python之线性回归

Love The Way You Lie 2022-05-19 02:44 274阅读 0赞

# 什么是线性回归 #

通过训练来得到线性方程系数ehta的过程叫做线性回归，这样我们可以预测某个x对应的y值

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# 线性 回归的数学表示 #

![70 1][]

# 线性回归中的最小二乘 法 #

通过对L求导使ehta最小

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# 线性回归的梯度下降算法 #

![70 3][]

![70 4][]

# 复习numpy的向量操作 #

#coding=gbk
    import numpy as np
    from numpy.linalg import inv
    from numpy import  dot
    from numpy import  mat
    
    #inv 求矩阵的逆
    #dot求矩阵点乘
    #mat是矩阵
    
    A = np.mat([1,1])
    #B = np.array([1,1])
    print('矩阵A:\n',A)
    print("A的转置:\n",A.T)
    #reshape给矩阵整形
    print(A.reshape(2,1))
    #print('B:\n',B)
    
    B = mat([[1,2],[2,3]])
    print(B.reshape(1,4))
    print("B的逆:\n",inv(B))
    print("矩阵B:\n",B)
    print("B的第一行:\n",B[0,:])
    print("B的第一列:\n",B[:,0])
    #A 1*2 B 2*2
    print('A和B点乘',dot(A,B))
    print("B点乘A的转置:\b",dot(B,A.T))

# 实现最小二乘法 #

#coding=gbk
    import numpy as np
    from numpy.linalg import inv
    from numpy import  dot
    from numpy import  mat
    
    #y=2x
    X = mat([1,2,3]).reshape(3,1)
    Y = 2*X
    
    #theta = (X'X)^-1X'Y
    
    theta = dot(dot(inv(dot(X.T,X)),X.T),Y)
    print (theta)

# 实现梯度下降 #

#coding=gbk
    import numpy as np
    from numpy.linalg import inv
    from numpy import  dot
    from numpy import  mat
    
    #y=2x
    X = mat([1,2,3]).reshape(3,1)
    Y = 2*X
    
    #theta = theta - alpha*(theta*X -Y)*X
    
    theta = 1.
    alpha = 0.1
    for i in range(100):
        theta = theta + np.sum(alpha * (Y - dot(X, theta))*X.reshape(1,3))/3.
    
    print (theta)

# 回归分析实战 #

## 随机数据生成代码,生成data.csv文件 ##

import random
    def Y(X1, X2, X3):
     return 0.65 * X1 + 0.70 * X2 - 0.55 * X3 + 1.95
    def Produce():
     filename = 'data.csv'
     with open(filename, 'w') as file:
      file.write('X1,X2,X3,Y\n')
      for i in range(200):
       random.seed()
       x1 = random.random() * 10
       x2 = random.random() * 10
       x3 = random.random() * 10
       y = Y(x1, x2, x3)
       try:
        file.write(str(x1) + ',' + str(x2) + ',' + str(x3) + ',' + str(y) + '\n')
       except Exception as e:
        print ('Write Error')
        print (str(e))
    
    Produce()

## 分别用最小二乘法和梯度下降法计算theta ##

#coding=gbk
    import numpy as np
    from numpy.linalg import inv
    from numpy import  dot
    from numpy import  mat
    import pandas as pd
    
    #获取X和Y向量
    dataset = pd.read_csv('data.csv')
    #print(dataset)
    temp = dataset.iloc[:,0:3]#取数据集中
    temp['x0'] = 1#截距
    
    X = temp.iloc[:,[3,0,1,2]]
    
    Y = dataset.iloc[:,3].values.reshape(200,1)
    #Y =  dataset.iloc[:,3].values.reshape(200,1)
    #print(Y)
    
    #最小二乘法
    theta = dot(dot(inv(dot(X.T,X)),X.T),Y)
    print("最小二乘法计算theta:\n",theta.reshape(4,1))
    
    #梯度下降法
    theta = np.array([1.,1.,1.,1.]).reshape(4,1)
    alpha = 0.02#学习率
    temp = theta #缓存，一块同步更新，很重要
    X0 = X.iloc[:,0].values.reshape(len(Y),1)
    X1 = X.iloc[:,1].values.reshape(len(Y),1)
    X2 = X.iloc[:,2].values.reshape(len(Y),1)
    X3 = X.iloc[:,3].values.reshape(len(Y),1)
    
    for i in range(10000):
        temp[0] = theta[0] + alpha*np.sum((Y-dot(X,theta))*X0)/200.
        temp[1] = theta[1] + alpha*np.sum((Y-dot(X,theta))*X1)/200.
        temp[2] = theta[2] + alpha*np.sum((Y-dot(X,theta))*X2)/200.
        temp[3] = theta[3] + alpha*np.sum((Y-dot(X,theta))*X3)/200.
        theta = temp
    print ("梯度下降法计算theta:",theta)

运行结果

![70 5][]

[70]: /images/20220519/1ccb489605f04de6935ad10cba0c419a.png
[70 1]: /images/20220519/49af6e8ea64743beb12a01f821155f59.png
[70 2]: /images/20220519/0d52c81aae4a40b0853696e5289dc0eb.png
[70 3]: /images/20220519/442f231dc8d64b37bcdedb6a43b35a80.png
[70 4]: /images/20220519/7edd1297123b488b8cee27ed3fedc25b.png
[70 5]: /images/20220519/57603b4921f343789020b0e7c366382a.png