Find closest value in array
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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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Accepted Answer
Joe S
on 10 Sep 2018
Edited: MathWorks Support Team
on 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.
6 Comments
David
on 5 Jul 2023
This just saved my night!
For anybody juse searching the index, faster variant:
[~,closestIndex] = min(abs(N-V));
More Answers (4)
Andrew Reibold
on 25 Aug 2014
Edited: Andrew Reibold
on 25 Aug 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.
4 Comments
reetu hooda
on 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
on 17 Feb 2018
reetu, if N is just a single number then you can do this
[minDistance, indexOfMin] = min(abs(V-N));
Image Analyst
on 25 Aug 2014
How about this:
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
3 Comments
Image Analyst
on 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)
John D'Errico
on 10 Jul 2024
Edited: John D'Errico
on 10 Jul 2024
To be honest, the easiest way is to use knnsearch. It works well in one dimension, as you have here, and it should be quite efficient.
V = [1375471092848936; 1375473388165900; 1375471320476780; 1375473388947681; 1375473392527002];
N = [1375473384440853; 1375471277856598; 1375473388165900; 1375471322465961; 1375471335206288];
help knnsearch
ids = knnsearch(N,V)
There is no need to look at differences, compute absolute values, etc. Just use the tool that is designed to solve your problem directly.
0 Comments
Korosh Agha Mohammad Ghasemi
on 25 Jun 2024
Moved: Voss
on 25 Jun 2024
% Example V and N vectors
V = [1375471092848936; 1375473388165900; 1375471320476780; 1375473388947681; 1375473392527002];
N = [1375473384440853; 1375471277856598; 1375473388165900; 1375471322465961; 1375471335206288];
W = 1e4; % Window size
% Initialize the closest times array
closest_times = zeros(size(V));
% Find the closest time in N for each time in V within the window
for i = 1:length(V)
% Calculate the absolute differences
diffs = abs(N - V(i));
% Find the indices within the window
within_window = diffs <= W;
if any(within_window)
% Find the closest time
[~, closest_idx] = min(diffs(within_window));
% Get the actual index in N
closest_times(i) = N(find(within_window, closest_idx, 'first'));
else
% No times within the window
closest_times(i) = NaN;
end
end
% Display the closest times
disp('Closest times:');
disp(closest_times);
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