Closest Points between two datasets without using pdist2

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Alberto Belvedere
Alberto Belvedere 2020-10-24
评论： Alberto Belvedere 2020-10-25
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
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Walter Roberson
Walter Roberson 2020-10-25
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.
Alberto Belvedere
Alberto Belvedere 2020-10-25
How could i use implicit expansion in this case? Furthermore, is it better to use a combination of find+min or to use sort twice (first by rows, then by columns) on the resultant matrix?

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采纳的回答

Walter Roberson
Walter Roberson 2020-10-25
Ran in:
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)
P = 4×5
40 34 26 5 32 8 2 10 13 0 16 10 10 1 8 68 50 58 13 36
●
A_idx_in_B = sum(cumprod(P ~= min(P,[],1),1),1)+1
A_idx_in_B = 1×5
2 2 2 3 2
●
B_idx_in_A = sum(cumprod(P ~= min(P,[],2),2),2)+1
B_idx_in_A = 4×1
4 5 4 4
●
This code resolves ties in favor of the first match.
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Walter Roberson
Walter Roberson 2020-10-25
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,:)

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更多回答（1 个）

Mitchell Thurston
Mitchell Thurston 2020-10-24
编辑：Mitchell Thurston 2020-10-24
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)
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Mitchell Thurston
Mitchell Thurston 2020-10-24
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.
Alberto Belvedere
Alberto Belvedere 2020-10-25
I'll try some minor tweaks tomorrow starting from your excellent job. Thank you so much for your time!

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