(newbie) N-uples from list of lists

[vd12005 at yahoo.fr]

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