Indexing structure without using scalars

2012 10 31
2 个回答
回答已采纳
3 次查看（30 天）

0 个投票

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 个评论
显示 无 隐藏 无

Matt J
Matt J 2012-10-31
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 2012-11-1
This was just an example structure. Data is different, and matrices are way larger.

请先登录，再进行评论。

请先登录，再回答此问题。

 采纳的回答

Sean Little
Sean Little 2012-10-31

0 个投票

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,:)

更多回答（1 个）

类别

在 帮助中心 和 File Exchange 中查找有关 Matrix Indexing 的更多信息

产品

标签

Community Treasure Hunt

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

Start Hunting!

Translated by