Find closest value in array
2 202 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Chiranjibi
le 25 Août 2014
Modifié(e) : John D'Errico
le 10 Juil 2024 à 18:40
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
0 commentaires
Réponse acceptée
Joe S
le 10 Sep 2018
Modifié(e) : MathWorks Support Team
le 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 commentaires
David
le 5 Juil 2023
This just saved my night!
For anybody juse searching the index, faster variant:
[~,closestIndex] = min(abs(N-V));
Plus de réponses (4)
Andrew Reibold
le 25 Août 2014
Modifié(e) : Andrew Reibold
le 25 Août 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 commentaires
reetu hooda
le 17 Fév 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
le 17 Fév 2018
reetu, if N is just a single number then you can do this
[minDistance, indexOfMin] = min(abs(V-N));
Image Analyst
le 25 Août 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 commentaires
Image Analyst
le 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
le 10 Juil 2024 à 18:36
Modifié(e) : John D'Errico
le 10 Juil 2024 à 18:40
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 commentaires
Korosh Agha Mohammad Ghasemi
le 25 Juin 2024 à 16:39
Déplacé(e) : Voss
le 25 Juin 2024 à 17:02
% 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);
0 commentaires
Voir également
Catégories
En savoir plus sur Characters and Strings dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!