Index slice of ND array of unknown dimension

12 visualizaciones (últimos 30 días)
Evan
Evan el 12 de Jun. de 2017
Editada: Stephen23 el 26 de Mzo. de 2020
Let's say I have a 3D array A, and I want to remove a slice:
A(i,:,:) = [];
But what if I don't know the dimension of the array beforehand? Just doing
A(i,:) = [];
Is valid, but reshapes the array into a 2D matrix.
I was thinking something like:
A(i,indices{:}) = [];
but I don't think you can put a colon operator in a cell.

Respuesta aceptada

Stephen23
Stephen23 el 12 de Jun. de 2017
Editada: Stephen23 el 12 de Jun. de 2017
If you do not know the size of the array before hand then you can call subsasgn directly. The function subsasgn is the actual function that MATLAB calls whenever you assign to an array using the indexing convenience operators () or {}, e.g. X(1) = 2, in just the same way that 1+2 is just a convenience operator for plus(1,2). We can also call subasgn directly, just like we could call plus if we so wished.
Here is a simple example of removing one row of A without hard-coding how many dimensions it has:
>> A = rand(5,4,3,2);
>> S.subs = repmat({':'},1,ndims(A));
>> S.subs{1} = 3; % the third row
>> S.type = '()';
>> B = subsasgn(A,S,[]);
>> size(B)
ans =
4 4 3 2
  4 comentarios
Daniel Plotnick
Daniel Plotnick el 23 de Mzo. de 2020
It looks like this has changed in more recent updates: I had been using this technique, but it failed when using it on a gpuArray. The included code below now works for gpuArrays and normal arrays:
A = gpuArray.rand(5,4,3,2);
v = repmat({':'},ndims(A),1);
v{1} = 3; % Do the third row
B = A;
B(v{:}) = []; % Note, you can replace this with a numeric to substitite instead of delete
size(B)
Stephen23
Stephen23 el 26 de Mzo. de 2020
Editada: Stephen23 el 26 de Mzo. de 2020
@Daniel Plotnick: you should put that as an Answer, so other users could vote for it.

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Matrix Indexing en Help Center y File Exchange.

Productos

Community Treasure Hunt

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

Start Hunting!

Translated by