AI 利用BP算法及Sigmoid函数，研究函数f(x)=2sinx-0.7的逼近问题-实验报告

import numpy as np
import matplotlib.pyplot as plt
import random
from pylab import *
mpl.rcParams['font.sans-serif'] = ['SimHei']
x = np.linspace(1, np.pi / 2, 10)
y = 2 * np.sin(x) - 0.7
x_test = np.random.random(10) * (np.pi / 2 - 1) + 1
y_test = 2 * np.sin(x_test) - 0.7
plt.scatter(x, y, marker='o')
plt.scatter(x_test, y_test, marker='+')
plt.title('训练集：红点，测试集：加号')
plt.show()
hide = 10  # 隐藏层神经元个数
W1 = np.random.random((hide, 1))  # 权重
B1 = np.random.random((hide, 1))   # 偏置
W2 = np.random.random((1, hide))
B2 = np.random.random((1, 1))
learningrate = 0.005 # 学习率
iteration = 10000  # 迭代次数
def sigmoid(x):
    return 1 / (1 + np.exp(-x))
E = np.zeros((iteration, 1))  # 存放误差
Y = np.zeros((10, 1))  # 存放训练结果
for k in range(iteration):
    temp = 0
    for i in range(10):
        hide_in = np.dot(x[i], W1) - B1
        hide_out = np.zeros((hide, 1))
        for j in range(hide):
            hide_out[j] = sigmoid(hide_in[j])
        y_out = np.dot(W2, hide_out) - B2
        Y[i] = y_out
        e = y_out - y[i]
        dB2 = -1 * learningrate * e
        dW2 = e * learningrate * np.transpose(hide_out)
        dB1 = np.zeros((hide, 1))
        for j in range(hide):
            dB1[j] = np.dot(np.dot(W2[0][j], sigmoid(hide_in[j])), (1 - sigmoid(hide_in[j])) * (-1) * e * learningrate)
        dW1 = np.zeros((hide, 1))
        for j in range(hide):
            dW1[j] = np.dot(np.dot(W2[0][j], sigmoid(hide_in[j])), (1 - sigmoid(hide_in[j])) * x[i] * e * learningrate)
        W1 = W1 - dW1
        B1 = B1 - dB1
        W2 = W2 - dW2
        B2 = B2 - dB2
        temp = temp + abs(e)
    E[k] = temp
plt.scatter(x, y, c='r')
plt.plot(x, Y)
plt.title('训练集：红点，训练结果：蓝线')
plt.show()
plt.scatter(x_test, y_test, c='y', marker='+')
plt.plot(x, Y)
plt.title('测试集：加号，训练结果：蓝线')
plt.show()
xx = np.linspace(0, iteration, iteration)
plt.plot(xx, E)
plt.title('训练误差随迭代次数趋势图')