Thank you David. I don't think that I could pass by rweibull function since I use uniform random variable to generate survival times having weibull distribution. Therefore, like you, I have not found any other solution to set the shape parameter. If I want to calculate hazard, I will need both scale and shape parameters.
Sorry, just forgot to reply to all when replying to your email. Unfortunately, nobody else has replied yet so I wonder if anybody else could be helpful. Cheers, --- On Sat, 11/14/09, David Winsemius <dwinsem...@comcast.net> wrote: > From: David Winsemius <dwinsem...@comcast.net> > Subject: Re: [R] survreg function in survival package > To: "carol white" <wht_...@yahoo.com> > Date: Saturday, November 14, 2009, 8:44 AM > It appears from a look at the str() > output from your survreg object that you did set the scale > parameter, at least the $scale value is set to 1 which is > not what happens when that procedure is employed without > that explicit setting. Does that mean that the coefficients > were the shape parameters? The help page for > survreg.distributions {survival} says that the scale = > 1/shape and that the intercept is =log(scale). > > ?survreg.distributions > ?survreg.object > > And this is found in the survreg example: > # There are multiple ways to parameterize a Weibull > distribution. The survreg > # function imbeds it in a general location-scale familiy, > which is a > # different parameterization than the rweibull function, > and often leads > # to confusion. > # survreg's scale = 1/(rweibull shape) > # survreg's intercept = log(rweibull scale) > # For the log-likelihood all parameterizations lead to the > same value. y <- rweibull(1000, shape=2, scale=5) > survreg(Surv(y)~1, dist="weibull") > I find that a bit confusing because it would seem that the > scale should not be a (pseudo-)random number. I was > especting to read that scale would be either pweibull(shape) > or qweibull(shape). Guess I will need to go back to my > textbooks when I have time, which sadly I do not have > today. > Given that you are asking this offlist, I am sending it > only to you which is not the optimal method for this > exchange. It means that neither one of us wiil get our > confusion and questions addressed by more knowledgeable > persons reading the r-help list. My suggestion is that you > copy this to the list. > --David > On Nov 13, 2009, at 9:02 AM, carol white wrote: > > > Thanks for your reply. > > > > which parameter presents the base line scale > parameter? How is it possible to set the shape parameter for > weibull in survreg? > > > > Many thanks > > > > --- On Fri, 11/13/09, David Winsemius <dwinsem...@comcast.net> > wrote: > > > >> From: David Winsemius <dwinsem...@comcast.net> > >> Subject: Re: [R] survreg function in survival > package > >> To: "carol white" <wht_...@yahoo.com> > >> Cc: r-h...@stat.math.ethz.ch > >> Date: Friday, November 13, 2009, 3:56 AM > >> > >> On Nov 13, 2009, at 3:17 AM, carol white wrote: > >> > >>> Hi, > >>> Is it normal to get intercept in the list of > >> covariates in the output of survreg function with > standard > >> error, z, p.value etc? Does it mean that intercept > was > >> fitted with the covariates? Does Value column > represent > >> coefficients or some thing else? > >>> > >> > >> Don't you need a baseline scale parameter for the > Weibull > >> function? You didn't offer the structure of your > dataframe, > >> but if it is the "standard" ovarian set, then the > rx coef is > >> just the difference between the scale parameter of > rx=2 from > >> that of rx=1, and similarly for ecog.ps. You would > not have > >> an estimate for rx=1 and ecog.ps=1 if you were not > given the > >> Intercept coef. > >> > >> In the future it would be good manners to indicate > what > >> grad school you are taking classes at. > >> > >> > >> --David > >> > >>> Regards, > >>> > >>> > ------------------------------------------------- > >>> tmp = survreg(Surv(futime, fustat) ~ ecog.ps + > rx, > >> ovarian, dist='weibull',scale=1) > >>>> summary(tmp) > >>> > >>> Call: > >>> survreg(formula = Surv(futime, fustat) ~ > ecog.ps + rx, > >> data = ovarian, > >>> dist = "weibull", > scale = 1) > >>> > >> Value Std. Error > >> z p > >>> (Intercept) 6.962 > >> 1.322 5.267 1.39e-07 > >>> ecog.ps -0.433 > >> 0.587 -0.738 4.61e-01 > >>> rx > >> 0.582 0.587 > >> 0.991 3.22e-01 > >>> > >>> Scale fixed at 1 > >>> > >>> Weibull distribution > >>> Loglik(model)= > -97.2 Loglik(intercept > >> only)= -98 > >>> Chisq= 1.67 on 2 > degrees of > >> freedom, p= 0.43 > >>> Number of Newton-Raphson Iterations: 4 > >>> n= 26 > -- > David Winsemius, MD > Heritage Laboratories > West Hartford, CT > > ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.