Asked by Ryan Magee
on 12 Jun 2013

Hey everyone, been banging my head against this for awhile and can't come up with something efficient.

I have a large matrix, roughly 1000x1000, and I want to discard all elements (or replace with 0) that don't appear in the matrix at least three times. Additionally, I'm trying to allow for an error range so that, say, 10.1 and 9.9 (some arbitrary interval) will count as "10" (but this is secondary to the original problem).

I'm guessing that my main issue is that rewriting/editing a matrix is computationally expensive. The only solution I came up with involved numel in a loop, which is dreadfully slow.

Thanks for looking, advice is appreciated!

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Answer by Roger Stafford
on 13 Jun 2013

Accepted answer

Actually I didn't need to find the inverse of permutation p. The following is one step shorter:

[B,p] = sort(A(:)); t = [true;diff(B)~=0;true]; q = cumsum(t(1:end-1)); t = diff(find(t))<3; A(p(t(q))) = 0;

Ryan Magee
on 15 Jun 2013

Roger Stafford
on 15 Jun 2013

In the previous version I first took the inverse of p with the line

p(p) = 1:length(p);

and subsequently did this

A(t(q(p))) = 0;

It only occurred to me later that the inverse operation is not needed if we do the last step this way:

A(p(t(q))) = 0;

Without that inverse operation this last order p(t(q)) is essential. It wouldn't work otherwise.

Answer by Roger Stafford
on 12 Jun 2013

Here's a modification of Azzi's code that avoids the 'ismember' call.

[B,~,p] = unique(A(:)); t = histc(A(:),B)<3; A(t(p)) = 0;

Answer by Roger Stafford
on 13 Jun 2013

This version uses the 'sort' function instead of 'unique' and 'histc'. Consequently it might be faster.

[B,p] = sort(A(:)); p(p) = 1:length(p); t = [true;diff(B)~=0;true]; q = cumsum(t); t = diff(find(t))<3; A(t(q(p))) = 0;

Answer by Azzi Abdelmalek
on 12 Jun 2013

Edited by Azzi Abdelmalek
on 12 Jun 2013

A=[1 2 1 1;1 2 3 1;3 3 3 3;3 0 0 1]; B=unique(A(:)); A(ismember(A(:),B(histc(A(:),B)<3)))=0

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