Indexing matrix using logicals
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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
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Respuestas (3)
Matt Fig
el 16 de Oct. de 2012
Editada: Matt Fig
el 16 de Oct. de 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 comentarios
Matt Fig
el 16 de Oct. de 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)
Sean de Wolski
el 16 de Oct. de 2012
Editada: Sean de Wolski
el 16 de Oct. de 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);
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Azzi Abdelmalek
el 16 de Oct. de 2012
Editada: Azzi Abdelmalek
el 16 de Oct. de 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 comentario
Matt Kindig
el 16 de Oct. de 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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