Working with a three dimensional matrix
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I have a matrix with three dimensions, and I need to take two dimensions through the use of cycle FOR.
For example:
Name variable Value
k <39x17x60>
And I need this:
a = 60;
for i = 1:60;
l = k...; %size of l <39x17>
xlswrite...;
end
Accepted Answer
More Answers (1)
Kenneth Eaton
on 24 Jan 2011
If you're asking how you would index k in your loop to get a 2-D submatrix, you would do the following:
l = k(:,:,i);
14 Comments
Paulo Silva
on 24 Jan 2011
I wasn't fast enough, you posted first :)
% k <39x17x60>
k=randn(39,17,60); %example data
%And I need this:
l=zeros(39,17); %preallocation of l
for i = 1:60
l=k(:,:,i); %size of l <39x17>
%xlswrite...;
end
Pepa
on 25 Jan 2011
Paulo Silva
on 25 Jan 2011
please explain exactly what doesn't work in your simple code?
Pepa
on 25 Jan 2011
Kenneth Eaton
on 25 Jan 2011
Pepa, I'm still not clear on exactly what error/problem you are having when you run your code. The two screenshots of the variables look fine. Note that kunn(:,:,1) and jaj are the same, there is just a scaling factor of 1.0e+003 used to display kunn.
Pepa
on 25 Jan 2011
Walter Roberson
on 25 Jan 2011
By the way, preallocation of jaj is not needed as you are overwritten its complete contents each trip through the loop.
Walter Roberson
on 25 Jan 2011
Pepa, put in kunn(:,:,X) *where* ?? Code example, please.
Pepa
on 25 Jan 2011
Oleg Komarov
on 25 Jan 2011
Why don't you apply what Kenneth suggested (not to mention Walter):
jaj = kunn(:,:,s)
And what's the problem, btw, can you elaborate, post any error message or unwanted result?
Pepa
on 25 Jan 2011
Pepa
on 25 Jan 2011
Oleg Komarov
on 25 Jan 2011
isequal(kunn(:,:,1),kunn(:,:,2)) should be 0, can you confirm?
Pepa
on 25 Jan 2011
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