Closest Points between two datasets without using pdist2
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Alberto Belvedere
il 24 Ott 2020
Commentato: Alberto Belvedere
il 25 Ott 2020
Hi, i have two matrices A, of size mx2, and B, of size nx2.
Each row of both matrices identifies a point in 2D space. What i want to do is to write a code, that does not involve loops and pdist2, as fast as possible, that tells me the indices of the row of both A and B such that the distance squared of the two points is the minimum one.
Example:
A=[5 6;
1 2;
3 4
1 8];
B=[3 0;
2 1;
4 1;
3 5;
1 2];
My function must be like [indA,indB]=function(A_matrix,B_matrix)
I want as output [2,5]=function(A,B)
I found a solution using for-loops but i really would like to find a solution using repmat that involves vectorization.
Thanks
4 Commenti
Walter Roberson
il 25 Ott 2020
repmat is slower than implicit expansion in many cases.
There are vectorized ways to get indices of the minimum, but they are not necessarily faster than using find() (would have to be tested) and would have problems with ties.
Risposta accettata
Walter Roberson
il 25 Ott 2020
A=[5 6;
1 2;
3 4
1 8];
B=[3 0;
2 1;
4 1;
3 5;
1 2];
P = permute(sum((A-permute(B,[3 2 1])).^2,2),[1 3 2]) %will be size(A,1) by size(B,1)
A_idx_in_B = sum(cumprod(P ~= min(P,[],1),1),1)+1
B_idx_in_A = sum(cumprod(P ~= min(P,[],2),2),2)+1
This code resolves ties in favor of the first match.
3 Commenti
Walter Roberson
il 25 Ott 2020
The idx_in variables are the indices of the minimum distance. For example the closest entry in B to A(4,:) is the 4th entry of A_idx_in_B which is 3, so A(4,:) is closest to B(3,:)
Più risposte (1)
Mitchell Thurston
il 24 Ott 2020
Modificato: Mitchell Thurston
il 24 Ott 2020
Came up with a solution:
[m,~] = size(A);
[n,~] = size(B);
A_rep = repmat(A,n,1);
B_rep = B';
B_rep = repmat(B_rep(:)',m,1);
dist = hypot( A_rep(:,1)-B_rep(:,1:2:end), A_rep(:,2)-B_rep(:,2:2:end) );
ind = find(dist == min(dist));
indB = floor((ind-1)./m)+1
indA = mod(ind-(indB-1)*m,n)
3 Commenti
Mitchell Thurston
il 24 Ott 2020
Not as far as I know, this is the method I've always used for cases like this. What is nice about this though is if there's a tie for the closest it will return all of those indicies of the tie.
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