Question

This question already has answers here:

Possible Duplicate:
item frequency count in python

速效项目

你们是如何发现一字在阵列中出现多少次?

我有一阵列,有大约5,000字案文,我想找到阵列中“帮助”一词的多少次。如何这样做?

阵列储存在X,因此,我的法典认为:

x = [...]
word = "help"

然后,我知道,用什么来使“帮助”的次数出现在x中。

感谢您的帮助!

Answer 1

>>> import collections
>>> print collections.Counter([ a ,  word ,  is ,  a ,  thing ,  that ,  is ,  countable ])
Counter({ a : 2,  is : 2,  word : 1,  that : 1,  countable : 1,  thing : 1})

http://docs.python.org/library/collections.html# Collections.Counter>Counter 。

Based on your edit, where each element in the list is a letter instead of the full word, then:

>>> import re
>>> letters = 
[ i ,  n ,  e ,  e ,  d ,  s ,  o ,  m ,  e ,  h ,  e ,  l ,  p ,  h ,  e ,  l ,  p ,  m ,  e ,  p ,  l ,  e ,  a ,  s ,  e ,  I ,  n ,  e ,  e ,  d ,  h ,  e ,  l ,  p ]
>>> len(re.findall( help , "".join(letters)))
3

Answer 2

As @sberry has depicted, Counter would server the purpose, but in case you are only searching a single word once and not interested to get the occurrence of all the words, you can use a simpler tool for the purpose

(我从树林中举出了例子)

既然有一言之词,你可以使用名单的<条码>(<>>><>>代码/代码>方法。

>>> list_of_words=[ a ,  word ,  is ,  a ,  thing ,  that ,  is ,  countable ]
>>> list_of_words.count( is )
2

你的评论表明,你可能有兴趣寻找一份特征清单。 ......

letters =
[ i ,  n ,  e ,  e ,  d ,  s ,  o ,  m ,  e ,  h ,  e ,  l ,  p ,  h ,  e ,  l ,  p ,  m ,  e ,  p ,  l ,  e ,  a ,  s ,  e ,  I ,  n ,  e ,  e ,  d ,  h ,  e ,  l ,  p ]

你们也可以利用通过压缩所有特性而形成的后遗体。

>>>   .join(letters).count( help )
3

如果这些词语被打碎,<代码>收集。反

>>> def count_words_in_jumbled(jumbled,word):
    jumbled_counter = collections.Counter(jumbled)
    word_counter = collections.Counter(word)
    return min(v /word_counter[k] for k,v in jumbled_counter.iteritems() if k in word)

>>> count_words_in_jumbled([ h , e , l , l , h , e , l , l , h , e , l ], hel )
3
>>> count_words_in_jumbled([ h , e , l , l , h , e , l , l , h , e , l ], hell )
2
>>> count_words_in_jumbled([ h , x , e , y , l , u , p ] , help )
1

Answer 3

nhelps = len(  .join(charlist).split( help )[1:]

友情链接