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))
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--
Simon Wood, Mathematical Science, University of Bath BA2 7AY UK
+44 (0)1225 386603 http://people.bath.ac.uk/sw283
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
