From christos.hatzis at nuverabio.com  Mon Aug 31 20:09:33 2009
From: christos.hatzis at nuverabio.com (Christos Hatzis)
Date: Mon, 31 Aug 2009 14:09:33 -0400
Subject: [Rd] Problem in matrix definition?
In-Reply-To: <1251741444.3976.9.camel@ubuntu>
References: <798421.77107.qm@web59715.mail.ac4.yahoo.com>
	<1251741444.3976.9.camel@ubuntu>
Message-ID: <17E6277E375B4F86B16FCB772452DBE6@headquarters.silicoinsights>

See details on subsetting
?"["
specifically the drop argument.

To prevent the effects of drop() use

L[k, 1:(r-1), drop=FALSE]

-Christos 

> -----Original Message-----
> From: r-devel-bounces at r-project.org 
> [mailto:r-devel-bounces at r-project.org] On Behalf Of Mathieu Ribatet
> Sent: Monday, August 31, 2009 1:57 PM
> To: Fabio Mathias Corr?a
> Cc: r-devel at r-project.org; Uwe Ligges
> Subject: Re: [Rd] Problem in matrix definition?
> 
> 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 ao?t 2009 ? 19:09 +0200, Fabio Mathias Corr?a a ?crit :
> > 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!
> > 
> >                F?bio Mathias Corr?a
> > Estat?stica e Experimenta??o Agropecu?ria/UFLA
> >                      Brazil
> > 
> > 
> >       
> > 
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[Rd] Problem in matrix definition?
Christos Hatzis