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? 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

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 > > ______________________________________________ > R-help at 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.David Winsemius, MD Heritage Laboratories West Hartford, CT

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 <dwinsemius at comcast.net> wrote:> From: David Winsemius <dwinsemius at comcast.net> > Subject: Re: [R] survreg function in survival package > To: "carol white" <wht_crl at 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 <dwinsemius at comcast.net> > wrote: > > > >> From: David Winsemius <dwinsemius at comcast.net> > >> Subject: Re: [R] survreg function in survival > package > >> To: "carol white" <wht_crl at yahoo.com> > >> Cc: r-help at 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 > >