Problem with the mean of index

% tr(i,j,ni) is  361*361*432
t_r = zeros(361,361,35);
ind=10:16;
for t=1:(ni/12)-1;
    if t<=34
        * *mean_tr* = mean(tr(i,j,ind));*
        t_r(i,j,t) = mean_tr;
        ind = ind  + 12;
    else
    end
end

mean_winter_sid = mean(t_r,3);

[Azzi Abdelmalek]

What is your question?

[Sophia]

why it is showing all zeros, values in tr(i,j,ni) are fine

[Star Strider]

Could it possibly be that:

mean(tr(i,j,ind))

are zero?

For example:

q = mean([-3:3])

q =
     0.0000e+000

[Sophia]

I tried

ind = 10:16;
mean_tr = mean(tr(i,j,ind));
*It is showing zeros, but this should not be the case*

Answer 1

[Azzi Abdelmalek]

You can calculate the mean without a for loop

A=rand(361,361,432);
mina=mean(A(:))  % the mean off all the matrix
min1=mean(A,1);  % dimension 1
min2=mean(A,2);  % dimension 2
min3=mean(A,3);  % dimension 3

[Sophia]

I need the mean for each winter, so 361*361*35 size So that i can use it later the way i want it to work

[Sophia]

I tried just this

ind = 10:16;
a = tr(i,j,ind);
It should be 361*361*7;
but it is showing 1*1*7, there is a problem in the ind part, i think

[Azzi Abdelmalek]

use

a = tr(1:361,1:361,ind);

[Sophia]

Its working , thanks Azzi

ind=10:16;
for t=1:(ni/12)-1;
    if t<=34
        a = tr(1:361,1:361,ind);
        mean_tr = mean(a,3);
        t_r(:,:,t) = mean_tr;
        ind = ind  + 12;
    else
    end
end

Answer 2

[Chad Greene]

Do the values of indices i and j ever change?

Can you describe in words what you are trying to do? I have a feeling it can be done much more simply and efficiently without loops.

[Sophia]

I have a 361*361*432 matrix that describes the 361*361 size monthly images for 36 years.. So i have 432 matrices, i am trying to find the average over just the winter months, say Oct to April (10:16)