and its constraints in terms of a matrix-vector problem where we minimize w.r.t. \( \lambda \) the following problem subject to \( \boldsymbol{y}^T\boldsymbol{\lambda}=0 \). Here we defined the vectors \( \boldsymbol{\lambda} =[\lambda_1,\lambda_2,\dots,\lambda_n] \) and
+\( \boldsymbol{y}=[y_1,y_2,\dots,y_n] \).
+
-Solving the above problem, yields the values of \( \lambda_i \).
+ Solving the above problem, yields the values of \( \lambda_i \).
To find the coefficients of your hyperplane we need simply to compute
+ With our vector \( \boldsymbol{w} \) we can in turn find the value of the intercept \( b \) (here in two dimensions) via resulting in or if we write it out in terms of the support vectors only, with \( N_s \) being their number, we have With our hyperplane coefficients we can use our classifier to assign any observation by simply using Below we discuss how to find the optimal values of \( \lambda_i \). Before we proceed however, we discuss now the so-called soft classifier.
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.
-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
+ 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
+ We introduce thus the so-called slack variables \( \boldsymbol{\xi} =[\xi_1,x_2,\dots,x_n] \) and
modify our previous equation
+ to 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 occur when \( \xi_i > 1 \). Thus bounding the total sum by some value \( C \) bounds in turn the total number of
misclassifications.
+
-This has in turn the consequences that we change our optmization problem to finding the minimum of
+ This has in turn the consequences that we change our optmization problem to finding the minimum of subject to with the requirement \( \xi_i\geq 0 \).
-Taking the derivatives with respect to \( b \) and \( \boldsymbol{w} \) we obtain
+ Taking the derivatives with respect to \( b \) and \( \boldsymbol{w} \) we obtain and and Inserting these constraints into the equation for \( {\cal L} \) we obtain the same equation as before 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
+ and
+
-The cases we have studied till now, were all characterized by two classes
+ 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.
+
-If our feature space is not easy to separate, as shown in the figure
+ 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.
+obtain a separation between the classes which is almost linear.
+
-The change of basis, from \( x\rightarrow z=\phi(x) \) leads to the same type of equations to be solved, except that
+ 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.
+
-
+
+
-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)
+ 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)
-With our new basis, the equations we solved earlier are basically the same, that is we have now (without the slack option for simplicity)
+ With our new basis, the equations we solved earlier are basically the same, that is we have now (without the slack option for simplicity) subject to the constraints \( \lambda_i\geq 0 \), \( \sum_i\lambda_iy_i=0 \), and for the support vectors 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
+ For the above example, the kernel reads
-We note that this is nothing but the dot product of the two original
+ 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
+ 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.
+
Using our definition of the kernel We can rewrite again the Lagrangian subject to the constraints \( \lambda_i\geq 0 \), \( \sum_i\lambda_iy_i=0 \) in terms of a convex optimization problem 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 \( 0\leq \lambda_i \leq C \).
+
-We can rewrite this (see the solutions below) in terms of a convex optimization problem of the type
+ We can rewrite this (see the solutions below) in terms of a convex optimization problem of the type 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 \( \boldsymbol{h} \).
+
-There are several popular kernels being used. These are
-
+ There are several popular kernels being used. These are and many other ones.
-An important theorem for us is Mercer's
+ 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
+
-So you can use \( K \) as a kernel since you know \( \phi \) exists, even if
-you don’t know what \( \phi \) is.
+ 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)
+ 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.
+
+
+
+
-A mathematical (quadratic) optimization problem, or just optimization problem, has the form
+ A mathematical (quadratic) optimization problem, or just optimization problem, has the form 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.
+
-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.
+ 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.
+ Convex optimization problems play a central role in applied mathematics and we recommend strongly Boyd and Vandenberghe's text on the topics.
-If we use Python as programming language and wish to venture beyond
+ 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 package CVXOPT and we need to import it together with numpy as
+ The functions we need are contained in the quadratic programming package CVXOPT and we need to import it together with numpy as
-
-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 t write your own optimizer.
-We remind ourselves about the general problem we want to solve
+ We remind ourselves about the general problem we want to solve
-Let us show how to perform the optmization using a simple case. Assume we want to optimize the following problem
+ Let us show how to perform the optmization using a simple case. Assume we want to optimize the following problem The minimization problem can be rewritten in terms of vectors and matrices as (with \( x \) and \( y \) being the unknowns) 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 We have collapsed all the inequalities into a single matrix \( \boldsymbol{G} \). We see also that our matrix is clearly positive semi-definite (all eigenvalues larger or equal zero).
Finally, the vector \( \boldsymbol{h} \) is defined as
+
-Since we don't have any equalities the matrix \( \boldsymbol{A} \) is set to zero
+ Since we don't have any equalities the matrix \( \boldsymbol{A} \) is set to zero
The following code solves the equations for us
-
+
+
+
-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
+ 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 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 \( 0\leq \lambda_i \leq C \).
+
code will be added
-
-
-
-
-
-The moons example
-The moons example
-from __future__ import division, print_function, unicode_literals
+
from __future__ import division, print_function, unicode_literals
import numpy as np
np.random.seed(42)
@@ -1173,14 +1251,26 @@ plt.figure(figsize=(11, r"$\gamma = {}, C = {}$".format(gamma, C), fontsize=16)
plt.show()
-
+Mathematical optimization of convex functions
-Mathematical optimization of convex functions
-
-
+How do we solve these problems?
-How do we solve these problems?
-
-import numpy
+
import numpy
import cvxopt
-
+A simple example
-A simple example
-
-# Import the necessary packages
+
# Import the necessary packages
import numpy
from cvxopt import matrix
from cvxopt import solvers
@@ -1292,14 +1397,26 @@ sol = solvers.qp(P,q,G,h)
# Extract optimal value and solution
sol[’x’]
sol[’primal objective’]
-
+Back to the more realistic cases
-Back to the more realistic cases
-
-Week 46: Support Vector Machines and Project 3
+Week 46: Gradient Boosting Summary and Support Vector Machines
+Nov 14, 2021
+
-
-Nov 12, 2020
-
-
-Overview of week 46
+Overview of week 46
-
+Note that on Thursday we will have only one lecture, from 1215pm to 1pm. There is no session 115pm-2pm.
-Geron's chapter 5. Chapter 12 (sections 12.1-12.3 are the most relevant ones) of Hastie et al contains also a good discussion.
-
+
+
+
+Friday
-Thursday
+
+
+
+Support Vector Machines, overarching aims
-Friday
-
-
-
-Support Vector Machines, overarching aims
-
-
+Hyperplanes and all that
-Hyperplanes and all that
-
-from sklearn import datasets
+
from sklearn import datasets
from sklearn.svm import SVC, LinearSVC
from sklearn.linear_model import SGDClassifier
from sklearn.preprocessing import StandardScaler
@@ -272,87 +412,96 @@ plt.legend(loc=
plt.axis([0, 5.5, 0, 2])
plt.show()
-
+What is a hyperplane?
-What is a hyperplane?
-
-
+A \( p \)-dimensional space of features
-A \( p \)-dimensional space of features
-
-The two-dimensional case
-The two-dimensional case
-
-
+Getting into the details
-Getting into the details
-
-
+First attempt at a minimization approach
-First attempt at a minimization approach
-
-
+Solving the equations
-Solving the equations
-
-
+Code Example
-Code Example
-
-
-
+
+
+Problems with the Simpler Approach
-Problems with the Simpler Approach
-
-
+A better approach
-A better approach
-
-
+A quick Reminder on Lagrangian Multipliers
-A quick Reminder on Lagrangian Multipliers
-
-
+Adding the Multiplier
-Adding the Multiplier
-
-
-Setting up the Problem
-In order to solve the above problem, we define the following Lagrangian function to be minimized
+
+Setting up the Problem
+
+
+The problem to solve
-The problem to solve
-
-
+The last steps
-The last steps
-
-
+A soft classifier
-A soft classifier
+
+Soft optmization problem
-Soft optmization problem
-
-
+Kernels and non-linearity
-Kernels and non-linearity
-
-import numpy as np
+
import numpy as np
import os
np.random.seed(42)
@@ -863,62 +1005,72 @@ plt.plot([-4.5<
plt.axis([-4.5, 4.5, -1, 17])
plt.subplots_adjust(right=1)
plt.show()
-
+The equations
-The equations
-
-
-
-The problem to solve
-Using our definition of the kernel We can rewrite again the Lagrangian
+The problem to solve
+
+Different kernels and Mercer's theorem
-Different kernels and Mercer's theorem
-
-
+
-
-The moons example
-The moons example
-from __future__ import division, print_function, unicode_literals
+
from __future__ import division, print_function, unicode_literals
import numpy as np
np.random.seed(42)
@@ -1178,14 +1328,26 @@ plt.figure(figsize.title(r"$\gamma = {}, C = {}$".format(gamma, C), fontsize=16)
plt.show()
-
+Mathematical optimization of convex functions
-Mathematical optimization of convex functions
-
-
+How do we solve these problems?
-How do we solve these problems?
-
-import numpy
+
import numpy
import cvxopt
-
+A simple example
-A simple example
-
-# Import the necessary packages
+
# Import the necessary packages
import numpy
from cvxopt import matrix
from cvxopt import solvers
@@ -1297,14 +1474,26 @@ sol = solvers.<
# Extract optimal value and solution
sol[’x’]
sol[’primal objective’]
-
+Back to the more realistic cases
-Back to the more realistic cases
-
-