R help - Jun 2015 - How to make new predictions from a GAM with a spline forced through the origin

Hi Gavan,

After running the code in your message, do something like this...

xp <- c(.2,.3) ## x values for prediction
Xp <- PredictMat(sm,data.frame(x=xp))[,-3] ## prediction matrix
yp <- predict(b,list(X=Xp,off=rep(.6,2)))  ## call predict.gam
points(xp,yp,col=2) ## plot points

... hope that's clear enough, let me know if not.

best,
Simon


On 29/05/15 13:12, Gavan McGrath wrote:
> Hi,
>
> I?m followed an example to fit a GAM with a spline forced through a point,
i.e. (0,0). This works fine from one of Simon?s examples however when it comes
to making a prediction from a new set of x values I?m a bit stumped.
>
> In the example below a smooth term is constructed and the basis and
penalties at x=0 are removed then the gam is fitted to a spline basis matrix X
using spline penalties.
>
> Can someone suggest a way that I can make predictions at new  x  values
based on the gam b below.
>
>
> Here is Simon Wood's example:
>
> library(mgcv)
> set.seed(0)
> n <- 100
> x <- runif(n)*4-1;x <- sort(x);
> f <- exp(4*x)/(1+exp(4*x));y <- f+rnorm(100)*0.1;plot(x,y)
> dat <- data.frame(x=x,y=y)
>
> ## Create a spline basis and penalty, making sure there is a knot
> ## at the constraint point, (0 here, but could be anywhere)
> knots <- data.frame(x=seq(-1,3,length=9)) ## create knots
> ## set up smoother...
> sm <- smoothCon(s(x,k=9,bs="cr"),dat,knots=knots)[[1]]
>
> ## 3rd parameter is value of spline at knot location 0,
> ## set it to 0 by dropping...
> X <- sm$X[,-3]        ## spline basis
> S <- sm$S[[1]][-3,-3] ## spline penalty
> off <- y*0 + .6       ## offset term to force curve through (0, .6)
>
> ## fit spline constrained through (0, .6)...
> b <- gam(y ~ X - 1 + offset(off),paraPen=list(X=list(S)))
> lines(x,predict(b))
>
>
>
> 	[[alternative HTML version deleted]]
>
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-- 
Simon Wood, Mathematical Science, University of Bath BA2 7AY UK
+44 (0)1225 386603               http://people.bath.ac.uk/sw283