could anyone help me to calculate the euclidean distance for the matrix.

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A= [1.8667 0.1553;
-0.0844 2.4322;
-0.3485 1.4434;
2.3628 0.6821]
I want to calculate the euclidean distance of first row with respect to second,third and fourth row.
Similarly i want to calculate the euclidean distance of second row with respect to first,third and fourth row and so on.
could anyone please help me onthis.

Accepted Answer

Stephen23
Stephen23 on 11 Sep 2019
Edited: Stephen23 on 11 Sep 2019
>> B = sqrt(sum(bsxfun(@minus,permute(A,[1,3,2]),permute(A,[3,1,2])).^2,3))
B =
0.00000 2.99851 2.56248 0.72363
2.99851 0.00000 1.02346 3.00859
2.56248 1.02346 0.00000 2.81615
0.72363 3.00859 2.81615 0.00000
Or use pdist (requires the Statistics Toolbox):
  1 Comment
jaah navi
jaah navi on 11 Sep 2019
i used pdist to calculate the euclidean distance with respect to the following code;
code:
PPP=[1.8667 0.1553;
-0.0844 2.4322;
-0.3485 1.4434;
2.3628 0.6821]
D = pdist(PPP)
f=sum(D)/numel(D)
indices = find(abs(D)<f)
D(indices) = []
when i execute the above code i can get the result as
D = [2.9985 2.5625 3.0086 2.8162]
but i want to display which two rows gives the above result.the expected output is
D=[(1,2) (1,3) (2,4) (3,4)]
Could you please help me on it.

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More Answers (5)

Christine Tobler
Christine Tobler on 11 Sep 2019
For MATLAB R2017b or later, you can use the vecnorm function for a simpler construction than the one involving sqrt, sum, and .^2:
vecnorm(permute(A,[1,3,2]) - permute(A,[3,1,2]), 2, 3)
This will give the exact same result as constructing two for-loops and computing B(i, j) = norm(x(i, :) - x(j, :)) individually for each combination.

Bruno Luong
Bruno Luong on 25 Sep 2019
Hi Christtine, very good initiative. I think at least 90% of use-case would be p=2.

Fabio Freschi
Fabio Freschi on 11 Sep 2019
Edited: Fabio Freschi on 11 Sep 2019
I have found that this is usually the fastest way, since the square of the binomial is unrolled
D = sqrt(abs(bsxfun(@plus,sum(A.*A,2),sum(A.*A,2).')-2*A*A'));
If you have two sets of points A and B
D = sqrt(abs(bsxfun(@plus,sum(A.*A,2),sum(B.*B,2).')-2*A*B'));
  3 Comments
Bruno Luong
Bruno Luong on 11 Sep 2019
Edited: Bruno Luong on 11 Sep 2019
The decompose method might have worst roundoff numerical issue:
A=[1e9 0;
1e9+1 0]
D = sqrt(abs(bsxfun(@plus,sum(A.*A,2),sum(A.*A,2).')-2*A*A'))
B = sqrt(sum(bsxfun(@minus,permute(A,[1,3,2]),permute(A,[3,1,2])).^2,3))
B is correct D is not.
It could even return complex numbers.
Fabio Freschi
Fabio Freschi on 11 Sep 2019
Good point @Bruno Luong
I used this function in a "safe" environment and never met this condition. In any case, thanks for pointing out

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Bruno Luong
Bruno Luong on 11 Sep 2019
I would put as comment of Christine's vecnorm soluton, but somehow Answers rejects it.
The vecnorm is slower than standard solution (Stephen's) in case of p = 2.
N = 10000;
A = rand(N,2);
tic
B = sqrt(sum((permute(A,[1,3,2])-permute(A,[3,1,2])).^2,3));
toc % Elapsed time is 1.249531 seconds.
tic
C = vecnorm(permute(A,[1,3,2]) - permute(A,[3,1,2]), 2, 3);
toc % Elapsed time is 4.590562 seconds.

Christine Tobler
Christine Tobler on 24 Sep 2019
Edited: Christine Tobler on 24 Sep 2019
Hi Bruno, I have the same problem where I can't comment on your answer, so adding another answer here.
That's a good point - vecnorm returns the exact same value as vecnorm, however it's not been optimized for performance as sum. Still, with all the additional overhead in the computation using sum, .^2 and sqrt, it would make sense for vecnorm to be faster here. I've made an enhancement request for this.

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