[R] Weibull survival modeling with covariate

[Göran Broström]

On Wed, Jun 08, 2005 at 08:39:21PM -0400, sms13+ at pitt.edu wrote:
> I was wondering if someone familiar 
> with survival analysis can help me with 
> the following.
> I would like to fit a Weibull curve, 
> that may be dependent on a covariate, 
> my dataframe "labdata" that has the 
> fields "cov", "time", and "censor".  Do 
> I do the following?
> wieb<-survreg(Surv(labdata$time, 
> labadata$censor)~labdata$cov, 
> dist="weibull")
> 
> This returns:
> >weib
> Call:
> survreg(formula = Surv(labdata$time, 
> labdata$censor) ~ labdata$cov,
>    dist = "weibull")
> 
> Coefficients:
> (Intercept) labdata$cov
> 8.091955112 0.001552897
> 
> Scale= 0.7532474
> 
> Loglik(model)= -12633.6 
> Loglik(intercept only)= -12734.8
>        Chisq= 202.41 on 1 degrees of 
> freedom, p= 0
> n= 5496
> 
> 
> I am not quite sure how to use the 
> output.  I see that it gives the Scale 
> parameter.  How do I find the Shape 
> paramater as a function of the 
> covariate?

You don't. The analysis is performed on the logs of durations, so scale is
transformed to location and shape to scale. For more intuitive output, use 
'weibreg' in package 'eha'. It can also handle left truncated data. But
only Weibull (and exponential) regression.
 
> 
> Thank you,
> Steven
> 
> ---------------------------------------
> -------------------------
> Steven Shechter
> PhD Candidate in Industrial Engineering
> University of Pittsburgh
> www.pitt.edu/~sms13
> 
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-- 
Göran Broström                    tel: +46 90 786 5223
Professor and Head
Department of Statistics          fax: +46 90 786 6614
Umeå University                   http://www.stat.umu.se/~goran.brostrom/
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