have a func(:,:,t) that I took the average of, how do I create just an array of those values?

So I have iterations of a 2-D temperature profile, in which I do

    Tavg = mean(mean(T(:,:,:)));

where T(i,j,t) is what I used.

Anyways, "Tavg" spits back a matrix of size [1 1 "#timesteps"]

Not sure why it's that size because when I type "Tavg" it gives me:

             Tavg(:,:,17280) =

            306.7141

an actual value. (note, that's one of thousands necessary for my script)

How would I take all those computed averages to be able to plot them against time? Seems trivial, but I can't figure it out.

Answer 1

[Cedric]

MEAN computes a mean along the 1st non-singleton dimension of its (1st) arg, unless you provide the dim as a second arg.

 >> A = rand(2)
 A =
    0.4218    0.7922
    0.9157    0.9595
 >> mean(A)
 ans =
    0.6687    0.8758

that was along dim 1. Now if A had been a 1x2 array (dim 1 is a singleton):

 >> A = rand(1,2)
 A =
    0.6557    0.0357
 >> mean(A)
 ans =
    0.3457

you see that the mean was performed over dim 2 (as dim 1 is a singleton dimension).

In your case, the inner MEAN performs a mean along dim 1 and reduces it to singleton. The outter MEAN performs a mean along dim 2 (1st non-singleton). You are left with a 3D array with two singleton dims, which is what is displayed..

 >> A = rand(2, 2, 3)
 A(:,:,1) =
    0.8491    0.6787
    0.9340    0.7577
 A(:,:,2) =
    0.7431    0.6555
    0.3922    0.1712
 A(:,:,3) =
    0.7060    0.2769
    0.0318    0.0462

 >> mean(A)                  % Along dim 1.
 ans(:,:,1) =
    0.8916    0.7182
 ans(:,:,2) =
    0.5677    0.4133
 ans(:,:,3) =
    0.3689    0.1615

 >> mean(mean(A))            % Along dim 1 and 2, but still 3D!
 ans(:,:,1) =
    0.8049
 ans(:,:,2) =
    0.4905
 ans(:,:,3) =
    0.2652

You can actually SQUEEZE singleton dimensions if/when needed:

 >> squeeze(mean(mean(A)))
 ans =
    0.8049
    0.4905
    0.2652

but you could also permute dimensions:

 >> mean(mean(permute(A, [3,1,2]), 3), 2)
 ans =
    0.8049
    0.4905
    0.2652

or

 >> mean(mean(permute(A, [1,3,2]), 3))
 ans =
    0.8049    0.4905    0.2652

So .. no, it was not completely trivial ;-)

[Jon]

Wow, thorough and perfect answer, Cedric.

Many thanks!

[Cedric]

My pleasure. Just have a look again at the 1st sentence please; I think that I reworded it while you were posting your comment, so you might not have seen the update!