How to find the closest data row index

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Input data (sort by time): M:
given data:
I want to find the most closer one in the input data
I use the below function:
[~,idx] = min(sum(bsxfun(@minus,M,V).^2,2));
But here, there two rows (duplicate data) row2 &row5, if this is the case, I want to pick the more recent row index i.e., row5 (but the present code giving row 2).

Accepted Answer

Cam Salzberger
Cam Salzberger on 18 Sep 2017
Hello Kanakaiah,
I think you're pretty much on the right track, though there's no need for bsxfun if you just want one index out. Since min will always return the index of the first element found, just flip it around before calling min:
dist = abs(A-V);
[~, revIdx] = min(dist(end:-1:1)); % Could also use flipud
idx = numel(A)-revIdx+1;
Cam Salzberger
Cam Salzberger on 18 Sep 2017
Edited: Cam Salzberger on 18 Sep 2017
Alright, so your "distance" in this case is the 2-norm of the vector with each row of M? In that case, you can just change how distance is calculated, but the rest of my code should still work. In R2016b and later, you can use implicit expansion to allow a little shortcut:
dist = sum((M-V).^2, 2);
Otherwise you have to get fancy with repmat:
Vexp = repmat(V, size(M, 1), 1);
dist = sum((M-Vexp).^2, 2);

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More Answers (1)

KL on 18 Sep 2017
Edited: KL on 18 Sep 2017
EDITED for 2D matrix,
A = abs(M-V);
ind = arrayfun(@(k) find(A(:,k)==min(A(:,k)),1,'last'),1:length(V))


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