How to find the closest data row index

Input data (sort by time): M:

1.0
1.2
3.1
4.0
1.2
1.0
4.8

given data:

V:1.2 
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));
  M(idx,:)

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).

Answer 1

[Cam Salzberger]

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

[Kanakaiah Jakkula]

Sir,

my input data is 3x10 matrix(i just cut down the data), original data is:

Hi, I have below matrix:

M:
1.5 2.1 5.1 4.0
1.0 2.9 3.6 6.9
1.0 2.9 3.6 6.9
1.1 2.3 2.8 3.9

V: 1.0 2.9 3.6 6.9

[Cam Salzberger]

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);

Answer 2

[KL]

EDITED for 2D matrix,

A = abs(M-V);
ind = arrayfun(@(k) find(A(:,k)==min(A(:,k)),1,'last'),1:length(V))