use a regex or not?

[Paul McGuire]

Here's a hybrid solution, using pyparsing to parse your input pattern
string (p), and transforming it into a regexp string.  Then uses
re.match using the regexp to process string (s), and then builds a
dictionary from the matched groups.

Download pyparsing at http://pyparsing.sourceforge.net.

-- Paul

===================
import re
from pyparsing import Word,OneOrMore

previousKeys = []
def createUpperKey(s,l,t):
    if t[0] not in previousKeys:
        ret = "(?P<%s>.*)" % t[0]
        previousKeys.append(t[0])
    else:
        ret = "(?P=%s)" % t[0]
    return ret

lowers = "abcdefghijklmnopqrstuvwxyz"
uppers = lowers.upper()
lowerLetter = Word(lowers,max=1)
upperLetter = Word(uppers,max=1).setParseAction(createUpperKey)
pattern = OneOrMore( lowerLetter | upperLetter )

def fill(s,p):
    # reset keys found in p-patterns
    del previousKeys[:]

    # create regexp by running pyparsing transform
    regexp = pattern.transformString(p) + "$"

    # create dict from re-matched groups
    match = re.match(regexp,s)
    if match:
        ret = dict( [ (k,match.group(k)) for k in previousKeys ] )
    else:
        ret = dict()
    return ret

tests = [
    ('ab','aA'),
    ('ab','Ab'),
    ('bb','Aa'),
    ('aa','Aa'),
    ('aa','Ab'),
    ('abb','aA'),
    ('aba','aAa'),
    ('abbba','aAa'),
    ('abbbaba','aAa'),
    ('abb','aAa'),
    ('abab','aAaA'),
    ('abac','aAaA'),
    ('abac','aAaB'),
    ]

for t in tests:
    print t,fill( *t )

================
Gives:
('ab', 'aA') {'A': 'b'}
('ab', 'Ab') {'A': 'a'}
('bb', 'Aa') {}
('aa', 'Aa') {'A': 'a'}
('aa', 'Ab') {}
('abb', 'aA') {'A': 'bb'}
('aba', 'aAa') {'A': 'b'}
('abbba', 'aAa') {'A': 'bbb'}
('abbbaba', 'aAa') {'A': 'bbbab'}
('abb', 'aAa') {}
('abab', 'aAaA') {'A': 'b'}
('abac', 'aAaA') {}
('abac', 'aAaB') {'A': 'b', 'B': 'c'}