intertools combination 2 vs 2 in a list python

Question

I need to create team of 2v2 people

here is my list of player

L=["P1","P2","P3","P4"]

import itertools

I know of to create the all 1v1 or 1v1v1 possible by using

>>> L=["P1","P2","P3","P4"]
>>> for p in itertools.combinations(L,2) :
...     print(p)
... 
('P1', 'P2')
('P1', 'P3')
('P1', 'P4')
('P2', 'P3')
('P2', 'P4')
('P3', 'P4')

or

>>> for p in itertools.combinations(L,3) :
...     print(p)
... 
('P1', 'P2', 'P3')
('P1', 'P2', 'P4')
('P1', 'P3', 'P4')
('P2', 'P3', 'P4')

but how to print all the 2V2 possible ?

I want all the combinaison possible in order to create match 2 vs2 . Exemple : P1&P2 vs P3&P4, P2&P3 vs P4&P1 etc — Grendel, Apr 02 '20 at 09:29
well I do not know the exact outut that is why I need a code for that.. — Grendel, Apr 02 '20 at 09:30
@Grendel could there be an odd number of people under consideration? — juanpa.arrivillaga, Apr 02 '20 at 09:37
Does this answer your question? [League fixture generator in python](https://stackoverflow.com/questions/11245746/league-fixture-generator-in-python) — Georgy, Apr 02 '20 at 10:14
Also, see [Grouping list combinations for round-robin tournament](https://stackoverflow.com/q/32358841/7851470) — Georgy, Apr 02 '20 at 10:15

user2390182 · Accepted Answer · 2020-04-02T09:43:53.587

2

You could do the following, rather naively but effectively:

for p in itertools.combinations(L,2) :
    o = tuple(x for x in L if x not in p)
    print(p, o)

('P1', 'P2') ('P3', 'P4')
('P1', 'P3') ('P2', 'P4')
('P1', 'P4') ('P2', 'P3')
('P2', 'P3') ('P1', 'P4')
('P2', 'P4') ('P1', 'P3')
('P3', 'P4') ('P1', 'P2')

For more players, you can do:

L = ["P1", "P2", "P3", "P4", "P5"]
for p in itertools.combinations(L,2) :
    o = [x for x in L if x not in p]
    for x in itertools.combinations(o, 2):
        print(p, x)

Which would you give a double round-robin ;) For a single round-robin:

for p in itertools.combinations(L,2) :
    o = [x for x in L if x > p[0] and x != p[1]]
    for x in itertools.combinations(o, 2):
        print(p, x)

('P1', 'P2') ('P3', 'P4')
('P1', 'P2') ('P3', 'P5')
('P1', 'P2') ('P4', 'P5')
('P1', 'P3') ('P2', 'P4')
('P1', 'P3') ('P2', 'P5')
('P1', 'P3') ('P4', 'P5')
('P1', 'P4') ('P2', 'P3')
('P1', 'P4') ('P2', 'P5')
('P1', 'P4') ('P3', 'P5')
('P1', 'P5') ('P2', 'P3')
('P1', 'P5') ('P2', 'P4')
('P1', 'P5') ('P3', 'P4')
('P2', 'P3') ('P4', 'P5')
('P2', 'P4') ('P3', 'P5')
('P2', 'P5') ('P3', 'P4')

edited Apr 02 '20 at 09:43

answered Apr 02 '20 at 09:34

user2390182

72,016
6
67
89

Ok and what if now I have a list of 5 players ? – Grendel Apr 02 '20 at 09:37
@Grendel Just loop over `combinations(o, 2)`. A better name for `o` would be `remaining`, or what I presume it stands for, `other` – Mateen Ulhaq Apr 02 '20 at 09:38
@Grendel so, order does not matter, e.g. `('P1', 'P2') ('P3', 'P4')` is not the same as `('P3', 'P4') ('P1', 'P2')` ? – juanpa.arrivillaga Apr 02 '20 at 09:42
heu yes it is the same indeed – Grendel Apr 02 '20 at 09:43
Added the single-round robin variant, without duplicates. – user2390182 Apr 02 '20 at 09:44

intertools combination 2 vs 2 in a list python

1 Answers1