[R] basehaz and newdata

[Austin, Matt]

I don't believe basehaz takes a newdata argument.

> args(basehaz)
function (fit, centered = TRUE)
NULL

It simply extracts info from the fitted object.  I think you want to use survfit method for coxph objects and then derive/extract what you need.

--Matt

Matt Austin
Global Statistical Leader, denosumab PMO
Biostatistics Director
Amgen, Inc


-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of mah
Sent: Monday, April 07, 2008 10:01 AM
To: r-help at r-project.org
Subject: Re: [R] basehaz and newdata

Thanks Roland, but using data frames does not resolve the issue.  See revised code and output below:

>  test1 <- data.frame(
+                 time=  c(4, 3,1,1,2,2,3),
+                 status=c(1,NA,1,0,1,1,0),
+                 x=     c(0, 2,1,1,1,0,0),
+                 sex=   c(0, 0,0,0,1,1,1)
+                 )
>  f1 <- coxph( Surv(time, status) ~ x + strata(sex), data=test1)
> #stratified model
>
>  test2 <- data.frame(
+                 time=  c(4, 3,1,1,2,2,3),
+                 status=c(1,NA,1,0,1,1,0),
+                 x=     c(1, 2,1,1,1,0,0),
+                 sex=   c(0, 0,0,0,1,1,1)
+                 )
>  f2 <- coxph( Surv(time, status) ~ x + strata(sex), data=test2)
> #stratified model
>
>  f1
Call:
coxph(formula = Surv(time, status) ~ x + strata(sex), data = test1)


  coef exp(coef) se(coef)     z    p
x 1.17      3.22     1.29 0.907 0.36

Likelihood ratio test=0.87  on 1 df, p=0.351  n=6 (1 observation deleted due to missingness)
>  f2
Call:
coxph(formula = Surv(time, status) ~ x + strata(sex), data = test2)


   coef exp(coef) se(coef)    z    p
x 0.896      2.45     1.42 0.63 0.53

Likelihood ratio test=0.38  on 1 df, p=0.535  n=6 (1 observation deleted due to missingness)
>  basehaz(f1, newdata=test2)
Error in basehaz(f1, newdata = test2) :
  unused argument(s) (newdata = list(time = c(4, 3, 1, 1, 2, 2, 3), status = c(1, NA, 1, 0, 1, 1, 0), x = c(1, 2, 1, 1, 1, 0, 0), sex = c(0, 0, 0, 0, 1, 1, 1)))
>

Mike

On Apr 7, 11:43 am, Roland Rau <roland.rproj... at gmail.com> wrote:
> Hi,
>
> just looked at it briefly and I don't know if it is the real cause for
> your problems. But 'data' as well as 'newdata' require a data.frame
> and not a list as input.
>
> Does this help?
>
> Best,
> Roland
>
>
>
>
>
> mah wrote:
> > I am unable to get thebasehazfunction to apply a proportional
> > hazards model to a new data frame.  I replicated my specific
> > situation with the example for coxph in the help, where I changed
> > the x value of the first record from 0 to 1.  Is there something
> > incorrect in the syntax that I am using?  Thanks in advance!
>
> > test1 <- list(time=  c(4, 3,1,1,2,2,3),
> >               status=c(1,NA,1,0,1,1,0),
> >               x=     c(0, 2,1,1,1,0,0),
> >               sex=   c(0, 0,0,0,1,1,1))
> > test2 <- list(time=  c(4, 3,1,1,2,2,3),
> >               status=c(1,NA,1,0,1,1,0),
> >               x=     c(1, 2,1,1,1,0,0),
> >               sex=   c(0, 0,0,0,1,1,1))
> > f1 <- coxph( Surv(time, status) ~ x + strata(sex), data=test1)
> > #stratified model
> > f2 <- coxph( Surv(time, status) ~ x + strata(sex), data=test2)
> > #stratified model
> >  f1
> > Call:
> > coxph(formula = Surv(time, status) ~ x + strata(sex), data = test1)
>
> >   coef exp(coef) se(coef)     z    p
> > x 1.17      3.22     1.29 0.907 0.36
>
> > Likelihood ratio test=0.87  on 1 df, p=0.351  n=6 (1 observation
> > deleted due to missingness)
> >> f2
> > Call:
> > coxph(formula = Surv(time, status) ~ x + strata(sex), data = test2)
>
> >    coef exp(coef) se(coef)    z    p
> > x 0.896      2.45     1.42 0.63 0.53
>
> > Likelihood ratio test=0.38  on 1 df, p=0.535  n=6 (1 observation
> > deleted due to missingness)
> >>basehaz(f1, newdata=test2)
> > Error inbasehaz(f1, newdata = test2) :
> >   unused argument(s) (newdata = list(time = c(4, 3, 1, 1, 2, 2, 3),
> > status = c(1, NA, 1, 0, 1, 1, 0), x = c(1, 2, 1, 1, 1, 0, 0), sex =
> > c(0, 0, 0, 0, 1, 1, 1)))
>
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