[R-sig-ME] Repeated measures with a non-linear time effect
albrechj at staff.uni-marburg.de
Thu Jul 21 22:47:22 CEST 2016
that was a misunderstanding. I meant you should code time as a factor and run a modified version of the model (see code below).
df <- data.frame(subject = rep(c("T1", "T2", "T3", "C1", "C2", "C3"), 3),
group = rep(c(rep("T", 3), rep("C", 3)), 3),
time = c(rep(1, 6), rep(2, 6), rep(3, 6)),
measure = c(0, 253, 155, 16, 232, 251, 1035, 1014, 760, 98, 239, 87, 371, 60, 47, 0, 260, 190),
col = rep(c(rep("red", 3), rep("blue", 3)), 3), stringsAsFactors = FALSE)
df$time <- as.factor(df$time)
xyplot(measure ~ time|subject, data = df)
# using the lme4 package
fit <- lmer(measure ~ time * group + (1|subject), data = df)
# using anova
fit <- aov(measure ~ time * group + Error(subject), data = df)
Both models are able to detect the interaction between treatment and time.
Hope this helps.
Jörg Albrecht, PhD
Institute of Nature Conservation
Polish Academy of Sciences
31-120 Krakow, Poland
> Am 21.07.2016 um 21:50 schrieb Daniel Rubi <daniel_rubi at ymail.com>:
> Hi Jörg,
> Unfortunately setting time as integer doesn't change anything whereas setting it as a factor means that the number of random effects will be equal to the number of samples.
> Perhaps this is an alternative:
> fit1 <- gls(measure~time*group,correlation=corSymm(form=~1|subject),weight=varIdent(form=~1|time),data=df)
> as it does capture the group and interaction effects but I think I need to contrast it with a null model and I'm not sure what would that null model be.
> Any idea?
> On Thursday, July 21, 2016 3:27 PM, Jörg Albrecht <albrechj at staff.uni-marburg.de> wrote:
> Hi Dan,
> most likely the model treats your time covariate as a continuous predictor. Try str(df), then time should appear as integer (int). If you specify time as a factor (with three levels: 1, 2, 3) the model will be able to estimate the time x treatment interaction separately for each time point. However, you still have to decide whether treating time as a factorial variable makes sense for your dataset.
> > Am 21.07.2016 um 20:13 schrieb Daniel Rubi via R-sig-mixed-models <r-sig-mixed-models at r-project.org <mailto:r-sig-mixed-models at r-project.org>>:
> > Hi,
> > I have repeated measures from two groups (treatment and control), three subjects in each, over three time points.
> > Here's the data in an R data.frame:df <- data.frame(subject=rep(c("T1","T2","T3","C1","C2","C3"),3), group=rep(c(rep("T",3),rep("C",3)),3), time=c(rep(1,6),rep(2,6),rep(3,6)), measure=c(0,253,155,16,232,251,1035,1014,760,98,239,87,371,60,47,0,260,190), col=rep(c(rep("red",3),rep("blue",3)),3), stringsAsFactors=F)
> > The plot shows the time x group interaction:
> > R code for producing the plot:
> > plot(df$time,df$measure,col=df$col,xlab="time",ylab="measure")
> > legend("topleft",legend=c("treatment","control"),col=c("red","blue"),pch=1)
> > My question is what model to use to capture the time x group interaction.
> > I thought:library(lmerTest)fit <- lmer(measure~time+group+time*group+(time|subject),data=df)
> > might do it.
> > But the summary of this model doesn't really capture that:> summary(fit)Linear mixed model fit by REMLt-tests use Satterthwaite approximations to degrees of freedom ['lmerMod']Formula: measure ~ time + group + time * group + (time | subject) Data: df
> > REML criterion at convergence: 210
> > Scaled residuals:
> > Min 1Q Median 3Q Max
> > -1.228 -0.448 -0.163 0.275 1.923
> > Random effects:
> > Groups Name Variance Std.Dev. Corr
> > subject (Intercept) 0.00e+00 0.00e+00
> > time 3.06e-16 1.75e-08 NaN
> > Residual 1.05e+05 3.25e+02
> > Number of obs: 18, groups: subject, 6
> > Fixed effects:
> > Estimate Std. Error df t value Pr(>|t|)
> > (Intercept) 168.89 286.35 13.78 0.59 0.56
> > time -8.17 132.55 13.78 -0.06 0.95
> > groupT 218.33 404.96 13.78 0.54 0.60
> > time:groupT 19.83 187.46 13.78 0.11 0.92
> > Correlation of Fixed Effects:
> > (Intr) time groupT
> > time -0.926
> > groupT -0.707 0.655
> > time:groupT 0.655 -0.707 -0.926
> > So my question is what model to use?
> > Thanks a lot,Dan
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