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我想使用 python 计算文件中所有二元组(相邻单词对)的出现次数。在这里,我正在处理非常大的文件,因此我正在寻找一种有效的方法。我尝试在文件内容上使用带有正则表达式 "\w+\s\w+" 的计数方法,但它并没有被证明是有效的。

例如,假设我想计算文件 a.txt 中的二元组数,该文件具有以下内容:

"the quick person did not realize his speed and the quick person bumped "

对于上述文件,二元组及其计数将是:

(the,quick) = 2
(quick,person) = 2
(person,did) = 1
(did, not) = 1
(not, realize) = 1
(realize,his) = 1
(his,speed) = 1
(speed,and) = 1
(and,the) = 1
(person, bumped) = 1

我在 Python 中遇到了一个 Counter 对象的示例,它用于计算 unigrams(单个单词)。它还使用正则表达式方法。

这个例子是这样的:

>>> # Find the ten most common words in Hamlet
>>> import re
>>> from collections import Counter
>>> words = re.findall('\w+', open('a.txt').read())
>>> print Counter(words)

上面代码的输出是:

[('the', 2), ('quick', 2), ('person', 2), ('did', 1), ('not', 1),
 ('realize', 1),  ('his', 1), ('speed', 1), ('bumped', 1)]

我想知道是否可以使用 Counter 对象来获取二元数。除了 Counter 对象或正则表达式之外的任何方法也将受到赞赏。

4

6 回答 6

53

一些itertools魔法:

>>> import re
>>> from itertools import islice, izip
>>> words = re.findall("\w+", 
   "the quick person did not realize his speed and the quick person bumped")
>>> print Counter(izip(words, islice(words, 1, None)))

输出:

Counter({('the', 'quick'): 2, ('quick', 'person'): 2, ('person', 'did'): 1, 
  ('did', 'not'): 1, ('not', 'realize'): 1, ('and', 'the'): 1, 
  ('speed', 'and'): 1, ('person', 'bumped'): 1, ('his', 'speed'): 1, 
  ('realize', 'his'): 1})

奖金

获取任何 n-gram 的频率:

from itertools import tee, islice

def ngrams(lst, n):
  tlst = lst
  while True:
    a, b = tee(tlst)
    l = tuple(islice(a, n))
    if len(l) == n:
      yield l
      next(b)
      tlst = b
    else:
      break

>>> Counter(ngrams(words, 3))

输出:

Counter({('the', 'quick', 'person'): 2, ('and', 'the', 'quick'): 1, 
  ('realize', 'his', 'speed'): 1, ('his', 'speed', 'and'): 1, 
  ('person', 'did', 'not'): 1, ('quick', 'person', 'did'): 1, 
  ('quick', 'person', 'bumped'): 1, ('did', 'not', 'realize'): 1, 
  ('speed', 'and', 'the'): 1, ('not', 'realize', 'his'): 1})

这也适用于惰性迭代和生成器。因此,您可以编写一个生成器,它逐行读取文件,生成单词,然后将其传递ngarms给惰性消耗,而无需读取内存中的整个文件。

于 2012-09-19T04:54:53.743 回答
14

怎么样zip()

import re
from collections import Counter
words = re.findall('\w+', open('a.txt').read())
print(Counter(zip(words,words[1:])))
于 2012-09-19T04:59:24.367 回答
5

您可以Counter像这样简单地使用任何 n_gram:

from collections import Counter
from nltk.util import ngrams 

text = "the quick person did not realize his speed and the quick person bumped "
n_gram = 2
Counter(ngrams(text.split(), n_gram))
>>>
Counter({('and', 'the'): 1,
         ('did', 'not'): 1,
         ('his', 'speed'): 1,
         ('not', 'realize'): 1,
         ('person', 'bumped'): 1,
         ('person', 'did'): 1,
         ('quick', 'person'): 2,
         ('realize', 'his'): 1,
         ('speed', 'and'): 1,
         ('the', 'quick'): 2})

对于 3 克,只需将 更改n_gram为 3:

n_gram = 3
Counter(ngrams(text.split(), n_gram))
>>>
Counter({('and', 'the', 'quick'): 1,
         ('did', 'not', 'realize'): 1,
         ('his', 'speed', 'and'): 1,
         ('not', 'realize', 'his'): 1,
         ('person', 'did', 'not'): 1,
         ('quick', 'person', 'bumped'): 1,
         ('quick', 'person', 'did'): 1,
         ('realize', 'his', 'speed'): 1,
         ('speed', 'and', 'the'): 1,
         ('the', 'quick', 'person'): 2})
于 2019-01-16T02:18:40.973 回答
3

从 开始Python 3.10,新pairwise函数提供了一种在连续元素对之间滑动的方法,这样您的用例就变成了:

from itertools import pairwise
import re
from collections import Counter

# text = "the quick person did not realize his speed and the quick person bumped "
Counter(pairwise(re.findall('\w+', text)))
# Counter({('the', 'quick'): 2, ('quick', 'person'): 2, ('person', 'did'): 1, ('did', 'not'): 1, ('not', 'realize'): 1, ('realize', 'his'): 1, ('his', 'speed'): 1, ('speed', 'and'): 1, ('and', 'the'): 1, ('person', 'bumped'): 1})

中间结果的详细信息:

re.findall('\w+', text)
# ['the', 'quick', 'person', 'did', 'not', 'realize', 'his', ...]
pairwise(re.findall('\w+', text))
# [('the', 'quick'), ('quick', 'person'), ('person', 'did'), ...]
于 2020-12-08T17:32:37.673 回答
1

自从提出这个问题并成功回答以来,已经有很长时间了。我受益于创建自己的解决方案的响应。我想分享它:

    import regex
    bigrams_tst = regex.findall(r"\b\w+\s\w+", open(myfile).read(), overlapped=True)

这将提供不被标点符号打断的所有二元组。

于 2014-11-28T00:12:53.267 回答
0

可以使用scikit-learn ( ) 中的CountVectorizer来生成二元组(或更一般地说,任何 ngram)。pip install sklearn

示例(使用 Python 3.6.7 和 scikit-learn 0.24.2 测试)。

import sklearn.feature_extraction.text

ngram_size = 2
train_set = ['the quick person did not realize his speed and the quick person bumped']

vectorizer = sklearn.feature_extraction.text.CountVectorizer(ngram_range=(ngram_size,ngram_size))
vectorizer.fit(train_set) # build ngram dictionary
ngram = vectorizer.transform(train_set) # get ngram
print('ngram: {0}\n'.format(ngram))
print('ngram.shape: {0}'.format(ngram.shape))
print('vectorizer.vocabulary_: {0}'.format(vectorizer.vocabulary_))

输出:

>>> print('ngram: {0}\n'.format(ngram)) # Shows the bi-gram count
ngram:   (0, 0) 1
  (0, 1)        1
  (0, 2)        1
  (0, 3)        1
  (0, 4)        1
  (0, 5)        1
  (0, 6)        2
  (0, 7)        1
  (0, 8)        1
  (0, 9)        2

>>> print('ngram.shape: {0}'.format(ngram.shape))
ngram.shape: (1, 10)
>>> print('vectorizer.vocabulary_: {0}'.format(vectorizer.vocabulary_))
vectorizer.vocabulary_: {'the quick': 9, 'quick person': 6, 'person did': 5, 'did not': 1, 
'not realize': 3, 'realize his': 7, 'his speed': 2, 'speed and': 8, 'and the': 0, 
'person bumped': 4}
于 2021-07-15T17:23:31.273 回答