Keep only elements that appear multiple times
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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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採用された回答
Roger Stafford
2013 年 6 月 13 日
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;
2 件のコメント
Roger Stafford
2013 年 6 月 15 日
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.
その他の回答 (4 件)
Roger Stafford
2013 年 6 月 12 日
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;
0 件のコメント
Roger Stafford
2013 年 6 月 13 日
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;
0 件のコメント
Azzi Abdelmalek
2013 年 6 月 12 日
編集済み: Azzi Abdelmalek
2013 年 6 月 12 日
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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