Reduce dimensionality using indices

Hi all.

I have what I tink should be a fairly simple question but I just can't work it out. In one of my scripts I have a three-dimensional array A like this:

A(:,:,1) =

       2     4     8
       5     6     3
      10     3     6

A(:,:,2) =

       7     6     3
       9     2     9
      10     2     3

A(:,:,3) =

       9     4     7
       3     2     5
      10     3     4

I also have a 2D matrix d which looks sth like this:

d =

       1     2     3
       1     1     3
       2     1     1

Now, what I want to reduce A to two dimensions using the third dimension elements indicated by d. The result should be:

AA =

       2     6     7
       5     6     5
      10     3     6

Surely there's a clever way to do this without loops? I fiddled around with a combination of meshgrid and d but that got ugly really quickly...

EDIT : I should have made clear that this is just an example and I don't necessarily know the dimensions of A. It will always be true that

size(A,1) == size(d,1)
size(A,2) == size(d,2)

but A may be 5x5x13.

Thanks so much for your help.

Answer 1

[Jos (10584)]

Convert sub indices into linear indices:

% some data
  A = ceil(10*rand(3,4,3)) % A has an arbitrary size
  d = ceil(size(A,3)*rand(size(A,1),size(A,2))) % d is the subindex in the 3rd dimension

% engine
  Nrc = size(d) ;
  [r,c] = ndgrid(1:Nrc(1), 1:Nrc(2)) ; %r,c, specify the subindices in the 1st and 2nd dimension
  ix = sub2ind(size(A),r, c, d) % convert to linear indices
  B = A(ix) % retrieve from A
% check
  x = 2 ; y = 3 ;
  A(x,y,d(x,y)) == B(x,y)

Answer 2

[George Papazafeiropoulos]

This is an alternative solution which does not contain external functions, thanks to Jos!

% initial data
A=ceil(10*rand(3,4,3))
d=ceil(size(A,3)*rand(size(A,1),size(A,2)))
% engine
dim1=size(A,1);
dim2=size(A,2);
linind=(1:dim1*dim2)';
linind=reshape(linind,dim1,dim2);
modind=linind+(d-1)*dim1*dim2;
AA=A(modind);
% check
x=2;
y=1;
A(x,y,d(x,y)) == AA(x,y)

Thanks Jos!

Answer 3

[George Papazafeiropoulos]

After specifying A...

d=[1 2 3;
    1 1 3;
    2 1 1];
B=zeros(3,3,3);
B(:,:,1)=(d==1);
B(:,:,2)=2*(d==2);
B(:,:,3)=3*(d==3);
AA=sum(A.*B,3)

[fromLondon]

Hi.

Thanks for the quick reply. I like the solution but I should have made clear that I don't necessarily know the size of A. I will change the question accordingly. My bad.

Answer 4

[fromLondon]

Perfect. Thank you both! :-)