Azzera filtri
Azzera filtri

Indexing structure without using scalars

1 visualizzazione (ultimi 30 giorni)
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 Commenti
Matt J
Matt J il 31 Ott 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 il 1 Nov 2012
This was just an example structure. Data is different, and matrices are way larger.

Accedi per commentare.

Risposta accettata

Sean Little
Sean Little il 31 Ott 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,:)

Più risposte (1)

Matt J
Matt J il 31 Ott 2012
tmp=cat(1,mm([1,2]).mtx);
a2=tmp(1)

Prodotti

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by