Python实现简单的梯度下降计算

import matplotlib.pyplot as plt
import math
import numpy as np

x = []
y = []
for i in np.linspace(0, 7):
    x.append(i)
    y.append(i**2 * math.sin(i))
plt.plot(x, y)
plt.show()

# w, b 为当前的参数值， sets为（x, y)的点
def average_error(w, b, sets):
    # 另总error值初始值为0
    total_error = 0
    # 对每个取得的set点进行迭代
    for i in range(0, len(sets)):
        # 得出所有的（x, y）点
        x = sets[i, 0]
        y = sets[i, 1]
        total_error += (y - (w*x + b)) ** 2
        # 将所有计算出的误差值的平方和累加起来
    return total_error / float(len(sets))
         # 返回累加出的平方和的均值

def gradient_compute(w_current, b_current, sets, learningrate):
    # w_current、b_current为当前值，sets为(x, y)点， learningrate为学习率
    w_gradient = 0
    b_gradient = 0
    # 两者初始梯度值为0
    n = float(len(sets))
    for i in range(0, len(sets)):
        x = sets[i, 0]
        y = sets[i, 1]

        b_gradient += -(2/n) * (y - ((w_current * x) + b_current))
        w_gradient += -(2/n) * x * (y - ((w_current * x) + b_current))
        # 依据梯度计算公式，列出上两式子
        # 直接对所有点的累加值做平均计算即为（2/n）
    new_w = w_current - learningrate * w_gradient
    new_b = b_current - learningrate * b_gradient
    return [new_b, new_w]
    # 返回新的w和b

def run_gradient(sets, b_start, w_start, learningrate, number):
    # number=迭代次数
    b = b_start
    w = w_start
    for i in range(number):
        w, b = gradient_compute(w, b, np.array(sets), learningrate)
        return [w, b]
        # 得到循环100次后，w和b的最优解