《机器学习》（周志华） 习题3.1-3.3个人笔记

3.1  试分析在什么情况下式（3.2）中不必考虑偏置项b.

其实从前面第一章开始的习题就有很多不会的，第二章更是只会做前两道，现在到第三章，发现第一题都不是很明了了。从我个人来看：f(x)=w'x+b中，x代表d维向量，w则是相应的权重向量，而b=b*x0可看做权重为b，x0=1为相应属性. 很明显，1与x中的xi线性无关，也就是说，f(x)实际上是d+1维空间，当所有示例的x0属性都只有一种取值时，这样的属性已经失去了作为分类属性的意义，因此，将其权重b设为0，表示将所有示例具有的同样的属性值的属性去掉。（感觉这样不对啊T_T）

3.2 试证明，对于参数w，对数几率回归的目标函数(3.18)是非凸的，但其对数似然函数(3.27)是凸的。

对目标函数(3.18)求取二阶导数可发现exp（w'x+b）-1不能确定正负号，因此是非凸的，而对数似然函数3.27经由式（3.31）推导可见是恒大于0的，因此是凸函数。

3.3 编程实现对数几率回归，并给出西瓜数据集3.0a上的结果。

# -*- coding: utf-8 -*-
import numpy as np
# the exercise is from p69 > 3.3
# the training_dataset is p89, assign to data(type=matrix).
# divide the set into 3 column: density, sugar, label.
# the indices are data[:,0], data[:,1], data[:,2] respectively.
# in this example, the number of  attributes equals to 2.
data = [[0.697,0.460,1],
[0.774,0.376,1],
[0.634,0.264,1],
[0.608,0.318,1],
[0.556,0.215,1],
[0.403,0.237,1],
[0.481,0.149,1],
[0.437,0.211,1],
[0.666,0.091,0],
[0.243,0.267,0],
[0.245,0.057,0],
[0.343,0.099,0],
[0.639,0.161,0],
[0.657,0.198,0],
[0.360,0.370,0],
[0.593,0.042,0],
[0.719,0.103,0]]
beta=np.array([1,1,1]).reshape((-1,1))  # initial beta is a column vector of [w1,w2,b]'

data = np.matrix(data)
beta = np.matrix(beta)
density, sugar, label = data[:,0], data[:,1], data[:,2]
# in the label list, set 'good' to 1 and 'bad' to '0'
x = np.c_[density,sugar,np.ones(len(sugar))].T  # initial x is a column vector of [x1,x2,1]'

def cal_l(beta,x,label):
# solve the l'(beta) and l''(beta) of data to l1 and l2 respectively
l1, l2 = 0, np.mat(np.zeros((3,3)))
# l1,l2 = np.zeros((1,3)), np.zeros((3,3))
for i in range(x.shape[1]):
l1 += x[:,i] * (np.exp(beta.T*x[:,i])/(1+np.exp(beta.T*x[:,i])) - label[i])
l2 += x[:,i]*x.T[i,:] * (np.exp(beta.T*x[:,i])/((1+np.exp(beta.T*x[:,i]))**2))[0,0]
return [l1,l2]

dist = 1  # carry out the distance between new_beta and beta.
while dist >= 0.01:
new_beta = beta - cal_l(beta,x,label)[1].I * cal_l(beta,x,label)[0]
dist = np.linalg.norm(new_beta-beta)
beta = new_beta
c = []  # save the logit regression result
for i in range(17):
c.append(1/(1+np.exp(-beta.T*x[:,i]))[0,0])
print(new_beta)
print(c)

result:
[[  3.15832966]
 [ 12.52119579]
 [ -4.42886451]]

[0.97159134201182584, 0.93840796737854693, 0.7066382101828117, 0.81353420973519985, 

0.50480582132703811, 0.45300555631425837, 0.26036934432276743, 0.39970315015130975,

0.23397722179395924, 0.42110689644219934, 0.050146188402258575, 0.10851898058397864,

0.40256730484729258, 0.53129773794877577, 0.79265049892320416, 0.11608022112650698, 

0.29559934850614572]

结果发现，17个示例中，正例和反例各有3/2个划分错误，错误率为5/17