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average sections of matrix

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Luke McLellan
Luke McLellan el 12 de Jul. de 2018
Comentada: Star Strider el 17 de Jul. de 2018
Hi there, I am trying to average sections of a matrix at a time. The matrix is 31x109. Initially I am trying to build another matrix from the average of the first 25 elements increments in the first row, then the second 25(beginning from the the last element in the previous average); this will later be applied to all rows. My current attempt is
for f = 1:[1:25:108]; Ave_PpIX(f,:) = PpIX_Dose_Matrix(1,1:f); end
ave_PpIX = mean(Ave_PpIX);
Am idea of the matrix is: [ave(1-25) ave(13-38) ave(25-50)...]
If anyone has had any previous experience doing this I would really appreciate any help.
Thnaks Luke
  1 comentario
Jan
Jan el 12 de Jul. de 2018
Why 108 in the loop, not 109? What should happen with the last chunk, which has less then 25 elements?
1:[a:b:c] is the same as 1:a.

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Star Strider
Star Strider el 12 de Jul. de 2018
I am not certain that I understand the result you want.
Try this:
M = rand(31, 109); % Original Matrix
Cols = [ones(1, fix(size(M,2)/25))*25 rem(size(M,2),25)]; % Create Column Sub-Sections
C = mat2cell(M, size(M,1), Cols); % Cell Array Of Sub-Sections
Cmean = cellfun(@(x)mean(x,2), C, 'Uni',0); % Row Means For Each Section
Mmean = cell2mat(Cmean); % Convert To Double Matrix Of Row Mean Vectors For Each Sub-Section
  9 comentarios
Luke McLellan
Luke McLellan el 17 de Jul. de 2018
I do believe that you are my knight in shining algorithm but I will have to return to the code next week now
Star Strider
Star Strider el 17 de Jul. de 2018
Thank you!

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Jan
Jan el 12 de Jul. de 2018
Editada: Jan el 13 de Jul. de 2018
PpIX_Dose_Matrix = rand(31, 109);
n = size(PpIX_Dose_Matrix, 2);
k = 0;
for f = 1:24:n
k = k + 1;
Ave_PpIX(k) = mean(PpIX_Dose_Matrix(1, f:f+24));
end
This can fail, when f+25 is outside the existing range. A test with if can catch this:
if f+25 <= n
...
  3 comentarios
Jan
Jan el 13 de Jul. de 2018
for f = 1:12:n
k = k + 1;
Ave_PpIX(k) = mean(PpIX_Dose_Matrix(1, f:f+24));
end
Luke McLellan
Luke McLellan el 16 de Jul. de 2018
Thanks again for your comments

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