一、代码实现
#!/usr/bin/python
# -*- coding utf-8 -*-
import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
import matplotlib as mpl
from sklearn import svm
from sklearn.model_selection import train_test_split
from sklearn.metrics import accuracy_score
def load_data():
path = 'E:\数据挖掘\Machine learning\[小象学院]机器学习课件\8.Regression代码\8.Regression\iris.data'
# 读取文件路径
data = pd.read_csv(path, header = None)
# 从data 读取数据, x为前4列的所有数据, y为第5列数据
x, y = data[range(4)], data[4]
# 返回字符类别的位置索引, 因y数组包含三类, 对应返回下标值
y = pd.Categorical(y).codes
# 取x的前两列数据, 一般SVM只做二特征分类, 多特征的转化为多个二特征分类再bagging?
x = x[[0, 1]]
# x = x[[0 ,2]]
return x, y
def classifier(x,y):
# 鸢尾花包含四个特征属性, 包含三类标签, 山鸢尾(0), 变色鸢尾(1), 维吉尼亚鸢尾(2)
iris_feature = u'花萼长度', u'花萼宽度', u'花瓣长度', u'花瓣宽度'
# 按 0.6 的比例，test_data 占40%， train_data 占60%, random_state随机数的种子, 1为产生相同随机数, 产生不同随机数
x_train, x_test, y_train, y_test = train_test_split(x, y, random_state=1, train_size=0.6)
# 使用SVM进行分类训练, 包含关键字, C, gamma, kernel
# kernel='linear'时，为线性核，C越大分类效果越好, kernel= 'rbf' 时（default）, 为高斯核
# gamma值越小，分类界面越连续；gamma值越大，分类界面越“散”，分类效果越好
# decision_function_shape = 'ovr' 时，为one vs rest, 即一个类别与其他类别进行划分，decision_function_shape = 'ovo'
# 为one vs one,即将类别两两之间进行划分，用二分类的方法模拟多分类的结果
clf = svm.SVC(C=0.8, kernel='rbf', gamma=20, decision_function_shape='ovr')
clf.fit(x_train, y_train.ravel())
# score函数返回返回该次预测的系数R2, 在（0, 1)之间、accuracy_score指的是分类准确率，即分类正确占所有分类的百分比
# recall_score 召回率 = 提取出的正确信息条数 / 样本中的信息条数
print(clf.score(x_train, y_train))
print('训练集准确率：', accuracy_score(y_train, clf.predict(x_train)))
print(clf.score(x_test, y_test))
print('测试集准确率：', accuracy_score(y_test, clf.predict(x_test)))
# decision_function()的功能: 计算样本点到分割超平面的函数距离, 每一列的值代表距离各类别的距离
print('decision_function:\n', clf.decision_function(x_train))
print('\npredict:\n', clf.predict(x_train))
# 画图
x1_min, x2_min = x.min() # 第0列的范围
x1_max, x2_max = x.max() # 第1列的范围
x1, x2 = np.mgrid[x1_min:x1_max:500j, x2_min:x2_max:500j] # 生成网格采样点
grid_test = np.stack((x1.flat, x2.flat), axis=1) # 测试点
# print 'grid_test = \n', grid_test
# Z = clf.decision_function(grid_test) # 样本到决策面的距离
# print Z
grid_hat = clf.predict(grid_test) # 预测分类值
grid_hat = grid_hat.reshape(x1.shape) # 使之与输入的形状相同
mpl.rcParams['font.sans-serif'] = [u'SimHei']
mpl.rcParams['axes.unicode_minus'] = False
cm_light = mpl.colors.ListedColormap(['#A0FFA0', '#FFA0A0', '#A0A0FF'])
cm_dark = mpl.colors.ListedColormap(['g', 'r', 'b'])
plt.figure(facecolor='w')
plt.pcolormesh(x1, x2, grid_hat, cmap=cm_light)
plt.scatter(x[0], x[1], c=y, edgecolors='k', s=50, cmap=cm_dark) # 样本
plt.scatter(x_test[0], x_test[1], s=120, facecolors='none', zorder=10) # 圈中测试集样本
plt.xlabel(iris_feature[0], fontsize=13)
plt.ylabel(iris_feature[1], fontsize=13)
plt.xlim(x1_min, x1_max)
plt.ylim(x2_min, x2_max)
plt.title(u'鸢尾花SVM二特征分类', fontsize=16)
plt.grid(b=True, ls=':')
plt.tight_layout(pad=1.5)
plt.show()
if __name__ == "__main__":
x, y = load_data()
classifier(x, y)
二、公式推导
\(在样本空间中任意点x到超平面(w,b)的距离可写为：\)
\[ r = \frac{|w^Tx + b|}{||w||} \]
\[推导如下：\\
取x_0为任意点x在超平面y= w^Tx + b的投影\\
wx_0 +b = 0 \Longrightarrow |w\vec {xx_0}| = |w\vec r|= ||w||r \\
另一方面:|w\vec{xx_0}| = |w(x_0 -x)|=|-b-wx|=|b+wx|\\
\therefore r = \frac{|w^Tx + b|}{||w||}\]
\[
\hat r=yf(x)=y(w^Tx + b)\\
\tilde r = ry = y\frac{|w^Tx + b|}{||w||}=\frac {\hat r}{||w||}\\
\\定义\hat r为函数间隔，\tilde r为几何间隔
\]
\[ L(\boldsymbol{w}, b, \boldsymbol{\alpha})=\frac{1}{2}\|\boldsymbol{w}\|^{2}+\sum_{i=1}^{m} \alpha_{i}\left(1-y_{i}\left(\boldsymbol{w}^{\top} \boldsymbol{x}_{i}+b\right)\right) \]
\[原问题为极小极大问题\min_{\boldsymbol{w,b}}\quad \max_{\boldsymbol{\alpha}}\quad L(w,b,\alpha)\\ 转化为极大极小问题\max_{\boldsymbol{\alpha}}\quad \min_{\boldsymbol{w,b}}\quad L(w,b,\alpha)\]
\[推导如下：\\ 目标函数：min\frac{1}{2}||w||^2\\ 约束条件：y_i(w^Tx_i + b) \geq 1\\ \therefore 对每个在y_i(w^Tx_i+b)-1的i乘以\alpha_i\\ \therefore L(\boldsymbol{w}, b, \boldsymbol{\alpha})=\frac{1}{2}\|\boldsymbol{w}\|^{2}+\sum_{i=1}^{m} \alpha_{i}\left(1-y_{i}\left(\boldsymbol{w}^{\top} \boldsymbol{x}_{i}+b\right)\right)\]
\[在其他的机器学习上述公式是L(\boldsymbol{w}, b, \boldsymbol{\alpha})=\frac{1}{2}\|\boldsymbol{w}\|^{2}-\sum_{i=1}^{m} \alpha_{i}\left(y_{i}\left(\boldsymbol{w}^{\top} \boldsymbol{x}_{i}+b\right)-1\right)，两者等价\]
\[
\begin{aligned}
w &= \sum_{i=1}^m\alpha_iy_i\boldsymbol{x}_i \\
0 &=\sum_{i=1}^m\alpha_iy_i
\end{aligned}
\]
\[推导如下：\\ \begin{aligned} L(\boldsymbol{w},b,\boldsymbol{\alpha}) &= \frac{1}{2}||\boldsymbol{w}||^2+\sum_{i=1}^m\alpha_i(1-y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)) \\ & = \frac{1}{2}||\boldsymbol{w}||^2+\sum_{i=1}^m(\alpha_i-\alpha_iy_i \boldsymbol{w}^T\boldsymbol{x}_i-\alpha_iy_ib)\\ & =\frac{1}{2}\boldsymbol{w}^T\boldsymbol{w}+\sum_{i=1}^m\alpha_i -\sum_{i=1}^m\alpha_iy_i\boldsymbol{w}^T\boldsymbol{x}_i-\sum_{i=1}^m\alpha_iy_ib \end{aligned}\]
\[\frac {\partial L}{\partial \boldsymbol{w}}=\frac{1}{2}\times2\times\boldsymbol{w} + 0 - \sum_{i=1}^{m}\alpha_iy_i \boldsymbol{x}_i-0= 0 \Longrightarrow \boldsymbol{w}=\sum_{i=1}^{m}\alpha_iy_i \boldsymbol{x}_i\]
\[\frac {\partial L}{\partial b}=0+0-0-\sum_{i=1}^{m}\alpha_iy_i=0 \Longrightarrow \sum_{i=1}^{m}\alpha_iy_i=0\]
\[
\begin{aligned}
\max_{\boldsymbol{\alpha}} & \sum_{i=1}^m\alpha_i - \frac{1}{2}\sum_{i = 1}^m\sum_{j=1}^m\alpha_i \alpha_j y_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j \\
s.t. & \sum_{i=1}^m \alpha_i y_i =0 \\
& \alpha_i \geq 0 \quad i=1,2,\dots ,m
\end{aligned}
\]
\(推导如下：\\计算拉格朗日函数，即将求得的两个公式代入\)
\[\begin{aligned} \min_{\boldsymbol{w},b} L(\boldsymbol{w},b,\boldsymbol{\alpha}) &=\frac{1}{2}\boldsymbol{w}^T\boldsymbol{w}+\sum_{i=1}^m\alpha_i -\sum_{i=1}^m\alpha_iy_i\boldsymbol{w}^T\boldsymbol{x}_i-\sum_{i=1}^m\alpha_iy_ib \\ &=\frac {1}{2}\boldsymbol{w}^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i-\boldsymbol{w}^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i+\sum _{i=1}^m\alpha_ i -b\sum _{i=1}^m\alpha_iy_i \\ & = -\frac {1}{2}\boldsymbol{w}^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i+\sum _{i=1}^m\alpha_i -b\sum _{i=1}^m\alpha_iy_i \end{aligned}\]
\[\begin{aligned}
\min_{\boldsymbol{w},b} L(\boldsymbol{w},b,\boldsymbol{\alpha}) &= -\frac {1}{2}\boldsymbol{w}^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i+\sum _{i=1}^m\alpha_i \\
&=-\frac {1}{2}(\sum_{i=1}^{m}\alpha_iy_i\boldsymbol{x}_i)^T(\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i)+\sum _{i=1}^m\alpha_i \\
&=-\frac {1}{2}\sum_{i=1}^{m}\alpha_iy_i\boldsymbol{x}_i^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i+\sum _{i=1}^m\alpha_i \\
&=\sum _{i=1}^m\alpha_i-\frac {1}{2}\sum_{i=1 }^{m}\sum_{j=1}^{m}\alpha_i\alpha_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j
\end{aligned}\]
\[
\begin{aligned}
& \min_{\boldsymbol{\alpha}}\frac{1}{2}\sum_{i = 1}^m\sum_{j=1}^m\alpha_i \alpha_j y_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j- \sum_{i=1}^m\alpha_i\\
& s.t. \sum_{i=1}^m \alpha_i y_i =0 \\
& \alpha_i \geq 0 \quad i=1,2,\dots ,m
\end{aligned}
\]
\[在原\max_{\boldsymbol{\alpha}}\quad \min_{\boldsymbol{w,b}}\quad L(w,b,\alpha)加负号，同样转化为约束最优化问题，为了求解最优解\alpha^*\]
\[
计算得到\\w^* = \sum_{i =1}^m{\alpha_i}^*y_ix_i\\
b^* = y_i -\sum_{i=1}^m{\alpha_i}^*y_ix_ix_j\\
分离得到超平面：\\
w^*x+ b^* =0\\
分类决策函数：\\
f(x) =sign(w^*x+b^*)
\]
\[引入松弛因子\xi_i的目标函数如下:\\\]
\[
\begin{aligned}
& \min_{\boldsymbol{w,b,\xi}}\frac{1}{2}||w||^2 +C\sum_{i = 1}^m\xi_i\\
s.t. & y_i(w.x_i+b)\geq1-\xi_i, i=1,2,\dots ,m \\
& \xi_i \geq 0 \quad i=1,2,\dots ,m
\end{aligned}
\]
\[同理如上式，构造拉格朗日函数L，再对w,b,\xi分别求偏导，再代入L\]
\[
\begin{aligned}
L(\boldsymbol{w},b,\boldsymbol{\alpha},\boldsymbol{\xi},\boldsymbol{\mu}) &= \frac{1}{2}||\boldsymbol{w}||^2+C\sum_{i=1}^m \xi_i+\sum_{i=1}^m \alpha_i(1-\xi_i-y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b))-\sum_{i=1}^m\mu_i \xi_i
\end{aligned}
\]
\[对w,b,\xi求偏导\]
\[
\begin{aligned}
w &= \sum_{i=1}^m\alpha_iy_i\boldsymbol{x}_i \\
0 &=\sum_{i=1}^m\alpha_iy_i\\
C & = a_i+\mu_i
\end{aligned}
\]
\[代入L\]
\[
\begin{aligned}
\min_{\boldsymbol{w},b,\boldsymbol{\xi}}L(\boldsymbol{w},b,\boldsymbol{\alpha},\boldsymbol{\xi},\boldsymbol{\mu}) &= \frac{1}{2}||\boldsymbol{w}||^2+C\sum_{i=1}^m \xi_i+\sum_{i=1}^m \alpha_i(1-\xi_i-y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b))-\sum_{i=1}^m\mu_i \xi_i \\
&=\frac{1}{2}||\boldsymbol{w}||^2+\sum_{i=1}^m\alpha_i(1-y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b))+C\sum_{i=1}^m \xi_i-\sum_{i=1}^m \alpha_i \xi_i-\sum_{i=1}^m\mu_i \xi_i \\
& = -\frac {1}{2}\sum_{i=1}^{m}\alpha_iy_i\boldsymbol{x}_i^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i+\sum _{i=1}^m\alpha_i +\sum_{i=1}^m C\xi_i-\sum_{i=1}^m \alpha_i \xi_i-\sum_{i=1}^m\mu_i \xi_i \\
& = -\frac {1}{2}\sum_{i=1}^{m}\alpha_iy_i\boldsymbol{x}_i^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i+\sum _{i=1}^m\alpha_i +\sum_{i=1}^m (C-\alpha_i-\mu_i)\xi_i \\
&=\sum _{i=1}^m\alpha_i-\frac {1}{2}\sum_{i=1 }^{m}\sum_{j=1}^{m}\alpha_i\alpha_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j
\end{aligned}
\]
\[再求\alpha极大\max\]
\[
\begin{aligned}
&\max_{\alpha}\sum _{i=1}^m\alpha_i-\frac {1}{2}\sum_{i=1 }^{m}\sum_{j=1}^{m}\alpha_i\alpha_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j\\
转化为\\
&\min_{\alpha}\frac {1}{2}\sum_{i=1 }^{m}\sum_{j=1}^{m}\alpha_i\alpha_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j-\sum _{i=1}^m\alpha_i\\
&s.t.\sum_{i=1}^m \alpha_i y_i=0 \\
&0 \leq\alpha_i \leq C \quad i=1,2,\dots ,m
\end{aligned}
\]
\[求最优解\alpha^*\]
\[
计算得到\\w^* = \sum_{i =1}^m{\alpha_i}^*y_ix_i\\
b^* = (\max_{i: y_i =1} w^*.x_i + \min_{i: y_i =-1} w^x* +x_i)/2\\
分离得到超平面：\\
w^*x+ b^* =0\\
分类决策函数：\\
f(x) =sign(w^*x+b^*)
\]
\[
\left\{\begin{array}{l}
{\alpha_{i}\left(f\left(\boldsymbol{x}_{i}\right)-y_{i}-\epsilon-\xi_{i}\right)=0} \\ {\hat{\alpha}_{i}\left(y_{i}-f\left(\boldsymbol{x}_{i}\right)-\epsilon-\hat{\xi}_{i}\right)=0} \\ {\alpha_{i} \hat{\alpha}_{i}=0, \xi_{i} \hat{\xi}_{i}=0} \\
{\left(C-\alpha_{i}\right) \xi_{i}=0,\left(C-\hat{\alpha}_{i}\right) \hat{\xi}_{i}=0}
\end{array}\right.
\]
\[推导如下：\\\]
\[
\left\{\begin{array}{l}2{f\left(\boldsymbol{x}_{i}\right)-y_{i}-\epsilon-\xi_{i} \leq 0 } \\ 3{y_{i}-f\left(\boldsymbol{x}_{i}\right)-\epsilon-\hat{\xi}_{i} \leq 0 } \\ 4{-\xi_{i} \leq 0} \\5{-\hat{\xi}_{i} \leq 0}6\end{array}\right.
\]
\[ \left\{\begin{array}{l} {\alpha_i\left(f\left(\boldsymbol{x}_{i}\right)-y_{i}-\epsilon-\xi_{i} \right) = 0 } \\ {\hat{\alpha}_i\left(y_{i}-f\left(\boldsymbol{x}_{i}\right)-\epsilon-\hat{\xi}_{i} \right) = 0 } \\ {-\mu_i\xi_{i} = 0 \Rightarrow \mu_i\xi_{i} = 0 } \\ {-\hat{\mu}_i \hat{\xi}_{i} = 0 \Rightarrow \hat{\mu}_i \hat{\xi}_{i} = 0 } \end{array}\right. \]
\[\because\begin{aligned}
\mu_i=C-\alpha_i \\
\hat{\mu}_i=C-\hat{\alpha}_i
\end{aligned}\]
\[
\left\{\begin{array}{l}
{\alpha_i\left(f\left(\boldsymbol{x}_{i}\right)-y_{i}-\epsilon-\xi_{i} \right) = 0 } \\
{\hat{\alpha}_i\left(y_{i}-f\left(\boldsymbol{x}_{i}\right)-\epsilon-\hat{\xi}_{i} \right) = 0 } \\
{(C-\alpha_i)\xi_{i} = 0 } \\
{(C-\hat{\alpha}_i) \hat{\xi}_{i} = 0 }
\end{array}\right.
\]
\[前面硬间隔与软间隔均处理线性问题，而对非线性问题需要将低维空间映射到高维空间，引入核函数\]
\[多项式核函数，高斯核函数，SMO算法，暂时还未完全理解，之后再补充\]
三、练习题目
3.1 给定三个数据点，正例点\(x_1 = (3, 3)^T\), \(x_2 = (4, 3)^T\),负例点\(x_3 = (1, 1)^T\)，求线性可分SVM
3.2 SVM可否用于多分类
3.3 SVM和Logistic回归的比较
3.4 核函数是什么？高斯核映射到无穷维是怎么回事？
3.5 如何理解SVM的损失函数？
3.6 使用高斯核函数，请描述SVM的参数C和 \(\sigma\) 对分类器的影响
3.7 比较感知机的对偶性形式与线性可分支持向量机的对偶形式
3-8 证明内积的正整数幂函数：
\[ K(x,z) = (x,z)^p\\ 是正定核函数，此处p为正整数，x,z为R \]
3.9 线性支持向量机还可定义为以下形式：
\[ \begin{aligned} \min_{\boldsymbol{w,b,\xi}}\quad \frac{1}{2}||w||^2+C\sum_{i=1}^N{\xi_i}^2\\ s.t.\quad y_i(\boldsymbol w.{x_i}+b)\geq 1-\xi_i, i= 1,2,...,N\\ {\xi}_i \geq 0, i=1, 2,...,N \end{aligned}\\ 求其对偶形式 \]
3.10 给定数据点，正例点\(x_1 = (1, 2)^T\), \(x_2 = (2, 3)^T\), \(x_3 = (1, 1)^T,\)负例点\(x_4 = (1, 1)^T\),\(x_5 = (1, 1)^T\),求最大间隔分离超平面和分类决策函数，并在图上画出分离超平面，间隔边界及支持向量
3.11 分析SVM对噪声敏感的原因
3.12 使用核技巧推广对数几率回归，产生核对率回归
3.13 给出式（6.52）的KKT条件
3.13 讨论线性判别分析与线性核支持向量机在何种条件等价
四、参看文献
[1] 《机器学习》 邹博
[2] 《SVM的三重境界》 July
[3] 《pumpkin-book》 Datawhale
[4] 《机器学习》周志华
[5] 《机器学习实战》Peter
[6] 《统计学习方法》李航，清华大学出版社，2012