机器学习作业（一）线性回归——Python(numpy)实现

2023年4月10日下午11:55 • 机器学习

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第1题

简述：设计一个5*5的单位矩阵。

1 import numpy as np
2 A = np.eye(5)
3 print(A)

运行结果：

机器学习作业（一）线性回归——Python(numpy)实现

第2题

简述：实现单变量线性回归。

 1 import numpy as np
 2 import matplotlib.pyplot as plt
 3 from mpl_toolkits.mplot3d import Axes3D
 4 
 5 #-----------------计算代价值函数-----------------------
 6 def computeCost(X, y, theta):
 7     m = np.size(X[:,0])
 8     J = 1/(2*m)*np.sum((np.dot(X,theta)-y)**2)
 9     return J
10 
11 
12 #----------------根据人口预测利润----------------------
13 #读取数据集中数据，第一列是人口数据，第二列是利润数据
14 data = np.loadtxt('ex1data1.txt',delimiter=",",dtype="float")
15 m = np.size(data[:,0])
16 # print(data)
17 
18 #------------------绘制样本点--------------------------
19 X = data[:,0:1]
20 y = data[:,1:2]
21 plt.plot(X,y,"rx")
22 plt.xlabel('Population of City in 10,000s')
23 plt.ylabel('Profit in $10,000s')
24 # plt.show()
25 
26 #-----------------梯度下降计算局部最优解----------------
27 #添加第一列1
28 one = np.ones(m)
29 X = np.insert(X,0,values=one,axis=1)
30 # print(X)
31 
32 #设置α、迭代次数、θ
33 theta = np.zeros((2,1))
34 iterations = 1500
35 alpha = 0.01
36 
37 #梯度下降，并显示线性回归
38 J_history = np.zeros((iterations,1))
39 for iter in range(0,iterations):
40     theta = theta - alpha/m*np.dot(X.T,(np.dot(X,theta)-y))
41     J_history[iter] = computeCost(X,y,theta)
42 plt.plot(data[:,0],np.dot(X,theta),'-')
43 plt.show()
44 # print(theta)
45 # print(J_history)
46 
47 #--------------------显示三维图------------------------
48 theta0 = np.linspace(-10,10,100)
49 theta1 = np.linspace(-1,4,100)
50 J_vals = np.zeros((np.size(theta0),np.size(theta1)))
51 for i in range(0,np.size(theta0)):
52     for j in range(0,np.size(theta1)):
53         t = np.asarray([theta0[i],theta1[j]]).reshape(2,1)
54         J_vals[i,j] = computeCost(X,y,t)
55 # print(J_vals)
56 J_vals = J_vals.T    #需要转置一下，否则轴会反
57 fig1 = plt.figure()
58 ax = Axes3D(fig1)
59 ax.plot_surface(theta0,theta1,J_vals,rstride=1,cstride=1,cmap=plt.get_cmap('rainbow'))
60 ax.set_xlabel('theta0')
61 ax.set_ylabel('theta1')
62 ax.set_zlabel('J')
63 plt.show()
64 
65 #--------------------显示轮廓图-----------------------
66 lines = np.logspace(-2,3,20)
67 plt.contour(theta0,theta1,J_vals,levels = lines)
68 plt.xlabel('theta0')
69 plt.ylabel('theta1')
70 plt.plot(theta[0],theta[1],'rx')
71 plt.show()

运行结果：

机器学习作业（一）线性回归——Python(numpy)实现

第3题

简述：实现多元线性回归。

 1 import numpy as np
 2 import matplotlib.pyplot as plt
 3 
 4 #-----------------计算代价值函数-----------------------
 5 def computeCost(X, y, theta):
 6     m = np.size(X[:,0])
 7     J = 1/(2*m)*np.sum((np.dot(X,theta)-y)**2)
 8     return J
 9 
10 #-------------------根据面积和卧室数量预测房价----------
11 #读取数据集中数据，第一列是面积数据，第二列是卧室数量，第三列是房价
12 data = np.loadtxt('ex1data2.txt',delimiter=",",dtype="float")
13 m = np.size(data[:,0])
14 # print(data)
15 X = data[:,0:2]
16 y = data[:,2:3]
17 
18 #----------------------均值归一化---------------------
19 mu = np.mean(X,0)
20 sigma = np.std(X,0)
21 X_norm = np.divide(np.subtract(X,mu),sigma)
22 one = np.ones(m)   #添加第一列1
23 X_norm = np.insert(X_norm,0,values=one,axis=1)
24 # print(mu)
25 # print(sigma)
26 # print(X_norm)
27 
28 #----------------------梯度下降-----------------------
29 alpha = 0.05
30 num_iters = 100
31 theta = np.zeros((3,1));
32 J_history = np.zeros((num_iters,1))
33 for iter in range(0,num_iters):
34     theta = theta - alpha/m*np.dot(X_norm.T,(np.dot(X_norm,theta)-y))
35     J_history[iter] = computeCost(X_norm,y,theta)
36 # print(theta)
37 x_col = np.arange(0,num_iters)
38 plt.plot(x_col,J_history,'-b')
39 plt.xlabel('Number of iterations')
40 plt.ylabel('Cost J')
41 plt.show()
42 
43 #----------使用上述结果对[1650,3]的数据进行预测--------
44 test1 = [1,1650,3]
45 test1[1:3] = np.divide(np.subtract(test1[1:3],mu),sigma)
46 price = np.dot(test1,theta)
47 print(price)        #输出预测结果[292455.63375132]
48 
49 #-------------使用正规方程法求解----------------------
50 one = np.ones(m)
51 X = np.insert(X,0,values=one,axis=1)
52 theta = np.dot(np.dot(np.linalg.pinv(np.dot(X.T,X)),X.T),y)
53 # print(theta)
54 price = np.dot([1,1650,3],theta)
55 print(price)       #输出预测结果[293081.46433497]