Nearest Neighbor Matching without Replacement

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Al
Al il 7 Dic 2016
Risposto: Bruno Luong il 25 Feb 2022
Hello there,
I am looking to match data in two vectors, x and y, based on shortest Euclidean distance. Each match should be unique; that is, numbers in vectors x and y cannot be matched twice. I have looked into knnsearch, but have not found anything that suggests the function works without replacement. Thank you!
  3 Commenti
Phillip
Phillip il 30 Mar 2018
It does not violate the shortest distance if it is removed from the set once matched (what Al is referring too when he says " without Replacement").
Peng Li
Peng Li il 31 Mar 2020
any update here? also was wondering if there is a way to do knn search without replacement.

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Risposte (2)

neuroDuck
neuroDuck il 25 Feb 2022
A late response, but for anyone that might come across this in future, I think a brute force approach should work, with the following code, assuming you don't have too many comparisons to go through:
[cIdx] = unique(knnsearch(x,y));
% brute force knnsearch to do without replacements
startingK = 2;
while length(cIdx)<length(y)
[cIdx] = unique(knnsearch(x,y,'k',startingK));
startingK = startingK + 1;
end

Bruno Luong
Bruno Luong il 25 Feb 2022
If you have R2019a release
x=rand(1,10)
x = 1×10
0.7999 0.9374 0.3693 0.3038 0.4504 0.9999 0.9559 0.9041 0.6531 0.1136
y=rand(1,10)
y = 1×10
0.2727 0.5885 0.7090 0.6540 0.6661 0.3712 0.2943 0.0732 0.8695 0.5349
C=abs(x(:)-y(:).');
M = matchpairs(C,max(C(:)));
px = M(:,1);
xm = x(px)
xm = 1×10
0.3693 0.6531 0.9999 0.7999 0.9559 0.4504 0.3038 0.1136 0.9374 0.9041
d = abs(xm-y)
d = 1×10
0.0966 0.0646 0.2910 0.1459 0.2898 0.0792 0.0094 0.0404 0.0679 0.3692

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