Machine Learning


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

Table of Contents

1. Linear Regression

Consider a linear regression.

$\text{Given} \; \begin{cases} x_{i} \; \text{: inputs} \\ y_{i} \; \text{: outputs} \end{cases}$ , Find $\theta_{0}$ and $\theta_{1}$

$$x= \begin{bmatrix} x_{1} \\ x_{2} \\ \vdots \\ x_{m} \end{bmatrix}, \qquad y= \begin{bmatrix} y_{1} \\ y_{2} \\ \vdots \\ y_{m} \end{bmatrix} \approx \hat{y}_{i} = \theta_{0} + \theta_{1}x_{i}$$
  • $ \hat{y}_{i} $ : predicted output

  • $ \theta = \begin{bmatrix} \theta_{0} \\ \theta_{1} \\ \end{bmatrix} $ : Model parameters

$$ \hat{y}_{i} = f(x_{i}\,; \theta) \; \text{ in general}$$
  • in many cases, a linear model is used to predict $y_{i}$
$$ \hat{y}_{i} = \theta_{0} + \theta_{1}x_{i} \; \quad \text{ such that }\quad \min\limits_{\theta_{0}, \theta_{1}}\sum\limits_{i = 1}^{m} (\hat{y}_{i} - y_{i})^2$$



1.1. Re-cast Problem as a Least Squares

  • For convenience, we define a function that maps inputs to feature vectors, $\phi$


$$\begin{array}{Icr}\begin{align*} \hat{y}_{i} & = \theta_0 + x_i \theta_1 = 1 \cdot \theta_0 + x_i \theta_1 \\ \\ & = \begin{bmatrix}1 & x_{i}\end{bmatrix}\begin{bmatrix}\theta_{0} \\ \theta_{1}\end{bmatrix} \\\\ & =\begin{bmatrix}1 \\ x_{i} \end{bmatrix}^{T}\begin{bmatrix}\theta_{0} \\ \theta_{1}\end{bmatrix} \\\\ & =\phi^{T}(x_{i})\theta \end{align*}\end{array} \begin{array}{Icr} \quad \quad \text{feature vector} \; \phi(x_{i}) = \begin{bmatrix}1 \\ x_{i}\end{bmatrix} \end{array}$$


$$\Phi = \begin{bmatrix}1 & x_{1} \\ 1 & x_{2} \\ \vdots \\1 & x_{m} \end{bmatrix}=\begin{bmatrix}\phi^T(x_{1}) \\\phi^T(x_{2}) \\\vdots \\\phi^T(x_{m}) \end{bmatrix} \quad \implies \quad \hat{y} = \begin{bmatrix}\hat{y}_{1} \\\hat{y}_{2} \\\vdots \\\hat{y}_{m}\end{bmatrix}=\Phi\theta$$


  • Optimization problem
$$\min\limits_{\theta_{0}, \theta_{1}}\sum\limits_{i = 1}^{m} (\hat{y}_{i} - y_{i})^2 =\min\limits_{\theta}\lVert\Phi\theta-y\rVert^2_2 \qquad \qquad \left(\text{same as} \; \min_{x} \lVert Ax-b \rVert_2^2 \right)$$



$$ \text{solution} \; \theta^* = (\Phi^{T}\Phi)^{-1}\Phi^{T} y $$

1.2. Solve using Linear Algebra

  • known as least square
$$ \theta = (A^TA)^{-1}A^T y $$
In [12]:
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
In [13]:
# data points in column vector [input, output]
x = np.array([0.1, 0.4, 0.7, 1.2, 1.3, 1.7, 2.2, 2.8, 3.0, 4.0, 4.3, 4.4, 4.9]).reshape(-1, 1)
y = np.array([0.5, 0.9, 1.1, 1.5, 1.5, 2.0, 2.2, 2.8, 2.7, 3.0, 3.5, 3.7, 3.9]).reshape(-1, 1)

plt.figure(figsize = (10,8))
plt.plot(x,y,'ko')
plt.title('Data', fontsize = 15)
plt.xlabel('X', fontsize = 15)
plt.ylabel('Y', fontsize = 15)
plt.axis('equal')
plt.grid(alpha = 0.3)
plt.xlim([0, 5])
plt.show()
In [14]:
m = y.shape[0]
#A = np.hstack([np.ones([m, 1]), x])
A = np.hstack([x**0, x])
A = np.asmatrix(A)

theta = (A.T*A).I*A.T*y

print('theta:\n', theta)
theta:
 [[0.65306531]
 [0.67129519]]
In [15]:
# to plot
plt.figure(figsize = (10, 8))
plt.title('Regression', fontsize = 15)
plt.xlabel('X', fontsize = 15)
plt.ylabel('Y', fontsize = 15)
plt.plot(x, y, 'ko', label = "data")

# to plot a straight line (fitted line)
xp = np.arange(0, 5, 0.01).reshape(-1, 1)
yp = theta[0,0] + theta[1,0]*xp

plt.plot(xp, yp, 'r', linewidth = 2, label = "regression")
plt.legend(fontsize = 15)
plt.axis('equal')
plt.grid(alpha = 0.3)
plt.xlim([0, 5])
plt.show()

1.3 Scikit-Learn


  • Machine Learning in Python
  • Simple and efficient tools for data mining and data analysis
  • Accessible to everybody, and reusable in various contexts
  • Built on NumPy, SciPy, and matplotlib
  • Open source, commercially usable - BSD license