1 Support Vector Machines

1.1 Example Dataset 1

%matplotlib inline
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sb
from scipy.io import loadmat
from sklearn import svm

大多数SVM的库会自动帮你添加额外的特征X₀已经θ₀，所以无需手动添加

mat = loadmat('./data/ex6data1.mat')
print(mat.keys())
# dict_keys(['__header__', '__version__', '__globals__', 'X', 'y'])
X = mat['X']
y = mat['y']

def plotData(X, y):
    plt.figure(figsize=(8,5))
    plt.scatter(X[:,0], X[:,1], c=y.flatten(), cmap='rainbow')
    plt.xlabel('X1')
    plt.ylabel('X2')
    plt.legend() 
plotData(X, y)

def plotBoundary(clf, X):
    '''plot decision bondary'''
    x_min, x_max = X[:,0].min()*1.2, X[:,0].max()*1.1
    y_min, y_max = X[:,1].min()*1.1,X[:,1].max()*1.1
    xx, yy = np.meshgrid(np.linspace(x_min, x_max, 500),
                         np.linspace(y_min, y_max, 500))
    Z = clf.predict(np.c_[xx.ravel(), yy.ravel()])
    Z = Z.reshape(xx.shape)
    plt.contour(xx, yy, Z)

models = [svm.SVC(C, kernel='linear') for C in [1, 100]]
clfs = [model.fit(X, y.ravel()) for model in models]

title = ['SVM Decision Boundary with C = {} (Example Dataset 1'.format(C) for C in [1, 100]]
for model,title in zip(clfs,title):
    plt.figure(figsize=(8,5))
    plotData(X, y)
    plotBoundary(model, X)
    plt.title(title)

可以从上图看到，当C比较小时模型对误分类的惩罚增大，比较严格，误分类少，间隔比较狭窄。

当C比较大时模型对误分类的惩罚增大，比较宽松，允许一定的误分类存在，间隔较大。

1.2 SVM with Gaussian Kernels

这部分，使用SVM做非线性分类。我们将使用高斯核函数。

为了用SVM找出一个非线性的决策边界，我们首先要实现高斯核函数。我可以把高斯核函数想象成一个相似度函数，用来测量一对样本的距离，(x ⁽ ʲ ⁾,y ⁽ ⁱ ⁾)

这里我们用sklearn自带的svm中的核函数即可。

1.2.1 Gaussian Kernel

def gaussKernel(x1, x2, sigma):
    return np.exp(- ((x1 - x2) ** 2).sum() / (2 * sigma ** 2))
gaussKernel(np.array([1, 2, 1]),np.array([0, 4, -1]), 2.)  # 0.32465246735834974

1.2.2 Example Dataset 2

mat = loadmat('./data/ex6data2.mat')
X2 = mat['X']
y2 = mat['y']

plotData(X2, y2)

sigma = 0.1
gamma = np.power(sigma,-2.)/2
clf = svm.SVC(C=1, kernel='rbf', gamma=gamma)
modle = clf.fit(X2, y2.flatten())
plotData(X2, y2)
plotBoundary(modle, X2)

1.2.3 Example Dataset 3

mat3 = loadmat('data/ex6data3.mat')
X3, y3 = mat3['X'], mat3['y']
Xval, yval = mat3['Xval'], mat3['yval']
plotData(X3, y3)

Cvalues = (0.01, 0.03, 0.1, 0.3, 1., 3., 10., 30.)
sigmavalues = Cvalues
best_pair, best_score = (0, 0), 0
for C in Cvalues:
    for sigma in sigmavalues:
        gamma = np.power(sigma,-2.)/2
        model = svm.SVC(C=C,kernel='rbf',gamma=gamma)
        model.fit(X3, y3.flatten())
        this_score = model.score(Xval, yval)
        if this_score > best_score:
            best_score = this_score
            best_pair = (C, sigma)
print('best_pair={}, best_score={}'.format(best_pair, best_score))
# best_pair=(1.0, 0.1), best_score=0.965

model = svm.SVC(C=1., kernel='rbf', gamma = np.power(.1, -2.)/2)
model.fit(X3, y3.flatten())
plotData(X3, y3)
plotBoundary(model, X3)

# 这我的一个练习画图的，和作业无关，给个画图的参考。
import numpy as np
import matplotlib.pyplot as plt
from sklearn import svm
# we create 40 separable points
np.random.seed(0)
X = np.array([[3,3],[4,3],[1,1]])
Y = np.array([1,1,-1])
# fit the model
clf = svm.SVC(kernel='linear')
clf.fit(X, Y)
# get the separating hyperplane
w = clf.coef_[0]
a = -w[0] / w[1]
xx = np.linspace(-5, 5)
yy = a * xx - (clf.intercept_[0]) / w[1]
# plot the parallels to the separating hyperplane that pass through the
# support vectors
b = clf.support_vectors_[0]
yy_down = a * xx + (b[1] - a * b[0])
b = clf.support_vectors_[-1]
yy_up = a * xx + (b[1] - a * b[0])
# plot the line, the points, and the nearest vectors to the plane
plt.figure(figsize=(8,5))
plt.plot(xx, yy, 'k-')
plt.plot(xx, yy_down, 'k--')
plt.plot(xx, yy_up, 'k--')
# 圈出支持向量
plt.scatter(clf.support_vectors_[:, 0], clf.support_vectors_[:, 1],
            s=150, facecolors='none', edgecolors='k', linewidths=1.5)
plt.scatter(X[:, 0], X[:, 1], c=Y, cmap=plt.cm.rainbow)
plt.axis('tight')
plt.show()
print(clf.decision_function(X))

[ 1. 1.5 -1. ]

2 Spam Classification

2.1 Preprocessing Emails

这部分用SVM建立一个垃圾邮件分类器。你需要将每个email变成一个n维的特征向量，这个分类器将判断给定一个邮件x是垃圾邮件(y=1)或不是垃圾邮件(y=0)。

take a look at examples from the dataset

with open('data/emailSample1.txt', 'r') as f:
    email = f.read()
    print(email)

> Anyone knows how much it costs to host a web portal ?
>
Well, it depends on how many visitors you're expecting.
This can be anywhere from less than 10 bucks a month to a couple of $100. 
You should checkout http://www.rackspace.com/ or perhaps Amazon EC2 
if youre running something big..
To unsubscribe yourself from this mailing list, send an email to:
groupname-unsubscribe@egroups.com

可以看到，邮件内容包含 a URL, an email address(at the end), numbers, and dollar amounts. 很多邮件都会包含这些元素，但是每封邮件的具体内容可能会不一样。因此，处理邮件经常采用的方法是标准化这些数据，把所有URL当作一样，所有数字看作一样。

例如，我们用唯一的一个字符串‘httpaddr'来替换所有的URL，来表示邮件包含URL，而不要求具体的URL内容。这通常会提高垃圾邮件分类器的性能，因为垃圾邮件发送者通常会随机化URL，因此在新的垃圾邮件中再次看到任何特定URL的几率非常小。

我们可以做如下处理：

  1. Lower-casing: 把整封邮件转化为小写。
  2. Stripping HTML: 移除所有HTML标签，只保留内容。
  3. Normalizing URLs: 将所有的URL替换为字符串 “httpaddr”.
  4. Normalizing Email Addresses: 所有的地址替换为 “emailaddr”
  5. Normalizing Dollars: 所有dollar符号($)替换为“dollar”.
  6. Normalizing Numbers: 所有数字替换为“number”
  7. Word Stemming(词干提取): 将所有单词还原为词源。例如，“discount”, “discounts”, “discounted” and “discounting”都替换为“discount”。
  8. Removal of non-words: 移除所有非文字类型，所有的空格(tabs, newlines, spaces)调整为一个空格.

%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
from scipy.io import loadmat
from sklearn import svm
import re #regular expression for e-mail processing
# 这是一个可用的英文分词算法(Porter stemmer)
from stemming.porter2 import stem
# 这个英文算法似乎更符合作业里面所用的代码，与上面效果差不多
import nltk, nltk.stem.porter

def processEmail(email):
    """做除了Word Stemming和Removal of non-words的所有处理"""
    email = email.lower()
    email = re.sub('<[^<>]>', ' ', email)  # 匹配<开头，然后所有不是< ,> 的内容，知道>结尾，相当于匹配<...>
    email = re.sub('(http|https)://[^\s]*', 'httpaddr', email )  # 匹配//后面不是空白字符的内容，遇到空白字符则停止
    email = re.sub('[^\s]+@[^\s]+', 'emailaddr', email)
    email = re.sub('[\$]+', 'dollar', email)
    email = re.sub('[\d]+', 'number', email) 
    return email

接下来就是提取词干，以及去除非字符内容。

def email2TokenList(email):
    """预处理数据，返回一个干净的单词列表"""
    # I'll use the NLTK stemmer because it more accurately duplicates the
    # performance of the OCTAVE implementation in the assignment
    stemmer = nltk.stem.porter.PorterStemmer()
    email = preProcess(email)
    # 将邮件分割为单个单词，re.split() 可以设置多种分隔符
    tokens = re.split('[ \@\$\/\#\.\-\:\&\*\+\=\[\]\?\!\(\)\{\}\,\'\"\>\_\<\;\%]', email)
    # 遍历每个分割出来的内容
    tokenlist = []
    for token in tokens:
        # 删除任何非字母数字的字符
        token = re.sub('[^a-zA-Z0-9]', '', token);
        # Use the Porter stemmer to 提取词根
        stemmed = stemmer.stem(token)
        # 去除空字符串‘'，里面不含任何字符
        if not len(token): continue
        tokenlist.append(stemmed)
    return tokenlist  

2.1.1 Vocabulary List(词汇表)

在对邮件进行预处理之后，我们有一个处理后的单词列表。下一步是选择我们想在分类器中使用哪些词，我们需要去除哪些词。

我们有一个词汇表vocab.txt，里面存储了在实际中经常使用的单词，共1899个。

我们要算出处理后的email中含有多少vocab.txt中的单词，并返回在vocab.txt中的index，这就我们想要的训练单词的索引。

def email2VocabIndices(email, vocab):
    """提取存在单词的索引"""
    token = email2TokenList(email)
    index = [i for i in range(len(vocab)) if vocab[i] in token ]
    return index

2.2 Extracting Features from Emails

def email2FeatureVector(email):
    """
    将email转化为词向量，n是vocab的长度。存在单词的相应位置的值置为1，其余为0
    """
    df = pd.read_table('data/vocab.txt',names=['words'])
    vocab = df.as_matrix()  # return array
    vector = np.zeros(len(vocab))  # init vector
    vocab_indices = email2VocabIndices(email, vocab)  # 返回含有单词的索引
    # 将有单词的索引置为1
    for i in vocab_indices:
        vector[i] = 1
    return vector

vector = email2FeatureVector(email)
print('length of vector = {}\nnum of non-zero = {}'.format(len(vector), int(vector.sum())))

length of vector = 1899
num of non-zero = 45

2.3 Training SVM for Spam Classification

读取已经训提取好的特征向量以及相应的标签。分训练集和测试集。

# Training set
mat1 = loadmat('data/spamTrain.mat')
X, y = mat1['X'], mat1['y']
# Test set
mat2 = scipy.io.loadmat('data/spamTest.mat')
Xtest, ytest = mat2['Xtest'], mat2['ytest']

clf = svm.SVC(C=0.1, kernel='linear')
clf.fit(X, y)

2.4 Top Predictors for Spam

predTrain = clf.score(X, y)
predTest = clf.score(Xtest, ytest)
predTrain, predTest

(0.99825, 0.989)

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