Classification: Perceptron


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




  • 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 loss function



$$ \mathscr{L}(\omega) = \sum_{n =1}^{m} \max \left\{ 0, -y_n \cdot \left(\omega^T x_n \right)\right\} $$

  • Loss $ = 0$ on examples where perceptron is correct, i.e., $y_n \cdot \left(\omega^T x_n \right) > 0$
  • Loss $ > 0$ on examples where perceptron misclassified, i.e., $y_n \cdot \left(\omega^T x_n \right) < 0$


Note: $\text{sign}\left(\omega^T x_n \right) \neq y_n$ is equivalent to $ y_n \cdot \left(\omega^T x_n \right) < 0$

3. Perceptron in Python


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



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

In [1]:
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
In [2]:
#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
In [3]:
C1 = np.where(g >= 1)
C0 = np.where(g < -1)
print(C1)

(array([ 3,  4,  7, 13, 17, 19, 22, 24, 27, 34, 36, 40, 42, 43, 44, 50, 59,
       63, 68, 71, 77, 78, 84, 88, 91, 92, 95, 96, 97, 98, 99],
      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, 0, 0], dtype=int64))
In [4]:
C1 = np.where(g >= 1)[0]
C0 = np.where(g < -1)[0]
print(C1.shape)
print(C0.shape)

(31,)
(43,)
In [5]:
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()
$$ \begin{align*} x &= \begin{bmatrix} \left(x^{(1)}\right)^T \\ \left(x^{(2)}\right)^T \\ \left(x^{(3)}\right)^T\\ \vdots \\ \left(x^{(m)}\right)^T \end{bmatrix} = \begin{bmatrix} 1 & x_1^{(1)} & x_2^{(1)} \\ 1 & x_1^{(2)} & x_2^{(2)} \\ 1 & x_1^{(3)} & x_2^{(3)}\\\vdots & \vdots & \vdots \\ 1 & x_1^{(m)} & x_2^{(m)}\end{bmatrix} \qquad y = \begin{bmatrix}y^{(1)} \\ y^{(2)} \\ y^{(3)}\\ \vdots \\ y^{(m)} \end{bmatrix} \end{align*}$$
In [6]:
X1 = np.hstack([np.ones([C1.shape[0],1]), x1[C1], x2[C1]])
X0 = np.hstack([np.ones([C0.shape[0],1]), x1[C0], x2[C0]])
X = np.vstack([X1, X0])

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

X = np.asmatrix(X)
y = np.asmatrix(y)
$$\omega = \begin{bmatrix} \omega_0 \\ \omega_1 \\ \omega_2\end{bmatrix}$$


$$\omega \leftarrow \omega + yx$$ where $(x, y)$ is a misclassified training point

In [7]:
w = np.ones([3,1])
w = np.asmatrix(w)

n_iter = y.shape[0]
for k in range(n_iter):
    for i in range(n_iter):
        if y[i,0] != np.sign(X[i,:]*w)[0,0]:
            w += y[i,0]*X[i,:].T

print(w)

[[-20.        ]
 [  5.87514833]
 [  9.87930425]]
$$ \begin{align*} g(x) &= \omega_0 + \omega^Tx = \omega_0 + \omega_1x_1 + \omega_2x_2 = 0 \\\\ \implies x_2 &= -\frac{\omega_1}{\omega_2} x_1 - \frac{\omega_0}{\omega_2} \end{align*} $$
In [8]:
x1p = np.linspace(0,8,100).reshape(-1,1)
x2p = - w[1,0]/w[2,0]*x1p - w[0,0]/w[2,0]

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 = 3, 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()

4. Perceptron using Scikit-Learn



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

In [9]:
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])])
In [10]:
from sklearn import linear_model

clf = linear_model.Perceptron(tol=1e-3)
clf.fit(X, np.ravel(y))
Out[10]:
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)
In [11]:
clf.predict([[3, -2]])
Out[11]:
array([-1.])
In [12]:
clf.predict([[6, 2]])
Out[12]:
array([1.])
In [13]:
clf.coef_
Out[13]:
array([[4.45077785, 8.31238814]])
In [14]:
clf.intercept_
Out[14]:
array([-17.])
$$ \begin{align*} g(x) &= \omega_0 + \omega^Tx = \omega_0 + \omega_1x_1 + \omega_2x_2 = 0 \\\\ \implies x_2 &= -\frac{\omega_1}{\omega_2} x_1 - \frac{\omega_0}{\omega_2} \end{align*} $$
In [15]:
w0 = clf.intercept_[0]
w1 = clf.coef_[0,0]
w2 = clf.coef_[0,1]
In [16]:
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()

5. 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



6. Vidoe Lectures

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