线性拟合最小二乘法Python实现

下面代码实现的是最小二乘法线性拟合，并且包含自己造的轮子与别人造的轮子的结果比较。

问题：对 y=2.5x+0.8y=2.5x+0.8直线附近的带有噪声的数据进行线性拟合，最终求出w,b的估计值。

最小二乘法基本思想是使得样本方差最小。

代码中self_func()函数为自定义拟合函数，skl_func()为调用scikit-learn中线性模块的函数。

import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression

n = 101

x = np.linspace(0,10,n)
noise = np.random.randn(n)
y = 2.5 * x + 0.8 + 2.0 * noise

def self_func(steps=100, alpha=0.01):
    w = 0.5
    b = 0
    alpha = 0.01
    for i in range(steps):
        y_hat = w*x + b
        dy = 2.0*(y_hat - y)
        dw = dy*x
        db = dy
        w = w - alpha*np.sum(dw)/n
        b = b - alpha*np.sum(db)/n
        e = np.sum((y_hat-y)**2)/n
        #print (i,'W=',w,'\tb=',b,'\te=',e)
    print ('self_func:\tW =',w,'\n\tb =',b)
    plt.scatter(x,y)
    plt.plot(np.arange(0,10,1), w*np.arange(0,10,1) + b, color = 'r', marker = 'o', label = 'self_func(steps='+str(steps)+', alpha='+str(alpha)+')')

def skl_func():
    lr = LinearRegression()
    lr.fit(x.reshape(-1,1),y)
    y_hat = lr.predict(np.arange(0,10,0.75).reshape(-1,1))
    print('skl_fun:\tW = %f\n\tb = %f'%(lr.coef_,lr.intercept_))
    plt.plot(np.arange(0,10,0.75), y_hat, color = 'g', marker = 'x', label = 'skl_func')
    
self_func(10000)
skl_func()
plt.legend(loc='upper left')
plt.show()

结果：

self_func:  W = 2.5648753825503197     b = 0.24527830841237772
skl_fun:     W = 2.564875                         b = 0.245278