Week 46: Support Vector Machines and Project 3. Start Principal Component Analysis

Overview of week 46

• Thursday: Support Vector Machines, discussion of project 3
• Video of lecture
• Friday: Support vector machines and start Principal Component Analysis

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.

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

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

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

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

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

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

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

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

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

and

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

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

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 \( \boldsymbol{w} \). 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

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 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.

Adding the Multiplier

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

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

Setting up the Problem

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

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

1. 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.
2. 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

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 \( \boldsymbol{y}=[y_1,y_2,\dots,y_n] \).

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

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.

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

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.

Soft optmization problem

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

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

Kernels and non-linearity

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()

The equations

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)

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 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.

The problem to solve

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

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 \( \boldsymbol{h} \).

Different kernels and Mercer's theorem

There are several popular kernels being used. These are

1. Linear: \( K(\boldsymbol{x},\boldsymbol{y})=\boldsymbol{x}^T\boldsymbol{y} \),
2. Polynomial: \( K(\boldsymbol{x},\boldsymbol{y})=(\boldsymbol{x}^T\boldsymbol{y}+\gamma)^d \),
3. Gaussian Radial Basis Function: \( K(\boldsymbol{x},\boldsymbol{y})=\exp{\left(-\gamma\vert\vert\boldsymbol{x}-\boldsymbol{y}\vert\vert^2\right)} \),
4. Tanh: \( K(\boldsymbol{x},\boldsymbol{y})=\tanh{(\boldsymbol{x}^T\boldsymbol{y}+\gamma)} \),

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

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.

The moons example

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()

Mathematical optimization of convex functions

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.

Convex optimization problems play a central role in applied mathematics and we recommend strongly Boyd and Vandenberghe's text on the topics.

How do we solve these problems?

If we use Python as programming language and wish to venture beyond scikit-learn, tensorflow and similar software which makes our lives so much easier, we need to dive into the wonderful world of quadratic programming. We can, if we wish, solve the minimization problem using say standard gradient methods or conjugate gradient methods. However, these methods tend to exhibit a rather slow converge. So, welcome to the promised land of quadratic programming.

The functions we need are contained in the quadratic programming library CVXOPT and we need to import it together with numpy as

import numpy
import cvxopt

This will make our life much easier. You don't need t0 write your own optimizer.

A simple example

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

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 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']
print(sol['x'] )

Back to the more realistic cases

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 \( 0\leq \lambda_i \leq C \).

Using the CVXOPT library, the matrix \( P \) would then be defined by the above matrix while the KKT conditions would all be collected by the matrix \( G \).

Basic ideas of the Principal Component Analysis (PCA)

The principal component analysis deals with the problem of fitting a low-dimensional affine subspace \( S \) of dimension \( d \) much smaller than the total dimension \( D \) of the problem at hand (our data set). Mathematically it can be formulated as a statistical problem or a geometric problem. In our discussion of the theorem for the classical PCA, we will stay with a statistical approach. Historically, the PCA was first formulated in a statistical setting in order to estimate the principal component of a multivariate random variable.

We have a data set defined by a design/feature matrix \( \boldsymbol{X} \) (see below for its definition)

• Each data point is determined by \( p \) extrinsic (measurement) variables
• We may want to ask the following question: Are there fewer intrinsic variables (say \( d < < p \)) that still approximately describe the data?
• If so, these intrinsic variables may tell us something important and finding these intrinsic variables is what dimension reduction methods do.

A good read is for example Vidal, Ma and Sastry.

Introducing the Covariance and Correlation functions

Before we discuss the PCA theorem, we need to remind ourselves about the definition of the covariance and the correlation function. These are quantities

Suppose we have defined two vectors \( \hat{x} \) and \( \hat{y} \) with \( n \) elements each. The covariance matrix \( \boldsymbol{C} \) is defined as

where for example

With this definition and recalling that the variance is defined as

we can rewrite the covariance matrix as

More on the covariance

The covariance takes values between zero and infinity and may thus lead to problems with loss of numerical precision for particularly large values. It is common to scale the covariance matrix by introducing instead the correlation matrix defined via the so-called correlation function

The correlation function is then given by values \( \mathrm{corr}[\boldsymbol{x},\boldsymbol{y}] \in [-1,1] \). This avoids eventual problems with too large values. We can then define the correlation matrix for the two vectors \( \boldsymbol{x} \) and \( \boldsymbol{y} \) as

In the above example this is the function we constructed using pandas.

Reminding ourselves about Linear Regression

In our derivation of the various regression algorithms like Ordinary Least Squares or Ridge regression we defined the design/feature matrix \( \boldsymbol{X} \) as

with \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \), with the predictors/features \( p \) refering to the column numbers and the entries \( n \) being the row elements. We can rewrite the design/feature matrix in terms of its column vectors as

with a given vector

Simple Example

With these definitions, we can now rewrite our \( 2\times 2 \) correlation/covariance matrix in terms of a moe general design/feature matrix \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \). This leads to a \( p\times p \) covariance matrix for the vectors \( \boldsymbol{x}_i \) with \( i=0,1,\dots,p-1 \)

The Correlation Matrix

and the correlation matrix

Numpy Functionality

The Numpy function np.cov calculates the covariance elements using the factor \( 1/(n-1) \) instead of \( 1/n \) since it assumes we do not have the exact mean values. The following simple function uses the np.vstack function which takes each vector of dimension \( 1\times n \) and produces a \( 2\times n \) matrix \( \boldsymbol{W} \)

which in turn is converted into into the \( 2\times 2 \) covariance matrix \( \boldsymbol{C} \) via the Numpy function np.cov(). We note that we can also calculate the mean value of each set of samples \( \boldsymbol{x} \) etc using the Numpy function np.mean(x). We can also extract the eigenvalues of the covariance matrix through the np.linalg.eig() function.

# Importing various packages
import numpy as np
n = 100
x = np.random.normal(size=n)
print(np.mean(x))
y = 4+3*x+np.random.normal(size=n)
print(np.mean(y))
W = np.vstack((x, y))
C = np.cov(W)
print(C)

Correlation Matrix again

The previous example can be converted into the correlation matrix by simply scaling the matrix elements with the variances. We should also subtract the mean values for each column. This leads to the following code which sets up the correlations matrix for the previous example in a more brute force way. Here we scale the mean values for each column of the design matrix, calculate the relevant mean values and variances and then finally set up the \( 2\times 2 \) correlation matrix (since we have only two vectors).

import numpy as np
n = 100
# define two vectors                                                                                           
x = np.random.random(size=n)
y = 4+3*x+np.random.normal(size=n)
#scaling the x and y vectors                                                                                   
x = x - np.mean(x)
y = y - np.mean(y)
variance_x = np.sum(x@x)/n
variance_y = np.sum(y@y)/n
print(variance_x)
print(variance_y)
cov_xy = np.sum(x@y)/n
cov_xx = np.sum(x@x)/n
cov_yy = np.sum(y@y)/n
C = np.zeros((2,2))
C[0,0]= cov_xx/variance_x
C[1,1]= cov_yy/variance_y
C[0,1]= cov_xy/np.sqrt(variance_y*variance_x)
C[1,0]= C[0,1]
print(C)

We see that the matrix elements along the diagonal are one as they should be and that the matrix is symmetric. Furthermore, diagonalizing this matrix we easily see that it is a positive definite matrix.

The above procedure with numpy can be made more compact if we use pandas.

Using Pandas

We whow here how we can set up the correlation matrix using pandas, as done in this simple code

import numpy as np
import pandas as pd
n = 10
x = np.random.normal(size=n)
x = x - np.mean(x)
y = 4+3*x+np.random.normal(size=n)
y = y - np.mean(y)
X = (np.vstack((x, y))).T
print(X)
Xpd = pd.DataFrame(X)
print(Xpd)
correlation_matrix = Xpd.corr()
print(correlation_matrix)

And then the Franke Function

We expand this model to the Franke function discussed above.

# Common imports
import numpy as np
import pandas as pd


def FrankeFunction(x,y):
	term1 = 0.75*np.exp(-(0.25*(9*x-2)**2) - 0.25*((9*y-2)**2))
	term2 = 0.75*np.exp(-((9*x+1)**2)/49.0 - 0.1*(9*y+1))
	term3 = 0.5*np.exp(-(9*x-7)**2/4.0 - 0.25*((9*y-3)**2))
	term4 = -0.2*np.exp(-(9*x-4)**2 - (9*y-7)**2)
	return term1 + term2 + term3 + term4


def create_X(x, y, n ):
	if len(x.shape) > 1:
		x = np.ravel(x)
		y = np.ravel(y)

	N = len(x)
	l = int((n+1)*(n+2)/2)		# Number of elements in beta
	X = np.ones((N,l))

	for i in range(1,n+1):
		q = int((i)*(i+1)/2)
		for k in range(i+1):
			X[:,q+k] = (x**(i-k))*(y**k)

	return X


# Making meshgrid of datapoints and compute Franke's function
n = 4
N = 100
x = np.sort(np.random.uniform(0, 1, N))
y = np.sort(np.random.uniform(0, 1, N))
z = FrankeFunction(x, y)
X = create_X(x, y, n=n)    

Xpd = pd.DataFrame(X)
# subtract the mean values and set up the covariance matrix
Xpd = Xpd - Xpd.mean()
covariance_matrix = Xpd.cov()
print(covariance_matrix)

We note here that the covariance is zero for the first rows and columns since all matrix elements in the design matrix were set to one (we are fitting the function in terms of a polynomial of degree \( n \)). We would however not include the intercept and wee can simply drop these elements and construct a correlation matrix without them by centering our matrix elements by subtracting the mean of each column.

Links with the Design Matrix

We can rewrite the covariance matrix in a more compact form in terms of the design/feature matrix \( \boldsymbol{X} \) as

To see this let us simply look at a design matrix \( \boldsymbol{X}\in {\mathbb{R}}^{2\times 2} \)

Computing the Expectation Values

If we then compute the expectation value

which is just

where we wrote $$\boldsymbol{C}[\boldsymbol{x}_0,\boldsymbol{x}_1] = \boldsymbol{C}[\boldsymbol{x}]$$ to indicate that this the covariance of the vectors \( \boldsymbol{x} \) of the design/feature matrix \( \boldsymbol{X} \).

It is easy to generalize this to a matrix \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \).

Towards the PCA theorem

We have that the covariance matrix (the correlation matrix involves a simple rescaling) is given as

Let us now assume that we can perform a series of orthogonal transformations where we employ some orthogonal matrices \( \boldsymbol{S} \). These matrices are defined as \( \boldsymbol{S}\in {\mathbb{R}}^{p\times p} \) and obey the orthogonality requirements \( \boldsymbol{S}\boldsymbol{S}^T=\boldsymbol{S}^T\boldsymbol{S}=\boldsymbol{I} \). The matrix can be written out in terms of the column vectors \( \boldsymbol{s}_i \) as \( \boldsymbol{S}=[\boldsymbol{s}_0,\boldsymbol{s}_1,\dots,\boldsymbol{s}_{p-1}] \) and \( \boldsymbol{s}_i \in {\mathbb{R}}^{p} \).

Assume also that there is a transformation \( \boldsymbol{S}^T\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{S}=\boldsymbol{C}[\boldsymbol{y}] \) such that the new matrix \( \boldsymbol{C}[\boldsymbol{y}] \) is diagonal with elements \( [\lambda_0,\lambda_1,\lambda_2,\dots,\lambda_{p-1}] \).

That is we have

since the matrix \( \boldsymbol{S} \) is not a data dependent matrix. Multiplying with \( \boldsymbol{S} \) from the left we have

and since \( \boldsymbol{C}[\boldsymbol{y}] \) is diagonal we have for a given eigenvalue \( i \) of the covariance matrix that

More on the PCA Theorem

In the derivation of the PCA theorem we will assume that the eigenvalues are ordered in descending order, that is \( \lambda_0 > \lambda_1 > \dots > \lambda_{p-1} \).

The eigenvalues tell us then how much we need to stretch the corresponding eigenvectors. Dimensions with large eigenvalues have thus large variations (large variance) and define therefore useful dimensions. The data points are more spread out in the direction of these eigenvectors. Smaller eigenvalues mean on the other hand that the corresponding eigenvectors are shrunk accordingly and the data points are tightly bunched together and there is not much variation in these specific directions. Hopefully then we could leave it out dimensions where the eigenvalues are very small. If \( p \) is very large, we could then aim at reducing \( p \) to \( l < < p \) and handle only \( l \) features/predictors.

Here is how we would proceed in setting up the algorithm for the PCA, see also discussion below here.

• Set up the datapoints for the design/feature matrix with the predictors/features \( p \) referring to the column numbers and the entries \( n \) being the row elements.
• Center the data by subtracting the mean value for each column.
• Compute then the covariance/correlation matrix.
• Find the eigenpairs of the covariance matrix with eigenvalues \( [\lambda_0,\lambda_1,\dots,\lambda_{p-1}] \) and eigenvectors \( [\boldsymbol{s}_0,\boldsymbol{s}_1,\dots,\boldsymbol{s}_{p-1}] \).
• Order the eigenvalue (and the eigenvectors accordingly) in order of decreasing eigenvalues.
• Keep only those \( l \) eigenvalues larger than a selected threshold value, discarding thus \( p-l \) features since we expect small variations in the data here.

Writing our own PCA code

We will use a simple example first with two-dimensional data drawn from a multivariate normal distribution with the following mean and covariance matrix (we have fixed these quantities but will play around with them below):

Note that the mean refers to each column of data. We will generate \( n = 10000 \) points \( X = \{ x_1, \ldots, x_N \} \) from this distribution, and store them in the \( 1000 \times 2 \) matrix \( \boldsymbol{X} \). This is our design matrix where we have forced the covariance and mean values to take specific values.

Implementing it

The following Python code aids in setting up the data and writing out the design matrix. Note that the function multivariate returns also the covariance discussed above and that it is defined by dividing by \( n-1 \) instead of \( n \).

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from IPython.display import display
n = 10000
mean = (-1, 2)
cov = [[4, 2], [2, 2]]
X = np.random.multivariate_normal(mean, cov, n)

Now we are going to implement the PCA algorithm. We will break it down into various substeps.

First Step

The first step of PCA is to compute the sample mean of the data and use it to center the data. Recall that the sample mean is

and the mean-centered data \( \bar{X} = \{ \bar{x}_1, \ldots, \bar{x}_n \} \) takes the form

When you are done with these steps, print out \( \mu_n \) to verify it is close to \( \mu \) and plot your mean centered data to verify it is centered at the origin! The following code elements perform these operations using pandas or using our own functionality for doing so. The latter, using numpy is rather simple through the mean() function.

df = pd.DataFrame(X)
# Pandas does the centering for us
df = df -df.mean()
# we center it ourselves
X_centered = X - X.mean(axis=0)

Scaling

Alternatively, we could use the functions we discussed earlier for scaling the data set. That is, we could have used the StandardScaler function in Scikit-Learn, a function which ensures that for each feature/predictor we study the mean value is zero and the variance is one (every column in the design/feature matrix). You would then not get the same results, since we divide by the variance. The diagonal covariance matrix elements will then be one, while the non-diagonal ones need to be divided by \( 2\sqrt{2} \) for our specific case.

Centered Data

Now we are going to use the mean centered data to compute the sample covariance of the data by using the following equation

where the data points \( x_i \in \mathbb{R}^p \) (here in this example \( p = 2 \)) are column vectors and \( x^T \) is the transpose of \( x \). We can write our own code or simply use either the functionaly of numpy or that of pandas, as follows

print(df.cov())
print(np.cov(X_centered.T))

Note that the way we define the covariance matrix here has a factor \( n-1 \) instead of \( n \). This is included in the cov() function by numpy and pandas. Our own code here is not very elegant and asks for obvious improvements. It is tailored to this specific \( 2\times 2 \) covariance matrix.

# extract the relevant columns from the centered design matrix of dim n x 2
x = X_centered[:,0]
y = X_centered[:,1]
Cov = np.zeros((2,2))
Cov[0,1] = np.sum(x.T@y)/(n-1.0)
Cov[0,0] = np.sum(x.T@x)/(n-1.0)
Cov[1,1] = np.sum(y.T@y)/(n-1.0)
Cov[1,0]= Cov[0,1]
print("Centered covariance using own code")
print(Cov)
plt.plot(x, y, 'x')
plt.axis('equal')
plt.show()

Exploring

Depending on the number of points \( n \), we will get results that are close to the covariance values defined above. The plot shows how the data are clustered around a line with slope close to one. Is this expected? Try to change the covariance and the mean values. For example, try to make the variance of the first element much larger than that of the second diagonal element. Try also to shrink the covariance (the non-diagonal elements) and see how the data points are distributed.

Diagonalize the sample covariance matrix to obtain the principal components

Now we are ready to solve for the principal components! To do so we diagonalize the sample covariance matrix \( \Sigma \). We can use the function np.linalg.eig to do so. It will return the eigenvalues and eigenvectors of \( \Sigma \). Once we have these we can perform the following tasks:

• We compute the percentage of the total variance captured by the first principal component
• We plot the mean centered data and lines along the first and second principal components
• Then we project the mean centered data onto the first and second principal components, and plot the projected data.
• Finally, we approximate the data as

where \( v_0 \) is the first principal component.

Collecting all Steps

Collecting all these steps we can write our own PCA function and compare this with the functionality included in Scikit-Learn.

The code here outlines some of the elements we could include in the analysis. Feel free to extend upon this in order to address the above questions.

# diagonalize and obtain eigenvalues, not necessarily sorted
EigValues, EigVectors = np.linalg.eig(Cov)
# sort eigenvectors and eigenvalues
#permute = EigValues.argsort()
#EigValues = EigValues[permute]
#EigVectors = EigVectors[:,permute]
print("Eigenvalues of Covariance matrix")
for i in range(2):
    print(EigValues[i])
FirstEigvector = EigVectors[:,0]
SecondEigvector = EigVectors[:,1]
print("First eigenvector")
print(FirstEigvector)
print("Second eigenvector")
print(SecondEigvector)
#thereafter we do a PCA with Scikit-learn
from sklearn.decomposition import PCA
pca = PCA(n_components = 2)
X2Dsl = pca.fit_transform(X)
print("Eigenvector of largest eigenvalue")
print(pca.components_.T[:, 0])

This code does not contain all the above elements, but it shows how we can use Scikit-Learn to extract the eigenvector which corresponds to the largest eigenvalue. Try to address the questions we pose before the above code. Try also to change the values of the covariance matrix by making one of the diagonal elements much larger than the other. What do you observe then?

Classical PCA Theorem

We assume now that we have a design matrix \( \boldsymbol{X} \) which has been centered as discussed above. For the sake of simplicity we skip the overline symbol. The matrix is defined in terms of the various column vectors \( [\boldsymbol{x}_0,\boldsymbol{x}_1,\dots, \boldsymbol{x}_{p-1}] \) each with dimension \( \boldsymbol{x}\in {\mathbb{R}}^{n} \).

The PCA theorem states that minimizing the above reconstruction error corresponds to setting \( \boldsymbol{W}=\boldsymbol{S} \), the orthogonal matrix which diagonalizes the empirical covariance(correlation) matrix. The optimal low-dimensional encoding of the data is then given by a set of vectors \( \boldsymbol{z}_i \) with at most \( l \) vectors, with \( l < < p \), defined by the orthogonal projection of the data onto the columns spanned by the eigenvectors of the covariance(correlations matrix).

The PCA Theorem

To show the PCA theorem let us start with the assumption that there is one vector \( \boldsymbol{s}_0 \) which corresponds to a solution which minimized the reconstruction error \( J \). This is an orthogonal vector. It means that we now approximate the reconstruction error in terms of \( \boldsymbol{w}_0 \) and \( \boldsymbol{z}_0 \) as

We are almost there, we have obtained a relation between minimizing the reconstruction error and the variance and the covariance matrix. Minimizing the error is equivalent to maximizing the variance of the projected data.

We could trivially maximize the variance of the projection (and thereby minimize the error in the reconstruction function) by letting the norm-2 of \( \boldsymbol{w}_0 \) go to infinity. However, this norm since we want the matrix \( \boldsymbol{W} \) to be an orthogonal matrix, is constrained by \( \vert\vert \boldsymbol{w}_0 \vert\vert_2^2=1 \). Imposing this condition via a Lagrange multiplier we can then in turn maximize

Taking the derivative with respect to \( \boldsymbol{w}_0 \) we obtain

meaning that

The direction that maximizes the variance (or minimizes the construction error) is an eigenvector of the covariance matrix! If we left multiply with \( \boldsymbol{w}_0^T \) we have the variance of the projected data is

If we want to maximize the variance (minimize the construction error) we simply pick the eigenvector of the covariance matrix with the largest eigenvalue. This establishes the link between the minimization of the reconstruction function \( J \) in terms of an orthogonal matrix and the maximization of the variance and thereby the covariance of our observations encoded in the design/feature matrix \( \boldsymbol{X} \).

The proof for the other eigenvectors \( \boldsymbol{w}_1,\boldsymbol{w}_2,\dots \) can be established by applying the above arguments and using the fact that our basis of eigenvectors is orthogonal, see Murphy chapter 12.2. The discussion in chapter 12.2 of Murphy's text has also a nice link with the Singular Value Decomposition theorem. For categorical data, see chapter 12.4 and discussion therein.

For more details, see for example Vidal, Ma and Sastry, chapter 2.

Geometric Interpretation and link with Singular Value Decomposition

For a detailed demonstration of the geometric interpretation, see Vidal, Ma and Sastry, section 2.1.2.

Principal Component Analysis (PCA) is by far the most popular dimensionality reduction algorithm. First it identifies the hyperplane that lies closest to the data, and then it projects the data onto it.

The following Python code uses NumPy’s svd() function to obtain all the principal components of the training set, then extracts the first two principal components. First we center the data using either pandas or our own code

import numpy as np
import pandas as pd
from IPython.display import display
np.random.seed(100)
# setting up a 10 x 5 vanilla matrix 
rows = 10
cols = 5
X = np.random.randn(rows,cols)
df = pd.DataFrame(X)
# Pandas does the centering for us
df = df -df.mean()
display(df)

# we center it ourselves
X_centered = X - X.mean(axis=0)
# Then check the difference between pandas and our own set up
print(X_centered-df)
#Now we do an SVD
U, s, V = np.linalg.svd(X_centered)
c1 = V.T[:, 0]
c2 = V.T[:, 1]
W2 = V.T[:, :2]
X2D = X_centered.dot(W2)
print(X2D)

PCA assumes that the dataset is centered around the origin. Scikit-Learn’s PCA classes take care of centering the data for you. However, if you implement PCA yourself (as in the preceding example), or if you use other libraries, don’t forget to center the data first.

Once you have identified all the principal components, you can reduce the dimensionality of the dataset down to \( d \) dimensions by projecting it onto the hyperplane defined by the first \( d \) principal components. Selecting this hyperplane ensures that the projection will preserve as much variance as possible.

W2 = V.T[:, :2]
X2D = X_centered.dot(W2)

PCA and scikit-learn

Scikit-Learn’s PCA class implements PCA using SVD decomposition just like we did before. The following code applies PCA to reduce the dimensionality of the dataset down to two dimensions (note that it automatically takes care of centering the data):

#thereafter we do a PCA with Scikit-learn
from sklearn.decomposition import PCA
pca = PCA(n_components = 2)
X2D = pca.fit_transform(X)
print(X2D)

After fitting the PCA transformer to the dataset, you can access the principal components using the components variable (note that it contains the PCs as horizontal vectors, so, for example, the first principal component is equal to

pca.components_.T[:, 0]

Another very useful piece of information is the explained variance ratio of each principal component, available via the \( explained\_variance\_ratio \) variable. It indicates the proportion of the dataset’s variance that lies along the axis of each principal component.

Back to the Cancer Data

We can now repeat the above but applied to real data, in this case our breast cancer data. Here we compute performance scores on the training data using logistic regression.

import matplotlib.pyplot as plt
import numpy as np
from sklearn.model_selection import  train_test_split 
from sklearn.datasets import load_breast_cancer
from sklearn.linear_model import LogisticRegression
cancer = load_breast_cancer()

X_train, X_test, y_train, y_test = train_test_split(cancer.data,cancer.target,random_state=0)

logreg = LogisticRegression()
logreg.fit(X_train, y_train)
print("Train set accuracy from Logistic Regression: {:.2f}".format(logreg.score(X_train,y_train)))
# We scale the data
from sklearn.preprocessing import StandardScaler
scaler = StandardScaler()
scaler.fit(X_train)
X_train_scaled = scaler.transform(X_train)
X_test_scaled = scaler.transform(X_test)
# Then perform again a log reg fit
logreg.fit(X_train_scaled, y_train)
print("Train set accuracy scaled data: {:.2f}".format(logreg.score(X_train_scaled,y_train)))
#thereafter we do a PCA with Scikit-learn
from sklearn.decomposition import PCA
pca = PCA(n_components = 2)
X2D_train = pca.fit_transform(X_train_scaled)
# and finally compute the log reg fit and the score on the training data	
logreg.fit(X2D_train,y_train)
print("Train set accuracy scaled and PCA data: {:.2f}".format(logreg.score(X2D_train,y_train)))

We see that our training data after the PCA decomposition has a performance similar to the non-scaled data.

Instead of arbitrarily choosing the number of dimensions to reduce down to, it is generally preferable to choose the number of dimensions that add up to a sufficiently large portion of the variance (e.g., 95%). Unless, of course, you are reducing dimensionality for data visualization — in that case you will generally want to reduce the dimensionality down to 2 or 3. The following code computes PCA without reducing dimensionality, then computes the minimum number of dimensions required to preserve 95% of the training set’s variance:

pca = PCA()
pca.fit(X)
cumsum = np.cumsum(pca.explained_variance_ratio_)
d = np.argmax(cumsum >= 0.95) + 1

You could then set \( n\_components=d \) and run PCA again. However, there is a much better option: instead of specifying the number of principal components you want to preserve, you can set \( n\_components \) to be a float between 0.0 and 1.0, indicating the ratio of variance you wish to preserve:

pca = PCA(n_components=0.95)
X_reduced = pca.fit_transform(X)

Incremental PCA

One problem with the preceding implementation of PCA is that it requires the whole training set to fit in memory in order for the SVD algorithm to run. Fortunately, Incremental PCA (IPCA) algorithms have been developed: you can split the training set into mini-batches and feed an IPCA algorithm one minibatch at a time. This is useful for large training sets, and also to apply PCA online (i.e., on the fly, as new instances arrive).

Randomized PCA

Scikit-Learn offers yet another option to perform PCA, called Randomized PCA. This is a stochastic algorithm that quickly finds an approximation of the first d principal components. Its computational complexity is \( O(m \times d^2)+O(d^3) \), instead of \( O(m \times n^2) + O(n^3) \), so it is dramatically faster than the previous algorithms when \( d \) is much smaller than \( n \).

Kernel PCA

The kernel trick is a mathematical technique that implicitly maps instances into a very high-dimensional space (called the feature space), enabling nonlinear classification and regression with Support Vector Machines. Recall that a linear decision boundary in the high-dimensional feature space corresponds to a complex nonlinear decision boundary in the original space. It turns out that the same trick can be applied to PCA, making it possible to perform complex nonlinear projections for dimensionality reduction. This is called Kernel PCA (kPCA). It is often good at preserving clusters of instances after projection, or sometimes even unrolling datasets that lie close to a twisted manifold. For example, the following code uses Scikit-Learn’s KernelPCA class to perform kPCA with an

from sklearn.decomposition import KernelPCA
rbf_pca = KernelPCA(n_components = 2, kernel="rbf", gamma=0.04)
X_reduced = rbf_pca.fit_transform(X)

Other techniques

There are many other dimensionality reduction techniques, several of which are available in Scikit-Learn.

Here are some of the most popular:

• Multidimensional Scaling (MDS) reduces dimensionality while trying to preserve the distances between the instances.
• Isomap creates a graph by connecting each instance to its nearest neighbors, then reduces dimensionality while trying to preserve the geodesic distances between the instances.
• t-Distributed Stochastic Neighbor Embedding (t-SNE) reduces dimensionality while trying to keep similar instances close and dissimilar instances apart. It is mostly used for visualization, in particular to visualize clusters of instances in high-dimensional space (e.g., to visualize the MNIST images in 2D).
• Linear Discriminant Analysis (LDA) is actually a classification algorithm, but during training it learns the most discriminative axes between the classes, and these axes can then be used to define a hyperplane onto which to project the data. The benefit is that the projection will keep classes as far apart as possible, so LDA is a good technique to reduce dimensionality before running another classification algorithm such as a Support Vector Machine (SVM) classifier discussed in the SVM lectures.