What I have so far does pretty much nothing
def dress_me(shirt, tie, suit): # if type(shirt) != list or type(tie) != list or type(suit) != list: # return None combinations = dress_me(shirt, tie, suit) for combo in combinations: print(combo)
What I have so far does pretty much nothing
def dress_me(shirt, tie, suit): # if type(shirt) != list or type(tie) != list or type(suit) != list: # return None combinations = dress_me(shirt, tie, suit) for combo in combinations: print(combo)
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)]
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)
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 :)
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.