How make regex that means "contains regex#1 but NOT regex#2" ??

[Reedick, Andrew]

> -----Original Message-----
> From: python-list-bounces+jr9445=att.com at python.org [mailto:python-
> list-bounces+jr9445=att.com at python.org] On Behalf Of Reedick, Andrew
> Sent: Tuesday, July 01, 2008 10:07 AM
> To: seberino at spawar.navy.mil; python-list at python.org
> Subject: RE: How make regex that means "contains regex#1 but NOT
> regex#2" ??
>  
> Match 'foo.*bar', except when 'not' appears between foo and bar.
> 
> 
> import re
> 
> s = 'fooAAABBBbar'
> print "Should match:", s
> m = re.match(r'(foo(.(?!not))*bar)', s);
> if m:
> 	print m.groups()
> 
> print
> 
> s = 'fooAAAnotBBBbar'
> print "Should not match:", s
> m = re.match(r'(foo(.(?!not))*bar)', s);
> if m:
> 	print m.groups()
> 
> 
> == Output ==
> Should match: fooAAABBBbar
> ('fooAAABBBbar', 'B')
> 
> Should not match: fooAAAnotBBBbar
> 


Fixed a bug with 'foonotbar'.  Conceptually it breaks down into:

	First_half_of_Regex#1(not
Regex#2)(any_char_Not_followed_by_Regex#2)*Second_half_of_Regex#1

However, if possible, I would make it a two pass regex.  Match on
Regex#1, throw away any matches that then match on Regex#2.  A two pass
is faster and easier to code and understand.  Easy to understand == less
chance of a bug.  If you're worried about performance, then a) a
complicated regex may or may not be faster than two simple regexes, and
b) if you're passing that much data through a regex, you're probably I/O
bound anyway.


import re

ss = ('foobar', 'fooAAABBBbar', 'fooAAAnotBBBbar', 'fooAAAnotbar',
'foonotBBBbar', 'foonotbar')

for s in ss:
	print s,
	m = re.match(r'(foo(?!not)(?:.(?!not))*bar)', s);
	if m:
		print m.groups()
	else:
		print


== output ==
foobar ('foobar',)
fooAAABBBbar ('fooAAABBBbar',)
fooAAAnotBBBbar
fooAAAnotbar
foonotBBBbar
foonotbar

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