Indexing matrix using logicals

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

Antworten (3)

Matt Fig
Matt Fig am 16 Okt. 2012
Bearbeitet: Matt Fig am 16 Okt. 2012

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% 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 Kommentare

What if
a=[10 13 14 15 16;11 12 15 16 17; 3 5 8 9 11]; % Array
Matt Fig
Matt Fig am 16 Okt. 2012
Bearbeitet: Matt Fig am 16 Okt. 2012
Interesting, what would you expect in that case, Azzi?
Perhaps [2 2 nan] would be most useful?
Azzi Abdelmalek
Azzi Abdelmalek am 16 Okt. 2012
I prefer [2 2 5],
Matt Fig
Matt Fig am 16 Okt. 2012
But from the OP: "with the goal of finding/indexing the first value to meet a threshold"
a(3,5) does not meet the threshold.
Azzi Abdelmalek
Azzi Abdelmalek am 16 Okt. 2012
Yes, but it's the nearest
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 am 16 Okt. 2012
Bearbeitet: Sean de Wolski am 16 Okt. 2012

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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 am 16 Okt. 2012
Bearbeitet: Azzi Abdelmalek am 16 Okt. 2012

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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 Kommentar

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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am 16 Okt. 2012

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