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
Respuesta aceptada
Más respuestas (1)
Kenneth Eaton
el 24 de En. de 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 comentarios
Paulo Silva
el 24 de En. de 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
el 25 de En. de 2011
Paulo Silva
el 25 de En. de 2011
please explain exactly what doesn't work in your simple code?
Pepa
el 25 de En. de 2011
Kenneth Eaton
el 25 de En. de 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
el 25 de En. de 2011
Walter Roberson
el 25 de En. de 2011
By the way, preallocation of jaj is not needed as you are overwritten its complete contents each trip through the loop.
Walter Roberson
el 25 de En. de 2011
Pepa, put in kunn(:,:,X) *where* ?? Code example, please.
Pepa
el 25 de En. de 2011
Oleg Komarov
el 25 de En. de 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
el 25 de En. de 2011
Pepa
el 25 de En. de 2011
Oleg Komarov
el 25 de En. de 2011
isequal(kunn(:,:,1),kunn(:,:,2)) should be 0, can you confirm?
Pepa
el 25 de En. de 2011
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