Using cell array for indexing

조회 수: 4 (최근 30일)
Lionel Pöffel
Lionel Pöffel 2021년 8월 2일
댓글: Lionel Pöffel 2021년 8월 2일
I have matrices of varying dimensionality I need to index.
So just writing A(i1,...,in) is not an option because n can vary in my application.
What is known to work for getting a single element out of A is using a cell array:
IndArray={i1,...,in}
Elem=A(IndArray{:})
However I'd like to do something like SubMatrix= A(i1,i2,...,im,:,:,...,:), where m<n. Is there a way to do that?
The following does not work:
IndArray={i1,...,im}
SubMatrix=A(IndArray{:})
Thanks in advance for any hints.

채택된 답변

Rik
Rik 2021년 8월 2일
I would suggest something like this:
IndArray={i1,...,im}
for dim=(numel(IndArray)+1):ndims(A)
IndArray{dim}=1:size(A,dim);
end
SubMatrix=A(IndArray{:})

추가 답변 (2개)

Stephen23
Stephen23 2021년 8월 2일
편집: Stephen23 2021년 8월 2일
No need for a loop, here is the simple and efficient MATLAB approach:
IndArray = {i1,..,im};
IndArray(1+end:ndims(A)) = {':'};
SubMatrix = A(IndArray{:})
How it works: MATLAB always allows a scalar element on the RHS to be allocated to any number of elements on the LHS. This applies to any array class: numeric, cell (as above), char, string, struct, etc.
  댓글 수: 1
Lionel Pöffel
Lionel Pöffel 2021년 8월 2일
Hello Stephen,
thanks for your answer. That is, of course, an additional improvement.

댓글을 달려면 로그인하십시오.


Lionel Pöffel
Lionel Pöffel 2021년 8월 2일
Hi Rik, that seems to work, thanks.
DGM posted an answer that surprisingly also seems to work but unfortunately deleted it (thanks to him/her as well).
For completeness I will note it here:
IndArray={i1,...,im}
for dim=(numel(IndArray)+1):ndims(A)
IndArray(dim)={':'};
end
SubMatrix=A(IndArray{:})
  댓글 수: 1
DGM
DGM 2021년 8월 2일
Sorry about that. I tried reading your other posts to try to figure out what exactly you were trying to do and had the feeling that I had misunderstood the intent.

댓글을 달려면 로그인하십시오.

카테고리

Help CenterFile Exchange에서 Matrix Indexing에 대해 자세히 알아보기

제품

Community Treasure Hunt

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

Start Hunting!

Translated by