Asked by jaah navi
on 11 Sep 2019

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.

Answer by Stephen Cobeldick
on 11 Sep 2019

Edited by Stephen Cobeldick
on 11 Sep 2019

Accepted Answer

>> 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):

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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Answer by 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.

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Answer by Fabio Freschi
on 11 Sep 2019

Edited by 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'));

Fabio Freschi
on 11 Sep 2019

Check

>> N = 10000;

>> A = rand(N,2);

>> tic; D = sqrt(abs(bsxfun(@plus,sum(A.*A,2),sum(A.*A,2).')-2*A*A')); toc

Elapsed time is 1.245305 seconds.

>> tic; B = sqrt(sum(bsxfun(@minus,permute(A,[1,3,2]),permute(A,[3,1,2])).^2,3)); toc

Elapsed time is 1.589360 seconds.

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
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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Answer by 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.

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Answer by Christine Tobler
on 24 Sep 2019

Edited by 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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Answer by Bruno Luong
on 25 Sep 2019

Hi Christtine, very good initiative. I think at least 90% of use-case would be p=2.

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