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
[EMAIL PROTECTED] 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
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