[R] Weibull and survival
Prof Brian Ripley
ripley at stats.ox.ac.uk
Tue Nov 22 09:15:43 CET 2005
On Tue, 22 Nov 2005, Stephen wrote:
> I have been asked to provide Weibull parameters from a paper using
> Kaplan Meir survival analysis.
>
> This is something I am not familiar with.
>
> The survival analysis in R works nicely and is the same as commercial
> software (only the graphs are superior in R).
>
> The Weibull does not and produces an error (see below).
>
> Any ideas why this error should occur?
Do you have zero survival times? They cannot occur for a Weibull, but
might as a result of rounding.
For a worked example of how to deal with these, see MASS4 p. 380.
The other possibility is that you have infinite survival times, but that
is unlikely to be correct ....
> My approach may be spurious.
>
> Code follows
>
> #The following works fine
>
>> surv.mod1 <- survfit( Surv(SURALL2, relall6==1)~randgrpc,
> type=c("kaplan-meier"),data=Dataset)
>
> #The following works produces the error below
>
>> surv.mod2 <- survreg( Surv(SURALL2, relall6 == 1)~randgrpc,
> data=Dataset, dist="weibull")
>
> Error in survreg(Surv(SURALL2, relall6 == 1) ~ randgrpc, data = Dataset,
> :
>
> Invalid survival times for this distribution
> PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
--
Brian D. Ripley, ripley at stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
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