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

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 1

[Stephen23]

>> 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):

https://www.mathworks.com/help/stats/pdist.html

[jaah navi]

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.

Answer 2

[Christine Tobler]

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.

Answer 3

[Bruno Luong]

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

Answer 4

[Fabio Freschi]

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]

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]

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]

Good point @Bruno Luong

I used this function in a "safe" environment and never met this condition. In any case, thanks for pointing out

Answer 5

[Bruno Luong]

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.

Answer 6

[Christine Tobler]

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.