[R] help with a regression problem

[Eik Vettorazzi]

Hi,
maybe working with a data.frame in long format is an option - then you
can use e.g. lmList and so on up to mixed models, depending on your
final goals of analyses (e.g. check for differential slopes).

vmat<-matrix(c("X1","X2","X3","X4","Y1","Y2","Y3","Y4"),nrow=2,byrow=T)
aa.l<-reshape(aa,idvar="ID",direction="long",varying=vmat,v.names=c("X","Y"))
library(nlme)
(ll<-lmList(Y~X|ID,aa.l,na.action=na.omit))
summary(ll)

cheers.

Am 01.08.2012 15:06, schrieb R Heberto Ghezzo, Dr:
> Hello,
> I have a big data frame where consecutive time dates and corresponding observed values for each subject (ID) are on a line. I want to compute the linear slope for each subject. I would like to use apply but I do
> not know how to express the corresponding function. An example using a loop follows 
> #
> # create dummy data set There are missing values
>  a <- c(1,2,3,4, 1,1,1,1, 2,2,3,3, 3,4,NA,4, 5,5,5,5,
>             2.1,2.2,2.3,2.4, 2.3,2.4,2.6,2.6, 2.5,2.6,2.9,3,
>             2.6,NA,3.2,4)
> a <- matrix(a, nr=4)
> aa <- as.data.frame(a)
> names(aa) <- c("ID","X1","X2","X3","X4","Y1","Y2","Y3","Y4")
> #
> #  I want the regression coefficientes of the Y on the X for each ID
> #
> sl <- rep(NA,4)
> for(i in 1:4) {
>   x1 <- a[i,2:5]
>   y1 <- a[i,6:9]
>   sl[i] <- lm(y1 ~ x1)$coef[2]
> }
> sl
> #
> #   I would like to use apply on the data.frame aa but with which function?
> #
> sl <- apply(aa,1,FUN) # FUN = ??
> #
> Thanks for any help
> 
> R.Heberto Ghezzo Ph.D.
> Montreal - Canada
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> and provide commented, minimal, self-contained, reproducible code.
> 


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Eik Vettorazzi

Department of Medical Biometry and Epidemiology
University Medical Center Hamburg-Eppendorf

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