Python / Numpy中的正规方程实现

Question

我编写了一些初学者代码来使用正规方程计算简单线性模型的系数。

# Modules
import numpy as np

# Loading data set
X, y = np.loadtxt('ex1data3.txt', delimiter=',', unpack=True)

data = np.genfromtxt('ex1data3.txt', delimiter=',')

def normalEquation(X, y):
    m = int(np.size(data[:, 1]))

    # This is the feature / parameter (2x2) vector that will
    # contain my minimized values
    theta = []

    # I create a bias_vector to add to my newly created X vector
    bias_vector = np.ones((m, 1))

    # I need to reshape my original X(m,) vector so that I can
    # manipulate it with my bias_vector; they need to share the same
    # dimensions.
    X = np.reshape(X, (m, 1))

    # I combine these two vectors together to get a (m, 2) matrix
    X = np.append(bias_vector, X, axis=1)

    # Normal Equation:
    # theta = inv(X^T * X) * X^T * y

    # For convenience I create a new, tranposed X matrix
    X_transpose = np.transpose(X)

    # Calculating theta
    theta = np.linalg.inv(X_transpose.dot(X))
    theta = theta.dot(X_transpose)
    theta = theta.dot(y)

    return theta

p = normalEquation(X, y)

print(p)

使用这里找到的小数据集：

http://www.lauradhamilton.com/tutorial-linear-regression-with-octave

我得到了系数：[-0.34390603; 0.2124426 ]，使用上面的代码而不是：[24.9660; 3.3058]。谁能帮助澄清我哪里出错了？

Answer 1

你可以像下面这样实现正规方程：

import numpy as np

X = 2 * np.random.rand(100, 1)
y = 4 + 3 * X + np.random.randn(100, 1)

X_b = np.c_[np.ones((100, 1)), X]  # add x0 = 1 to each instance
theta_best = np.linalg.inv(X_b.T.dot(X_b)).dot(X_b.T).dot(y)

X_new = np.array([[0], [2]])
X_new_b = np.c_[np.ones((2, 1)), X_new]  # add x0 = 1 to each instance
y_predict = X_new_b.dot(theta_best)
y_predict

Answer 2

这假设 X 是一个 m × n+1 维矩阵，其中 x_0 始终 = 1，而 y 是一个 m 维向量。

import numpy as np

step1 = np.dot(X.T, X)
step2 = np.linalg.pinv(step1)
step3 = np.dot(step2, X.T)
theta = np.dot(step3, y) # if y is m x 1.  If 1xm, then use y.T

Answer 3

您的实现是正确的。您只交换了X 和y（仔细查看它们如何定义x 和y），这就是您得到不同结果的原因。

调用normalEquation(y, X) 给[ 24.96601443 3.30576144] 应该是这样的。

Answer 4

这是一行中的正规方程：

theta = np.dot(np.linalg.inv(np.dot(X.T,X)),np.dot(X.T,Y))