Cleaner way to calculate mean

I have an array with 18 columns - each column is a cell array itself of varying lengths. I am trying to perform basic calculations with this, finding the mean, but all the cells are divided by different cells. So cell 4/cell 1. Cell 6/3

A = [21x8 cell  46x8 cell  13x8 cell  11x8 cell  27x8 cell  4x8 cell  10x8 cell  77x8 cell  173x8 cell]

I was going to write something similar to what is below, but that would require a lot of repeating lines. Is there a cleaner/more efficient way to write this?

for k = 1:size(A)
                for h = 1:length(A)
                      B = length(A{k,h});
                      avg.A = (B(a,6)/C(a,3))*100;
                      avg.B = (B(a,5)/C(a,2))*100;
                      avg.C = (B(a,4)/C(a,1))*100;
                      avg.D = (B(a,9)/C(a,3))*100;
                      avg.E = (B(a,8)/C(a,2))*100;
                      avg.F = (B(a,7)/C(a,1))*100;

                  end
              end

...and so forth. As you can see, this will produce 18 different lines of code which looks terrible.

[dpb]

Well, no, "I don't see", at least completely... :)

What is a in

(B(a,6)/C(a,3))*100;
(B(a,5)/C(a,2))*100;
(B(a,4)/C(a,1))*100;
(B(a,9)/C(a,3))*100;
(B(a,8)/C(a,2))*100;
(B(a,7)/C(a,1))*100;

?

Since there's an apparent symmetry in the second subscripts as paired with each other, what's the general rule that determines those?

B=length(A{k,h});

will be a constant so how can it have subscript (a,N).

What is the actual arithmetic computation you're trying to accomplish?

Show us a (small) sample A and then the desired output from it and how got it from the input. It shouldn't take but a small subset of the real problem with small(ish) arrays to illustrate.

Answer 1

[Image Analyst]

I have no idea what B, C, and "a" are, but this will get you the mean of all cells of A

for k = 1 : length(A)
    thisCellsContents = A{k}; % Extract double array.
    meansOfA(k) = mean(thisCellsContents(:));
end

You could probably do it more compactly, though more cryptically, with cellfun().