机器学习(周志华)-python编程练习-习题3-3

 

习题3.3  编程实现对率回归，并给出西瓜数据集 3.0α 上的结果.

数据集3.0α

sn	density	suger_ratio	good_melon
1	0.697	0.46	1
2	0.774	0.376	1
3	0.634	0.264	1
4	0.608	0.318	1
5	0.556	0.215	1
6	0.403	0.237	1
7	0.481	0.149	1
8	0.437	0.211	1
9	0.666	0.091	0
10	0.243	0.267	0
11	0.245	0.057	0
12	0.343	0.099	0
13	0.639	0.161	0
14	0.657	0.198	0
15	0.36	0.37	0
16	0.593	0.042	0
17	0.719	0.103	0

 

对应的python实现代码

#encoding:UTF-8
import csv
import numpy as np
from math import *
from numpy.linalg import *
from matplotlib import pyplot as plt

#牛顿迭代法实现对数机率回归

old_ml=0 #初始似然函数值
n=0 
beta=np.mat([[0],[0],[1]]) #初始参数[w1;w2;b]  

#读取数据集
wm_data_30a = csv.reader(open('../data_set/watermelon_data_set_30a.csv','r'))

xi_d = np.mat(np.zeros((17,3)))
yi_d = np.mat(np.zeros((17,1)))

#遍历数据集
sn=0
for stu in wm_data_30a:
	#[sn x1 x2 y]
	if(stu[0].isdigit()==True):
		xi_d[sn,:] = np.mat([float(stu[1]),float(stu[2]),1])
		yi_d[sn,0] = float(stu[3])
		sn = sn+1
#print('xi_d=\n',xi_d)
#print('yi_d=\n',yi_d)
#print(wm_data_30a)
iter=0
xi = np.mat(np.zeros((3,1)))
while(iter<10):
	sn=0
	#遍历数据
	print('iter=',iter)
	new_ml=0
	for idx in range(17):#0~16
		#print('idx=',idx)
				
		xi = xi_d[idx,:].T #3*1的矩阵
		yi=yi_d[idx,0]
		#print(yi)
		#计算当前ml函数
		# ln(1+beta*xi) - yi*beta*xi
		new_ml = new_ml + log(1+e**det(beta.T*xi)) - yi*det(beta.T*xi)
	print('new ml=',new_ml)
	if abs(new_ml-old_ml)<0.001:
		print(new_ml-old_ml)
		print(beta)
		break

	#牛顿迭代法更新参数
	old_ml = new_ml
	dml_beta_1 = np.mat([[0],[0],[0]])#3*1矩阵
	dml_beta_2 = np.mat(np.zeros((3,3))) #3*3矩阵
	#遍历数据
	for idx in range(17):#0~16			
		xi = xi_d[idx,:].T 
		yi=yi_d[idx,0]
		#print(det(beta.T*xi))
		p1_xi = 1-1/(1+e**(det(beta.T*xi)))
		#计算一阶导
		dml_beta_1 = dml_beta_1 - xi*(yi-p1_xi) #矩阵
		#计算二阶导
		dml_beta_2 = dml_beta_2 + xi*(xi.T)*p1_xi*(1-p1_xi)
		#print('Neton iter',dml_beta_2)
	beta = beta - (dml_beta_2.I)*dml_beta_1
	print('beta=',beta)
	iter=iter+1

#%画出散点图以及计算出的直线
#%逐点画  分别表示是否好瓜

for idx in range(17):
	if yi_d[idx,0]==1:
		plt.plot(xi_d[idx,0],xi_d[idx,1],'+r')
	else:
		plt.plot(xi_d[idx,0],xi_d[idx,1],'ob')
#画出回归线
ply=-(0.1*beta[0,0]+beta[2,0])/beta[1,0];
pry=-(0.9*beta[0,0]+beta[2,0])/beta[1,0];

px=[0.1,0.9]
py=[ply,pry]

plt.plot(px,py)


plt.xlabel('density')
plt.ylabel('suger ratio')
plt.title('logistic function regression')
plt.show()

运算结果

beta=

[[ 3.15832738]
 [12.52119012]
 [-4.42886222]]