Refit a (merMod) Model with a Different Response
Refit a model, possibly after modifying the response vector. This makes use of the model representation and directly goes to the optimization.
refit(object, newresp, ...)
## S3 method for class 'merMod'
refit(object, newresp = NULL, rename.response = FALSE,
maxit = 100, ...)object |
a fitted model, usually of class
|
newresp |
an (optional) numeric vector providing the new
response, of the same length as the original response (see
|
rename.response |
when refitting the model, should the name of
the response variable in the formula and model frame be replaced with
the name of |
maxit |
scalar integer, currently only for GLMMs: the maximal number of Pwrss update iterations. |
... |
optional additional parameters. For the |
Refit a model, possibly after modifying the response vector. This
could be done using update(), but the refit()
approach should be faster because it bypasses the creation of the
model representation and goes directly to the optimization step.
Setting rename.response = TRUE may be necessary if one
wants to do further operations (such as update) on the fitted
model. However, the refitted model will still be slightly different
from the equivalent model fitted via update; in particular, the
terms component is not updated to reflect the new response
variable, if it has a different name from the original.
If newresp has an na.action attribute, then it is
assumed that NA values have already been removed from the numeric
vector; this allows the results of simulate(object) to
be used even if the original response vector contained NA values.
Otherwise, the length of newresp must be the same as the
original length of the response.
an object like x, but fit to a different response vector Y.
update.merMod for more flexible and extensive model
refitting;
refitML for refitting a REML fitted model with maximum
likelihood (‘ML’).
## Ex. 1: using refit() to fit each column in a matrix of responses -------
set.seed(101)
Y <- matrix(rnorm(1000),ncol=10)
## combine first column of responses with predictor variables
d <- data.frame(y=Y[,1],x=rnorm(100),f=rep(1:10,10))
## (use check.conv.grad="ignore" to disable convergence checks because we
## are using a fake example)
## fit first response
fit1 <- lmer(y ~ x+(1|f), data = d,
control= lmerControl(check.conv.grad="ignore",
check.conv.hess="ignore"))
## combine fit to first response with fits to remaining responses
res <- c(fit1,lapply(as.data.frame(Y[,-1]), refit, object=fit1))
## Ex. 2: refitting simulated data using data that contain NA values ------
sleepstudyNA <- sleepstudy
sleepstudyNA$Reaction[1:3] <- NA
fm0 <- lmer(Reaction ~ Days + (1|Subject), sleepstudyNA)
## the special case of refitting with a single simulation works ...
ss0 <- refit(fm0, simulate(fm0))
## ... but if simulating multiple responses (for efficiency),
## need to use na.action=na.exclude in order to have proper length of data
fm1 <- lmer(Reaction ~ Days + (1|Subject), sleepstudyNA, na.action=na.exclude)
ss <- simulate(fm1, 5)
res2 <- refit(fm1, ss[[5]])Please choose more modern alternatives, such as Google Chrome or Mozilla Firefox.