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I have tried to use itertools to compute all combinations of a list ['a', 'b', 'c'] using combinations_with_replacement with repeating elements. The problem is in the fact that the indices seem to be used to distinguish the elements:

Return r length subsequences of elements from the input iterable allowing individual elements to be repeated more than once.

Combinations are emitted in lexicographic sort order. So, if the input iterable is sorted, the combination tuples will be produced in sorted order.

Elements are treated as unique based on their position, not on their value. So if the input elements are unique, the generated combinations will also be unique.

Sot this code snippet: 

import itertools

for item in itertools.combinations_with_replacement(['a','b','c'], 3): 
    print (item)

results in this output:

('a', 'a', 'a')
('a', 'a', 'b')
('a', 'a', 'c')
('a', 'b', 'b')
('a', 'b', 'c')
('a', 'c', 'c')
('b', 'b', 'b')
('b', 'b', 'c')
('b', 'c', 'c')
('c', 'c', 'c')

And what I need is the combination set to contain elements like: ('a', 'b', 'a') which seem to be missing. How to compute the complete combination set?

tmaric
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2 Answers2

14

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')
>>>
  • 2
    Right. `itertools.combinations_with_replacement` sorts the elements in each tuple, so you'll only get `(a, a, b)`, never `(a, b, a)` – dano May 23 '14 at 16:09
  • This gives permutations. – Supertech Apr 17 '20 at 18:55
  • @Supertech There are only 6 permutations for a 3-element list, and anything with a duplicate element is not one of them. – Mew Dec 30 '22 at 01:43
2

For such small sequences you could use no 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')]
Stefan van den Akker
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