Mean of every n number of doubles in a cell

I have a cell with 990 doubles of 300 by 300 matrices. Lets call it T

I want to create a new cell with 26 doubles of 300 by 300 matrices each of which are the mean values of every 39 doubles from T (the 26th double of the new cell will be the mean of the final 15 doubles of T). i.e. (39 X 25) + (15 X 1) = 990.

Kindly help me with the code for this.

[Dyuman Joshi]

How do you want to group them?

#1 - [1-39], [40-78], [79-117], ...

#2 - [1,39,78,117, ...], [2, 40, 79, 118, ...], [3, 41, 80, 119, ...]

[Adnan Habib]

Hi Dyuman Joshi, I want the first one 1-39, 40-78 etc.

Answer 1

[Jan]

I've reduced the test data size to 30x30 matrices - use the original sizes for your implementation:

T = squeeze(num2cell(rand(30, 30, 990), 1:2));  % Some test data, {990 x 1} cell

nT = numel(T);
R  = cell(26, 1);
iR = 0;
for iT = 1:39:nT                  % Initial index of this block
    fT  = min(nT, iT + 38);       % Final index of this block
    tmp = 0;
    for k = iT:fT                 % Accumulate elements of T
        tmp = tmp + T{k};
    end
    
    iR    = iR + 1;               % Next output
    R{iR} = tmp / (fT - iT + 1);  % Mean value
end

Alternative approach:

nT = numel(T);

% Create [1, 40, 79, ...] with the last element is nT+1:
w  = 39;
iT = 1:w:nT;
iT(end + (iT(end) == nT)) = nT + 1;  % Consider nT is divisable by w

nR = numel(iT) - 1;
R2 = cell(nR, 1);
for iR = 1:nR
    fT  = iT(iR + 1) - 1;       % Final index of this block
    tmp = 0;
    for k = iT(iR):fT           % Accumulate elements of T
        tmp = tmp + T{k};
    end
    R2{iR} = tmp / (fT - iT(iR) + 1);  % Mean value
end

[Adnan Habib]

Thanks a lot Jan. The first one works. I didn't even try the alternative approach after that.