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Looping through 3rd dimension

Asked by ellis langdon on 13 Feb 2019
Latest activity Commented on by Matt J
on 14 Feb 2019
Accepted Answer by Jan
Hi
I am currently working on a project with temperature data. The data is stored in a 128 x 64 x 2160 matrix called T.
I have calculated that summer is between index 30:48 and then +72 for the consecutive summer years.
I want to loop through the third dimension and stored this in a new variable ST.
Please see below code I have already made, i'm not too sure where to go after this.
ST=NaN*ones(128,64,30) %Setting up grid
x=0;
for i=30:48;
x=x+1;
ST(:,:,x) = T(:,:,[i:i]); %Looping through third dimension 30:48
end
ST should essentially be 128 x 64 x 30 or 128 x 64 x 540 matrix
Any help would be great, thank you!

  3 Comments

I think you mean the final matrix should be 128 x 64 x 570. Your summer data has 19 elements times 30 years = 570.
Sorry, it should actually have 18 elements so the final grid should be 128 x 64 x 540. However, if I modify the code for the ammount of elements I need, should that work?
Thank you
Yes.

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2 Answers

Answer by Jan
on 13 Feb 2019
Edited by Jan
on 14 Feb 2019
 Accepted Answer

Your loop has 19 iterations. It is not clear, why you expect that the output has a size of 30 or 540 in the 3rd dimension.
[i:i] is the same as the much simpler: i .
A simplified version of your code:
ST = T(:, :, 30:48)
No loop needed. But I do not see, how this could produce a size of 30 or 540.
By the way, this can be simplified also:
ST=NaN*ones(128,64,30)
% Better:
ST = NaN(128, 64, 30)
But it is not useful here, so better omit it.
[EDITED] After the clarifications:
ST = NaN(128, 64, 18, 30); % Pre-allocate
ini = 30;
len = 18;
for k = 1:30
ST(:, :, :, k) = T(:, :, ini:ini + len - 1);
ini = ini + 72;
end
ST = reshape(ST, 128, 64, 540);
% Or:
v = 1:72;
match = (30 <= v & v < 48);
ST = T(:, :, repmat(match, 1, 30));

  2 Comments

I am trying to put all of the summer values into the new grid (ST). The summer values are between 30 and 48, which increase annually by 72.
For example the loop will put all the values between 30:48 into ST(:,:,x), after the loop has reached 48, it should go back and increase everyhing by 72 until the ST array has been filled.
I hope that makes sense.
ellis langdon commented:
Thank you Jan, that seemed to do the trick! :)

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Answer by Matt J
on 13 Feb 2019
Edited by Matt J
on 13 Feb 2019

ST=reshape( T, 128 ,64,72,30);
ST=reshape( ST(:,:,30:48,:), 128, 64,[]);

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