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Indexing structure without using scalars

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Milos
Milos el 31 de Oct. de 2012
Consider following structure:
mm(1,1).no=1;
mm(2,1).no=2;
mm(3,1).no=3;
mm(3,1).mtx=[3;3;3];
mm(2,1).mtx=[2;2;2];
mm(1,1).mtx=[1;1;1];
a1 = cat(1, mm([1,2]).no)
a1 =
1
2
a2 = cat(1, mm([1,2]).mtx(1))
Scalar index required for this type of multi-level indexing.
Is there a work-around for this type of indexing?
  2 comentarios
Matt J
Matt J el 31 de Oct. de 2012
If your mtx and no data re all the same size, as in this example, it makes more sense to hold them in a scalar struct
mm.no=[1;2;3];
mm.mtx=[1 2 3; 1 2 3; 1 2 3];
Then you can easily extract pieces that you need
a1=mm.no(1:2);
a2=mm.mtx(:,1);
Milos
Milos el 1 de Nov. de 2012
This was just an example structure. Data is different, and matrices are way larger.

Iniciar sesión para comentar.

Respuesta aceptada

Sean Little
Sean Little el 31 de Oct. de 2012
Unfortunately, Matlab does not support multi-level indexing like other languages do. You are probably going to have to create an intermediate variable.
a1 = [mm(1:2).mtx];
a2 = a1(1,:)

Más respuestas (1)

Matt J
Matt J el 31 de Oct. de 2012
tmp=cat(1,mm([1,2]).mtx);
a2=tmp(1)

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