You seem to be suggesting a different approach, one I hadn't thought of: explicitly test series of pairs, rather than the whole remaining string at each point, and do this just once starting at 0, and once starting at 1. That sounds as though it would work, though the regex would have to be called in a different way so as to seek non-overlapping patterns (rather than the elaborate precautions I've taken to seek overlapping ones) -- I'm not yet sure quite how, and I'm not yet clear that it's any more efficient and/or elegant than what I've got now. Hm -- lots to think about here. Thank you.
Charles Hartman Professor of English, Poet in Residence http://cherry.conncoll.edu/cohar http://villex.blogspot.com
pat = sre.compile('(x[x/])+')
(longest, startlongest) = max([(fnd.end()-fnd.start(), fnd.start()) for
i in range(len(marks))
for fnd in pat.finditer(marks,i)])
If I'm understanding that correctly, the only way for you to get different
best matches are at offsets 0 and 1; offset 2 will yield the same matches
as 0, with the possibility of excluding the first two characters -- i. e.
any different matches should be guaranteed to be shorter. Therefore
... for i in range(2) ...
instead of
... for i in range(len(marks)) ...
should be sufficient.
Peter
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