简单线性回归的Python实现

import numpy as np
import pandas as pd
from numpy.linalg import inv
from numpy import dot

iris = pd.read_csv('iris.csv')
# 拟合线性模型： Sepal.Length ~ x0 + x1*Sepal.Width + x2*Petal.Length + x3*Petal.Width

# 正规方程法
temp = iris.iloc[:, 1:4]
temp['x0'] = 1
X = temp.iloc[:, [3, 0, 1, 2]]
Y = iris.iloc[:, 4]
Y = Y.values.reshape(len(iris), 1)
theta_n = dot(dot(inv(dot(X.T, X)), X.T), Y)  # theta = (X'X)^(-1)X'Y
print(theta_n)

# 批量梯度下降法
theta_g = np.array([1., 1., 1., 1.])  # 初始化theta
theta_g = theta_g.reshape(4, 1)
alpha = 0.1
temp = theta_g
X0 = X.iloc[:, 0].values.reshape(150, 1)
X1 = X.iloc[:, 1].values.reshape(150, 1)
X2 = X.iloc[:, 2].values.reshape(150, 1)
X3 = X.iloc[:, 3].values.reshape(150, 1)
J = pd.Series(np.arange(800, dtype=float))
for i in range(5000):
# theta j := theta j + alpha*(yi - h(xi))*xi
#若划去/150.会导致报错溢出，以调低学习率弥补则会得到很差的效果
temp[0] = theta_g[0] + alpha * np.sum((Y - dot(X, theta_g)) * X0)/150.
temp[1] = theta_g[1] + alpha * np.sum((Y - dot(X, theta_g)) * X1)/150.
temp[2] = theta_g[2] + alpha * np.sum((Y - dot(X, theta_g)) * X2)/150.
temp[3] = theta_g[3] + alpha * np.sum((Y - dot(X, theta_g)) * X3)/150.
J[i] = 0.5 * np.sum((Y - dot(X, theta_g)) ** 2)  # 计算损失函数值
#J[i] = 0.5 * np.square(np.sum(Y - dot(X, theta_g)))  # 计算损失函数值
theta_g = temp  # 更新theta

print(theta_g)
J.plot(ylim=[0, 50])