How to measure the distance between 2 sets of points.

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Jim Lehane
Jim Lehane 2013-3-4
编辑： udara darshana panamulle arachchige 2018-12-10
I was wondering how I measure the distance between 2 sets of coordinates.
I have 2 matrices where each matrix (X and Y) have the same number of rows (m x 2). I calculated out the distance from every point in X to every point in Y using the pdist2 tool. This gave me a m x m matrix of distances.
I want the minimum distance between X and Y where each coordinate is in a 1-to-1 ratio between X and Y. I am not sure how to separate the distances out of the pdist2 tool to discrete sets.
Does anyone know how to do this or where I can find the answer?
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Matt J
Matt J 2013-3-4
Clarify what you mean by "is in a 1-to-1 ratio between X and Y".
Jim Lehane
Jim Lehane 2013-3-4
编辑：Jim Lehane 2013-3-4
Every point in X corresponds with 1 point in Y. So that there are no duplicates. (ie X1 to Y3; X2 to Y1; X3 to Y2...)

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Matt J
Matt J 2013-3-4
编辑：Matt J 2013-3-4
m=3;
Q=perms(1:m).';
P=repmat((1:m).',1,size(Q,2));
idx=sub2ind([m,m],P,Q);
result = min(sum(DistMatrix(idx)))
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udara darshana panamulle arachchige
udara darshana panamulle arachchige 2018-12-10
编辑：udara darshana panamulle arachchige 2018-12-10
I have a same qustion but data set size is bit larger A(15,3) and B(105,3)
using this method gives a error due to exceeding maximum array size.
Is there any other way to do this Matt J??
Q=perms(1:15).'; %% here I get the error
P=repmat((1:103).',1,size(Q,2));
idx=sub2ind([15,103],P,Q);
result = min(sum(DistMatrix(idx)));
ps: my qustion is same with Jim Lehane

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Azzi Abdelmalek
Azzi Abdelmalek 2013-3-4
编辑：Azzi Abdelmalek 2013-3-4
M1 & M2 are your two matrices
M1=rand(10,2);
M2=rand(10,2)
dist=sqrt((M2(:,1)-M1(:,1)).^2+(M2(:,2)-M1(:,2)).^2)
min_dist=min(dist)
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Jim Lehane
Jim Lehane 2013-3-4
For example, say I have 2 matrices X and Y with 3 coordinates each. The matrix below represents the pdist2 result of those with the distances.
Y1 Y2 Y3
X1 2 5 7
X2 4 1 6
X3 3 8 9
I want the minimum combined distance of each unique combination of distance:
X1,Y1 + X2,Y2 + X3,Y3 =
X1,Y1 + X2,Y3 + X3,Y2 =
X1,Y2 + X2,Y1 + X3,Y3 =
X1,Y2 + X2,Y3 + X3,Y1 =
X1,Y3 + X2,Y1 + X3,Y2 =
X1,Y3 + X2,Y3 + X3,Y1 =
In this instance there are a total of 6 possible combination of coordinates (3!).

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