List.Combinations


#1

I would have expected item at index no. 10 to be [2][1]. No?

This happens at the end of each combination, dropping the items as the list goes higher.

ListCombinations


#2

List.Permutations should give you more or less the behaviour you’re expecting (except for combinations like 1+1, 2+2 etc.) as it takes into account the order of items in each sublist. List.Combinations will “only” give you unique combinations of list elements.


#3

If I under you correctly - if [2][1] already exists in a list, List.Combination will not return a [1][2], but the List.Permutations will. Correct?

 


#4

Bewarned that using list.permutations on any list larger than 6 or 7 will suffer greatly in terms of performance since List.Permutations generates a list of length n!. That number gets big fast… and your computer will feel the pain. To test, just create a simple list 1…10…1, then run list.permutation on it. I have 32Gb of RAM and a 2.7Ghz processor, and Dynamo works for over 1o minutes when I do this before giving me a list of 3.6 million permutations.