6

我一直在研究网络上的不同来源并尝试了各种方法，但只能找到如何计算唯一单词而不是唯一短语的频率。我到目前为止的代码如下：

import collections
import re
wanted = set(['inflation', 'gold', 'bank'])
cnt = collections.Counter()
words = re.findall('\w+', open('02.2003.BenBernanke.txt').read().lower())
for word in words:
    if word in wanted:
        cnt [word] += 1
print (cnt)

如果可能的话，我还想统计一下本文中使用“中央银行”和“高通胀”这两个词的次数。感谢您提供的任何建议或指导。

4

3 回答 3

2

首先，这就是我将如何生成cnt你所做的（以减少内存开销）

def findWords(filepath):
  with open(filepath) as infile:
    for line in infile:
      words = re.findall('\w+', line.lower())
      yield from words

cnt = collections.Counter(findWords('02.2003.BenBernanke.txt'))

现在，关于您关于短语的问题：

from itertools import tee
phrases = {'central bank', 'high inflation'}
fw1, fw2 = tee(findWords('02.2003.BenBernanke.txt'))   
next(fw2)
for w1,w2 in zip(fw1, fw2)):
  phrase = ' '.join([w1, w2])
  if phrase in phrases:
    cnt[phrase] += 1

希望这可以帮助

于 2013-11-12T04:05:57.977 回答
1

要计算一个小文件中几个短语的字面出现次数：

with open("input_text.txt") as file:
    text = file.read()
n = text.count("high inflation rate")

有一个nltk.collocations模块提供了识别经常连续出现的单词的工具：

import nltk
from nltk.tokenize import word_tokenize, sent_tokenize
from nltk.collocations import BigramCollocationFinder, TrigramCollocationFinder

# run nltk.download() if there are files missing
words = [word.casefold() for sentence in sent_tokenize(text)
         for word in word_tokenize(sentence)]
words_fd = nltk.FreqDist(words)
bigram_fd = nltk.FreqDist(nltk.bigrams(words))
finder = BigramCollocationFinder(word_fd, bigram_fd)
bigram_measures = nltk.collocations.BigramAssocMeasures()
print(finder.nbest(bigram_measures.pmi, 5))
print(finder.score_ngrams(bigram_measures.raw_freq))

# finder can be constructed from words directly
finder = TrigramCollocationFinder.from_words(words)
# filter words
finder.apply_word_filter(lambda w: w not in wanted)
# top n results
trigram_measures = nltk.collocations.TrigramAssocMeasures()
print(sorted(finder.nbest(trigram_measures.raw_freq, 2)))
于 2013-11-12T13:05:52.507 回答
0

假设文件不大 - 这是最简单的方法

for w1, w2 in zip(words, words[1:]):
    phrase = w1 + " " + w2
    if phrase in wanted:
        cnt[phrase] += 1
print(cnt)
于 2013-11-12T04:28:39.937 回答