Re: a RegEx puzzle

[Charles Hartman]

Thanks -- not only for the code, which does almost exactly what I need 
to do, but for the reminder (thanks also to Jeremy Bowers for this!) to 
prefer simple solutions. I was, of course, so tied up in getting my 
nifty one-liner right that I totally lost sight of how 
straightforwardly the job could be done; and now that I've got it, I've 
also got room to tune it. For instance,  your code keeps the first 
"longest" match if several are equal in length; my program will I think 
do slightly better if I keep the last "longest" instead, and changing 
that required changing > into >=, which even I can't screw up.

Thanks to everyone who's helped on this. Makes me wish I were going to 
pycon.

Charles Hartman
Professor of English, Poet in Residence
http://cherry.conncoll.edu/cohar
http://villex.blogspot.com
Kent Johnson wrote:
It's pretty simple to put re.search() into a loop where subsequent 
searches start from the character after where the previous one 
matched. Here is a solution that uses a general-purpose longest match 
function:

import re
# RE solution
def longestMatch(rx, s):
    ''' Find the longest match for rx in s.
        Returns (start, length) for the match or (None, None) if no 
match found.
    '''

    start = length = current = 0
    while True:
        m = rx.search(s, current)
        if not m:
            break
        mStart, mEnd = m.span()
        current = mStart + 1
        if (mEnd - mStart) > length:
            start = mStart
            length = mEnd - mStart
    if length:
        return start, length
    return None, None
pairsRe = re.compile(r'(x[x/])+')
for s in [ '/xx/xxx///', '//////xx//' ]:
    print s, longestMatch(pairsRe, s)
--
http://mail.python.org/mailman/listinfo/python-list