[R] Frailty and coxph
Thomas Lumley
tlumley at u.washington.edu
Tue Oct 8 18:36:30 CEST 2002
On Tue, 8 Oct 2002, [iso-8859-1] Göran Broström wrote:
>
> Does someone know the rules by which 'coxph' returns 'frail', the
> predicted frailty terms? In my test function:
>
> -----------------------------------------------
> fr <- function(){
> #testing(frailty terms in 'survival'
> require(survival)
> dat <- data.frame(exit = 1:6,
> event = rep(1, 6),
> x = rep(c(0, 1), 3),
> id = rep(1:2, rep(3, 2))
> )
> fit1 <- coxph(Surv(exit, event) ~ x +
> frailty(id, dist = "gaussian"), data = dat)
> return(fit$frail)
> }
> -----------------------------------------------
>
> the result is 'NULL', but with 'real data', I usually get the predicted
> frailties. The help pages doesn't even mention the component 'frail', but
> in the code I can see that 'frail' is reurned if 'nfrail > 0'. And
> (from 'coxpenal.fit.s'):
The frailties are always returned. However, they aren't always returned
in the same place. They are returned in $frail if they are computed by
the sparse algorithm described in help(frailty), and in $coef if the full
Newton-Raphson algorithm is used.
By default, then, frailties on more than 5 individuals will be in $frail,
fewer than five in $coef. In your example we have
> coef(fit1)
x gauss:1 gauss:2
-0.1143062 1.4837374 -1.4837374
which are the missing frailties.
You can use frailty(id, dist="gaussian", sparse=TRUE) to force the
frailties to be returned in $frail or sparse=FALSE to force them to be in
$coef.
-thomas
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