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[QUOTE="David C. Fox, post: 1734442"]
Dennis is correct: alphabetizing both the pattern and the target strings 
is way to do this.  I would use a slightly different regex, though, 
constructed as follows:

import re

def char_counts(s):
     """returns a dictionary indicating how many times a given character
     appears in the string s
     """
     d = {}
     for char in s:
         d[char] = d.get(char, 0) + 1 # current count, or zero, plus one
     return d

def char_subset(s):
     """given a string s, returns a regular expression which matches a
     sorted character string containing a subset of the same characters
     (and with no more occurences of each character than in the original
     string)
     """
     counts = char_counts(s)
     l = []
     chars = counts.keys()
     chars.sort()
     for char in chars:
         r = '%s{0,%d}' % (char, counts[char])
         l.append(r)
         # regex fragment matching count or fewer occurances os that many
     l.append('$') # make sure that we match against the full string
     return ''.join(l)

def sorted_string(s):
     """given a string s, return a new one which all the characters
     sorted
     """
     l = list(s)
     l.sort()
     return ''.join(l)

Then, you can compare a given target string, t, against the string in 
yipee with:

r = char_subset(yipee)
t_sorted = sorted_string(t)
if re.match(r, t_sorted):
     print 'found a match: %s' %t

David
[/QUOTE]

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