如何在Python正则表达式后附加否定后置列表？

Question

I'm trying to split a paragraph into sentences using regex split and I'm trying to use the second answer posted here: a Regex for extracting sentence from a paragraph in python

But I have a list of abbreviations that I don't want to end the sentence on even though there's a period. But I don't know how to append it to that regular expression properly. I'm reading in the abbreviations from a file that contains terms like Mr. Ms. Dr. St. (one on each line).

Answer 1

Short answer: You can't, unless all lookbehind assertions are of the same, fixed width (which they probably aren't in your case; your example contained only two-letter abbreviations, but Mrs. would break your regex).

This is a limitation of the current Python regex engine.

Longer answer:

You could write a regex like (?s)(?<!.Mr|Mrs|.Ms|.St)\\. , padding each alternating part of the lookbehind assertion with as many . s as needed to get all of them to the same width. However, that would fail in some circumstances, for example when a paragraph begins with Mr. .

Anyway, you're not using the right tool here. Better use a tool designed for the job, for example the Natural Language Toolkit .

If you're stuck with regex (too bad!), then you could try and use a findall() approach instead of split() :

(?:(?:\b(?:Mr|Ms|Dr|Mrs|St)\.)|[^.])+\.\s*

would match a sentence that ends in . (optionally followed by whitespace) and may contain no dots unless preceded by one of the allowed abbreviations.

>>> import re
>>> s = "My name is Mr. T. I pity the fool who's not on the A-Team."
>>> re.findall(r"(?:(?:\b(?:Mr|Ms|Dr|Mrs|St)\.)|[^.])+\.\s*", s)
['My name is Mr. T. ', "I pity the fool who's not on the A-Team."]

Answer 2

I don't directly answer your question, but this post should contain enough information for you to write a working regex for your problem.

You can append a list of negative look-behinds. Remember that look-behinds are zero-width, which means that you can put as many look-behinds as you want next to each other, and you are still look-behind from the same position. As long as you don't need to use "many" quantifier (eg * , + , {n,} ) in the look-behind, everything should be fine (?).

So the regex can be constructured like this:

(?<!list )(?<!of )(?<!words )(?<!not )(?<!allowed )(?<!to )(?<!precede )pattern\w+

It is a bit too verbose. Anyway, I write this post just to demonstrate that it is possible to look-behind on a list of fixed string.

Example run:

>>> s = 'something patterning of patterned crap patternon not patterner, not allowed patternes to patternsses, patternet'
>>> re.findall(r'(?<!list )(?<!of )(?<!words )(?<!not )(?<!allowed )(?<!to )(?<!precede )pattern\w+', s)
['patterning', 'patternon', 'patternet']

There is a catch in using look-behind, though. If there are dynamic number of spaces between the blacklisted text and the text matching the pattern, the regex above will fail. I really doubt there exists a way to modify the regex so that it works for the case above while keeping the look-behinds . (You can always replace consecutive spaces into 1, but it won't work for more general cases).