Displaying 2 results from an estimated 2 matches for "randgrpc".
2005 Nov 24
4
Survreg Weibull lambda and p
Hi All,
I have conducted the following survival analysis which appears to be OK
(thanks BRipley for solving my earlier problem).
> surv.mod1 <- survreg( Surv(timep1, relall6)~randgrpc, data=Dataset,
dist="weibull", scale = 1)
> summary(surv.mod1)
Call:
survreg(formula = Surv(timep1, relall6) ~ randgrpc, data = Dataset,
dist = "weibull", scale = 1)
Value Std. Error z p
(Intercept) 7.36 0.259 28.42 1.27e-177
randg...
2005 Nov 22
3
Weibull and survival
...ame 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?
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,
:...