[R] command in survival package
llei at bccrc.ca
Mon Jan 23 18:33:25 CET 2006
Thank you guys.
But I tried the commands and I still get:
 time status x
<0 rows> (or 0-length row.names)
> esf.fit <- survfit(Surv(aml1$weeks,status) ~ 1)
Error in Surv(aml1$weeks, status) : Time variable is not numeric
In addition: Warning message:
is.na() applied to non-(list or vector) in: is.na(time)
which still looks confusing. Or are they should be applied in s-plus
instead of R?
Thanks a lot!
From: Marc Schwartz [mailto:MSchwartz at mn.rr.com]
Sent: Friday, January 20, 2006 4:49 PM
To: Thomas Lumley
Cc: Linda Lei; R-help at stat.math.ethz.ch
Subject: Re: [R] command in survival package
On Fri, 2006-01-20 at 15:27 -0800, Thomas Lumley wrote:
> On Fri, 20 Jan 2006, Linda Lei wrote:
> > Hi there,
> > I have a question about one command sentence when I follow the
> > in the book of "Survival analysis in S":
> > > aml1<-aml[aml$group==1]
> If this is really what the book says you should probably complain to
> author. However, if it says
> then you show type that instead.
> It is hard to be sure, because you don't say where the `aml' data set
> comes from. It can't be the one in the survival package as this
> have a variable called "group"
> > Thus, I couldn't keep going on the next command:
> > esf.fit<-survfit(Surv(aml1,status)~1).
> This looks implausible as well. I would have expected something like
> esf.fit<-survfit(Surv(time,status)~1, data=aml1)
Correct on both accounts Thomas. The text in the book on page 26 appears
as Linda posted.
However, there is an errata document here:
which shows the corrected text as:
aml1 <- aml[aml$group==1,] # Maintained group only
esf.fit <- survfit(Surv(aml1$weeks,status) ~ 1)
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