Thank you ver much.
Is there is a function in R which is doing penalized cubic regression, say
spl.plr(), that if I have weeks = 1:9 I can use somthing like pp <-
spl.plr(weeks,c(1,3,5,7)) and for 8 and 9 will be linear? Is rcs()
library(Design) doing this?
Many thanks,
Samuel
--- On Tue, 22/6/10, Joris Meys <jorism...@gmail.com> wrote:
From: Joris Meys <jorism...@gmail.com>
Subject: Re: [R] glm
To: "Samuel Okoye" <samu...@yahoo.com>
Cc: r-help@r-project.org
Date: Tuesday, 22 June, 2010, 9:50
On Tue, Jun 22, 2010 at 1:00 AM, Samuel Okoye <samu...@yahoo.com> wrote:
> Hi,
>
> I have the following data
>
> data1 <- data.frame(count = c(0,1,1,2,4,5,13,16,14), weeks = 1:9,
> treat=c(rep("1mg",3),rep("5mg",3),rep("10mg",3)))
> and I am using
>
> library(splines)
>
> to fit
>
> glm.m <- glm(count~weeks)+as.factor(treat),family=poisson,data=data1)
>
> and I am interested in predicting the count variale for the weeks 10, 11 and
> 12 with treat 10mg and 15mg.
bad luck for you.
newdat <-data.frame(
weeks=rep(10:12,each=2),
treat=rep(c("5mg","10mg"),times=3)
)
preds <- predict(glm.m,type="response",newdata=newdat,se.fit=T)
cbind(newdat,preds)
gives as expected :
Warning message:
In bs(weeks, degree = 3L, knots = numeric(0), Boundary.knots = c(1L, :
some 'x' values beyond boundary knots may cause ill-conditioned bases
weeks treat fit se.fit residual.scale
1 10 5mg 5.934881 5.205426 1
2 10 10mg 12.041639 9.514347 1
3 11 5mg 4.345165 6.924663 1
4 11 10mg 8.816168 15.805171 1
5 12 5mg 2.781063 8.123436 1
6 12 10mg 5.642667 18.221007 1
Watch the standard errors on the predicted values. No, you shouldn't
predict outside your data space, especially when using splines. And
when interested in 15mg, well, you shouldn't put treatment as a factor
to start with.
Cheers
Joris
--
Joris Meys
Statistical consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
tel : +32 9 264 59 87
joris.m...@ugent.be
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