使用朴素贝叶斯进行交叉验证时解析不出文档

原代码

def textParse(bigString):
    import re
    listOfTokens =re.split(r'\W*',bigString)
    return [tok.lower() for tok in listOfTokens if len(tok) > 2]
def spamTest():
    docList=[]; classList=[]; fullText=[]
    for i in range(1,25):
        #❶ （以下 七行） 导入 并解 析 文本 文件
        wordList = textParse(open('email/spam/%d.txt'%i).read())
        docList.append(wordList)
        fullText.extend(wordList)
        classList.append(1)
        wordList = textParse(open('email/spam/%d.txt'%i).read())
        docList.append(wordList)
        fullText.extend(wordList)
        classList.append(0)
    vocabList = createVocabList(docList)
    trainingSet = list(range(26)); testSet=[]
    #❷（ 以下 四行） 随机 构建 训练 集
    for i in range(10):
        randIndex = int(random.uniform(0,len(trainingSet)))
        testSet.append(trainingSet[randIndex])
        del(trainingSet[randIndex])
    trainMat=[]; trainClasses =[]
    for docIndex in trainingSet:
        trainMat.append(setOfWords2Vec(vocabList,docList[docIndex]))
        trainClasses.append(classList[docIndex])
    p0V,p1V,pSpam = trainNB0(array(trainMat),array(trainClasses))
    errorCount = 0
    #❸（ 以下 四行） 对 测试 集 分类
    for docIndex in testSet:
        wordVector = setOfWords2Vec(vocabList,docList[docIndex])
        if classifyNB(array(wordVector),p0V,p1V,pSpam)!= classList[docIndex]:
            errorCount += 1
    print('the error rate is:',float(errorCount)/len(testSet))
    return wordList,docList,classList,fullText,vocabList,testSet,trainingSet

终端输出，解析结果是空的

>>> wordList,docList,classList,fullText,vocabList,testSet,trainingSet=bayes.spamTest()
the error rate is: 0.7
>>> wordList
[]
>>> docList
[[], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], []]
>>> classList
[1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0]
>>> fullText
[]

https://www.jianshu.com/p/94ec02bcd663

兄弟，我也遇到你这样的问题了，在py3中，不过py2是没有问题的。所以py3中正则化匹配改为如下：listOfTokens=re.split(r'\W+',bigString)
可正常解析。详见https://blog.csdn.net/LaoYuanPython/article/details/100045507