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Indexing matrix using logicals

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Matt H
Matt H on 16 Oct 2012
I'm trying to index a large matrix, with the goal of finding/indexing the first value to meet a threshold. Right now I'm doing it in a loop, but it's rather slow.
For example: a=[10 13 14 15 16;... 11 12 15 16 17;... 3 5 8 9 12]; threshold=11.5;
I want it to return: ans=[2;2;5];
I've tried screwing with find, but all I can seem to get it to return is: ans=[4;5;7;8;10;11;13;14;15];
Thanks, - Matt

Answers (3)

Matt Fig
Matt Fig on 16 Oct 2012
Edited: Matt Fig on 16 Oct 2012
% Given:
a=[10 13 14 15 16;11 12 15 16 17; 3 5 8 9 12]; % Array
T = 11.5; % Threshold.
% The approach:
L = mod(findstr(reshape(a.',1,numel(a))<T,[1,0]),size(a,2))+1
  6 Comments
Azzi Abdelmalek
Azzi Abdelmalek on 16 Oct 2012
Yes, but it's the nearest
Matt Fig
Matt Fig on 16 Oct 2012
To Matt H,
If you want to fill the value with nan (or zero, just replace the nan with 0 in below code) when no member of the row matches, you could use this:
% This array has no element meet the
% threshold in the third row.
a = [10 13 14 15 16;11 12 15 16 17; 3 5 8 9 11];
% So we use NaN as the filler.
[I,J] = find(a>11.5);
A = accumarray(I,J,[size(a,1),1], @min, NaN)

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Sean de Wolski
Sean de Wolski on 16 Oct 2012
Edited: Sean de Wolski on 16 Oct 2012
If you can guarantee that each row in A has at least one value greater than the threshold, you can use the index output from max():
a=[10 13 14 15 16; 11 12 15 16 17; 3 5 8 9 12];
thresh = 11.5;
[~,index] = max(a>thresh,[],2);

Azzi Abdelmalek
Azzi Abdelmalek on 16 Oct 2012
Edited: Azzi Abdelmalek on 16 Oct 2012
a=[10 13 14 15 16;11 12 15 16 17; 3 5 8 9 12];
[n,m]=size(a)
b=a<11.5
out=arrayfun(@(x) max(find(b(x,:)==1)),1:n)+1
out(out>m)=m
%or
out(out>m)=nan
  1 Comment
Matt Kindig
Matt Kindig on 16 Oct 2012
Another way, just for fun (assuming that each row is sorted, which is assumed if we are interested in the "first" match):
n = size(a,2);
b = a > thresh;
first = n-sum(b,2)+1;
%if first > n, then not found-- set to NaN;
first( first > n)= NaN;

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