Mean of every n number of doubles in a cell

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Adnan Habib 2023-2-28

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https://ww2.mathworks.cn/matlabcentral/answers/1920125-mean-of-every-n-number-of-doubles-in-a-cell

评论： Adnan Habib 2023-2-28

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.

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Dyuman Joshi 2023-2-28

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 2023-2-28

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

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Answer 1

Jan 2023-2-28

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1920125-mean-of-every-n-number-of-doubles-in-a-cell#answer_1181645

编辑：Jan 2023-2-28

在 MATLAB Online 中打开

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