Python机器学习之随机梯度下降法的实现

import numpy as np

m = 10000

x = np.random.random(size=m)
y = x*3 + 4 + np.random.normal(size=m)

def dj_sgd(theta, x_i, y_i): # 传入一个样本，获取对应的梯度
    return x_i.T.dot(x_i.dot(theta)-y_i)*2 # MSE

def sgd(X_b, y, initial_theta, n_iters): # 求出整个theta的函数
    def learning_rate(i_iter):
        t0 = 5
        t1 = 50
        return t0/(i_iter+t1)
    theta = initial_theta
    i_iter = 1
    
    while i_iter <= n_iters:
        index = np.random.randint(0, len(X_b))
        x_i = X_b[index]
        y_i = y[index]
        gradient = dj_sgd(theta, x_i, y_i) # 求导数
        theta = theta - gradient*learning_rate(i_iter) # 求步长
        i_iter += 1
    return theta

def dj_sgd(theta, x_i, y_i): # 传入一个样本，获取对应的梯度
    return x_i.T.dot(x_i.dot(theta)-y_i)*2 # MSE

def sgd(X_b, y, initial_theta, n_iters): # 求出整个theta的函数
    def learning_rate(i_iter):
        t0 = 5
        t1 = 50
        return t0/(i_iter+t1)
    theta = initial_theta
    m = len(X_b)
    
    for cur_iter in range(n_iters): # 每一次循环都把样本打乱，n_iters的代表整个样本看几轮
        random_indexs = np.random.permutation(m)
        X_random = X_b[random_indexs]
        y_random = y[random_indexs]
        for i in range(m):
            theta = theta - learning_rate(cur_iter*m+i) * (dj_sgd(theta, X_random[i], y_random[i]))
        return theta