Python正则表达式拆分但将正则表达式匹配的结尾部分放回字符串中？

Question

I'd like to find a regex expression that can break up paragraphs (long strings, no newline characters to worry about) into sentences with the simple rule that an of {., ?, !} followed by a whitespace and then a capital letter should be the end of the sentence (I realize this is not a good rule for real life).

I've got something partly working, but it doesn't quite do the job:

line = 'a b c FFF! D a b a a FFF. gegtat FFF. A'
matchObj = re.split(r'(.*?\sFFF[\.|\?|\!])\s[A-Z]', line)
print (matchObj)

prints

['', 'a b c FFF!', '', ' a b a a FFF. gegtat FFF.', '']

whereas I'd like to get:

['a b c FFF!', 'D a b a a FFF. gegtat FFF.']

So two questions.

• Why are there empty members ( '' ) in the results?

• I understand why the D gets cut out from the split result - it's part of the first search. How can I structure my search differently so that the capital letter coming after the punctuation is put back so it can be included with the next sentence? In this case, how can I get D to turn up in the second element of the split result?

I know I could accomplish this with some sort of for-loop just peeling off the first result, adding back the capital letter and then doing it all over again, but this seems not-so-Pythonic. If regex is not the way to go here, is there something that still avoids the for loop?

Thanks for any suggestions.

Answer 1

1. To solve the first problem (empty strings in the result returned by split() ), use either findall() or finditer() :

    >>> re.findall(r'(.*?\\sFFF[\\.|\\?|\\!])\\s[AZ]', line) ['abc FFF!', ' abaa FFF. gegtat FFF.'] 

   You were seeing empty strings in the output because that's what split() is supposed to do: split the input string using the matched groups as delimiters.

2. For the second problem (the missing D from the output), use a lookahead assertion (?=...) :

    >>> re.findall(r'(.*?\\sFFF[\\.|\\?|\\!])\\s(?=[AZ])', line) ['abc FFF!', 'D abaa FFF. gegtat FFF.'] 

   Lookaheads, negative lookaheads, lookbehinds and negative lookbehinds are four kinds of assertions that you can use to say "match this group only if followed/preceded by group, but don't consume the string".

3. Reading carefully your expression, it seems you have misunderstood the syntax of the [...] operator. It seems you want to match one of . , ? and ! .

   If that is the case, then you can rewrite [\\.|\\?|\\!] as [.?!] :

    >>> re.findall(r'(.*?\\sFFF[.?!])\\s(?=[AZ])', line) ['abc FFF!', 'D abaa FFF. gegtat FFF.'] 

   [.?!] is not only more compact, but is also more correct: with [\\.|\\?|\\!] you were also matching the | character (so that 'abc FFF|' were a valid match)!