python正则表达式中的多个负面回顾断言？

Question

I'm new to programming, sorry if this seems trivial: I have a text that I'm trying to split into individual sentences using regular expressions. With the .split method I search for a dot followed by a capital letter like

"\. A-Z"

However I need to refine this rule in the following way: The . (dot) may not be preceeded by either Abs or S . And if it is followed by a capital letter ( AZ ), it should still not match if it is a month name, like January | February | March January | February | March January | February | March .

I tried implementing the first half, but even this did not work. My code was:

"( (?<!Abs)\. A-Z) | (?<!S)\. A-Z) ) "

Answer 1

First, I think you may want to replace the space with \\s+ , or \\s if it really is exactly one space (you often find double spaces in English text).

Second, to match an uppercase letter you have to use [AZ] , but AZ will not work (but remember there may be other uppercase letters than AZ ...).

Additionally, I think I know why this does not work. The regular expression engine will try to match \\. [AZ] \\. [AZ] if it is not preceeded by Abs or S . The thing is that, if it is preceeded by an S , it is not preceeded by Abs , so the first pattern matches. If it is preceeded by Abs , it is not preceeded by S , so the second pattern version matches. In either way one of those patterns will match since Abs and S are mutually exclusive.

The pattern for the first part of your question could be

(?<!Abs)(?<!S)(\. [A-Z])

or

(?<!Abs)(?<!S)(\.\s+[A-Z])

(with my suggestion)

That is because you have to avoid | , without it the expression now says not preceeded by Abs and not preceeded by S . If both are true the pattern matcher will continue to scan the string and find your match.

To exclude the month names I came up with this regular expression:

(?<!Abs)(?<!S)(\.\s+)(?!January|February|March)[A-Z]

The same arguments hold for the negative look ahead patterns.

Answer 2

I'm adding a short answer to the question in the title, since this is at the top of Google's search results:

The way to have multiple differently-lengthed negative lookbehinds is to chain them together like this:

"(?<!1)(?<!12)(?<!123)example"

This would match example 2example and 3example but not 1example 12example or 123example .

Answer 3

Use nltk punkt tokenizer . It's probably more robust than using regex.

>>> import nltk.data
>>> text = """
... Punkt knows that the periods in Mr. Smith and Johann S. Bach
... do not mark sentence boundaries.  And sometimes sentences
... can start with non-capitalized words.  i is a good variable
... name.
... """
>>> sent_detector = nltk.data.load('tokenizers/punkt/english.pickle')
>>> print '\n-----\n'.join(sent_detector.tokenize(text.strip()))
Punkt knows that the periods in Mr. Smith and Johann S. Bach
do not mark sentence boundaries.
-----
And sometimes sentences
can start with non-capitalized words.
-----
i is a good variable
name.

Answer 4

Use nltk or similar tools as suggested by @root.

To answer your regex question:

import re
import sys

print re.split(r"(?<!Abs)(?<!S)\.\s+(?!January|February|March)(?=[A-Z])",
               sys.stdin.read())

Input

First. Second. January. Third. Abs. Forth. S. Fifth.
S. Sixth. ABs. Eighth

Output

['First', 'Second. January', 'Third', 'Abs. Forth', 'S. Fifth',
 'S. Sixth', 'ABs', 'Eighth']

Answer 5

You can use Set [].

'(?<![1,2,3]example)'

This would not match 1example, 2example, 3example.