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
>
>
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