# Create all permutations of 1x13 vector used for Leave-p-out Cross-Validation

2 views (last 30 days)
Filip Wadman on 23 Feb 2022
Commented: Filip Wadman on 23 Feb 2022
Hi,
I am trying to develop a sound quality metric and for that I need to find all possible combination of my vector A=[1:1:13] values to pick out 11 set for training and 2 sets for validation to later use the regress function. Perms(A) is to large to be used and I can't use nchoosek(A,13). I want to creata a matrice which is 13x78 since that should be all possible permutation.
Any ideas?
If you need better clarifications please let me know.

Akira Agata on 23 Feb 2022
Though this generates 78x11 (instead of 13x78 which you expected), the result contains all possible combination.
% Create all possible combination of 11 elements out ouf 1~13
comb = nchoosek(1:13, 11);
% Show the result
disp(comb)
1 2 3 4 5 6 7 8 9 10 11 1 2 3 4 5 6 7 8 9 10 12 1 2 3 4 5 6 7 8 9 10 13 1 2 3 4 5 6 7 8 9 11 12 1 2 3 4 5 6 7 8 9 11 13 1 2 3 4 5 6 7 8 9 12 13 1 2 3 4 5 6 7 8 10 11 12 1 2 3 4 5 6 7 8 10 11 13 1 2 3 4 5 6 7 8 10 12 13 1 2 3 4 5 6 7 8 11 12 13 1 2 3 4 5 6 7 9 10 11 12 1 2 3 4 5 6 7 9 10 11 13 1 2 3 4 5 6 7 9 10 12 13 1 2 3 4 5 6 7 9 11 12 13 1 2 3 4 5 6 7 10 11 12 13 1 2 3 4 5 6 8 9 10 11 12 1 2 3 4 5 6 8 9 10 11 13 1 2 3 4 5 6 8 9 10 12 13 1 2 3 4 5 6 8 9 11 12 13 1 2 3 4 5 6 8 10 11 12 13 1 2 3 4 5 6 9 10 11 12 13 1 2 3 4 5 7 8 9 10 11 12 1 2 3 4 5 7 8 9 10 11 13 1 2 3 4 5 7 8 9 10 12 13 1 2 3 4 5 7 8 9 11 12 13 1 2 3 4 5 7 8 10 11 12 13 1 2 3 4 5 7 9 10 11 12 13 1 2 3 4 5 8 9 10 11 12 13 1 2 3 4 6 7 8 9 10 11 12 1 2 3 4 6 7 8 9 10 11 13 1 2 3 4 6 7 8 9 10 12 13 1 2 3 4 6 7 8 9 11 12 13 1 2 3 4 6 7 8 10 11 12 13 1 2 3 4 6 7 9 10 11 12 13 1 2 3 4 6 8 9 10 11 12 13 1 2 3 4 7 8 9 10 11 12 13 1 2 3 5 6 7 8 9 10 11 12 1 2 3 5 6 7 8 9 10 11 13 1 2 3 5 6 7 8 9 10 12 13 1 2 3 5 6 7 8 9 11 12 13 1 2 3 5 6 7 8 10 11 12 13 1 2 3 5 6 7 9 10 11 12 13 1 2 3 5 6 8 9 10 11 12 13 1 2 3 5 7 8 9 10 11 12 13 1 2 3 6 7 8 9 10 11 12 13 1 2 4 5 6 7 8 9 10 11 12 1 2 4 5 6 7 8 9 10 11 13 1 2 4 5 6 7 8 9 10 12 13 1 2 4 5 6 7 8 9 11 12 13 1 2 4 5 6 7 8 10 11 12 13 1 2 4 5 6 7 9 10 11 12 13 1 2 4 5 6 8 9 10 11 12 13 1 2 4 5 7 8 9 10 11 12 13 1 2 4 6 7 8 9 10 11 12 13 1 2 5 6 7 8 9 10 11 12 13 1 3 4 5 6 7 8 9 10 11 12 1 3 4 5 6 7 8 9 10 11 13 1 3 4 5 6 7 8 9 10 12 13 1 3 4 5 6 7 8 9 11 12 13 1 3 4 5 6 7 8 10 11 12 13 1 3 4 5 6 7 9 10 11 12 13 1 3 4 5 6 8 9 10 11 12 13 1 3 4 5 7 8 9 10 11 12 13 1 3 4 6 7 8 9 10 11 12 13 1 3 5 6 7 8 9 10 11 12 13 1 4 5 6 7 8 9 10 11 12 13 2 3 4 5 6 7 8 9 10 11 12 2 3 4 5 6 7 8 9 10 11 13 2 3 4 5 6 7 8 9 10 12 13 2 3 4 5 6 7 8 9 11 12 13 2 3 4 5 6 7 8 10 11 12 13 2 3 4 5 6 7 9 10 11 12 13 2 3 4 5 6 8 9 10 11 12 13 2 3 4 5 7 8 9 10 11 12 13 2 3 4 6 7 8 9 10 11 12 13 2 3 5 6 7 8 9 10 11 12 13 2 4 5 6 7 8 9 10 11 12 13 3 4 5 6 7 8 9 10 11 12 13
Filip Wadman on 23 Feb 2022
This worked very well. Thank you!
Now the next questions arises. For my vector A=1:13 we had another vector B=1:13 with values corresponding to each value in vector A which I now need to use with the regress function regress(A,B). Is it possible to find which element in my B vector which belongs to my new combTrain that we have created?
When I run my regress functions now it complains that the arrays don't match which is obvoius but to run it I need the right values from B belonging to A to receive a correct result.