Re: [R] survfit number of variables != number of variable names

-BEGIN PGP SIGNED MESSAGE- Hash: SHA1 Hi Terry, I attached a small data set to this email. This is what I get (I restricted the formula to avoid NA's): surv = with(small, Surv(time=absence, event=(censored==FALSE))) (cox.s = coxph(surv ~ bucket*(today) + strata(activity), data =

Re: [R] survfit number of variables != number of variable names

Hi! In answer to: I noticed that you were using what might be called an externally created Surv object. I have a memory that Terry Therneau has criticized that practice. I cannot remember if it was in exactly this situation but I might ask if setting up the model as: cox =

Re: [R] survfit number of variables != number of variable names

Hi! It seems the data file wasn't transmit. Please find a copy in attachment. Best, ge On 11/19/2012 09:02 AM, Terry Therneau-2 [via R] wrote: I can't reproduce the problem. Tell us what version of R and what version of the survival package. Create a reproducable example. I don't know if

[R] survfit number of variables != number of variable names

This works ok: cox = coxph(surv ~ bucket*(today + accor + both) + activity, data = data) fit = survfit(cox, newdata=data[1:100,]) but using strata leads to problems: cox.s = coxph(surv ~ bucket*(today + accor + both) + strata(activity), data = data) fit.s = survfit(cox.s,

[R] coxph reference hazard rate

Hi, In the following results I interpret exp(coef) as the factor that multiplies the base hazard rate if the corresponding variable is TRUE. For example, when the bucket is ks008 and fidelity = 3, then the rate, compared to the base rate h_0(t), is h(t) = 0.200 h_0(t). My question is then, to