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Data Analysis and Machine Learning: Support Vector Machines

Morten Hjorth-Jensen, Department of Physics, University of Oslo and Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University

Date: Nov 26, 2019

Support Vector Machines, overarching aims

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.

Hyperplanes and all that

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.

In [1]:

%matplotlib inline

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()

What is a hyperplane?

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.

A $p$-dimensional space of features

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.

The two-dimensional case

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.

Getting into the details

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).

First attempt at a minimization approach

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.

Solving 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


\boldsymbol{w} \leftarrow \boldsymbol{w} +\eta \frac{\partial C}{\partial \boldsymbol{w}},

where \eta is our by now well-known learning rate.

Code Example

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.

Problems with the Simpler Approach

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

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.

A quick Reminder on 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.

Adding the Multiplier

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.

Setting up the Problem

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.

If \lambda_i > 0, then y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)=1 and we say that x_i is on the boundary.
If y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)> 1, we say x_i is not on the boundary and we set \lambda_i=0.

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.

The problem to solve

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].

The last steps

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.

A 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,


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.

Soft optmization problem

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.