Mean of each Value in a Cell Array

Hello,

I havea cell A (107x3), with each 14x8 double.

I want to create a cell A_mean 1x3 that just contain three 14x8 doubles, that are the mean of all the value.

I mean, A_mean{1,1}(1,1) should be the mean(A{1,1}(1,1),A{2,1}(1,1)...A{107,1}(1,1)), and the same for all the others numbers.

I really don't know how can I do it..

Thanks in advance :)

Answer 1

[Voss]

% a cell A (107x3), with each 14x8 double
A = arrayfun(@(~)randn(14,8),zeros(107,3),'UniformOutput',false)

A = 107×3 cell array
    {14×8 double}    {14×8 double}    {14×8 double}
    {14×8 double}    {14×8 double}    {14×8 double}
    {14×8 double}    {14×8 double}    {14×8 double}
    {14×8 double}    {14×8 double}    {14×8 double}
    {14×8 double}    {14×8 double}    {14×8 double}
    {14×8 double}    {14×8 double}    {14×8 double}
    {14×8 double}    {14×8 double}    {14×8 double}
    {14×8 double}    {14×8 double}    {14×8 double}
    {14×8 double}    {14×8 double}    {14×8 double}
    {14×8 double}    {14×8 double}    {14×8 double}
    {14×8 double}    {14×8 double}    {14×8 double}
    {14×8 double}    {14×8 double}    {14×8 double}
    {14×8 double}    {14×8 double}    {14×8 double}
    {14×8 double}    {14×8 double}    {14×8 double}
    {14×8 double}    {14×8 double}    {14×8 double}
    {14×8 double}    {14×8 double}    {14×8 double}

% I want to create a cell A_mean 1x3 that just contain three 
% 14x8 doubles, that are the mean of all the value
N = size(A,2);
A_mean = cell(1,N);
for ii = 1:N
    A_mean{ii} = mean(cat(3,A{:,ii}),3);
end
disp(A_mean);

    {14×8 double}    {14×8 double}    {14×8 double}

[Fabio Taccaliti]

Thanks a lot Voss! This it is what I was looking for!

If I would like to do the average of only just the last 50 doubles of the cell A, I have to do

N = size(A,2);
A_mean = cell(1,N);
for ii = 1:N
    A_mean{ii} = mean(cat(3,A{length(A)-50:length(A),ii}),3);
end
disp(A_mean);

Right?

[Voss]

That will be the average of the last 51 (!) 14x8 matrices in each column of A, but yeah.

Using end is better than using length(A), so do A{end-49:end,ii} to get the last 50.

[Fabio Taccaliti]

Ohh you are right!! Thanks a lot for the help!

[Voss]

You're welcome!