Asked by Rafael Schwarzenegger
on 2 Nov 2017

Hi, I would like to ask how I could sum over a n dimensional matrix except one dimension, so the output is a vector. The ndims are not known in forehand. The summation is giving always a vector. (In my case a marginal pdf in statistics). Something like sum(K(:) except i-th dimension)

The best in a cyclus ( ndims not known).

For example having matrix K, the sums would be [6 22], [12 16], [10 18]

K(:,:,1) =

0 1

2 3

K(:,:,2) =

4 5

6 7

Answer by Stephen Cobeldick
on 2 Nov 2017

Edited by Stephen Cobeldick
on 2 Nov 2017

Accepted Answer

Here is one very simple way:

>> d = 3; % dimension

>> v = 1:ndims(K);

>> v([1,d]) = v([d,1]);

>> s = sum(reshape(permute(K,v),size(K,d),[]),2)

s =

6

22

It works by swapping the first and the desired dimension, then reshaping so that all trailing dimensions are reduced to one. The sum is then trivially along each row.

Note that the method I show here also has the significant advantage that it does not create any variables with copies of the data from the input array: this will be important for scaling to larger input arrays. Also note that slow and complex arrayfun or cellfun are not required for this task!

Stephen Cobeldick
on 2 Nov 2017

Just change the dimension d. It is on the very first line of code, I even labeled it for you.

" It gives [6 22]. "

No, it gives all of the answers that you showed in your question. How do I know this? Because I tested it. You just need to change the dimension value d. It would be trivial to do this in a loop, if you need to.

Rafael Schwarzenegger
on 2 Nov 2017

Oh yeah, that is correct. (I thought first, it is the total dimension.) Thank you very much!

Stephen Cobeldick
on 2 Nov 2017

@Rafael Schwarzenegger: the dimensions of the input array are automatically measured using ndims(K).

Sign in to comment.

Answer by Cedric Wannaz
on 2 Nov 2017

Here is one way, but there is probably a simpler approach:

buffer = arrayfun(@(k) permute(KL, circshift(1:ndims(KL), k-1)), 1:ndims(KL), 'UniformOutput', false) ;

s = cellfun(@(M) sum(reshape(M, [], size(M,2))), buffer, 'UniformOutput', false) ;

where s is a cell array of sum vectors.

EDIT : I won't have time, but maybe you can see how dimensions work with SHIFTDIM and understand how to have the outputs ordered as needed:

buffer = arrayfun(@(k) shiftdim(KL, k-1), 1:ndims(KL), 'UniformOutput', false) ;

s = cellfun(@(M) sum(reshape(M, [], size(M,2))), buffer, 'UniformOutput', false) ;

PS : you'll have to test whether it really does what you need. The approach is based on the fact that MATLAB reads memory column first

Stephen Cobeldick
on 2 Nov 2017

@Rafael Schwarzenegger: See my answer for a simpler, neater, and much more efficient solution.

Cedric Wannaz
on 2 Nov 2017

Rafael Schwarzenegger
on 2 Nov 2017

@Cedric Wannaz: Thank you anyway.

Sign in to comment.

Opportunities for recent engineering grads.

Apply Today
## 3 Comments

## KL (view profile)

## Direct link to this comment

https://www.mathworks.com/matlabcentral/answers/364673-sum-of-matrix-omitting-one-dimension#comment_500128

## Cedric Wannaz (view profile)

## Direct link to this comment

https://www.mathworks.com/matlabcentral/answers/364673-sum-of-matrix-omitting-one-dimension#comment_500135

## Rafael Schwarzenegger (view profile)

## Direct link to this comment

https://www.mathworks.com/matlabcentral/answers/364673-sum-of-matrix-omitting-one-dimension#comment_500168

Sign in to comment.