Combinations algorithm without enumeration

Asked Nov 04 '22 at 04:31

Active Nov 04 '22 at 04:51

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here is how you generate combinations :

   from itertools import combinations_with_replacement, combinations
   list(combinations([1,2,3,4,5],r=3))
   [(1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 3, 4), (1, 3, 5), (1, 4, 5), (2, 3, 4), (2, 3, 5), (2, 4, 5), (3, 4, 5)]

Is there a way to get the Nth combination without enumerating all the previous items.

  combinations([1,2,3,4,5],r=3, nth=3)   ==>  (1,3,4)

The important thing is not to enumerate so that I can call for large numbers f.e.

  combinations(range(100),r=5, nth=3000000)   ==>  .....

I'm interested of the Algorithm/Source rather than callable function.

if it is possible at all, because I want to modify it for my needs

asked Nov 04 '22 at 04:31

sten

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does this answer your question : https://stackoverflow.com/questions/1776442/nth-combination – Deepak Tripathi Nov 04 '22 at 04:47
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Also if you can use `more_itertools` then you can use `more_itertools.nth_combination` this method is there. – Deepak Tripathi Nov 04 '22 at 04:48
@Mark you are correct. But in my first comment, there is already a discussion on this topic. – Deepak Tripathi Nov 04 '22 at 04:54
on first glance both seem like possible answers.. have to read through ... thanks – sten Nov 04 '22 at 17:50

Combinations algorithm without enumeration

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