Dear Fabio,
The problem is that L[k,1:(r-1)] is not anymore a matrix but a vector.
Hence when you do t(L[k,1:(r-1)]) you get a matrix with only one row
while I think you expected one column instead. You can see this feature
with a simpler example like the following one:
x <- runif(10)
is.vector(x)
dim(t(x))##1,10
dim(t(t(x)))##10,1
Getting back to your code, you probably want that L[k,1:(r-1)] is a
matrix with r-1 columns. This is possible by substituting
t(L[k,1:(r-1)]) for t(t(L[k,1:(r-1)])) or matrix(L[k,1:(r-1)], nrow = 1)
Maybe somebody else will find a more elegant fix for your problem.
However, I tried and it did solve your issue.
Cheers,
Mathieu
Le lundi 31 août 2009 à 19:09 +0200, Fabio Mathias Corrêa a écrit :
> The problem is that arrays are the same size. The only difference is how they
> were built!
>
> The way to build the matrix should not influence in the outcome!
>
> Below is the commented code in MatLab!
>
> Neural Information Processing - Letters and Reviews Vol.8, No.2, August 2005
>
> function Y = geninv(G)
> % Returns the Moore-Penrose inverse of the argument
> % Transpose if m < n
> [m,n]=size(G); transpose=false;
> if m<n
> transpose=true;
> A=G*G';
> n=m;
> else
> A=G'*G;
> end
> % Full rank Cholesky factorization of A
> dA=diag(A); tol= min(dA(dA>0))*1e-9;
> L=zeros(size(A));
> r=0;
> for k=1:n
> r=r+1;
> L(k:n,r)=A(k:n,k)-L(k:n,1:(r-1))*L(k,1:(r-1))';
> % Note: for r=1, the substracted vector is zero
> if L(k,r)>tol
> L(k,r)=sqrt(L(k,r));
> if k<n
> L((k+1):n,r)=L((k+1):n,r)/L(k,r);
> end
> else
> r=r-1;
> end
> end
> L=L(:,1:r);
> % Computation of the generalized inverse of G
> M=inv(L'*L);
> if transpose
> Y=G'*L*M*M*L';
> else
> Y=L*M*M*L'*G';
> end
>
>
> Thanks!
>
> Fábio Mathias Corrêa
> Estatística e Experimentação Agropecuária/UFLA
> Brazil
>
>
>
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