I'm familiar w/ relevel but that doesn't seem to help here (e.g., tried setting the reference level to 3 but the second model w/ the subset argument doesn't seem to estimate the desired model).
Possibly another contrasts statement needs to be provided prior to the second model, this time w/ a reduced number of levels, but that seems to introduce problems w.r.t. the total number of levels prior to subsetting. Is the only way around this to define a new dataframe for the subset and define the contrasts for that dataframe accordingly? -----Original Message----- From: Francisco J. Zagmutt [mailto:[EMAIL PROTECTED] Sent: Friday, July 06, 2007 6:40 PM To: Afshartous, David Cc: [email protected] Subject: Re: maintaining specified factor contrasts when subsetting in lmer See ?relevel Regards, Francisco Afshartous, David wrote: > All, > > I'm using lmer for some repeated measures data and have specified > the contrasts for a time factor such that say time 3 is the base. This > works fine. However, when > I next use the subset argument to remove the last two time values, the > output indicates that > the specified contrast is not maintained (see below). I can solve this > by creating a new dataframe > for the subset of interest and redefining the constrasts, but I was > wondering if there is a direct method that allows me to continue w/ > the subset argument? (perhaps via supplying a contrast argument to > lmer directly, but this doesn't seem possible based on the defintion > of this argument in model.matrix.default). > > Thanks, > Dave > > > z <- rnorm(24, mean=0, sd=1) > time <- factor(paste("Time-", rep(1:6, 4), sep="")) Patient <- > rep(1:4, each = 6) drug <- factor(rep(c("D", "P"), each = 6, times = > 2)) ## P = placebo, D = Drug dat.new <- data.frame(time, drug, z, > Patient) > > ### specify the contrast as time 3: > contrasts(dat.new$time) <- contr.treatment(6, base=3) > dimnames(contrasts(dat.new$time))[[1]] <- > as.character(levels(dat.new$time)) > dimnames(contrasts(dat.new$time))[[2]] <- > as.character(levels(dat.new$time)[-3]) > > fm1 <- lmer(z ~ drug + time + (1 | Patient), data = dat.new ) Fixed > effects: > Estimate Std. Error t value > (Intercept) -0.182774 0.464014 -0.39390 > drugP -0.281103 0.352309 -0.79789 > timeTime-1 0.150505 0.606462 0.24817 > timeTime-2 0.612016 0.606462 1.00916 > timeTime-4 0.775342 0.606462 1.27847 > timeTime-5 0.093741 0.606462 0.15457 > timeTime-6 0.452442 0.606462 0.74604 > > ## time 3 is the base as specified > > fm2 <- lmer(z ~ drug + time + (1 | Patient), data = dat.new, subset = > dat.new$time !="Time-6" & dat.new$time != "Time-5") > > Fixed effects: > Estimate Std. Error t value > (Intercept) 0.052975 0.500675 0.10581 > drugP -0.451593 0.447818 -1.00843 > timeTime-2 0.461511 0.633310 0.72873 > timeTime-3 -0.150505 0.633310 -0.23765 > timeTime-4 0.624837 0.633310 0.98662 > > ### time 3 no longer the base; was expecting to see the fixed effects > for time-1, time-2, and time-4, w/ Intercept ### representing time-3 > > > > [[alternative HTML version deleted]] > > ______________________________________________ > [email protected] mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > ______________________________________________ [email protected] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
