I need help coming up with a function in python that can take 3 arguements as lists and give me all combinations of there elements [duplicate]

Answer 1

Or, for completeness, in the generator-fashion without an extra function:

import itertools

for combination in itertools.product(shirts, ties, suits):
    whatever_you_want_to_do(combination)

Answer 2

Use itertools.product:

def dress_me(shirt, tie, suit):
    if type(shirt) != list or type(tie) != list or type(suit) != list:
        return None
    return list(itertools.product(shirt, tie, suit))

Demo:

>>> dress_me([1,2,3],[4,5,6],[7,8,9])
[(1, 4, 7), (1, 4, 8), (1, 4, 9), (1, 5, 7), (1, 5, 8), (1, 5, 9), (1, 6, 7), (1, 6, 8), (1, 6, 9), (2, 4, 7), (2, 4, 8), (2, 4, 9), (2, 5, 7), (2, 5, 8), (2, 5, 9), (2, 6, 7), (2, 6, 8), (2, 6, 9), (3, 4, 7), (3, 4, 8), (3, 4, 9), (3, 5, 7), (3, 5, 8), (3, 5, 9), (3, 6, 7), (3, 6, 8), (3, 6, 9)]

Answer 3

Since it looks like you were trying to come up with a recursive solution, here is the general form of it:

def all_perms(thing):
    if len(thing) <=1:
        yield thing
    else:
        for perm in all_perms(thing[1:]):
            for i in range(len(perm)+1):
                yield perm[:i] + thing[0:1] + perm[i:]

This works for most kinds of iterables. Demo:

In [5]: list(all_perms(('shirt','tie','suit')))
Out[5]: 
[('shirt', 'tie', 'suit'),
 ('tie', 'shirt', 'suit'),
 ('tie', 'suit', 'shirt'),
 ('shirt', 'suit', 'tie'),
 ('suit', 'shirt', 'tie'),
 ('suit', 'tie', 'shirt')]

Recursion is difficult to understand at first, but the general form is:

if simplest_case:
    return simplest_case
else:
    #recurse

In this case, return is replaced by yield in order to make a generator, which is more memory-friendly. You still shouldn't expect this to be the best-performing solution, but I'm including it for completeness since "USE ITERTOOLS" doesn't end up teaching you much other than itertools is cool.

Answer 4

def dress_me(shirt, tie, suit):
    all_combinations = []
    for s in shirt:
        for t in tie:
            for su in suit:
                all_combinations.append((s,t,su))
    return all_combination

Maybe there is a more pythonic way to do it :)