Removing far points between vectors

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Daniel ziv
Daniel ziv on 15 May 2017
Edited: KL on 15 May 2017
Hello, i have 2 very large vectors which represent points in a 3d image, im interested in removing all points from the first vector that are farther then 'distance' from the second vector. for example: let A=(1 1 1, 10 10 10, 20 20 20) and b be (1 1 1, 2 2 2, 3 3 3), and distance be 5, only point 1 1 1 in a is closer then 5 to a certain point in b therefore the result should be C=(1 1 1)
  2 Comments
Jan
Jan on 15 May 2017
Edited: Jan on 15 May 2017
It is a benefit to use standard Matlab syntax in this forum, although we can guess how the matrix A might look like.
The distance between [1,1,1] and [3,3,3] is less then 5 also. What exactly does "to a certain point" mean here?
Daniel ziv
Daniel ziv on 15 May 2017
im sorry about my matlab ileteracy, certain point means any point. like the example if in the example b would also have a point (9,9,9), then the distance would be less then 5 and the point 10 10 10 in A would also be valid and C=[1 1 1; 10 10 10;]

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Accepted Answer

KL
KL on 15 May 2017
Edited: KL on 15 May 2017
A=[1 1 1; 10 10 10; 20 20 20]
B = [1 1 1; 2 2 2; 3 3 3]
C = A(sqrt(sum((A-B).^2,2))<=5,:) %edited
  3 Comments
Daniel ziv
Daniel ziv on 15 May 2017
im sorry A and B do not have the same size
KL
KL on 15 May 2017
Edited: KL on 15 May 2017
As Jan says, the above works only for the matrices of same size. If you want to compare matrices of different sizes and with all the elements in matrix B, then use the following one.
Arep = kron(A,ones(size(B,1),1));
Brep = repmat(B,size(A,1),1);
dAB = sqrt(sum((Arep-Brep).^2,2));
d = mean(reshape(dAB,fliplr(size(A))))
C = A(d<=5,:)

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