py itertools?

Parker xenoszh at gmail.com
Sun Dec 20 06:21:46 EST 2009


>>> a = 'qwerty'
>>> b = '^%&$#'
>>> c = [(x,y) for x in a for y in b]
>>> c
[('q', '^'), ('q', '%'), ('q', '&'), ('q', '$'), ('q', '#'), ('w',
'^'), ('w', '%'), ('w', '&'), ('w', '$'), ('w', '#'), ('e', '^'),
('e', '%'), ('e', '&'), ('e', '$'), ('e', '#'), ('r', '^'), ('r',
'%'), ('r', '&'), ('r', '$'), ('r', '#'), ('t', '^'), ('t', '%'),
('t', '&'), ('t', '$'), ('t', '#'), ('y', '^'), ('y', '%'), ('y',
'&'), ('y', '$'), ('y', '#')]


This one is better and simple.




On Dec 19, 12:48 pm, Chris Rebert <c... at rebertia.com> wrote:
> On Sat, Dec 19, 2009 at 2:54 AM, mattia <ger... at gmail.com> wrote:
> > Hi all, I need to create the permutation of two strings but without
> > repeat the values, e.g. 'ab' for me is equal to 'ba'. Here is my
> > solution, but maybe the python library provides something better:
>
> >>>> def mcd(a, b):
> > ...     if b == 0:
> > ...         return a
> > ...     else:
> > ...         return mcd(b, a % b)
> > ...
> >>>> def mcm(a, b):
> > ...     return int((a * b) / mcd(a, b))
> > ...
> >>>> s1 = 'abc'
> >>>> s2 = 'wt'
> >>>> m = mcm(len(s1), len(s2))
> >>>> set(zip(s1*m, s2*m))
> > {('a', 'w'), ('a', 't'), ('b', 'w'), ('c', 't'), ('b', 't'), ('c', 'w')}
>
> > Any help?
>
> Surprised you didn't think of the seemingly obvious approach:
>
> def permute_chars(one, two):
>     for left in set(one):
>         for right in set(two):
>             yield (left, right)
>
> >>> list(permute_chars('abc', 'wt'))
>
> [('a', 'w'), ('a', 't'), ('b', 'w'), ('b', 't'), ('c', 'w'), ('c', 't')]
>
> Cheers,
> Chris
> --http://blog.rebertia.com




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