打印包含和排除停用词的10个最常出现的文字

use*_*809 10 python nltk word-frequency find-occurrences

我从这里得到了我的改变的问题.我有以下代码:

from nltk.corpus import stopwords
>>> def content_text(text):
    stopwords = nltk.corpus.stopwords.words('english')
    content = [w for w in text if w.lower() in stopwords]
    return content
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如何打印 1)最常出现的文字1)包括和2)不包括停用词?

igo*_*shi 15

nltk中有一个FreqDist函数

import nltk
allWords = nltk.tokenize.word_tokenize(text)
allWordDist = nltk.FreqDist(w.lower() for w in allWords)

stopwords = nltk.corpus.stopwords.words('english')
allWordExceptStopDist = nltk.FreqDist(w.lower() for w in allWords if w not in stopwords)    
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提取10个最常见的:

mostCommon= allWordDist.most_common(10).keys()
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  • 您应该询问带有小写字符串的停用词。来自:`allWordExceptStopDist = nltk.FreqDist(w.lower() for w in allWords if w not in stopwords) ` To: `allWordExceptStopDist = nltk.FreqDist(w.lower() for w in allWords if w.lower() not在停用词中)` (2认同)

Pad*_*ham 5

不确定在is stopwords函数中,我想它需要是in 但你可以使用 Counterdict withmost_common(10)来获得 10 个最常见的:

from collections import Counter
from string import punctuation


def content_text(text):
    stopwords = set(nltk.corpus.stopwords.words('english')) # 0(1) lookups
    with_stp = Counter()
    without_stp  = Counter()
    with open(text) as f:
        for line in f:
            spl = line.split()
            # update count off all words in the line that are in stopwrods
            with_stp.update(w.lower().rstrip(punctuation) for w in spl if w.lower() in stopwords)
               # update count off all words in the line that are not in stopwords
            without_stp.update(w.lower().rstrip(punctuation)  for w in spl if w  not in stopwords)
    # return a list with top ten most common words from each 
    return [x for x in with_stp.most_common(10)],[y for y in without_stp.most_common(10)]
wth_stop, wthout_stop = content_text(...)
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如果您传入一个 nltk 文件对象,只需对其进行迭代:

def content_text(text):
    stopwords = set(nltk.corpus.stopwords.words('english'))
    with_stp = Counter()
    without_stp  = Counter()
    for word in text:
        # update count off all words in the line that are in stopwords
        word = word.lower()
        if word in stopwords:
             with_stp.update([word])
        else:
           # update count off all words in the line that are not in stopwords
            without_stp.update([word])
    # return a list with top ten most common words from each
    return [k for k,_ in with_stp.most_common(10)],[y for y,_ in without_stp.most_common(10)]

print(content_text(nltk.corpus.inaugural.words('2009-Obama.txt')))
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nltk 方法包括标点符号,因此可能不是您想要的。