python源码，朴素贝叶斯实现多分类
机器学习实战中，朴素贝叶斯那一章节只实现了二分类，网上大多数博客也只是照搬书上的源码，没有弄懂实现的根本。在此梳理了一遍朴素贝叶斯的原理，实现了5分类的例子，仅供参考！
from numpy import *
'''
贝叶斯公式 p(ci|w) = p(w|ci)*p(ci) / p(w)
即比较两类别分子大小，把结果归为分子大的一类
p(w|ci)条件概率，即在类别1或0下，w（词频）出现的概率（词频/此类别总词数即n/N）
'''
# 取得DataSet中不重复的word
def createVocabList(dataSet):
vocabSet = set([])#使用set创建不重复词表库
for document in dataSet:
vocabSet = vocabSet | set(document) #创建两个集合的并集
return list(vocabSet)
'''
我们将每个词的出现与否作为一个特征，这可以被描述为词集模型(set-of-words model)。
在词集中，每个词只能出现一次。
'''
def setOfWords2Vec(vocabList, inputSet):
returnVec = [0]*len(vocabList)#创建一个所包含元素都为0的向量
#遍历文档中的所有单词，如果出现了词汇表中的单词，则将输出的文档向量中的对应值设为1
for word in inputSet:
if word in vocabList:
returnVec[vocabList.index(word)] = 1
else: print("the word: %s is not in my Vocabulary!" % word)
return returnVec
'''
如果一个词在文档中出现不止一次，这可能意味着包含该词是否出现在文档中所不能表达的某种信息,
这种方法被称为词袋模型(bag-of-words model)。
在词袋中，每个单词可以出现多次。
为适应词袋模型，需要对函数setOfWords2Vec稍加修改，修改后的函数称为bagOfWords2VecMN
'''
def bagOfWords2Vec(vocabList, inputSet):
returnVec = [0]*len(vocabList)
for word in inputSet:
if word in vocabList:
returnVec[vocabList.index(word)] += 1
return returnVec
def countX(aList,el):
count = 0
for item in aList:
if item == el:
count += 1
return count
def trainNB0(trainMatrix,trainCategory):
'''
trainMatrix:文档矩阵
trainCategory:每篇文档类别标签
'''
numTrainDocs = len(trainMatrix)
numWords = len(trainMatrix[0])
pAbusive0 = countX(trainCategory,0) / float(numTrainDocs)
pAbusive1 = countX(trainCategory,1) / float(numTrainDocs)
pAbusive2 = countX(trainCategory,2) / float(numTrainDocs)
pAbusive3 = countX(trainCategory,3) / float(numTrainDocs)
pAbusive4 = countX(trainCategory,4) / float(numTrainDocs)
#初始化所有词出现数为1，并将分母初始化为2，避免某一个概率值为0
p0Num = ones(numWords); p1Num = ones(numWords)
p2Num = ones(numWords)
p3Num = ones(numWords)
p4Num = ones(numWords)
p0Denom = 2.0; p1Denom = 2.0 ;p2Denom = 2.0
p3Denom = 2.0; p4Denom = 2.0
for i in range(numTrainDocs):
# 1类的矩阵相加
if trainCategory[i] == 1:
p1Num += trainMatrix[i]
p1Denom += sum(trainMatrix[i])
if trainCategory[i] == 2:
p2Num += trainMatrix[i]
p2Denom += sum(trainMatrix[i])
if trainCategory[i] == 3:
p3Num += trainMatrix[i]
p3Denom += sum(trainMatrix[i])
if trainCategory[i] == 4:
p4Num += trainMatrix[i]
p4Denom += sum(trainMatrix[i])
if trainCategory[i] == 0:
p0Num += trainMatrix[i]
p0Denom += sum(trainMatrix[i])
#将结果取自然对数，避免下溢出，即太多很小的数相乘造成的影响
p4Vect = log(p4Num/p4Denom)
p3Vect = log(p3Num/p3Denom)
p2Vect = log(p2Num/p2Denom)
p1Vect = log(p1Num/p1Denom)#change to log()
p0Vect = log(p0Num/p0Denom)#change to log()
return p0Vect,p1Vect,p2Vect,p3Vect,p4Vect,pAbusive0,pAbusive1,pAbusive2,pAbusive3,pAbusive4
def classifyNB(vec2Classify,p0Vec,p1Vec,p2Vec,p3Vec,p4Vec,pClass0,pClass1,pClass2,pClass3,pClass4):
p1 = sum(vec2Classify * p1Vec) + log(pClass1)
p2 = sum(vec2Classify * p2Vec) + log(pClass2)
p3 = sum(vec2Classify * p3Vec) + log(pClass3)
p4 = sum(vec2Classify * p4Vec) + log(pClass4)
p0 = sum(vec2Classify * p0Vec) + log(pClass0)
## print(p0,p1,p2,p3,p4)
return [p0,p1,p2,p3,p4].index(max([p0,p1,p2,p3,p4]))
if __name__ == "__main__":
dataset = [['my','dog','has','flea','problems','help','please'],
['maybe','not','take','him','to','dog','park','stupid'],
['my','dalmation','is','so','cute','I','love','him'],
['stop','posting','stupid','worthless','garbage'],
['mr','licks','ate','my','steak','how','to','stop','him'],
['quit','buying','worthless','dog','food','stupid'],
['i','love','you'],
['you','kiss','me'],
['hate','heng','no'],
['can','i','hug','you'],
['refuse','me','ache'],
['1','4','3'],
['5','2','3'],
['1','2','3']]
# 0，1，2，3，4分别表示不同类别
classVec = [0,1,0,1,0,1,2,2,4,2,4,3,3,3]
print("正在创建词频列表")
myVocabList = createVocabList(dataset)
print("正在建词向量")
trainMat = []
for postinDoc in dataset:
trainMat.append(setOfWords2Vec(myVocabList,postinDoc))
print("开始训练")
p0V,p1V,p2V,p3V,p4V,pAb0,pAb1,pAb2,pAb3,pAb4 = trainNB0(array(trainMat),array(classVec))
# 输入的测试案例
tmp = ['love','you','kiss','you']
thisDoc = array(setOfWords2Vec(myVocabList,tmp))
flag = classifyNB(thisDoc,p0V,p1V,p2V,p3V,p4V,pAb0,pAb1,pAb2,pAb3,pAb4)
print('flag is',flag)如有帮助点个赞吧~