Classification

By Prof. Seungchul Lee
http://iai.postech.ac.kr/
Industrial AI Lab at POSTECH

Table of Contents

1. Classification

• where $y$ is a discrete value
  • develop the classification algorithm to determine which class a new input should fall into

• start with binary class problems
  • Later look at multiclass classification problem, although this is just an extension of binary classification

• We could use linear regression
  • Then, threshold the classifier output (i.e. anything over some value is yes, else no)
  • linear regression with thresholding seems to work

• We will learn
  • perceptron
  • logistic regression

2. Perceptron

• For input $x = \begin{bmatrix}x_1\\ \vdots\\ x_d \end{bmatrix}\;$ 'attributes of a customer'

• weights $\omega = \begin{bmatrix}\omega_1\\ \vdots\\ \omega_d \end{bmatrix}$

$$\begin{align*} \text{Approve credit if} \; & \sum\limits_{i=1}^{d}\omega_ix_i > \text{threshold}, \\ \text{Deny credit if} \; & \sum\limits_{i=1}^{d}\omega_ix_i < \text{threshold}. \end{align*}$$

$$h(x) = \text{sign} \left(\left( \sum\limits_{i=1}^{d}\omega_ix_i \right)- \text{threshold} \right) = \text{sign}\left(\left( \sum\limits_{i=1}^{d}\omega_ix_i \right)+ \omega_0\right)$$

• Introduce an artificial coordinate $x_0 = 1$:

$$h(x) = \text{sign}\left( \sum\limits_{i=0}^{d}\omega_ix_i \right)$$

• In a vector form, the perceptron implements

$$h(x) = \text{sign}\left( \omega^T x \right)$$

• Sign function

$$ \text{sgn}(x) = \begin{cases} 1, &\text{if }\; x < 0\\ 0, &\text{if }\; x = 0\\ -1, &\text{if }\; x > 0 \end{cases} $$

• Hyperplane

  • Separates a D-dimensional space into two half-spaces
  • Defined by an outward pointing normal vector $\omega$
  • $\omega$ is orthogonal to any vector lying on the hyperplane
  • Assume the hyperplane passes through origin, $\omega^T x = 0$ with $x_0 = 1$

• Sign with respect to a line

$$ \begin{align*} \omega = \begin{bmatrix}\omega_1 \\ \omega_2 \end{bmatrix}, \quad x = \begin{bmatrix} x_1 \\ x_2\end{bmatrix} &\implies g(x) = \omega_0 + \omega_1 x_1 + \omega_2 x_2 = \omega_0 + \omega^T x\\\\ \omega = \begin{bmatrix}\omega_0 \\ \omega_1 \\ \omega_2 \end{bmatrix}, \quad x = \begin{bmatrix} 1 \\ x_1 \\ x_2\end{bmatrix} &\implies g(x) = \omega_0 \cdot 1 + \omega_1 x_1 + \omega_2 x_2 = \omega^T x \end{align*} $$

• If $\vec p$ and $\vec q$ are on the decision line

$$ \begin{align*} g\left(\vec p\right) = g\left(\vec q\right) = 0 \quad & \Rightarrow \quad \omega_0 + \omega^T \vec p = \omega_0 + \omega^T \vec q = 0 \\ & \Rightarrow \quad \omega^T \left( \vec p- \vec q \right) = 0 \end{align*} $$

• Goal: to learn the hyperplane $g_{\omega}(x)=0$ using the training data

• How to find $\omega$

  • All data in class 1 ($y = +1$) $$g(x) > 0$$
  • All data in class 0 ($y = -1$) $$g(x) < 0$$

2.1. Perceptron Algorithm

The perceptron implements

$$h(x) = \text{sign}\left( \omega^Tx \right)$$

Given the training set

$$(x_1, y_1), (x_2, y_2), \cdots, (x_N, y_N) \quad \text{where } y_i \in \{-1,1\}$$

1) pick a misclassified point

$$ \text{sign}\left(\omega^Tx_n \right) \neq y_n$$

2) and update the weight vector

$$\omega \leftarrow \omega + y_nx_n$$

Why perceptron updates work ?

• Let's look at a misclassified positive example ($y_n = +1$)

  • perceptron (wrongly) thinks $\omega_{old}^T x_n < 0$

• updates would be $$ \begin{align*}\omega_{new} &= \omega_{old} + y_n x_n = \omega_{old} + x_n \\ \\ \omega_{new}^T x_n &= (\omega_{old} + x_n)^T x_n = \omega_{old}^T x_n + x_n^T x_n \end{align*}$$

• Thus $\omega_{new}^T x_n$ is less negative than $\omega_{old}^T x_n$

2.2. Iterations of Perceptron

1. Randomly assign $\omega$
   
   
2. One iteration of the PLA (perceptron learning algorithm) $$\omega \leftarrow \omega + yx$$ where $(x, y)$ is a misclassified training point
   
   
3. At iteration $i = 1, 2, 3, \cdots,$ pick a misclassified point from $$(x_1,y_1),(x_2,y_2),\cdots,(x_N, y_N)$$
   
   
4. and run a PLA iteration on it
   
   
5. That's it!

Summary

2.3. Perceptron in Python

$$g(x) = \omega_0 + \omega^Tx = \omega_0 + \omega_1x_1 + \omega_2x_2 = 0$$

$$ \begin{align*} \omega &= \begin{bmatrix} \omega_0 \\ \omega_1 \\ \omega_2\end{bmatrix}\\ \\ x &= \begin{bmatrix} \left(x^{(1)}\right)^T \\ \left(x^{(2)}\right)^T \\ \left(x^{(3)}\right)^T\\ \vdots \\ \left(x^{(m)}\right)^T \end{bmatrix} = \begin{bmatrix} 1 & x_1^{(1)} & x_2^{(1)} \\ 1 & x_1^{(2)} & x_2^{(2)} \\ 1 & x_1^{(3)} & x_2^{(3)}\\\vdots & \vdots & \vdots \\ 1 & x_1^{(m)} & x_2^{(m)}\end{bmatrix}, \qquad y = \begin{bmatrix}y^{(1)} \\ y^{(2)} \\ y^{(3)}\\ \vdots \\ y^{(m)} \end{bmatrix} \end{align*}$$

import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline

#training data gerneration
m = 100
x1 = 8*np.random.rand(m, 1)
x2 = 7*np.random.rand(m, 1) - 4

g = 0.8*x1 + x2 - 3

C1 = np.where(g >= 1)
C0 = np.where(g < -1)
print(C1)

(array([ 5,  7, 12, 17, 18, 19, 32, 35, 43, 44, 46, 48, 49, 50, 51, 52, 55,
       57, 59, 60, 65, 72, 73, 77, 84, 88, 89, 94, 98], dtype=int64), array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0], dtype=int64))

C1 = np.where(g >= 1)[0]
C0 = np.where(g < -1)[0]
print(C1.shape)
print(C0.shape)

(29,)
(39,)

plt.figure(figsize=(10, 8))
plt.plot(x1[C1], x2[C1], 'ro', alpha = 0.4, label = 'C1')
plt.plot(x1[C0], x2[C0], 'bo', alpha = 0.4, label = 'C0')
plt.title('Linearly Separable Classes', fontsize = 15)
plt.legend(loc = 1, fontsize = 15)
plt.xlabel(r'$x_1$', fontsize = 15)
plt.ylabel(r'$x_2$', fontsize = 15)
plt.show()

$$ \begin{align*} x &= \begin{bmatrix} \left(x^{(1)}\right)^T \\ \left(x^{(2)}\right)^T \\ \left(x^{(3)}\right)^T\\ \vdots \\ \left(x^{(m)}\right)^T \end{bmatrix} = \begin{bmatrix} x_1^{(1)} & x_2^{(1)} \\ x_1^{(2)} & x_2^{(2)} \\ x_1^{(3)} & x_2^{(3)}\\ \vdots & \vdots \\ x_1^{(m)} & x_2^{(m)}\end{bmatrix} \qquad y = \begin{bmatrix}y^{(1)} \\ y^{(2)} \\ y^{(3)}\\ \vdots \\ y^{(m)} \end{bmatrix} \end{align*}$$

X1 = np.hstack([x1[C1], x2[C1]])
X0 = np.hstack([x1[C0], x2[C0]])
X = np.vstack([X1, X0])

y = np.vstack([np.ones([C1.shape[0],1]), -np.ones([C0.shape[0],1])])

from sklearn import linear_model

clf = linear_model.Perceptron(tol=1e-3)
clf.fit(X, np.ravel(y))

Perceptron(alpha=0.0001, class_weight=None, early_stopping=False, eta0=1.0,
           fit_intercept=True, max_iter=1000, n_iter_no_change=5, n_jobs=None,
           penalty=None, random_state=0, shuffle=True, tol=0.001,
           validation_fraction=0.1, verbose=0, warm_start=False)

clf.predict([[3, -2]])

array([-1.])

clf.predict([[6, 2]])

array([1.])

clf.coef_

array([[3.32453187, 7.36364365]])

clf.intercept_

array([-14.])

$$ \begin{align*} g(x) &= \omega_0 + \omega^Tx = \omega_0 + \omega_1x_1 + \omega_2x_2 = 0 \\\\ \implies x_2 &= -\frac{\omega_1}{\omega_2} x_1 - \frac{\omega_0}{\omega_2} \end{align*} $$

w0 = clf.intercept_[0]
w1 = clf.coef_[0,0]
w2 = clf.coef_[0,1]

x1p = np.linspace(0,8,100).reshape(-1,1)
x2p = - w1/w2*x1p - w0/w2

plt.figure(figsize=(10, 8))
plt.plot(x1[C1], x2[C1], 'ro', alpha = 0.4, label = 'C1')
plt.plot(x1[C0], x2[C0], 'bo', alpha = 0.4, label = 'C0')
plt.plot(x1p, x2p, c = 'k', linewidth = 4, label = 'perceptron')
plt.xlim([0, 8])
plt.xlabel('$x_1$', fontsize = 15)
plt.ylabel('$x_2$', fontsize = 15)
plt.legend(loc = 1, fontsize = 15)
plt.show()

2.4. The Best Hyperplane Separator?

• Perceptron finds one of the many possible hyperplanes separating the data if one exists

• Of the many possible choices, which one is the best?

• Utilize distance information

• Intuitively we want the hyperplane having the maximum margin

• Large margin leads to good generalization on the test data
  • we will see this formally when we cover Support Vector Machine

3. Logistic Regression

• Logistic regression is a classification algorithm - don't be confused

3.1. Distance from a Line

$$\omega = \begin{bmatrix}\omega_1 \\ \omega_2\end{bmatrix}, \, x = \begin{bmatrix}x_1\\x_2\end{bmatrix} \; \implies g(x) = \omega_0 + \omega^Tx = \omega_0 + \omega_1x_1 + \omega_2x_2 $$

• For any vector of $x$

$$ \; h = \frac{g(x)}{\lVert \omega \rVert} \implies\; \text{orthogonal signed distance from the line}$$

3.2. Using all Distances

• Perceptron: make use of sign of data

• We want to use distance information of all data points $\rightarrow$ logistic regression

• Basic idea: to find the decision boundary (hyperplane) of $g(x)=\omega^T x =0$ such that maximizes $\prod_i \lvert h_i \rvert$
  • Inequality of arithmetic and geometric means $$ \frac{h_1+h_2}{2} \geq \sqrt{h_1 h_2} $$ and that equality holds if and only if $h_1 = h_2$

• Roughly speaking, this optimization of $\max \prod_i \lvert h_i \rvert$ tends to position a hyperplane in the middle of two classes

$$h = \frac{g(x)}{\lVert \omega \rVert} = \frac{\omega^T x}{\lVert \omega \rVert} \sim \omega^T x$$

• We link or squeeze $(-\infty, +\infty)$ to $(0,1)$ for several reasons:

• If $\sigma(z)$ is the sigmoid function, or the logistic function
  
  $$ \sigma(z) = \frac{1}{1+e^{-z}} \implies \sigma \left(\omega^T x \right) = \frac{1}{1+e^{-\omega^T x}}$$
  
  
  • Logistic function always generates a value between 0 and 1
  • Crosses 0.5 at the origin, then flattens out

# plot a sigmoid function

import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline

z = np.linspace(-4,4,100)
s = 1/(1 + np.exp(-z))

plt.figure(figsize=(10,2))
plt.plot(z, s)
plt.xlim([-4, 4])
plt.axis('equal')
plt.grid(alpha = 0.3)
plt.show()

• Benefit of mapping via the logistic function

  • monotonic: same or similar optimziation solution
  • continuous and differentiable: good for gradient descent optimization
  • probability or confidence: can be considered as probability

  $$P\left(y = +1 \mid x\,;\omega\right) = \frac{1}{1+e^{-\omega^T x}} \;\; \in \; [0,1]$$
  

  • Often we do note care about predicting the label $y$

  • Rather, we want to predict the label probabilities $P\left(y \mid x\,;\omega\right)$

    • the probability that the label is $+1$ $$P\left(y = +1 \mid x\,;\omega\right)$$
    • the probability that the label is $0$ $$P\left(y = 0 \mid x\,;\omega\right) = 1 - P\left(y = +1 \mid x\,;\omega\right)$$

• Goal: we need to fit $\omega$ to our data

3.3. Logistic Regression using Scikit-Learn

$$ \begin{align*} \omega &= \begin{bmatrix} \omega_1 \\ \omega_2\end{bmatrix}, \qquad \omega_0, \qquad x = \begin{bmatrix} x_1 \\ x_2\end{bmatrix}\\ \\ X &= \begin{bmatrix} \left(x^{(1)}\right)^T \\ \left(x^{(2)}\right)^T \\ \left(x^{(3)}\right)^T \\ \vdots\end{bmatrix} = \begin{bmatrix} x_1^{(1)} & x_2^{(1)} \\ x_1^{(2)} & x_2^{(2)} \\ x_1^{(3)} & x_2^{(3)} \\ \vdots & \vdots \\\end{bmatrix}, \qquad y = \begin{bmatrix} y^{(1)}\\ y^{(2)} \\y^{(3)} \\ \vdots \end{bmatrix} \end{align*} $$

# data generation

m = 100
w = np.array([[-6], [2], [1]])
X = np.hstack([np.ones([m,1]), 4*np.random.rand(m,1), 4*np.random.rand(m,1)])

w = np.asmatrix(w)
X = np.asmatrix(X)

y = 1/(1 + np.exp(-X*w)) > 0.5 

C1 = np.where(y == True)[0]
C0 = np.where(y == False)[0]

y = np.empty([m,1])
y[C1] = 1
y[C0] = 0

plt.figure(figsize = (10,8))
plt.plot(X[C1,1], X[C1,2], 'ro', alpha = 0.3, label = 'C1')
plt.plot(X[C0,1], X[C0,2], 'bo', alpha = 0.3, label = 'C0')
plt.xlabel(r'$x_1$', fontsize = 15)
plt.ylabel(r'$x_2$', fontsize = 15)
plt.legend(loc = 1, fontsize = 12)
plt.axis('equal')
plt.xlim([0,4])
plt.ylim([0,4])
plt.show()

X = X[:,1:3]

X.shape

(100, 2)

from sklearn import linear_model

clf = linear_model.LogisticRegression(solver='lbfgs')
clf.fit(X,np.ravel(y))

LogisticRegression(C=1.0, class_weight=None, dual=False, fit_intercept=True,
                   intercept_scaling=1, l1_ratio=None, max_iter=100,
                   multi_class='warn', n_jobs=None, penalty='l2',
                   random_state=None, solver='lbfgs', tol=0.0001, verbose=0,
                   warm_start=False)

clf.coef_

array([[3.31856236, 1.24254905]])

clf.intercept_

array([-9.07267516])

w0 = clf.intercept_[0]
w1 = clf.coef_[0,0]
w2 = clf.coef_[0,1]

xp = np.linspace(0,4,100).reshape(-1,1)
yp = - w1/w2*xp - w0/w2

plt.figure(figsize = (10,8))
plt.plot(X[C1,0], X[C1,1], 'ro', alpha = 0.3, label = 'C1')
plt.plot(X[C0,0], X[C0,1], 'bo', alpha = 0.3, label = 'C0')
plt.plot(xp, yp, 'g', linewidth = 4, label = 'Logistic Regression')
plt.title('Logistic Regression')
plt.xlabel(r'$x_1$', fontsize = 15)
plt.ylabel(r'$x_2$', fontsize = 15)
plt.legend(loc = 1, fontsize = 12)
plt.axis('equal')
plt.xlim([0,4])
plt.ylim([0,4])
plt.show()

4. Multiclass Classification

• Generalization to more than 2 classes is straightforward
  • one vs. all (one vs. rest)
  • one vs. one

• Using the soft-max function instead of the logistic function (refer to UFLDL Tutorial)
  • see them as probability

$$P\left(y = k \mid x\,;\omega\right) = \frac{\exp{\left( \omega_k^T x \right) }}{\sum_k \exp{\left(\omega_k^T x \right)}} \in [0,1]$$

• We maintain a separator weight vector $\omega_k$ for each class $k$

• Note that the softmax function is often used in deep learning

5. Non-linear Classification

5.1. Kernel

• Often we want to capture nonlinear patterns in the data
  • Nonlinear regression: input and output relationship may not be linear
  • Nonlinear classification: classes may note be separable by a linear boundary

• Linear models (e.g. linear regression, linear SVM) are note just rich enough

• Kernels: make linear model work in nonlinear settings
  • By mapping data to higher dimensions where it exhibits linear patterns
  • Apply the linear model in the new input feature space
  • Mapping $=$ changing the feature representation

5.2. Classifying non-linear separable data

• Consider the binary classification problem
  • Each example represented by a single feature $x$
  • No linear separator exists for this data

• Now map each example as $x \rightarrow \{x,x^2\}$

• Data now becomes linearly separable in the new representation

• Linear in the new representation $=$ nonlinear in the old representation

• Let's look at another example
  • Each example defined by a two features $x=\{x_1, x_2\}$
  • No linear separator exists for this data

• Now map each example as $x=\{x_1, x_2\} \rightarrow z=\{x_1^2,\sqrt{2}x_1x_2,x_2^2\}$
  • Each example now has three features (derived from the old represenation)

• Data now becomes linear separable in the new representation

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