I have a list [1, 2, 3]. I want a function that takes in the list and another number, the length.
f([1, 2, 3], 4) = [
[1, 1, 1, 1],
[1, 1 , 1, 2],
[1, 1, 1, 3],
[1, 1, 2, 1],
[1, 1, 3, 1],
#and so on...
]
Maybe itertools has an answer?
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Moon Cheesez
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You can check it here: http://stackoverflow.com/questions/464864/python-code-to-pick-out-all-possible-combinations-from-a-list – raymelfrancisco Jul 21 '15 at 14:49
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"maybe itertools has an answer" have you even spent some time checking itertools? – DevLounge Jul 21 '15 at 14:53
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Can you be more specific in your requirement? Should two items in the result be included if they have the same elements in a different order? Is the next item `[1,1,2,1]` or `[1,1,2,2]`? In total, does the result have 15 items or 81? – Robᵩ Jul 21 '15 at 15:06
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Sorry for the unclear post – Moon Cheesez Jul 21 '15 at 15:08
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The intended result should contain 81. – Moon Cheesez Jul 21 '15 at 15:10
1 Answers
6
itertools.combinations_with_replacement is the function you seek.
In [17]: i=itertools.combinations_with_replacement((1,2,3), 4)
In [18]: next(i)
Out[18]: (1, 1, 1, 1)
In [19]: next(i)
Out[19]: (1, 1, 1, 2)
In [20]: next(i)
Out[20]: (1, 1, 1, 3)
In [21]: next(i)
Out[21]: (1, 1, 2, 2)
In [22]:
If you want the set of all combinations, including items which differ only in order, try this:
# Modified from itertools.combinations_with_replace example
# from the python doc.
import itertools
import pprint
def odometer(iterable, r):
pool = tuple(iterable)
n = len(pool)
for indices in itertools.product(range(n), repeat=r):
yield tuple(pool[i] for i in indices)
pprint.pprint (list(odometer([1,2,3], 4)))
Robᵩ
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Is there a possible way to get all the possibilities in one list? – Moon Cheesez Jul 21 '15 at 14:58
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