R help - Dec 2009 - model matrix with a spline

Hi all

I want to get the design matrix for a model, evaluated at a single value.
For example, if I pass in a data frame with a=2, b=2, y=3, and my
model is y ~ a+b+a:b, then I would like to get
the values 3, 2, 2, 4 out. I can do this with:

tmp <- model.matrix(fit, data=mydata)

or

tmp <- predict(fit, newdata=mydata, type="terms")

However, if the fit had a smoothing spline component, this fails. It
seems like the prediction function is trying to reevaluate the basis
for the spline, and as there is only one row in the new data, it can't
do that.

Is there a way I can get the value of the already-created spline? And
is there a simple way to do this programmatically so I don't need to
check each term of the formula individually?

Cheers,

David Hugh-Jones
Post-doctoral Researcher
Max Planck Institute of Economics, Jena
http://davidhughjones.googlepages.com

On Dec 19, 2009, at 2:22 AM, David Hugh-Jones wrote:
> Hi all
>
> I want to get the design matrix for a model, evaluated at a single  
> value.
> For example, if I pass in a data frame with a=2, b=2, y=3, and my
> model is y ~ a+b+a:b, then I would like to get
> the values 3, 2, 2, 4 out. I can do this with:
>
> tmp <- model.matrix(fit, data=mydata)
>
> or
>
> tmp <- predict(fit, newdata=mydata, type="terms")
>
> However, if the fit had a smoothing spline component, this fails. It
> seems like the prediction function is trying to reevaluate the basis
> for the spline, and as there is only one row in the new data, it can't
> do that.
>
> Is there a way I can get the value of the already-created spline? And
> is there a simple way to do this programmatically so I don't need to
> check each term of the formula individually?
Probably.

But you have not given any details about your modeling functions, so  
"probably" is about all that can be said. Parametric splines would  
have a greater likelihood of having a compact representation which is  
what you are probably seeking. Harrell's packages are the ones I have  
the most experience with and his Function function gives a very quick  
answer.

-- 

David Winsemius, MD
Heritage Laboratories
West Hartford, CT