I'm assuming that you meant to write
s <- smooth.spline(x, y)
You can look at the code for the predict method for smooth.spline by
typing
getAnywhere(predict.smooth.spline)
If a new value is provided (i.e., the 500 in your example) then the
predict method for smooth.spline.fit is applied to the spline object's fit
(i.e., s$fit in your example).
You can look at the code for the predict method for smooth.spline.fit by
typing
getAnywhere(predict.smooth.spline.fit)
Jean
Jesper wrote on 08/25/2011 03:03:09 AM:
>
> Hi,
>
> I want to use the result from smooth.spline outside R.
>
> I take my data ,which is 180 point stored in x and y
>
> s <- smooth(x,y)
>
> I can know use to e.g. find the interpolated value at e.g. x=500
>
> predict (s,500)
>
> My problem is, that i don't know how to implement the predict function.
I
> have looked at literature, but i cannot connect the output of the
> smooth.spline() to an actual spline model..
>
> Has anyone tried it?
>
> Br
>
> Jesper
>
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