# Find closest value in array

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Chiranjibi il 25 Ago 2014
Commentato: David il 5 Lug 2023
I have two vector(which are time stamps) like,
V N
1375471092848936 1375473384440853
1375473388165900 1375471277856598
1375471320476780 1375473388165900
1375473388947681 1375471322465961
1375473392527002 1375471335206288
.................. ..................
My goal is to find closest time in N with respect to V (i.e. find time in N which is nearly equal with V). My frame is W = 1e4, furthermore V should lies between N-W and N+W. So how do I get closest time through MATLAB? Any help would be appreciated.
Thanks
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### Risposta accettata

Joe S il 10 Set 2018
Modificato: MathWorks Support Team il 27 Nov 2018
To compute the closest value in a vector “N” for each element of “V”, try the following code with example vectors “N” and “V”:
V = randi(10,[5 1])
N = randi(10,[5 1])
A = repmat(N,[1 length(V)])
[minValue,closestIndex] = min(abs(A-V))
closestValue = N(closestIndex)
Note that if there is a tie for the minimum value in each column, MATLAB chooses the first element in the column.
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Adrian Aquino Arriaga il 8 Ott 2021
Great answer. I would only use dot-apostrophe (.') instead of only apostrophe to make sure you are taking the non-conjugate transpose. Of course, this is only relevant if you are working with complex numbers.
David il 5 Lug 2023
This just saved my night!
For anybody juse searching the index, faster variant:
[~,closestIndex] = min(abs(N-V));

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### Più risposte (3)

Andrew Reibold il 25 Ago 2014
Modificato: Andrew Reibold il 25 Ago 2014
This finds the value in N which is closest to the V value I am calling.
N = [1990 1998 2001 2004 2001]
V = [2000 2011 2010 2001 1998]
[c index] = min(abs(N-V(1)))
In this case Im looking for the closest value to 'V(1)' which is 2000. It should return the 3rd or 5th value of N which is 2001.
Note: 'index' is the index of the closest value. If two are the same, like in this example with two different '2001's, it will return the index of the first one.
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reetu hooda il 17 Feb 2018
if N is just a decimal number and it is to be searched in a matrix V(containing decimal numbers). how would the code change?
Image Analyst il 17 Feb 2018
reetu, if N is just a single number then you can do this
[minDistance, indexOfMin] = min(abs(V-N));

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Image Analyst il 25 Ago 2014
clc;
% Sample data
numberOfRows = 5;
V = rand(numberOfRows, 1)
N = rand(numberOfRows, 1)
% Find min distance
minDistance = inf;
for ni = 1 : numberOfRows
for vi = 1 : numberOfRows
distances(vi, ni) = abs(N(ni) - V(vi));
if distances(vi, ni) < minDistance
minNRow = ni;
minVRow = vi;
minDistance = distances(vi, ni);
end
end
end
% Report to command window:
distances
fprintf('Closest distance is %f which occurs between row %d of N and row %d of V\n',...
minDistance, minNRow, minVRow);
In the command window:
V =
0.5309
0.6544
0.4076
0.8200
0.7184
N =
0.9686
0.5313
0.3251
0.1056
0.6110
distances =
0.4378 0.0005 0.2057 0.4252 0.0801
0.3142 0.1231 0.3293 0.5488 0.0435
0.5610 0.1237 0.0825 0.3020 0.2033
0.1487 0.2886 0.4948 0.7144 0.2090
0.2503 0.1870 0.3932 0.6127 0.1074
Closest distance is 0.000470 which occurs between row 2 of N and row 1 of V
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Image Analyst il 2 Nov 2017
You can try this:
% Sample data
numberOfRows = 5;
V = rand(numberOfRows, 1)
N = rand(numberOfRows, 1)
% Find min distance
distances = pdist2(V, N)
[minDistance, index] = min(distances(:))
[minVRow, minNRow] = ind2sub(size(distances), index)
fprintf('The closest distance is %f which occurs between\nrow %d of V (%f) and\nrow %d of N (%f)\n',...
minDistance, minVRow, V(minVRow), minNRow, N(minNRow));
% Double-check / Prove it
V(minVRow) - N(minNRow)
Image Analyst il 10 Nov 2017
What's wrong with a for loop? And what is ni and vi?

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Hi,
I have a matrix A of size [30x36] and B of size [38x36]. How do I find the closest pairs?
FYI- Those are the HOG Descriptors of block size 16x16 and cell size 8x8. I am trying to find the closest HOG descriptors.
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