通过jupyter notebook用python实现线性回归

1.今天小关要讲的是用python实现线性回归

2.来吧，展示：

#python实现线性回归
import numpy as np
import pandas as pd
from numpy.linalg import inv
from numpy import dot
from sklearn.model_selection import train_test_split
import matplotlib.pyplot as plt
from sklearn import linear_model
# 最小二乘法
def lms(x_train,y_train,x_test):
  theta_n = dot(dot(inv(dot(x_train.T, x_train)), x_train.T), y_train) # theta = (X'X)^(-1)X'Y
  #print(theta_n)
  y_pre = dot(x_test,theta_n)
  mse = np.average((y_test-y_pre)**2)
  #print(len(y_pre))
  #print(mse)
  return theta_n,y_pre,mse
#梯度下降算法
def train(x_train, y_train, num, alpha,m, n):
  beta = np.ones(n)
  for i in range(num):
    h = np.dot(x_train, beta)       # 计算预测值
    error = h - y_train.T         # 计算预测值与训练集的差值
    delt = 2*alpha * np.dot(error, x_train)/m # 计算参数的梯度变化值
    beta = beta - delt
    #print('error', error)
  return beta
if __name__ == "__main__":
    #iris.csv保存为自己的本地路径
  iris = pd.read_csv('E:\新加卷（E）\python学习\iris.csv')
  iris['Bias'] = float(1)
  x = iris[['Sepal.Width', 'Petal.Length', 'Petal.Width', 'Bias']]
  y = iris['Sepal.Length']
  x_train, x_test, y_train, y_test = train_test_split(x, y, test_size=0.2, random_state=5)
  t = np.arange(len(x_test))
  m, n = np.shape(x_train)
  # Leastsquare
  theta_n, y_pre, mse = lms(x_train, y_train, x_test)
  # plt.plot(t, y_test, label='Test')
  # plt.plot(t, y_pre, label='Predict')
  # plt.show()
  # GradientDescent
  beta = train(x_train, y_train, 1000, 0.001, m, n)
  y_predict = np.dot(x_test, beta.T)
  # plt.plot(t, y_predict)
  # plt.plot(t, y_test)
  # plt.show()
  # sklearn
  regr = linear_model.LinearRegression()
  regr.fit(x_train, y_train)
  y_p = regr.predict(x_test)
  print(regr.coef_,theta_n,beta)
  l1,=plt.plot(t, y_predict)
  l2,=plt.plot(t, y_p)
  l3,=plt.plot(t, y_pre)
  l4,=plt.plot(t, y_test)
  plt.legend(handles=[l1, l2,l3,l4 ], labels=['GradientDescent', 'sklearn','Leastsquare','True'], loc='best')
  plt.show()

　　分享iris.csv下载地址：    http://yy.jb51.net:81/201812/yuanma/iris.csv(jb51.net).rar

3.打印结果：

 4.代码有点长，希望耐心分析，不懂的地方可以私信我，下面有联系方式

希望能帮到大家，问你们要一个赞，你们会给吗，谢谢大家
版权声明：本文版权归作者（@攻城狮小关）和CSDN共有，欢迎转载，但未经作者同意必须保留此段声明，且在文章页面明显位置给出原文连接，否则保留追究法律责任的权利。
大家写文都不容易，请尊重劳动成果~ 
交流加Q：1909561302
博客园地址https://www.cnblogs.com/guanguan-com/