python combinations of multiple list

Question

Is there any pythonic method to generate combinations between multiple list? (similar to Cartesian product but more complicated)

Example:

a = [1, 2, 3]
b = [4, 5, 6]
c = [7, 8, 9]
# ...
# there are more than 3 lists

Expected output:

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

Update:

Thanks for the quick reply~!!

To clarify the question:

The result are all non-repeated combinations of Cartesian product of list a, b, c.

It can be done by another ugly method:

1) Generate the whole list of Cartesian product

from itertools import product, combinations, chain
t = list(product(a, b, c))

2) Using combinations to generate all possible results

p = list(combinations(t, 3))

3) Filter the repeated conditions

cnt = len(list(chain(a, b, c)))
f = [x for x in p if len(set(chain(*x))) == cnt]

Update2:

Expected result generated by ugly method:

((1, 4, 7), (2, 5, 8), (3, 6, 9))
((1, 4, 7), (2, 5, 9), (3, 6, 8))
((1, 4, 7), (2, 6, 8), (3, 5, 9))
((1, 4, 7), (2, 6, 9), (3, 5, 8))
((1, 4, 8), (2, 5, 7), (3, 6, 9))
((1, 4, 8), (2, 5, 9), (3, 6, 7))
((1, 4, 8), (2, 6, 7), (3, 5, 9))
((1, 4, 8), (2, 6, 9), (3, 5, 7))
((1, 4, 9), (2, 5, 7), (3, 6, 8))
((1, 4, 9), (2, 5, 8), (3, 6, 7))
((1, 4, 9), (2, 6, 7), (3, 5, 8))
((1, 4, 9), (2, 6, 8), (3, 5, 7))
((1, 5, 7), (2, 4, 8), (3, 6, 9))
((1, 5, 7), (2, 4, 9), (3, 6, 8))
((1, 5, 7), (2, 6, 8), (3, 4, 9))
((1, 5, 7), (2, 6, 9), (3, 4, 8))
((1, 5, 8), (2, 4, 7), (3, 6, 9))
((1, 5, 8), (2, 4, 9), (3, 6, 7))
((1, 5, 8), (2, 6, 7), (3, 4, 9))
((1, 5, 8), (2, 6, 9), (3, 4, 7))
((1, 5, 9), (2, 4, 7), (3, 6, 8))
((1, 5, 9), (2, 4, 8), (3, 6, 7))
((1, 5, 9), (2, 6, 7), (3, 4, 8))
((1, 5, 9), (2, 6, 8), (3, 4, 7))
((1, 6, 7), (2, 4, 8), (3, 5, 9))
((1, 6, 7), (2, 4, 9), (3, 5, 8))
((1, 6, 7), (2, 5, 8), (3, 4, 9))
((1, 6, 7), (2, 5, 9), (3, 4, 8))
((1, 6, 8), (2, 4, 7), (3, 5, 9))
((1, 6, 8), (2, 4, 9), (3, 5, 7))
((1, 6, 8), (2, 5, 7), (3, 4, 9))
((1, 6, 8), (2, 5, 9), (3, 4, 7))
((1, 6, 9), (2, 4, 7), (3, 5, 8))
((1, 6, 9), (2, 4, 8), (3, 5, 7))
((1, 6, 9), (2, 5, 7), (3, 4, 8))
((1, 6, 9), (2, 5, 8), (3, 4, 7))

do you want an iterator ? then itertools.product(*args)is what you are looking for. just place your lists as arg. — Marvin Taschenberger
– Marvin Taschenberger, Commented Jul 25, 2017 at 16:02
Your question isn't completely clear. How many lines of output do you expect for that input data? If you want the Cartesian product of each permutation of a, b, and c, that'd result in 216 lines. — PM 2Ring
– PM 2Ring, Commented Jul 25, 2017 at 16:28
I've written code for @PM2Ring's suggestion. If that's the case then please reword the question. — Antti Haapala
– Antti Haapala, Commented Jul 25, 2017 at 17:00

Antti Haapala · Accepted Answer · 2017-07-26 01:46:17Z

2

What you want are not combinations but indeed permutations. 3 elements have 6 permutations, a Cartesian product of 2 sets of permutations has 36. PM 2Ring originally suspected that you want all 3 of these permuted since your question didn't tell otherwise. If the code in your question produces the desired output, it means you want b and c permuted but not a. Initially I wrote code that calculated the permutations for all of a, b and c. However, since a doesn't need to be permuted, we'll just wrap it in a list. This gets us very close to the desired output:

import itertools as it

a = [1, 2, 3]
b = [4, 5, 6]
c = [7, 8, 9]

for i in it.product([tuple(a)], it.permutations(b), it.permutations(c)):
    print(i)

The output is 36 lines that start with

((1, 2, 3), (4, 5, 6), (7, 8, 9))
((1, 2, 3), (4, 5, 6), (7, 9, 8))
((1, 2, 3), (4, 5, 6), (8, 7, 9))

It is already almost the same format that you want but with indexes transposed so o[x][y] would match o[y][x] of your desired output. We use some zip magic to transpose them. As a plus, this function now works for any number of arguments:

import itertools as it

a = [1, 2, 3]
b = [4, 5, 6]
c = [7, 8, 9]

def funnyperms(first, *rest):
    for i in it.product([first], *(it.permutations(j) for j in rest)):
        yield tuple(zip(*i))

for i in funnyperms(a, b, c):
    print(i)

The output is

((1, 4, 7), (2, 5, 8), (3, 6, 9))
((1, 4, 7), (2, 5, 9), (3, 6, 8))
((1, 4, 8), (2, 5, 7), (3, 6, 9))
((1, 4, 8), (2, 5, 9), (3, 6, 7))
((1, 4, 9), (2, 5, 7), (3, 6, 8))
((1, 4, 9), (2, 5, 8), (3, 6, 7))
((1, 4, 7), (2, 6, 8), (3, 5, 9))
((1, 4, 7), (2, 6, 9), (3, 5, 8))
((1, 4, 8), (2, 6, 7), (3, 5, 9))
((1, 4, 8), (2, 6, 9), (3, 5, 7))
((1, 4, 9), (2, 6, 7), (3, 5, 8))
((1, 4, 9), (2, 6, 8), (3, 5, 7))
((1, 5, 7), (2, 4, 8), (3, 6, 9))
((1, 5, 7), (2, 4, 9), (3, 6, 8))
((1, 5, 8), (2, 4, 7), (3, 6, 9))
((1, 5, 8), (2, 4, 9), (3, 6, 7))
((1, 5, 9), (2, 4, 7), (3, 6, 8))
((1, 5, 9), (2, 4, 8), (3, 6, 7))
((1, 5, 7), (2, 6, 8), (3, 4, 9))
((1, 5, 7), (2, 6, 9), (3, 4, 8))
((1, 5, 8), (2, 6, 7), (3, 4, 9))
((1, 5, 8), (2, 6, 9), (3, 4, 7))
((1, 5, 9), (2, 6, 7), (3, 4, 8))
((1, 5, 9), (2, 6, 8), (3, 4, 7))
((1, 6, 7), (2, 4, 8), (3, 5, 9))
((1, 6, 7), (2, 4, 9), (3, 5, 8))
((1, 6, 8), (2, 4, 7), (3, 5, 9))
((1, 6, 8), (2, 4, 9), (3, 5, 7))
((1, 6, 9), (2, 4, 7), (3, 5, 8))
((1, 6, 9), (2, 4, 8), (3, 5, 7))
((1, 6, 7), (2, 5, 8), (3, 4, 9))
((1, 6, 7), (2, 5, 9), (3, 4, 8))
((1, 6, 8), (2, 5, 7), (3, 4, 9))
((1, 6, 8), (2, 5, 9), (3, 4, 7))
((1, 6, 9), (2, 5, 7), (3, 4, 8))
((1, 6, 9), (2, 5, 8), (3, 4, 7))

Storing these into a set and comparing with the values produced by your method proves that they have identical output:

print(set(funnyperms(a, b, c)) == set(f))

prints True, Q.E.D.

edited Jul 26, 2017 at 1:46

answered Jul 25, 2017 at 19:46

Antti Haapala

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3 Comments

Vincent Li Over a year ago

Thanks for your answer! It's a pretty nice method, but the result is different (elements from input should be separated into different). I updated the question with expected result. Is there any algorithm to produce that result?

Antti Haapala Over a year ago

what do you mean? As I say there, my second code produces a set of tuples that is identical to what you provided...

Vincent Li Over a year ago

Oops, sorry I missed your second code. This result is exactly what I wanted. That's awesome!! Thank you so much for you answer!

pylang · Accepted Answer · 2017-07-25 16:38:02Z

2

You can use product from itertools for all combinations

>>> from itertools import product
>>> a = [1, 2, 3]
>>> b = [4, 5, 6]
>>> c = [7, 8, 9]
>>> A = [a,b,c]
>>> prod = list(product(*A))
>>> print(prod)

Expected output:

[(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)]

edited Jul 25, 2017 at 16:38

pylang

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answered Jul 25, 2017 at 16:16

dheeraj .A

1,1177 silver badges6 bronze badges

2 Comments

Akavall Over a year ago

This does not look like the output that OP is expecting.

pylang Over a year ago

I think you have it if you add list(itertools.permutations(prod, 3)) for the final result.

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