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[ACCEPTED]-How to use itertools to compute all combinations with repeating elements?-itertools

Accepted answer
Score: 12

It sounds like you want itertools.product:

>>> from itertools import product
>>> for item in product(['a', 'b', 'c'], repeat=3):
...     print item
...
('a', 'a', 'a')
('a', 'a', 'b')
('a', 'a', 'c')
('a', 'b', 'a')
('a', 'b', 'b')
('a', 'b', 'c')
('a', 'c', 'a')
('a', 'c', 'b')
('a', 'c', 'c')
('b', 'a', 'a')
('b', 'a', 'b')
('b', 'a', 'c')
('b', 'b', 'a')
('b', 'b', 'b')
('b', 'b', 'c')
('b', 'c', 'a')
('b', 'c', 'b')
('b', 'c', 'c')
('c', 'a', 'a')
('c', 'a', 'b')
('c', 'a', 'c')
('c', 'b', 'a')
('c', 'b', 'b')
('c', 'b', 'c')
('c', 'c', 'a')
('c', 'c', 'b')
('c', 'c', 'c')
>>>

0

Score: 2

For such small sequences you could use no 1 itertools at all:

abc = ("a", "b", "c")

print [(x, y, z) for x in abc for y in abc for z in abc]
# output:
[('a', 'a', 'a'),
 ('a', 'a', 'b'),
 ('a', 'a', 'c'),
 ('a', 'b', 'a'),
 ('a', 'b', 'b'),
 ('a', 'b', 'c'),
 ('a', 'c', 'a'),
 ('a', 'c', 'b'),
 ('a', 'c', 'c'),
 ('b', 'a', 'a'),
 ('b', 'a', 'b'),
 ('b', 'a', 'c'),
 ('b', 'b', 'a'),
 ('b', 'b', 'b'),
 ('b', 'b', 'c'),
 ('b', 'c', 'a'),
 ('b', 'c', 'b'),
 ('b', 'c', 'c'),
 ('c', 'a', 'a'),
 ('c', 'a', 'b'),
 ('c', 'a', 'c'),
 ('c', 'b', 'a'),
 ('c', 'b', 'b'),
 ('c', 'b', 'c'),
 ('c', 'c', 'a'),
 ('c', 'c', 'b'),
 ('c', 'c', 'c')]
Source: stackoverflow.com

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