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# How to vectorize this code to eliminate nested For loops

Asked by DaveF on 18 Jan 2013

Would like to know how this code be vectorized:

```X=zeros(nn);
for i=1:nn
for j=1:nn
X(i,j)=x(i)-x(j);
end
end
```

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Answer by Sean de Wolski on 18 Jan 2013
```X = bsxfun(@minus,x(1:nn).',x(1:nn));
```

## 1 Comment

Cedric Wannaz on 18 Jan 2013
` bsxfun(@minus,x(1:nn),x(1:nn).')`
Answer by Jan Simon on 18 Jan 2013
Edited by Jan Simon on 19 Jan 2013

This cannot compete with the BSXFUN approach, but it is 32% faster than the original loop:

```nn = numel(x);
X  = zeros(nn);
for i = 1:nn
for j = i+1:nn
a      = x(i) - x(j);
X(i,j) = a;
X(j,i) = -a;
end
end
```

Using a temporary variable for x(i), which is updated in the outer loop only, is slightly slower.

Cedric Wannaz on 19 Jan 2013

Well then this one would be more efficient:

``` X  = zeros(nn);
for ii = 1:nn
X(:,ii) = x - x(ii);
end```

It almost competes with bsxfun for nn=1e4.

Jan Simon on 19 Jan 2013

@Cedric: When you post this as an answer, you get my vote.

Your version does not exploit the symmetry as the other approaches. But this modification is much slower:

```X=zeros(nn);
for ii = 1:nn
a       = x(ii:nn) - x(ii);
X(ii:nn, ii) = a;
X(ii, ii:nn) = -a;
end
```

Obviously the overhead for explicit indexing is too high.

Cedric Wannaz on 19 Jan 2013

@Jan : thank you! Knowing that I would get your vote is enough, so I'll keep that as a comment ;-)

I just profiled your last version and it is very interesting indeed, because I would never have expected such an overhead. It might even lead me to review my code in a few projects, so thank you for the comment!

Answer by Jan Simon on 19 Jan 2013

And a C-Mex, which has the same speed as BSXFUN for a [1 x 1e3] vector under Matlab R2009a/64, Win7, MSVC 2008:

```#include "mex.h"
```
```void mexFunction(int nlhs, mxArray *plhs[], int nrhs, const mxArray *prhs[]) {
double *X, *Y;
mwSize i, j, n;
```
```     n = mxGetNumberOfElements(prhs[0]);
X = mxGetPr(prhs[0]);```
```     plhs[0] = mxCreateDoubleMatrix(n, n, mxREAL);
Y       = mxGetPr(plhs[0]);```
```     for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
*Y++ = X[i] - X[j];
}
}
}```

Answer by Azzi Abdelmalek on 18 Jan 2013
Edited by Azzi Abdelmalek on 18 Jan 2013
```x=[1 3 4 10 20 30];
X=cell2mat(arrayfun(@(y) y-x,x','un',0))
```

%or

```X=bsxfun(@minus,repmat(x',1,numel(x)),x)
```

Cedric Wannaz on 18 Jan 2013

Or

` [X, Xt] = meshgrid(x, x) ;  Xt-X`

More efficient than cell2mat/arrayfun and less efficient than bsxfun. So the latter (bsxfun) should be the winner ;) But Sean's version without repmat beats them all.

Sean de Wolski on 18 Jan 2013

Well the for-loops will smoke arrayfun/cell2mat like a Lambo would smoke bicycle...

bsxfun will be faster than the for-loops some of the time; meshgrid will only be faster if you need to reuse the gridded matrices.

Answer by Matt J on 19 Jan 2013
``` y=exp(x(1:nn));
X=log( y * (1./y.')  );```