Multiple negative lookbehind assertions in python regex?

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 (A-Z), it should still not match if it is a month name, like January | February | March.

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

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

I suggest replacing the space with \s+ (or \s if it really needs to be exactly one space). And matching an uppercase letter would be [A-Z] (you forgot the brackets). — hochl
– hochl, Commented Oct 2, 2012 at 11:13

hochl · Accepted Answer · 2012-10-02 11:43:09Z

24

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 [A-Z], but A-Z will not work (but remember there may be other uppercase letters than A-Z ...).

Additionally, I think I know why this does not work. The regular expression engine will try to match \. [A-Z] 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.

edited Oct 2, 2012 at 11:43

answered Oct 2, 2012 at 11:16

hochl

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2 Comments

jhiro009 Over a year ago

I'm new to multiple look-behinds. It looks like (?<!Abs)(?<!S) does the same as (?<!Abs|S). Is there any advantage to either (beyond personal preference on brevity/readability)?

Joel Wigton Over a year ago

@jhiro009 Yes, when you lump them together with the OR (pipe) operator, regex requires them to be a fixed-width pattern, so Abs and S are incompatible. You'd have to use the former case, in this situation.

Niko Fohr · Accepted Answer · 2020-04-14 09:51:41Z

10

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.

edited Apr 14, 2020 at 9:51

Niko Fohr

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answered Jul 12, 2019 at 8:13

Nathan Wailes

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Comments

root · Accepted Answer · 2012-10-02 11:13:47Z

1

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.

answered Oct 2, 2012 at 11:13

root

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Comments

jfs · Accepted Answer · 2012-10-02 11:41:00Z

1

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']

answered Oct 2, 2012 at 11:41

jfs

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Comments

qibin chen · Accepted Answer · 2020-06-11 08:43:48Z

-2

You can use Set [].

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

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

answered Jun 11, 2020 at 8:43

qibin chen

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