matrix with different randperm rows

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Jeremy
Jeremy on 17 Sep 2014
Commented: Jeremy on 18 Sep 2014
Hi,
Say I have vector of N integers, 1 through N. Is there a quick way, without loops, to create a matrix where each row is a different randperm of the vector? My N can get large (up to 100).
Thanks

Accepted Answer

Andrei Bobrov
Andrei Bobrov on 17 Sep 2014
Edited: Andrei Bobrov on 17 Sep 2014
[~, out] = sort(rand(M,N),2);
  3 Comments
Jeremy
Jeremy on 18 Sep 2014
Works orders of magnitude faster for large group sizes :)

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More Answers (3)

Guillaume
Guillaume on 17 Sep 2014
cell2mat(arrayfun(@(dummy) randperm(n), 1:m, 'UniformOutput', false)');
Whether or not that can be said to be without a loop is debatable as arrayfun is effectively looping over the array 1:m

Mikhail
Mikhail on 17 Sep 2014
  4 Comments
Image Analyst
Image Analyst on 17 Sep 2014
In what universe, or decade, is 100 considered large? You actually mean 100, right, like ten times ten, not 100 million or 100 billion or something?
Jeremy
Jeremy on 17 Sep 2014
100 is large when you talk about some types of group coordination. I need to select k individuals from a group of N to interact at different time steps. What I've been doing is generating all combinations and randomly selecting ones to fill an interaction matrix, but that combination matrix gets big fast.

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Joseph Cheng
Joseph Cheng on 17 Sep 2014
Well there is a randperm() function that will give you a random permuation. How to get it as a MxN matrix without a loop will take some more time to think about it.
  1 Comment
Joseph Cheng
Joseph Cheng on 17 Sep 2014
Edited: Joseph Cheng on 17 Sep 2014
ok well it is possible to do it without a loop in 2ish lines. Use the function arrayfun() or cellfun() on an array of N*ones(M,1).

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