阿里巴巴做网站联系人,seo免费培训,如何创造一个网站,做废铁在哪个网站推广梯度下降的直线拟合
实现说明
给定若干个x,yx, yx,y并且求得一个最佳的yaxby ax byaxb#xff0c;也就是二元一次方程组的解。
先放上给定的散点#xff0c;以及求得的线性回归的直线的图片。 我个人认为#xff0c;这里的梯度优化#xff0c;就是通过一个关键式子los…梯度下降的直线拟合
实现说明
给定若干个x,yx, yx,y并且求得一个最佳的yaxby ax byaxb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loss∑(axb−y)2loss \sum(ax b - y) ^{2}loss∑(axb−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∂a2x(axb−y)\frac{\partial{loss}}{\partial{a}} 2x(ax b - y)∂a∂loss2x(axb−y)
∂loss∂b2(axb−y)\frac{\partial{loss}}{\partial{b}} 2(ax b - y)∂b∂loss2(axb−y)
由此我们每次梯度下降更新的aa−∂loss∂a∗learning_ratea a - \frac{\partial{loss}}{\partial{a}} * learning\_rateaa−∂a∂loss∗learning_rate
同样的每次梯度下降更新的bb−∂loss∂b∗learning_rateb b - \frac{\partial{loss}}{\partial{b}} * learning\_ratebb−∂b∂loss∗learning_rate
然后通过这个迭代更新去得到最优损失的losslossloss同时a,ba, ba,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(Mys {0}, {1} Standards {2}, {3}.format(a, b, 0.487713, 3.0308))print()# for i in range(len(data)):# print(Datas y : {0} Mys y : {1} Standards 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)
