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I am writing a negative lookbehind assertion expression in Python which performs the following function to parse a plain text file:

Does not match anything followed after http://********** ; but will match the pattern when it is not inside a http://* link 

Example:
http://www.test.com/aa4   cd6
bx2 vq9 
yu9 http://www.bh9.com/cj3

Matches: cd6,bx2,vq9 and yu9

So I tried regexps like 

r'(?<!http://(.*))([a-z][a-z][0-9])'
r'(?<!http://*)([a-z][a-z][0-9])'

They did not work.

How to add .* or do similar opearion inside negative look behind assertion regex in Python.

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2 Answers 2

up vote 2 down vote accepted

Problem: Lookbehind does not allow pattern whose length is not fixed.

Quick hack: Perhaps the following regexp does the job?

r'(?<![./])[a-z][a-z][0-9]'

It works like this: 

>>> str = """http://www.test.com/aa4
... bx2 vq9 
... http://www.bh9.com/cj3
... """
>>> re.findall(r'(?<![./])[a-z][a-z][0-9]',str)
['bx2', 'vq9']

Or - as another solution - use a regexp matching urls to cut off all urls in your string and then search for r'[a-z][a-z][0-9]'

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How does [./] perform the functionality? –  thinkcool Jul 15 '11 at 8:12
    
@thinkcool: [./] matches a dot or a slash. I.e. the regexp presented in my answer matches all words [a-z][a-z][0-9] which do not follow a dot or a slash. –  phynfo Jul 15 '11 at 8:45
    
what happens when the pattern just follows a [.] or [./] It will not match .fd3 . Is there any foolproof solution? –  thinkcool Jul 15 '11 at 13:22
    
I am not able to understand your newly edited answer. Can you tell me how to proceed with this? –  thinkcool Jul 17 '11 at 21:42
up vote 1 down vote

That not possible. Python allows only fixed length lookbehinds. That means no quantifier inside the lookbehind.

See here the feature list on egular-expressions.info

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