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

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几率回归

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=
,xi_d)
#print(yi_d=
,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]]