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Indexing matrix using logicals

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Matt H
Matt H 2012-10-16
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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回答（3 个）

Matt Fig
Matt Fig 2012-10-16
编辑：Matt Fig 2012-10-16
% 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
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Matt Fig
Matt Fig 2012-10-16
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 2012-10-16
编辑：Sean de Wolski 2012-10-16
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 2012-10-16
编辑：Azzi Abdelmalek 2012-10-16
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 个评论
Matt Kindig
Matt Kindig 2012-10-16
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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