Quoting Bruno Desthuilliers <[EMAIL PROTECTED]>: > [EMAIL PROTECTED] a écrit : > > Hi, > > I want to generate all non-empty substrings of a string of length > >=2. > > Also, > > each substring is to be paired with 'string - substring' part and vice > > versa. > > Thus, ['abc'] gives me [['a', 'bc'], ['bc', 'a'], ['ab', 'c'], ['c', > > 'ab'], ['b', 'ac'], ['ac', 'b']] etc. > > Similarly, 'abcd' should give me [['a', 'bcd'], ['bcd', 'a'], ['abc', > > 'd'], ['d', 'abc'], ['b', 'acd'], ['acd', 'b'],['c', 'abd'], ['abd', > 'c'], > > ['ab', 'cd'], ['cd', 'ab'], ['bc', 'ad'], ['ad', 'bc'], ['ac', > > 'bd'],['bd','ac']] > > > > I've tried the following but i cant prevent duplicates and i'm > missing > > some substrings: > > > > > >>>>colocn = 'abcd' > >>>>k = 4 > >>>>for i in range(k-1): > > > > for j in range(1,k): > > rule1 = [colocn[i:i+j],colocn[:i]+colocn[i+j:]] > > rule2 = [colocn[:i]+colocn[i+j:],colocn[i:i+j]] > > rules.append(rule1) > > rules.append(rule2) > > > >>>>rules > > > > [['a', 'bcd'], ['bcd', 'a'], ['ab', 'cd'], ['cd', 'ab'], ['abc', 'd'], > > ['d', 'abc'], ['b', 'acd'], ['acd', 'b'], ['bc', 'ad'], ['ad', 'bc'], > > ['bcd', 'a'], ['a', 'bcd'], ['c', 'abd'], ['abd', 'c'], ['cd', 'ab'], > > ['ab', 'cd'], ['cd', 'ab'], ['ab', 'cd']] > > > > > > Any ideas?? > > Algorithmic problem. > > First, you need to get all permutations. This is a known algorithm, that > you'll find examples of on the net. Then for each permutation, list > possibles 'pair-splits'. > > Here's a (probably sub-optimal, but it's getting late here) possible > implementation in functional style: > > def rotate(lst): > yield lst > max = len(lst) > for i in range(1, max): > yield lst[i:] + lst[:-(max-i)] > > def permute(lst): > if len(lst) > 2: > for rotated in rotate(lst): > head, tail = rotated[0], rotated[1:] > for permuted in permute(tail): > yield [head] + permuted > elif len(lst) == 2: > yield lst > yield lst[::-1] > else: > yield lst > > def splits(lst): > for i in range(1, len(lst)): > yield lst[0:i], lst[i:] > > def allsplits(lst): > for permuted in permute(lst): > for pair in splits(permuted): > yield pair > > def listsubstrings(thestr): > format = lambda pair: (''.join(pair[0]), ''.join(pair[1])) > return [format(list(pair)) for pair in allsplits(list(thestr))] > > > print listsubstrings("abcd") Thanks Bruno. I wanted to avoid permutations as it would take more time, but i guess will have to deal with them now :(( Also this one gives each rule twice...i've to search for some double counting...
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > -- > http://mail.python.org/mailman/listinfo/python-list > ---------------------------------------------------------------- This message was sent using IMP, the Internet Messaging Program. -- http://mail.python.org/mailman/listinfo/python-list