# Preserving the shape of the indices vector when indexing into another vector.

2 views (last 30 days)
Federico Miretti on 16 Oct 2020
Edited: Ameer Hamza on 22 Oct 2020
Suppose i have a vector of values A
A = [5 10 15 20 25];
And a vector of indeces that may have any number of singleton leading dimensions, e.g.
idx = zeros(1,1,3);
idx(:) = 1:3
idx(:,:,1) =
1
idx(:,:,2) =
2
idx(:,:,3) =
I want to get the elements of A specified by the subscripts in idx, and I want the output B to have the same dimensions as idx. In this example, what I want to get is
B(:,:,1) =
5
B(:,:,2) =
10
B(:,:,3) =
15
Instead of the default indexing behaviour which would get me:
B = A(idx)
B =
5 10 15
I know that i could obtain this by first initializing B to have the same size of idx, i.e.
B = zeros(size(idx));
B(:) = A(idx);
However, in my application, I am developing a toolbox that must be able to work with user-created functions. I can act on idx, which will be fed as an input to the user-created function, but the rest is up to the user and I cannot expect him to write these two lines of code.
Is there any way to achieve this?
Federico Miretti on 21 Oct 2020
@Stephen Colbeck: I understand that that's the default behaviour for indexing vectors with vectors. My problem is, I need the exact opposite: I need the number of elements of C to be the number of elements in B with the orientation of B.
I know this may not be possible at all, but I still thought it was worth trying here on Answers.

Ameer Hamza on 16 Oct 2020
Edited: Ameer Hamza on 16 Oct 2020
This can be one of the way
B = reshape(A(idx), size(idx));
Although, I wonder is why MATLAB does not follow normal behavior for indexing A(idx). Maybe there is a good reason, or maybe this is an oversight.
Ameer Hamza on 22 Oct 2020
I don't think that overloading subsindex will help here. subsindex only returns the indexes, but as we already saw, the shape of indexes have no effect on the output of indexing when idx is 1x1x3.