similar to: how to plot hazard function for coxph model?

I?m trying to set up proportional hazard model that is stratified with respect to covariate 1 and has an interaction between covariate 1 and another variable, covariate 2. Both variables are categorical. In the following, I try to illustrate the two problems that I?ve encountered, using the lung dataset. The first problem is the warning: To me, it seems that there are too many dummies

Dear R users, My apologies for the simple question, as I'm starting to learn the concepts behind the Cox PH model. I was just experimenting with the survival and rms packages for this. I'm simply trying to obtain the expected survival time (as opposed to the probability of survival at a given time t). I can't seem to find an option from the "type" argument in the predict

First, here is your message as it appears on R-help. On 10/14/2012 05:00 AM, r-help-request@r-project.org wrote: > I?m trying to set up proportional hazard model that is stratified with > respect to covariate 1 and has an interaction between covariate 1 and > another variable, covariate 2. Both variables are categorical. In the > following, I try to illustrate the two problems that

Hi there, I have a question on how to extract the linear term in the penalized spline in coxph. Here is a sample code: n=100 set.seed(1) x=runif(100) f1 = cos(2*pi*x) hazard = exp(f1) T = 0 for (i in 1:100) { T[i] = rexp(1,hazard[i]) } C = runif(n)*4 cen = T<=C y = T*(cen) + C*(1-cen) data.tr=cbind(y,cen,x) fit=coxph(Surv(data.tr[,1],

It seems to me that R returns the unpenalized log-likelihood for the ratio likelihood test when ridge regression Cox proportional model is implemented. Is this as expected? In the example below, if I am not mistaken, fit$loglik[2] is unpenalized log-likelihood for the final estimates of coefficients. I would expect to get the penalized log-likelihood. I would like to check if this is as expected.

Hello everyone, I tried to use coxph.detail() to get the hazard function. But a warning messge always returns to me, even in the example provided by its help document: > ?coxph.detail > fit <- coxph(Surv(futime,fustat) ~ age + rx + ecog.ps, ovarian, x=TRUE) > fitd <- coxph.detail(fit) Warning message: data length [37] is not a sub-multiple or multiple of the number of rows

Dear All: I have some questions about the output of coxph. Below is the input and output: ---------------------------------------- > coxph(formula = Surv(futime, fustat) ~ age + rx + ecog.ps, data = + ovarian, x = TRUE) Call: coxph(formula = Surv(futime, fustat) ~ age + rx + ecog.ps, data = ovarian, x = TRUE) coef exp(coef) se(coef) z p age 0.147 1.158

Dear all, I think the coxph and survfit.coxph can give the cumulative hazard of cox model. But is there any method to calculate the hazard Lambda(t)=lambda_0(t)*exp{beta*X(t)}? Any suggestion will be great help. Thank you very much! Koshihaku -- View this message in context: http://r.789695.n4.nabble.com/How-to-get-the-hazard-of-coxph-not-cumulative-hazard-tp3873516p3873516.html Sent from the R

My question is, how to compute hazard function(H(t)) after building the coxph model. I even aware of the terminology that differs from hazard function(H(t)) and the hazard rate(h(t)). Here onward I wish to calculate both. Here what I have done in two different methods; ##########################################################################################

Dear R-users, I computed a simple coxph model and plotted survival over time with associated confidence intervals for 2 covariate levels (males and females). M1 <- coxph(survobject~sex, data=surv) M1 survsex <- survfit(survobject~sex,data=surv) summary(survsex) plot(survsex, conf.int=T, col=c("black","red"), lty = c(1,2), lwd=c(1,2), xlab="Time",

Hi I am estimating the following coxph function with stratification and frailty?where each person had multiple events. m<-coxph(Surv(dtime1,status1)~gender+cage+uplf+strata(enum)+frailty(id),xmodel) ? > head(xmodel) id enum dtime status gender cage uplf 1 1008666 1 2259.1412037 1 MA 0.000 0 2 1008666 2 36.7495023 1 MA 2259.141 0 3 1008666

Hello everybody. First of all, I would like to present myself. I'm a french student in public health and I like statistics though I'm not that good in mathematics (but I try to catch up). I've discovered R recently while trying to find a statistical program in order to avoid rebooting my computer under windows when I need to do some statistical work. And here is my first question.

I was looking at how the model.frame method for lm works and comparing it to my own for coxph. The big difference is that I try to retain xlevels and predvars information for a new model frame, and lm does not. I use a call to model.frame in predict.coxph, which is why I went that route, but never noted the difference till now (preparing for my course in Nashville). Could someone shed light

Hello, is there a R package that provides a log rank trend test for survival data in >=3 treatment groups? Or are there any comparable trend tests for survival data in R? Thanks a lot Markus -- Dipl. Inf. Markus Kreuz Universitaet Leipzig Institut fuer medizinische Informatik, Statistik und Epidemiologie (IMISE) Haertelstr. 16-18 D-04107 Leipzig Tel. +49 341 97 16 276 Fax. +49 341 97 16

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 what case does the base hazard rate correspond to? I would expect the reference to be the first

What is the best way to report the standard error when publishing Kaplan-Meier plots? In my field (Vascular Surgery), practitioners loosely refer to the "10% error" cutoff as the point at which to stop drawing the KM curve. I am interpreting this as the *standard error of the cumulative hazard*, although I'm having a difficult time finding some guidelines about this (perhaps I am

Hello users, I would like to obtain a survival curve from a Cox model that is smooth and does not have zero differences due to no events for those particular days. I have: > sum((diff(surv))==0) [1] 18 So you can see 18 days where the survival curve did not drop due to no events. Is there a way to ask survfit to fit a nice spline for the survival?? Note: I tried survreg and it did not

-------- Original Message -------- Subject: Re: How to obtain nonparametric baseline hazard estimates in the gamma frailty model? Date: Mon, 04 Nov 2013 17:27:04 -0600 From: Terry Therneau <therneau.terry at mayo.edu> To: Y <yuhanusa at gmail.com> The cumulative hazard is just -log(sfit$surv). The hazard is essentially a density estimate, and that is much harder. You'll notice

Since I read the list in digest form (and was out ill yesterday) I'm late to the discussion. There are 3 steps for predicting survival, using a Cox model: 1. Fit the data fit <- coxph(Surv(time, status) ~ age + ph.ecog, data=lung) The biggest question to answer here is what covariates you wish to base the prediction on. There is the usual tradeoff between too few (leave out something

On 2/25/20 9:13 PM, Jason Wang wrote: > > On 2020/2/26 ??12:51, Suman Anna wrote: >> Hi Jason, >> >> On 2/24/20 11:39 PM, Jason Wang wrote: >>> On 2020/2/25 ??5:26, Suman Anna wrote: >>>> The functions vring_new_virtqueue() and __vring_new_virtqueue() are >>>> used >>>> with split rings, and any allocations within these