meanMatrix = (matrix1 + matrix2 + matrix3 + .... etc...... + matrix 30 + matrix31)/31;

Yes, concatenate the matrices into a 3-D array and use mean(x,3) x = randn(4,4,10); mean(x,3) and std(x,[],3) To concatenate, you can use cat() x = randn(4,4); y = randn(4,4); z = cat(3,x,y); mean(z,3) std(z,[],3)

AVG = (R + G)/2; It works well. No need of a "for" loop.

If I read your question to the letter, you have 12263 recordings of pupil sizes. Each recording lasted 784 ms. Recoring frequency varied between 1000 Hz and (1134/784)*1000 Hz, resulting in different number of samples between recordings ... In this case: resample each recording to a the same timestampes using INTERP1 (or resample or something along these lines); concatenate all cells using CAT and get the mean MyFun = @(x) interp1(linspace(1,784,numel(x)), x, 1:784) ; pupil_resampled = cellfun(MyFun, pupil, 'un', 0) ; pupil_mean = mean(cat(2, pupil_resampled), 2) If you have a single sample frequency but sample durations vary, try concatenating using, for instance, PADCAT, which pads shorter arrays with NaNs and use nanmean pupil_padded = padcat(pupil{:}) ; pupil_mean = nanmean(pupil_padded, 2) PADCAT can be found on the File Exchange: https://uk.mathworks.com/matlabcentral/fileexchange/22909-padcat-varargin You might want to tweak the dimension arguments for mean and nanmean depending on the oriention of your pupil cells (all row or column vectors)

R = mean(reshape(X(1, 1, :, :, :), 1, []))

% make dummy data N = 8 ; for i = 1:N C{i} = rand(2,2,3) ; end % get the mean N = length(C) ; [m,n,p] = size(C{1}) ; themean = zeros(m,n,p) ; for i = 1:N themean = themean+C{i} ; end themean = themean/N ; Can be obtained without loop, using Cell2mat, reshape etc.

Hi, if you have Statistics Toolbox, you can use |nanmean| as follows: matrixA = [2 3 5 ; 3 NaN 1 ; 2 4 3]; matrixB = [3 4 NaN ; 1 2 5 ; NaN 3 5]; matrixC = [NaN 2 3 ; 2 5 3 ; 1 2 1]; allData = cat(3,matrixA,matrixB,matrixC); nanmean(allData,3);

As a general strategy, you could generate a large default array using zeros() based on the largest matrix. Then assign all the data into the large nan (or zeros) array and then manually calculate the average. Simple Example: A_tmp = rand([3,4]); B_tmp = rand([7,4]); C_tmp = rand([5,4]); defArry = zeros([max([size(A_tmp,1) size(B_tmp,1) size(C_tmp,1)]),4]); A = defArry; A(1:size(A_tmp,1),1:4) = A_tmp; B = defArry; B(1:size(B_tmp,1),1:4) = B_tmp; C = defArry; C(1:size(C_tmp,1),1:4) = C_tmp; D = (A + B + C) / 3

You would do:

c = (a+b)/2

This will give you the desired result.