how to do average over third dimension?

Hi all,

I have multiple 3D matrices with the dimension of 21*41*24 for each which 24 (third dimension) is 24 hours.

I wanted to calculate 6 hours average for the third dimension.

Where I am mistaken?

NHours=size(ucom,3);
NDays=NHours/6;
out   = nanmean(permute(reshape(ucom,[6,NDays]),[2,1,3]),3);

[KSSV]

Read about mean there is option to get mean in any dimension.

[Lilya]

OK..Thank you :)

Answer 1

[Star Strider]

Taking a slightly shorter example, try something like this —

% ucom = randi(9,21,41,24);
ucom = randi(9, 3, 4, 24);
sz_ucom = size(ucom);
ucomr = reshape(ucom,sz_ucom(1),sz_ucom(2),[],6);
out = nanmean(ucomr,4)

out = 
out(:,:,1) =

    5.6667    5.5000    4.8333    3.1667
    5.8333    4.5000    5.6667    4.0000
    5.8333    5.1667    5.0000    4.5000


out(:,:,2) =

    3.8333    6.0000    4.5000    6.0000
    6.0000    5.1667    5.6667    2.5000
    3.1667    6.0000    5.3333    5.3333


out(:,:,3) =

    3.8333    3.6667    6.3333    5.3333
    4.1667    6.0000    5.5000    5.6667
    3.6667    5.0000    4.1667    6.3333


out(:,:,4) =

    5.5000    5.5000    4.1667    2.6667
    3.1667    3.6667    4.8333    4.8333
    7.0000    3.8333    5.5000    6.3333

The ‘out’ variable now has 4 blocks of 6-hour averages

.

[Lilya]

Much appreciated Sir Strider :)

It works perfectly!

[Star Strider]

As always, my pleasure!

.

[Lilya]

Can i know to to add multiple 3d to new 3d one?

ex: the 'out' matrix should be 365 for each day in a year. The size of each 'out' is (21*41*4).

I want to have a final matrix contains all 'out' to get (21*41*1460), where 1460 is 4*365

I tried 'cat' function but it doesn't take all 'out', just only the last one.

Thanks in advance.

[Star Strider]

I am not certain that I understand what you want to do, or the matrices you want to .

It might be easier to concatenate the ‘ucom’ matrices in the third dimension, the do the resshape operation and calcuiate the mean.

Example —

% ucom = randi(9,21,41,24);
ucom1 = randi(9, 3, 4, 24);
ucom2 = randi([11 19], 3, 4, 24);
ucom12 = cat(3, ucom1, ucom2);
sz_ucom = size(ucom12)

sz_ucom = 1×3

3 4 48

ucom12r = reshape(ucom12,sz_ucom(1),sz_ucom(2),[],6);
sz_ucom12r = size(ucom12r)

sz_ucom12r = 1×4

3 4 8 6

out = nanmean(ucom12r,4)

out = 
out(:,:,1) =

    9.5000   12.1667   10.1667   10.6667
    9.3333   10.1667   11.3333    9.3333
   10.5000    9.5000    9.1667   10.5000


out(:,:,2) =

    9.6667   10.3333    9.3333   10.3333
   10.0000    9.5000    9.1667    8.6667
   10.1667    8.8333   10.8333   11.6667


out(:,:,3) =

    9.8333   10.3333    9.0000   12.5000
    9.6667   10.6667   10.3333    8.8333
   10.3333    9.0000   12.3333    9.8333


out(:,:,4) =

   10.0000    8.6667   10.6667   11.6667
    9.8333    8.8333    9.3333    9.3333
   10.8333   11.1667   10.1667   10.5000


out(:,:,5) =

   10.3333    9.6667    9.3333   10.3333
   11.6667   10.1667   10.3333    8.6667
   10.3333    9.8333    9.1667    8.8333


out(:,:,6) =

   11.1667   11.0000    9.3333   10.6667
    8.1667   11.0000    9.0000   10.0000
   11.5000    8.1667    9.5000   11.5000


out(:,:,7) =

   10.5000   11.1667    8.5000    8.3333
   10.8333   11.3333    7.6667   12.0000
    9.5000   10.6667   10.8333   10.3333


out(:,:,8) =

   10.3333   10.8333    8.8333   11.1667
   11.1667    8.6667   10.8333    9.0000
    9.5000   10.8333   10.8333    9.5000

size(out)

ans = 1×3

3 4 8

It would likely be necessary to do the concatenation in a loop with 365 matrices.

That is the only way I can think of to approach this.

.

[Lilya]

Exactly!

Thank you very, very much for your response.

result = [];
nfiles = dir('*.nc');
for i=1:length(nfiles)
    for j = 1:1460
        fname = nfiles(i).name; 
        ucom            = ncread(fname,'U10M');
        sz_ucom = size(ucom);
        ucomr = reshape(ucom,sz_ucom(1),sz_ucom(2),[],6);
        out = nanmean(ucomr,4);
        result(:,:,j) = out;  
    end
end

I've used that but this error appears

Unable to perform assignment because the indices on the left side are not compatible with the size of the right side.

[Star Strider]

As always, my pleasure!

In each iteration, ‘out’ contains multiples of 4 elements in the 3rd dimension, so a single index reference is not going to work.

Rather than using an indexing scheme (that may not work well in all situations), I would just save ‘out’ to a cell array —

result{j} = out;

Recover the matrices from it later.

.

[Lilya]

It works!!

Much much appreciated.

[Star Strider]

As always, my pleasure!

(Sorry for the delay — I was off doing other things.)

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