Python: Finding combinations

Question

I have a list, to make it easier to understand it's structure I'll write it out like so:

mylist = [[["a","b","c","d"]...]...]

Where the ... means the previous list is repeated (although the values inside may change)

An example list would be:

mylist = [[["a","b","c","d"], ["e","f","g","h"]], [["i", "j", "k", "l"]], [["m","n","o","p"], ["q","r","s","t"]]]

My current method is:

mylist2 = []
for a in mylist[0]:
    for b in mylist[1]:
        for c in mylist[2]:
            mylist2.append([a,b,c])

However this is very long, especially since in my actual code it goes on up to for x in mylist[35]

Is there a better way for me to write this code?

.. this is very long.. - did you mean it takes a long time to execute? Will you be using mylist2 more than once? — wwii
– wwii, Commented Oct 17, 2019 at 22:31
@wwii I mean the code is very long. There are nested for loops in for loops in for loops... Around 35 times. That's 35 tabs before code! Surely there's a shorter and faster way to do this — Rlz
– Rlz, Commented Oct 17, 2019 at 22:35
And the variable mylist2 is returned from the function as soon as it is found — Rlz
– Rlz, Commented Oct 17, 2019 at 22:36
Did you look through the Python docs for something that might work for you? — wwii
– wwii, Commented Oct 17, 2019 at 22:39

Mr_U4913 · Accepted Answer · 2019-10-17 23:05:34Z

3

your code

%%timeit
mylist2 = []
for a in mylist[0]:
    for b in mylist[1]:
        for c in mylist[2]:
            mylist2.append([a,b,c])
# 809 ns ± 18.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

use itertools

import itertools
%%timeit
list(itertools.product(*mylist))
# 528 ns ± 11.4 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

answered Oct 17, 2019 at 23:05

Mr_U4913

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2 Comments

ShadowRanger Over a year ago

Minor difference in behavior: The latter makes a list of three-tuples, where the former is making three-lists. If mutability is required, the latter can be efficiently made to match the former by mapping to list: list(map(list, itertools.product(*mylist))). But yes, this is the obvious way to go here, and especially useful when mylist might have a variable number of sub-lists.

Mr_U4913 Over a year ago

@ShadowRanger Thank you for your suggestion.

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