线性回归的梯度下降和正规方程组求解

 1  # coding:utf-8
 2 import matplotlib.pyplot as plt
 3 import numpy as np
 4
 5 def dataN(length):
 6     x = np.zeros(shape = (length,2))
 7     y = np.zeros(shape = length)
 8     for i in range(0,length):
 9       x[i][0] = 1
10       x[i][1] = i
11       y[i] = (i + 25) + np.random.uniform(0,1) *10
12     return x,y
13
14 def alphA(x,y): #选取前20次迭代cost最小的alpha
15     c=float("inf")
16     for k in range(1,1000):
17             a=1.0/k**3
18             f=gD(x,y,20,a)[1][-1]
19             if f>c:
20                 break
21             c=f
22             alpha=a
23     return alpha
24
25 def gD(x,y,iter,alpha):#梯度下降
26     theta=np.ones(2)
27     cost=[]
28     for i in range(iter):
29         hypothesis = np.dot(x,theta)
30         loss = hypothesis - y
31         cost.append(np.sum(loss ** 2))
32         gradient = np.dot(x.transpose(),loss)
33         theta = theta -alpha * gradient
34     return theta,cost
35
36 def sgD(x,y,iter,alpha):#随机梯度下降
37     theta=np.ones(2)
38     cost=[]
39
40     for i in range(iter):
41         l=0
42         for j in range(0,len(y)):
43             loss=theta[1]*x[j][1]+theta[0]-y[j]
44             theta[1]=theta[1]-alpha*x[j][1]*loss
45             theta[0]=theta[0]-alpha*loss
46             l=l+loss**2
47         cost.append(l)
48     return theta,cost
49
50 def eQ(x,y):#正则方程组
51    x=np.matrix(x)
52    y=np.matrix(y).T
53    a=np.dot(x.T,x).I
54    b=np.dot(a,x.T)
55    c=np.dot(b,y)
56    return c
57
58 def exP(x,y):#一元线性回归拟合方程
59     xmean=np.mean(x)
60     ymean=np.mean(y)
61     b=np.sum([(x1-xmean)*(y1-ymean)for x1,y1 in zip(x,y)])/np.sum([(x1-xmean)**2 for x1 in x])
62     a=ymean-b*xmean
63     return a,b
64
65 length=100
66 iter=50000
67 x,y=dataN(length)
68 #theta,cost=sgD(x,y,iter,alphA(x,y))  #[ 30.24623439   0.99707473]
69 theta,cost=gD(x,y,iter,alphA(x,y))  #[ 30.17228028   0.99806093]
70 print theta
71 print eQ(x,y) #[[ 30.20320097][  0.99759475]]
72 print exP(x[:,1],y) #与正则方程组结果一致
73
74 plt.figure(1)
75 plt.plot(range(iter),cost)
76 plt.figure(2)
77 plt.plot(x[:,1],y,‘b.‘)
78 plt.plot([0,length],[theta[0],theta[0]+length*theta[1]])
79 plt.show()