最小二乘法实现的线性回归误差逆向传递的简单演示

Posted 2021-01-04 lycsdhr

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

x_data = np.arange(100,)
print(x_data)
k0 = np.random.random()
b0 = np.random.random()

y_data = x_data*k0+np.random.random(100,)*3
print()
print(y_data)
# plt.ion()
plt.figure(figsize=(12,9))
plt.scatter(x_data,y_data,label=‘true data‘)
# 假设  y = k * x + b

def wucha(x_真值,y_真值,k,b):
    y_预测 = x_真值*k + b
    误差 = y_真值 - y_预测
    return abs(误差.sum())


def tiaocan(x_真值,y_真值,k,b,步长):
    s1 = wucha(x_真值,y_真值,k+步长,b+步长)
    s2 = wucha(x_真值,y_真值,k-步长,b-步长)
    s3 = wucha(x_真值,y_真值,k+步长,b-步长)
    s4 = wucha(x_真值,y_真值,k-步长,b+步长)
    s = wucha(x_真值,y_真值,k,b)
    s_list = list((s1,s2,s3,s4,s))
    print(s_list)
    s_list.sort()
    min = s_list[0]
    print(s_list)
    if min == s1:
        return True,True
    elif min == s2:
        return False,False
    elif min == s3:
        return True,False
    elif min == s4:
        return False,True
    else:
        return None


if __name__ == ‘__main__‘:

    #运行100次
    k = np.random.random()
    b = np.random.random()
    k1 = k
    b1 = b
    print(‘随机初始化的k,b初始值:‘,k1,b1)
    for i in range(1000000):
        n = 0.01
        s = tiaocan(x_data,y_data,k1,b1,n)
        print(s)
        if s:
            if s[0]:
                k1+=n
            else:
                k1-=n
            if s[1]:
                b1+=n
            else:
                b1-=n
        else:
            break
        print(‘第{}次运行k,b {}  {}‘.format(i,k1,b1))
        lines = plt.plot(x_data,x_data*k1+b1,color=‘black‘)

    print(‘初始化的k0 b0 真值为:‘,k0,b0*3)
    print(‘k,b的初始化随机值为:‘,k,b)
    print(‘一共运行了{}次‘.format(i+1))
    print(‘k,b的预测值为‘,k1,b1)
    plt.plot(x_data,x_data*k+b,color=‘green‘,label=‘first‘)
    plt.plot(x_data,k1*x_data+b1,color=‘red‘,label=‘final‘)
    plt.legend()
    # plt.ioff()
    plt.show()

运行以上代码，可以实现拟合随机生成的数据，拟合过后的线为红色，拟合前的为绿色，拟合过程中的线为黑色。为了便于理解，有些参数我使用了中文，Python3默认编码UTF8