[R] Simulation from a model fitted by survreg.

Spencer Graves spencer.graves at pdf.com
Wed Jul 28 16:15:45 CEST 2004

      Please check the documentation, e.g., Venables and Ripley (2002) 
Modern Applied Statistics with S, p. 360.  The Weibull shape parameter 
is the reciprocal of the scale parameter 0.598 in your printout, so 
shape = 1/0.598 = 1.672;  see also Meeker & Escobar (1998) Statistical 
Methods for Reliability Data (Wiley). 

      Does this answer the question?  You can get coefficients with 
coef(mod1).  Also, have you looked at attributes(summary(mod1))?  If 
"mod1" follows the old S3 standard, attributes may give you a list of 
names you can access via summary(mod1)$whateverthenameis (or via 
summary(mod1)[["whateverthenameis"]]).  If "mod1" follows the new S4 
standard, then getSlots(mod1) and getSlots(summary(mod1)) will give you 
the names of the slots and their classes, which can then be accessed via 
mod1 at nameofslotofinterest.  Sorry, I don't have time to construct an 
example myself, but I've done this kind of thing many times. 

      hope this helps. 
      spencer graves    

Sixten Borg wrote:

>Dear list,
>I would like to simulate individual survival times from a model that has been fitted using the survreg procedure (library survival). Output shown below.
>My plan is to extract the shape and scale arguments for use with rweibull() since my error terms are assumed to be Weibull, but it does not make any sense. The mean survival time is easy to predict, but I would like to simulate individual survival times.
>I am probably missing something completely obvious. Any hints or advice are appreciated.
>survreg(formula = Surv(tid, study$first.event.death) ~ regim + 
>    age + stadium2, data = study, dist = "weibull")
>              Value Std. Error      z        p
>(Intercept) 11.6005     0.7539 15.387 2.01e-53
>regimposto  -0.1350     0.1558 -0.867 3.86e-01
>age         -0.0362     0.0102 -3.533 4.11e-04
>stadium2ii  -0.0526     0.2794 -0.188 8.51e-01
>Log(scale)  -0.5148     0.1116 -4.615 3.93e-06
>Scale= 0.598 
>Weibull distribution
>Loglik(model)= -680.7   Loglik(intercept only)= -689.2
>        Chisq= 16.87 on 3 degrees of freedom, p= 0.00075 
>Number of Newton-Raphson Iterations: 8 
>n=1183 (4 observations deleted due to missing)
>         _              
>platform i386-pc-mingw32
>arch     i386           
>os       mingw32        
>system   i386, mingw32  
>major    1              
>minor    8.1            
>year     2003           
>month    11             
>day      21             
>language R              
>R-help at stat.math.ethz.ch mailing list
>PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html

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