R help - Jun 2006 - getting the smoother matrix from smooth.spline

Can anyone tell me the trick for obtaining the smoother matrix from
smooth.spline when there are non-unique values for x. I have the following code
but, of course, it only works when all values of x are unique.
   
  ## get the smoother matrix (x having unique values
smooth.matrix = function(x, df){
 n = length(x);
 A = matrix(0, n, n);
 for(i in 1:n){
       y = rep(0, n); y[i]=1;
       yi = smooth.spline(x, y, df=df)$y;
       A[,i]= yi;
}
 (A+t(A))/2;
}

   
  Thanks for any assistance,
  Gregory

 		
---------------------------------

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>>>>> "Gregory" == Gregory Gentlemen
<gregory_gentlemen at yahoo.ca>
>>>>>     on Sat, 24 Jun 2006 14:41:37 -0400 (EDT) writes:

    Gregory> Can anyone tell me the trick for obtaining the
    Gregory> smoother matrix from smooth.spline when there are
    Gregory> non-unique values for x. I have the following code
    Gregory> but, of course, it only works when all values of x
    Gregory> are unique.

   >>    ## get the smoother matrix (x having unique values
   >>    smooth.matrix = function(x, df){
   >>      n = length(x);
   >>      A = matrix(0, n, n);
   >>      for(i in 1:n){
   >>        y = rep(0, n); y[i]=1;
   >>        yi = smooth.spline(x, y, df=df)$y;
   >>        A[,i]= yi;
   >>      }
   >>      (A+t(A))/2;
   >>    }

{ All the extraneous ";" at the end of lines make the above
  unncessarily ugly  (and potentially even slightly inefficient).}
   
Package 'sfsmisc' has had a  function  hatMat()  which returns
the hat matrix aka smoother matrix, in a slightly more general
way -- you can use it also for other smoothers, see the examples
in  help(hatMat).  The smoother defaults to smooth.spline(), so
you can directly use it, e.g.,
    hatMat(x, df=5)

Why do you think your code or hatMat()  would not work for
non-unique x-values?


    Gregory> Thanks for any assistance,
    Gregory> Gregory

    Gregory> ---------------------------------

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[please do read the posting guide, and hence get rid of the line above ! ]

    Gregory> ______________________________________________
    Gregory> R-help at stat.math.ethz.ch mailing list
    Gregory> https://stat.ethz.ch/mailman/listinfo/r-help
    Gregory> PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html

smooth.matrix = function(x, df){
  n = length(x);
  A = matrix(0, n, n);
  for(i in 1:n){
        y = rep(0, n); y[i]=1;
        yi = predict(smooth.spline(x, y, df=df),x)$y;
        A[,i]= yi;
}
  (A+t(A))/2;
}

>- Simon Wood, Mathematical Sciences, University of Bath, Bath BA2 7AY
>-             +44 (0)1225 386603         www.maths.bath.ac.uk/~sw283/

On Sat, 24 Jun 2006, Gregory Gentlemen wrote:
> Can anyone tell me the trick for obtaining the smoother matrix from
smooth.spline when there are non-unique values for x. I have the following code
but, of course, it only works when all values of x are unique.
>
>  ## get the smoother matrix (x having unique values
> smooth.matrix = function(x, df){
> n = length(x);
> A = matrix(0, n, n);
> for(i in 1:n){
>       y = rep(0, n); y[i]=1;
>       yi = smooth.spline(x, y, df=df)$y;
>       A[,i]= yi;
> }
> (A+t(A))/2;
> }
>
>
>  Thanks for any assistance,
>  Gregory
>
>
> ---------------------------------
>
> ---------------------------------
> Get a sneak peak at messages with a handy reading pane.
> 	[[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
>

Perhaps this could be developed into a spline smooth method
for model.matrix and included in R.

On 6/30/06, Simon Wood <sw283 at maths.bath.ac.uk>
wrote:
> smooth.matrix = function(x, df){
>  n = length(x);
>  A = matrix(0, n, n);
>  for(i in 1:n){
>        y = rep(0, n); y[i]=1;
>        yi = predict(smooth.spline(x, y, df=df),x)$y;
>        A[,i]= yi;
> }
>  (A+t(A))/2;
> }
>
>
> >- Simon Wood, Mathematical Sciences, University of Bath, Bath BA2 7AY
> >-             +44 (0)1225 386603         www.maths.bath.ac.uk/~sw283/
>
>
> On Sat, 24 Jun 2006, Gregory Gentlemen wrote:
>
> > Can anyone tell me the trick for obtaining the smoother matrix from
smooth.spline when there are non-unique values for x. I have the following code
but, of course, it only works when all values of x are unique.
> >
> >  ## get the smoother matrix (x having unique values
> > smooth.matrix = function(x, df){
> > n = length(x);
> > A = matrix(0, n, n);
> > for(i in 1:n){
> >       y = rep(0, n); y[i]=1;
> >       yi = smooth.spline(x, y, df=df)$y;
> >       A[,i]= yi;
> > }
> > (A+t(A))/2;
> > }
> >
> >
> >  Thanks for any assistance,
> >  Gregory
> >
> >
> > ---------------------------------
> >
> > ---------------------------------
> > Get a sneak peak at messages with a handy reading pane.
> >       [[alternative HTML version deleted]]
> >
> > ______________________________________________
> > R-help at stat.math.ethz.ch mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
> >
>
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
>