how to use bsxfun or arrayfun to build the following matrix

Hi, there.

I want to use bsxfun or arrayfun to replace the following for loop.

for s=1:S
      New_P(s,:)=P(s,:,U0(s));
  end;

for example,

if P(：,：,1)=[1,2;3,4]; 
   P(:,:,2)=[5,6;7,8];
   U0=[1;2];

then the final result is New_P=[1,2;7,8];

Thank you in advance!

[Adam]

Is there any particular reason why you want to use arrayfun or bsxfun? arrayfun is generally slower than a for loop and is also more suited to 1d arrays than multidimensional arrays. It can certainly be used for multidimensional arrays, but it effectively just interprets it as a 1d array anyway using linear indexing on the array.

I'm not aware of a syntax of bsxfun that would be useful for this, though maybe it is. I use it for certain specific tasks so haven't used all its capabilities.

[Jason]

thank you for your comments. because when the size of the matrix is very large, the for loop is generally very unefficiency, so i want use a vectorized expression.

Answer 1

[Guillaume]

Neither arrayfun nor bsxfun are useful for this. sub2ind is:

P = cat(3, [1 2;3 4], [5 6;7 8]);
U0 = [1;2];
New_P = P(sub2ind(size(P), ...
   repmat((1:size(P, 1))', 1, size(P, 2)), ... rows where to pick the elements
   repmat(1:size(P, 2), size(P, 1), 1), ... columns where to pick the elements
   repmat(U0, 1, size(P, 2)))) %pages where to pick the elements

[Guillaume]

Note that you could generate the rows and columns matrix with ngrid if you prefer:

[rows, cols] = ndgrid(1:size(P, 1), 1:size(P, 2));
new_P = P(sub2ind(size(P), rows, cols, repmat(U0, 1, size(P, 2))))