Find closest value in array
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Chiranjibi
il 25 Ago 2014
Commentato: Image Analyst
il 8 Feb 2025 alle 18:00
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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Joe S
il 14 Nov 2024
Modificato: MathWorks Support Team
il 14 Nov 2024
Questo/a risposta è stato/a segnalato/a da Walter Roberson
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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Più risposte (5)
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.
4 Commenti
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));
Image Analyst
il 25 Ago 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 Commenti
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)
John D'Errico
il 10 Lug 2024
Modificato: John D'Errico
il 10 Lug 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.
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Steven Lord
il 4 Feb 2025 alle 19:04
As of release R2021b, you don't need to use abs on the input to the min function. Use the 'ComparisonMethod' option to tell MATLAB to take the minimum value by absolute value.
V = randi(100,[5 1])
N = randi(100,[5 1])
[minDifference,closestIndex] = min(N-V.', [], ComparisonMethod = "abs")
closestValue = N(closestIndex)
4 Commenti
Steven Lord
il 8 Feb 2025 alle 16:43
For the case you described you probably want to use the 'rows' option for ismember. [This assumes the row you're searching for is in the matrix you're searching, not 'close to' a row in the matrix. In that case you'd want to use ismembertol with ByRows=true.]
% Define variables to search and search for.
allValues = [1,20; 2,30; 3, 40; 4, 50; 5, 60]
sought = [3, 40];
[~, whichRow] = ismember(sought, allValues, 'rows')
matchpairs is intended to solve a different problem. It solves the combinatorial optimization assignment problem. It is intended to associate the elements of two different sets along with a cost or value for each association to minimize cost or maximize value, matching pairs of elements from those sets. One example is if you have workers and tasks, and each worker is more or less skilled with different tasks or is more or less satisfied with those tasks. You may want to assign each task to a worker maximizing the overall skill or the job satisfaction of the group of workers with their assignments.
Korosh Agha Mohammad Ghasemi
il 25 Giu 2024
Spostato: Voss
il 25 Giu 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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