Question

如果一封信重复的话,是否有办法删除该信?

Answer 1

如果我正确理解你的问题,你可以定期表示:

import re
re.sub(r (.)1+ , r 1 ,  aardvarrk )

这倒塌了所有性质相同的顺序,请见 ardvark 。

As for the implementation of your spell checker, I suggest "collapsing" all words that have repeating characters in sequence in your dictionary and keeping that in a dictionary (data structure), where the key is the collapsed word and the value is the original word (or possibly a set of original words):

{
  aple :  apple ,
  acord :  accord 
  halo : set([ hallo ,  halo ])
}

现在,在你分析你们的意见时,每一字:

您的行文正确无误。如果是,它就忽视了它。 (例如:投入为<代码>人。) 它是在文字清单上。在这里做不到。
如果有的话,“栏目”是指:
1. It exists in your word list. If it does, replace it. (eg.: computerr becomes computer . Now you just replace it with the original word in your list).
2. A key exists in your dictionary. If it does, replace it with the word associated with that key. (eg: aaapppleee become aple . Now you look up aple in your word list. It s not there. Now look in your dictionary for the key aple . If it is there. Replace it with its value, apple .)

我所看到的唯一问题是,可能把“col脚”二字改为“口号”。这意味着,你必须使用<条码><>>。

www.un.org/spanish/ga/president 现在,你不得不决定由谁取代。可以通过计算。

Answer 2

http://docs.python.org/library/difflib.html 标准图书馆:

import difflib

words = open( /usr/share/dict/words ).read().split()

difflib.get_close_matches( aaaappplllee , words, 3, 0.5)
[ appalled ,  apple ,  appellate ]

difflib.get_close_matches( aaardvarrk , words, 3, 0.5)
[ aardvark ,  aardvarks , "aardvark s"]

Answer 3

Here is a solution that will allow you to iterate over all versions of the string with different combinations of repeated letters:

from itertools import product, groupby

# groups == [ aaaa ,  ppp ,  lll ,  ee ]
groups = [  .join(g) for c, g in groupby( aaaappplllee )]

# lengths is an iterator that will return all combinations of string lengths to  
# use for each group, starting with [4, 3, 3, 2] and ending with [1, 1, 1, 1]
lengths = product(*[range(x, 0, -1) for x in map(len, groups)])

# Using the lengths from the previous line, this is a generator that yields all
# combinations of the original string with duplicate letters removed
words = (  .join(groups[i][:v] for i, v in enumerate(x)) for x in lengths)

>>> for word in words:
...   print word
... 
aaaappplllee
aaaapppllle
aaaapppllee
aaaappplle
aaaappplee
aaaappple
...
apple
aplllee
apllle
apllee
aplle
aplee
aple

这不是找到正确措辞的最有效办法,但它符合OP的原始方法。

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