Sum every n rows of matrix

Question

Is there any way that I can sum up columns values for each group of three rows in a matrix?
I can sum three rows up in a manual way.

For example

% matrix is the one I wanna store the new data.
% data is the original dataset.
matrix(1,1:end) = sum(data(1:3, 1:end))
matrix(2,1:end) = sum(data(4:6, 1:end))
...

But if the dataset is huge, this wouldn't work.
Is there any way to do this automatically without loops?

Answer 1

Here are four other ways:

1. The obligatory for-loop:

   % for-loop over each three rows
   matrix = zeros(size(data,1)/3, size(data,2));
   counter = 1;
   for i=1:3:size(data,1)
       matrix(counter,:) = sum(data(i:i+3-1,:));
       counter = counter + 1;
   end
   

2. Using mat2cell for tiling:

   % divide each three rows into a cell
   matrix = mat2cell(data, ones(1,size(data,1)/3)*3);
   
   % compute the sum of rows in each cell
   matrix = cell2mat(cellfun(@sum, matrix, 'UniformOutput',false));
   

3. Using third dimension (based on this):

   % put each three row into a separate 3rd dimension slice
   matrix = permute(reshape(data', [], 3, size(data,1)/3), [2 1 3]);
   
   % sum rows, and put back together
   matrix = permute(sum(matrix), [3 2 1]);
   

4. Using accumarray:

   % build array of group indices [1,1,1,2,2,2,3,3,3,...]
   idx = floor(((1:size(data,1))' - 1)/3) + 1;
   
   % use it to accumulate rows (appliead to each column separately)
   matrix = cell2mat(arrayfun(@(i)accumarray(idx,data(:,i)), 1:size(data,2), ...
       'UniformOutput',false));
   

Of course all the solution so far assume that the number of rows is evenly divisble by 3.

Answer 2

This one-liner reshapes so that all the values needed for a particular cell are in a column, does the sum, and then reshapes the back to the expected shape.

reshape(sum(reshape(data, 3, [])), [], size(data, 2))

The naked 3 could be changed if you want to sum a different number of rows together. It's on you to make sure the number of rows in each group divides evenly.