如何用Python编程实现梯度下降算法？

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pythonPython代码示例：使用梯度下降法求解y=2*x1 + x2 + 3

import numpy as np

目标函数def objective_function(x): return 2*x[0] + x[1] + 3

梯度函数def gradient_function(x): return np.array([2, 1])

梯度下降法def gradient_descent(x0, learning_rate, epochs): x=x0 for _ in range(epochs): grad=gradient_function(x) x=x - learning_rate * grad return x

初始参数x0=np.array([0, 0])learning_rate=0.01epochs=1000

求解solution=gradient_descent(x0, learning_rate, epochs)print(求解结果：, solution)

本文实例为大家分享了python实现梯度下降法的具体代码，供大家参考，具体内容如下

使用工具：Python（x,y） 2.6.6
运行环境：Windows10

问题：求解y=2*x1+x2+3，即使用梯度下降法求解y=a*x1+b*x2+c中参数a，b，c的最优值（监督学习）

训练数据：

x_train=[1, 2], [2, 1],[2, 3], [3, 5], [1,3], [4, 2], [7, 3], [4, 5], [11, 3], [8, 7]

y_train=[7, 8, 10, 14, 8, 13, 20, 16, 28,26]

测试数据：

x_test = [1, 4],[2, 2],[2, 5],[5, 3],[1,5],[4, 1]

# -*- coding: utf-8 -*- """ Created on Wed Nov 16 09:37:03 2016 @author: Jason """ import numpy as np import matplotlib.pyplot as plt # y=2 * (x1) + (x2) + 3 rate = 0.001 x_train = np.array([[1, 2], [2, 1],[2, 3], [3, 5], [1, 3], [4, 2], [7, 3], [4, 5], [11, 3], [8, 7] ]) y_train = np.array([7, 8, 10, 14, 8, 13, 20, 16, 28, 26]) x_test = np.array([[1, 4],[2, 2],[2, 5],[5, 3],[1, 5],[4, 1]]) a = np.random.normal() b = np.random.normal() c = np.random.normal() def h(x): return a*x[0]+b*x[1]+c for i in range(100): sum_a=0 sum_b=0 sum_c=0 for x, y in zip(x_train, y_train): for xi in x: sum_a = sum_a+ rate*(y-h(x))*xi sum_b = sum_b+ rate*(y-h(x))*xi #sum_c = sum_c + rate*(y-h(x)) *1 a = a + sum_a b = b + sum_b c = c + sum_c plt.plot([h(xi) for xi in x_test]) print(a) print(b) print(c) result=[h(xi) for xi in x_train] print(result) result=[h(xi) for xi in x_test] print(result) plt.show()

运行结果：

结论：线段是在逐渐逼近的，训练数据越多，迭代次数越多就越逼近真实值。

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