梯度下降：求线性回归

梯度下降的直线拟合

实现说明

给定若干个 $x, y$ 并且求得一个最佳的 $y = a x + b$ ，也就是二元一次方程组的解。

先放上给定的散点，以及求得的线性回归的直线的图片。

我个人认为，这里的梯度优化，就是通过一个关键式子 $\sum(ax + b - y) ^{2}$ ，通过求解这个凹函数来得到他的最小值从而实现整个线性回归方程的最优解，具体的实现就如下分析。

$∂loss∂a=2x(ax+b−y)\frac{\partial{loss}}{\partial{a}} = 2x(ax + b - y)$

$∂loss∂b=2(ax+b−y)\frac{\partial{loss}}{\partial{b}} = 2(ax + b - y)$

由此我们每次梯度下降更新的 $a=a−∂loss∂a∗learning_ratea = a - \frac{\partial{loss}}{\partial{a}} * learning\_rate$

同样的每次梯度下降更新的 $b=b−∂loss∂b∗learning_rateb = b - \frac{\partial{loss}}{\partial{b}} * learning\_rate$

然后通过这个迭代更新去得到最优损失的 $l o s s$ ，同时 $a, b$ ，也会在这个时候更新为最优值

PY‘S CODE

import torch
import numpy as np
import matplotlib.pyplot as pltx1 = np.array([1.1, 2.4, 2.4, 3.1, 2.2, 4.42, 5.43, 4.5, 5.28, 7.35, 10, 8.27, 12.6, 12.8, 9.69, 15.0, 13.69])
y1 = np.array([2.5, 1.7, 3, 4.0, 5.2, 6.53, 7.33, 8.7, 4.2, 5.8, 6.05, 8.05, 7.41, 8.89, 10.12, 9.72, 10.83])def calc_error(a, b, data):sum = 0for i in range(len(data)):x, y = data[i][0], data[i][1]sum += (a * x + b - y) ** 2return sum / (float)(len(data))def gradient_step(now_a, now_b, data, learning_rate):gradient_a, gradient_b = 0, 0for i in range(len(data)):x, y = data[i][0], data[i][1]gradient_a += 2 * x * (now_a * x + now_b - y)gradient_b += 2 * (now_a * x + now_b - y)gradient_a /= len(data)#取导数的平均值gradient_b /= len(data)new_a = now_a - learning_rate * gradient_anew_b = now_b - learning_rate * gradient_breturn [new_a, new_b]def algorithm(start_a, start_b, data, learning_rate, iterator_num):a, b = start_a, start_bfor i in range(iterator_num):a, b = gradient_step(a, b, data, learning_rate)return [a, b]def run():# 1.1, 2.4, 2.4, 3.1, 2.2, 4.42, 5.43, 4.5, 5.28, 7.35, 10, 8.27, 12.6, 12.8, 9.69, 15.0, 13.69# 2.5, 1.7, 3, 4.0, 5.2, 6.53, 7.33, 8.7, 4.2, 5.8, 6.05, 8.05, 7.41, 8.89, 10.12, 9.72, 10.83data = np.array([[1.100000, 2.500000], [2.400000, 1.700000], [2.400000, 3.000000],[3.100000, 4.000000], [2.200000, 5.200000], [4.420000, 6.530000],[5.430000, 7.330000], [4.500000, 8.700000], [5.280000, 4.200000],[7.350000, 5.800000], [10.000000, 6.050000], [8.270000, 8.050000],[12.600000, 7.410000], [12.800000, 8.890000], [9.690000, 10.120000],[15.000000, 9.720000], [13.690000, 10.830000]])a, b = 0, 0# for i in range(1, 6):#通过改变迭代次数，对比其答案，#     iterator_num = 10 ** i#     print("iterator_num is {0}".format(iterator_num))#     print("befor a:{0}, b:{1}, error{2}".format(a, b, calc_error(a, b, data)))#     a, b = algorithm(a, b, data, 0.0001, iterator_num)#     print("after a:{0}, b:{1}, error{2}".format(a, b, calc_error(a, b, data)))#     print("")a, b = algorithm(a, b, data, 0.001, 100000)#选了一个稍优的迭代次数，print("My's {0}, {1} Standard's {2}, {3}".format(a, b, 0.487713, 3.0308))print("")# for i in range(len(data)):#     print("Data's y : {0} My's y : {1} Standard's y : {2}".format(data[i][1], a * data[i][0] + b, 0.487713 * data[i][0] + 3.0308))#     print("")return [a, b]if __name__ == "__main__":plt.scatter(x1, y1, color = "red", label = "point")a, b = run()x = x1y = a * x + bplt.plot(x, y, label = "line")plt.legend(loc = "best")plt.show()# print("heloo, word")