Regression


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



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

Note




$$\begin{array}{Icr} \text{input} \\ x_{i} \end{array} \quad \rightarrow \quad \begin{array}{Icr} \text{feature} \\ \begin{bmatrix} 1 \\ x_{i} \end{bmatrix} \end{array} \quad \rightarrow \quad \begin{array}{Icr} \text{predicted output} \\ \hat{y}_{i} \end{array}$$


$$\begin{array}{Icr} \begin{bmatrix} 1 & x_{1} \\1 & x_{2}\\\vdots & \vdots\\1 & x_{m}\end{bmatrix}\begin{bmatrix}\theta_0\\\theta_1\end{bmatrix}=\begin{bmatrix}y_{1} \\y_{2} \\\vdots \\y_{m}\end{bmatrix} \\ \begin{array}{Icr} \uparrow \\ \vec{A}_1 \end{array} \;\; \begin{array}{Icr} \uparrow \\ \vec{A}_2 \end{array} \quad \begin{array}{Icr} \uparrow \\ \vec{x} \end{array} \quad\quad \;\; \begin{array}{Icr} \uparrow \\ \vec{B} \end{array} \end{array} \quad \begin{array}{Icr} \quad \text{over-determined or} \\ \quad \text{projection} \end{array}$$


$$A(= \Phi) = \left[ \vec{A}_1 \;\vec{A}_2 \right]$$

1.2. Solve Optimizaton in Linear Regression

1.2.1. Use Linear Algebra

  • known as least square
$$ \theta = (A^TA)^{-1}A^T y $$
In [1]:
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
In [2]:
# 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 [3]:
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 [4]:
# 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()

1.2.2. Use Gradient Descent



$$\min_{\theta} ~ \lVert \hat y - y \rVert_2^2 = \min_{\theta} ~ \lVert A\theta - y \rVert_2^2$$



$$ \begin{align*} f &= (A\theta-y)^T(A\theta-y) = (\theta^TA^T-y^T)(A\theta-y) \\ &= \theta^TA^TA\theta - \theta^TA^Ty - y^TA\theta + y^Ty \\\\ \nabla f &= A^TA\theta + A^TA\theta - A^Ty - A^Ty = 2(A^TA\theta - A^Ty) \end{align*} $$


$$ \theta \leftarrow \theta - \alpha \nabla f $$

In [5]:
theta = np.random.randn(2,1)
theta = np.asmatrix(theta)

alpha = 0.001

for _ in range(1000):
    df = 2*(A.T*A*theta - A.T*y)
    theta = theta - alpha*df

print (theta)
[[0.65242559]
 [0.6714885 ]]
In [6]:
# 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()

1.2.3. Use CVXPY Optimization



$$\min_{\theta} ~ \lVert \hat y - y \rVert_2 = \min_{\theta} ~ \lVert A\theta - y \rVert_2$$

In [7]:
import cvxpy as cvx

theta2 = cvx.Variable([2, 1])
obj = cvx.Minimize(cvx.norm(A*theta2-y, 2))
cvx.Problem(obj,[]).solve()

print('theta:\n', theta2.value)
theta:
 [[0.65306531]
 [0.67129519]]

By the way, do we have to use only $L_2$ norm? No.

  • Let's use $L_1$ norm



$$ \min_{\theta} ~ \lVert \hat y - y \rVert_1 = \min_{\theta} ~ \lVert A\theta - y \rVert_1 $$

In [8]:
theta1 = cvx.Variable([2, 1])
obj = cvx.Minimize(cvx.norm(A*theta1-y, 1))
cvx.Problem(obj).solve()

print('theta:\n', theta1.value)
theta:
 [[0.62276377]
 [0.69128229]]
In [9]:
# to plot data
plt.figure(figsize = (10, 8))
plt.title('$L_1$ and $L_2$ Regression', fontsize = 15)
plt.xlabel('X', fontsize = 15)
plt.ylabel('Y', fontsize = 15)
plt.plot(x, y, 'ko', label = 'data')

# to plot straight lines (fitted lines)
xp = np.arange(0, 5, 0.01).reshape(-1, 1)
yp1 = theta1.value[0,0] + theta1.value[1,0]*xp
yp2 = theta2.value[0,0] + theta2.value[1,0]*xp

plt.plot(xp, yp1, 'b', linewidth=2, label = '$L_1$')
plt.plot(xp, yp2, 'r', linewidth=2, label = '$L_2$')
plt.legend(fontsize = 15)
plt.axis('equal')
plt.xlim([0, 5])
plt.grid(alpha = 0.3)
plt.show()

$L_1$ norm also provides a decent linear approximation.

What if outliers exist?

  • Fit with the different norms
  • Discuss the result
  • It is important to understand what makes them different.
In [10]:
# add outliers
x = np.vstack([x, np.array([0.5, 3.8]).reshape(-1, 1)])
y = np.vstack([y, np.array([3.9, 0.3]).reshape(-1, 1)])

A = np.hstack([x**0, x])
A = np.asmatrix(A)
In [11]:
A.shape
Out[11]:
(15, 2)
In [12]:
plt.figure(figsize = (10, 8))
plt.plot(x, y, 'ko', label = 'data')
plt.axis('equal')
plt.xlim([0, 5])
plt.grid(alpha = 0.3)
plt.show()
In [13]:
theta2 = cvx.Variable([2, 1])
obj2 = cvx.Minimize(cvx.norm(A*theta2-y, 2))
prob2 = cvx.Problem(obj2).solve()
In [14]:
# to plot straight lines (fitted lines)
plt.figure(figsize = (10, 8))
plt.plot(x, y, 'ko', label = 'data')
xp = np.arange(0, 5, 0.01).reshape(-1,1)
yp2 = theta2.value[0,0] + theta2.value[1,0]*xp

plt.plot(xp, yp2, 'r', linewidth = 2, label = '$L_2$')
plt.axis('equal')
plt.xlim([0, 5])
plt.legend(fontsize = 15, loc = 5)
plt.grid(alpha = 0.3)
plt.show()
In [15]:
theta1 = cvx.Variable([2, 1])
obj1 = cvx.Minimize(cvx.norm(A*theta1-y, 1))
prob1 = cvx.Problem(obj1).solve()
In [16]:
# to plot straight lines (fitted lines)
plt.figure(figsize = (10, 8))
plt.plot(x, y, 'ko', label = 'data')
xp = np.arange(0, 5, 0.01).reshape(-1,1)
yp1 = theta1.value[0,0] + theta1.value[1,0]*xp

plt.plot(xp, yp1, 'b', linewidth = 2, label = '$L_1$')
plt.axis('equal')
plt.xlim([0, 5])
plt.legend(fontsize = 15, loc = 5)
plt.grid(alpha = 0.3)
plt.show()
In [17]:
# to plot data
plt.figure(figsize = (10, 8))
plt.plot(x, y, 'ko', label = 'data')
plt.title('$L_1$ and $L_2$ Regression w/ Outliers', fontsize = 15)
plt.xlabel('X', fontsize = 15)
plt.ylabel('Y', fontsize = 15)

# to plot straight lines (fitted lines)
xp = np.arange(0, 5, 0.01).reshape(-1,1)
yp1 = theta1.value[0,0] + theta1.value[1,0]*xp
yp2 = theta2.value[0,0] + theta2.value[1,0]*xp

plt.plot(xp, yp1, 'b', linewidth = 2, label = '$L_1$')
plt.plot(xp, yp2, 'r', linewidth = 2, label = '$L_2$')
plt.axis('scaled')
plt.xlim([0, 5])
plt.legend(fontsize = 15, loc = 5)
plt.grid(alpha = 0.3)
plt.show()

Think about what makes them different.

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
In [18]:
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)
In [19]:
from sklearn import linear_model
In [20]:
reg = linear_model.LinearRegression()
reg.fit(x, y)
Out[20]:
LinearRegression(copy_X=True, fit_intercept=True, n_jobs=None,
         normalize=False)
In [21]:
reg.coef_
Out[21]:
array([[0.67129519]])
In [22]:
reg.intercept_
Out[22]:
array([0.65306531])
In [23]:
# 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)
plt.plot(xp, reg.predict(xp), 'r', linewidth = 2, label = "regression")
plt.legend(fontsize = 15)
plt.axis('equal')
plt.grid(alpha = 0.3)
plt.xlim([0, 5])
plt.show()

2. Multivariate Linear Regression

(= linear regression for multivariate data)


$$ \hat{y} = \theta_0 + \theta_{1}x_1 + \theta_{2}x_2 $$

$$$$ $$$$

$$\phi \left(x^{(i)}\right) = \begin{bmatrix}1\\x^{(i)}_{1}\\x^{(i)}_{2} \end{bmatrix}$$

In [24]:
# for 3D plot
from mpl_toolkits.mplot3d import Axes3D
In [25]:
# y = theta0 + theta1*x1 + theta2*x2 + noise

n = 200
x1 = np.random.randn(n, 1)
x2 = np.random.randn(n, 1)
noise = 0.5*np.random.randn(n, 1);

y = 2 + 1*x1 + 3*x2 + noise

fig = plt.figure(figsize = (10, 8))
ax = fig.add_subplot(1, 1, 1, projection = '3d')
ax.set_title('Generated Data', fontsize = 15)
ax.set_xlabel('$X_1$', fontsize = 15)
ax.set_ylabel('$X_2$', fontsize = 15)
ax.set_zlabel('Y', fontsize = 15)
ax.scatter(x1, x2, y, marker = '.', label = 'Data')
#ax.view_init(30,30)
plt.legend(fontsize = 15)
plt.show()


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



$$\implies \theta^{*} = (\Phi^T \Phi)^{-1} \Phi^T y$$
In [26]:
#% matplotlib qt5

A = np.hstack([np.ones((n, 1)), x1, x2])
A = np.asmatrix(A)
theta = (A.T*A).I*A.T*y

X1, X2 = np.meshgrid(np.arange(np.min(x1), np.max(x1), 0.5), 
                     np.arange(np.min(x2), np.max(x2), 0.5))
YP = theta[0,0] + theta[1,0]*X1 + theta[2,0]*X2

fig = plt.figure(figsize = (10, 8))
ax = fig.add_subplot(1, 1, 1, projection = '3d')
ax.set_title('Regression', fontsize = 15)
ax.set_xlabel('$X_1$', fontsize = 15)
ax.set_ylabel('$X_2$', fontsize = 15)
ax.set_zlabel('Y', fontsize = 15)
ax.scatter(x1, x2, y, marker = '.', label = 'Data')
ax.plot_wireframe(X1, X2, YP, color = 'k', alpha = 0.3, label = 'Regression Plane')
#ax.view_init(30,30)
plt.legend(fontsize = 15)
plt.show()