CODEWITHSUNDEEP
pythonpython-itertoolscombinatorics
iago-lito
iago-lito

Reputation: 3218

Iterate over every possible n-sampling of a list

If I understand itertools "combinatoric iterators" doc correctly, the idea is to provide a set of standard functions for every common combinatory iteration.

But I miss one today. I need to iterate over every ordered combinations of items with repetitions.

combination_with_replacement('abcd', 4) yields

('a', 'a', 'a', 'a')
('a', 'a', 'a', 'b')
('a', 'a', 'a', 'c')
('a', 'a', 'a', 'd')
('a', 'a', 'b', 'b')
('a', 'a', 'b', 'c')
('a', 'a', 'b', 'd')
('a', 'a', 'c', 'c')
('a', 'a', 'c', 'd')
... etc

but (even though the results are sorted tuples), these combinations are not ordered.

I expect more results from an ideal ordered_combination_with_replacement('abcd', 4) for I need to distinguish between

('a', 'a', 'a', 'a')
('a', 'a', 'a', 'b')
('a', 'a', 'b', 'a')
('a', 'b', 'a', 'a')
('b', 'a', 'a', 'a')
('a', 'a', 'a', 'c')
('a', 'a', 'c', 'a')
... etc

In other words: order matters today.

Does itertool provide such an iteration? Why not, or why have I missed it?
What's a standard way to iterate over those?
Do I need to write this generic iterator myself?

Upvotes: 1

Views: 91

Answers (1)

Roy2012
Roy2012

Reputation: 12523

To wrap up some of the comments, there are (at least) two ways to do that: 

itertools.combinations_with_replacement("abcd", 4)

and

itertools.product("abcd", repeat=4)

Both produce the required results of: 

[('a', 'a', 'a', 'a'),
 ('a', 'a', 'a', 'b'),
 ('a', 'a', 'a', 'c'),
 ('a', 'a', 'a', 'd'),
 ('a', 'a', 'b', 'a'),
 ...

Upvotes: 1

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