Bugs item #1597014, was opened at 2006-11-15 14:27
Message generated for change (Comment added) made by ctimmerman
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Category: Regular Expressions
Group: Python 2.4
Status: Closed
Resolution: Invalid
Priority: 5
Private: No
Submitted By: Cees Timmerman (ctimmerman)
Assigned to: Gustavo Niemeyer (niemeyer)
Summary: Can't exclude words before capture group
Initial Comment:
Python 2.4.3 (#2, Oct 6 2006, 07:52:30)
[GCC 4.0.3 (Ubuntu 4.0.3-1ubuntu5)] on linux2
Tried:
>>> re.findall(r'(?!def)\b(\S+)\(', "def bla(): dof blu()")
>>> re.findall(r'(?:def){0}\b(\S+)\(', "def bla(): dof blu()")
Result:
['bla', 'blu']
Expected:
['blu']
Why doesn't (?!) work like it does here?:
>>> re.findall(r'\b(\S+): (?!bad)', "bob: bad; suzy: good")
['suzy']
Wouldn't it be nice if (^) worked?
>>> re.findall(r'\b(\S+): (^bad)', "bob: bad; suzy: good")
[]
[^()] does, sorta. Also not before a capture group:
>>> re.findall(r'\b(\S+): [^(bad)]', "bob: bad; suzy: good")
['suzy']
>>> re.findall(r'[^(def)]\b(\S+)\(', "def bla(): dof blu()")
['bla', 'blu']
>>> re.findall(r'[^(def)] (\S+)\(', "def bla(): dof blu()")
[]
>>> re.findall(r'(^def) (\S+)\(', "def bla(): dof blu()")
[('def', 'bla')]
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>Comment By: Cees Timmerman (ctimmerman)
Date: 2006-11-17 11:35
Message:
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Btw, thanks for the explanation, and I think you meant (?<[!=]
My final pattern:
>>> re.findall(r'(?<!def)[\W]+([.\w]+)\(', "def bla():
[ryawry.aet().blu()]")
['ryawry.aet', 'blu']
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Comment By: Cees Timmerman (ctimmerman)
Date: 2006-11-17 10:50
Message:
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Originator: YES
I tried (^ because [^ works. (^ doesn't seem to do anything. To match ^
inside () you need to use (\^), anyway.
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Comment By: Georg Brandl (gbrandl)
Date: 2006-11-15 17:20
Message:
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What you want is
>>> re.findall(r'(?<!def)\s(\S+)\(', "def bla(): dof blu()")
\b doesn't match a space, it matches an end of word. And to do a
lookbehind assertion, use (?<[!]...)
Why on earth do you expect that (^...) works?
[^(bad)] is something entirely different, it's a character class excluding
(, b, a, d and ).
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