(newbie) N-uples from list of lists

[bonono at gmail.com]

Interesting, I found a reduce one liner(just one step further of
Raymond's) easiest to understand and match my thinking about what the
problem is about.

That once again tell me that different people think and approach the
problem differently. It is possible to talk about one "fastest" way,
but many times there isn't such a thing of one "best" way.

Martin Miller wrote:
> FWIW, I found Steven Taschuk's solution easiest to understand regarding
> the question posed in your original post -- namely how to solve the
> problem non-recursively with generators -- because it was similar to my
> own thinking about how to do it -- but suspect that Raymond Hettinger's
> is the likely the "best" (as is usually the case ;-).
>
> Best,
> -Martin
>
>
> vd12005 at yahoo.fr wrote:
> > great thanks to all.
> >
> > actually i have not seen it was a cross product... :) but then there
> > are already few others ideas from the web, i paste what i have found
> > below...
> >
> > BTW i was unable to choose the best one, speaking about performance
> > which one should be prefered ?
> >
> > ### --------------------------------------------------
> >
> > ### from title: variable X procuct - [(x,y) for x in list1 for y in
> > list2]
> > ### by author:  steindl fritz
> > ### 28 mai 2002
> > ### reply by:   Jeff Epler
> >
> > def cross(l=None, *args):
> >     if l is None:
> >         # The product of no lists is 1 element long,
> >         # it contains an empty list
> >         yield []
> >         return
> >     # Otherwise, the product is made up of each
> >     # element in the first list concatenated with each of the
> >     # products of the remaining items of the list
> >     for i in l:
> >         for j in cross(*args):
> >             yield [i] + j
> >
> > ### reply by:   Raymond Hettinger
> >
> > def CartesianProduct(*args):
> >     ans = [()]
> >     for arg in args:
> >         ans = [ list(x)+[y] for x in ans for y in arg]
> >     return ans
> >
> > """
> > print CartesianProduct([1,2], list('abc'), 'do re mi'.split())
> > """
> >
> > ### from:
> > http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/159975
> > ### by: Raymond Hettinger
> >
> > def cross(*args):
> >     ans = [[]]
> >     for arg in args:
> >         ans = [x+[y] for x in ans for y in arg]
> >     return ans
> >
> > ### from:
> > http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/159975
> > ### by: Steven Taschuk
> > """
> > Iterator version, Steven Taschuk, 2003/05/24
> > """
> > def cross(*sets):
> >     wheels = map(iter, sets) # wheels like in an odometer
> >     digits = [it.next() for it in wheels]
> >     while True:
> >         yield digits[:]
> >         for i in range(len(digits)-1, -1, -1):
> >             try:
> >                 digits[i] = wheels[i].next()
> >                 break
> >             except StopIteration:
> >                 wheels[i] = iter(sets[i])
> >                 digits[i] = wheels[i].next()
> >         else:
> >             break