Re: creation of a permutation matrix (WAS: Re: remove for-loop from following code)

[c.]

On 6 Nov 2011, at 13:30, Juan Pablo Carbajal wrote:

> 
> Hi,
> 
> Indeed, you get the permutation matrix if you use K=max(Y). If you
> just want a submatrix out of the full permutation matrix, I think the
> only solution to save storage is using a sparse matrix. right?
> 

or a a rectangular diagonal matrix times a square permutation matrix:

>> P = eye (4) ([1 3 2 4], :)
P =

Permutation Matrix

   1   0   0   0
   0   0   1   0
   0   1   0   0
   0   0   0   1

>> C = diag (ones (3, 1), 3, 4)
C =

Diagonal Matrix

   1   0   0   0
   0   1   0   0
   0   0   1   0

>> whos P C
Variables in the current scope:

   Attr Name        Size                     Bytes  Class
   ==== ====        ====                     =====  ===== 
        P           4x4                         16  double
        C           3x4                         24  double

Total is 28 elements using 40 bytes

>> 

though one then must be careful to compute C * P * X as C * (P * X) and not (C 
* P) * X
to avoid creating a full matrix again 

c.