Question

我想用python计算文件中所有bigrams（一对相邻单词）的出现次数。在这里，我正在处理非常大的文件，所以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count方法，但它没有被证明是有效的。

e.g。假设我想从文件a.txt中计算出双字母组的数量，该文件包含以下内容：

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

对于上述文件，bigram集及其计数将为：

(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对象的例子，它用于计算unigrams（单个单词）。它还使用正则表达式方法。

示例如下：

>>> # 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)

上述代码的输出为：

[('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对象获取bigrams的数量。除了Counter对象或正则表达式之外的任何方法也将受到赞赏。

Answer 1

一些itertools魔术：

>>> 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})

<强>加成

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})

这也适用于懒惰的迭代和生成器。因此，您可以编写一个生成器，逐行读取文件，生成单词，并将其传递给ngarms以便懒惰地使用，而无需读取内存中的整个文件。

Answer 2

zip()怎么样？

import re
from collections import Counter
words = re.findall('\w+', open('a.txt').read())
print(Counter(zip(words,words[1:])))

Answer 3

随着即将到来的findMaxSum(Node) ^{Java passes parameters by value, but every object variable in Java implicitly references the actual object}，新的release schedule函数提供了一种在成对的连续元素之间滑动的方法，从而使您的用例简单变成：

Python 3.10

中间结果的详细信息：

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})

Answer 4

自问这个问题并成功回复以来已经有很长时间了。我从创建自己的解决方案的响应中受益。我想分享一下：

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

这将提供所有不会被标点符号打断的双字母组。

Answer 5

对于任何n_gram，您都可以简单地使用Counter，例如：

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n_gram更改为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})

Answer 6

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

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}

使用python计算文件中的双字节（一对两个单词）

6 个答案: