词频计数奇怪结果

Question

当对我的语料库进行词频统计时，结果似乎不准确（不是我感觉最频繁的单词，频率计数只有一两个），有些结果显示“as over\xe2”和“\xad”。有人能帮忙吗？在

def toptenwords(mycorpus):
  mywords = mycorpus.words()
  nocapitals = [word.lower() for word in mywords] 
  filtered = [word for word in nocapitals if word not in stoplist]
  nopunctuation= [s.translate(None, string.punctuation) for s in filtered] 
  woordcounter = {}
     for word in nopunctuation:
        if word in wordcounter:
          woordcounter[word] += 1
        else:
          woordcounter[word] = 1
    frequentwords = sorted(wordcounter.iteritems(), key = itemgetter(1), reverse = True)
    top10 = frequentwords[:10]
    woord1 = frequentwords[1]
    woord2 = frequentwords[2]
    woord3 = frequentwords[3]
    woord4 = frequentwords[4]
    woord5 = frequentwords[5]
    woord6 = frequentwords[6]
    woord7 = frequentwords[7]
    woord8 = frequentwords[8]
    woord9 = frequentwords[9]
    woord10 = frequentwords[10]
    print "De 10 meest frequente woorden zijn: ", woord1, ",", woord2, ",", woord3, ",",    woord4, ",", woord5, ",", woord6, ",", woord7, ",", woord8, ",", woord9, "en", woord10

代码最初是荷兰语，这是未翻译的代码：

^{pr2}$

Answer 1

使用collections.Counter。它非常适合计算（散列）项的频率，并且它有一个most_common方法，可以返回前十个最频繁的项，而无需您自己编写逻辑：

import string
import collections


def topNwords(mywords, N = 10, stoplist = set(), filtered = set()):
    # mywords = mycorpus.words()
    nocapitals = [word.lower() for word in mywords]
    filtered = [word for word in nocapitals if word not in stoplist]
    nopunctuation = [s.translate(None, string.punctuation) for s in filtered]
    woordcounter = collections.Counter(nopunctuation)
    top_ten = [word for word, freq in woordcounter.most_common(N)]
    return top_ten


top_ten = topNwords('This is a test. It is only a test. In case of a real emergency'.split(), N = 10)
print("De 10 meest frequente woorden zijn: {w}".format(w = ', '.join(top_ten)))