Summing within an array to change the size

I have an array that has dimensions of 111x46x2. I want to sum the values in the first 3x3 block to become the first value of the next matrix repeating this until the last column is summed together. The dimensions of the new matrix should be 37x16x2.

I could make the dimensions of the original matrix 111x48x2 if that makes the calculations easier and more efficient. The 9x9 blocks will have values only in certain locations such as [ 0 0 #; # 0 0; 0 # 0]; and the last column will have values in each row.

[Sean de Wolski]

46 or 48?

[Matthew Vincent]

It would be easiest if I could keep the original array at 46 and the last "block" would be a 3x1 column but it would be possible to change it to 48 just inconvenient in the code.

[Sean de Wolski]

blockproc below is smart enough to handle either of those scenarios :)

[Walter Roberson]

Please read the guide to tags and retag this Question; see http://www.mathworks.co.uk/matlabcentral/answers/43073-a-guide-to-tags

Answer 1

[Sean de Wolski]

If you have the Image Processing Toolbox:

x = rand(111,46,2);
y = blockproc(x,[3 3],@(s)sum(sum(s.data,1),2));

Or if you don't:

z = convn(x,ones(3),'valid');
z = z(1:3:end,1:3:end,:); %may need to crop differently depending on last edges

[Matthew Vincent]

I think that the convn function will work for me. Have other things I need to incorporate before I can test it and be sure. Thanks

Answer 2

[Matt Fig]

Here is one way to do it:

A = magic(6);
cellfun(@(x) sum(x(:)),mat2cell(A,2*ones(1,3),2*ones(1,3)))

[Matthew Vincent]

Is there anyway to expand this to be useful for an array input? I attempted to use num2cell but am not very familiar with that function

[Sean de Wolski]

mat2cell is painfully slow for anything at all large. I recommend avoiding it:

x = rand(999,999,3);
tic,mat2cell(x,ones(1,333)*3,ones(1,333)*3,3);toc

Elapsed time is 7.549038 seconds.

[Matt Fig]

What do you mean 'array input'? A is an array....

[Matt Fig]

MAT2CELL is slow for large inputs, indeed. What are you MathWorkers doing with your time?! ;-)

Answer 3

[Jan]

You can use the fast FEX: BlockMean. Either modify the code or multiply the results accordingly.

A Matlab version:

x = rand(111, 46, 2);

S = size(X);
M = S(1) - mod(S(1), 3);
N = S(2) - mod(S(2), 3);

% Cut and reshape input such that the 1st and 3rd dimension have the lengths V
% and W:
XM = reshape(X(1:M, 1:N, :), 3, M / 3, 3, N / 3, []);

Y = squeeze(sum(sum(XM, 1), 3));

[Andrei Bobrov]

with two reshape

s = size(x);
t = mod(-s(1:2),[3 3]);
xx = [x zeros(s(1),t(2),s(3))];
xx = [xx; zeros(t(1),s(2)+t(2),s(3))];
s2 = s + [t, 0];
y = reshape(sum(sum(reshape(xx,3, s2(1)/3, 3,[]),3)),s2(1)/3,s2(2)/3,[]);

[Jan]

Difference between Andrei's and my solution:

• RESHAPE instead of SQUEEZE (which calls RESHAPE internally)
• Andrei appends zeros, I remove the not matching lines on the bottom.