similar to: censoring

how do l programme the logrank test. l am trying to compare 2 survival curves --------------------------------- [[alternative HTML version deleted]]

How can I get the Log - Rank p value to be output? The chi square value can be output, so I was thinking if I can also have the degrees of freedom output I could generate the p value, but can't see how to find df either. > (survtest <- survdiff(Surv(time, cens) ~ group, data = surv,rho=0)) Call: survdiff(formula = Surv(time, cens) ~ group, data = surv, rho = 0) N Observed

Hello, I am having some trouble with the 'censoring-adjusted C-index' by Uno et al, in the package survAUC. The relevant function is UnoC. The question has to do with what happens when I specify a time point t for the upper limit of the time range under consideration (we want to avoid using the right-end tail of the KM curve). Copying from the example in the help file: TR <-

Hi all, Using: R version 2.8.1 Patched (2009-03-07 r48068) on OSX (10.5.6) with survival version: Version: 2.35-3 Date: 2009-02-10 I get the following using the first example in ?summary.survfit: > summary( survfit( Surv(futime, fustat)~1, data=ovarian)) Call: survfit(formula = Surv(futime, fustat) ~ 1, data = ovarian) time n.risk n.event survival

Hola Estoy tratando de correr un survival analysis usando un Cox regression model. Tengo una duda respecto a la organizacion del script. Tengo una variable que es -tamano del individuo- y quiero ver si hay diferencia en sobrevivencia respecto a tamano. Como diseno de campo los tamanos fueron ubicados de forma aleatoria en bloques al azar. Cuado planteo el script tengo algo como:

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,

Dear R users, I am trying to use "muhaz" and "plot.muhaz" functions in "muhaz" package to estimate and plot hazard funciton. However function "muhaz" always gives error message "Error in Surv(times, delta) : object "times" not found". I could not even run their sample codes in the user's manual as follows: data(ovarian)

Dear all, I am using R 3.4.3 on Windows 10. I am preparing some teaching materials and I'm having trouble matching the by-hand version with the R code. I have fitted a Cox model - let's use the ovarian data as an example: library(survival) data(ovarian) ova_mod <- coxph(Surv(futime,fustat)~age+rx,data=ovarian) If I want to make predict survival for a new set of individuals at 100

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

Hi, I'm using natural cubic splines from splines::ns() in survival regression (regressing inter-arrival times of patients to a queue on queue size). The queue size fluctuates between 3600 and 3900. I would like to be able to run predict.survreg() for sizes <3600 and >3900 by assuming that the rate for <3600 is the same as for 3600 and that for >4000 it's the same as for

Hi I am trying to understand how to get the validate() function in Design to work with the subset option. I tried this: ovarian.cph=cph(Surv(futime, fustat) ~ age+factor(ecog.ps)+strat(rx), time.inc=1000, x=T, y=T, data=ovarian) validate(ovarian.cph) #fine when no subset is used, but the following two don't work: > validate(ovarian.cph, subset=ovarian$ecog.ps==2) Error in

Dear all, I am confused with the output of survfit.coxph. Someone said that the survival given by summary(survfit.coxph) is the baseline survival S_0, but some said that is the survival S=S_0^exp{beta*x}. Which one is correct? By the way, if I use "newdata=" in the survfit, does that mean the survival is estimated by the value of covariates in the new data frame? Thank you very much!

Hello there, I am doing analysis on survival data. How do I pick out a probability of survival at a chosen landmark time(for example, 3 years, 4 years) from the result of "survfit"? or any other functions? As I know, the result of 'survfit' only have the probabilities for all event or cencor time. Thank you very much Lisa Wang Princess Margaret Hospital Toronto, Ca tel:

Hello, I have questions regarding penalized Cox regression using survival package (functions coxph() and ridge()). I am using R 2.8.0 on Ubuntu Linux and survival package version 2.35-4. Question 1. Consider the following example from help(ridge): > fit1 <- coxph(Surv(futime, fustat) ~ rx + ridge(age, ecog.ps, theta=1), ovarian) As I understand, this builds a model in which `rx' is

Hello, Can anyone suggest a simple way to generate a Kaplan-Meier plot with 2 survfit objects, just like this one:? https://drive.google.com/file/d/1fEcpdIdE2xYtA6LBQN9ck3JkL6-goabX/view?usp=sharing Suppose I have 2 survfit objects: fit1 is for the curve on the left (survtime has been truncated to the cutoff line: year 5), fit2 is for the curve on the right (minimum survival time is at the

Dear UseR, I do not know if this a problem with me, my data or cph/survest in package design. The example below works with a standard data set, but not with my data, but I cannot locate the problem. Note that I am using an older package of survival to avoid a problem with the newly renamed function in survival meeting Design. Dieter # First, check standard example to make sure library(Design)

l'm trying to find the "survdiff" package/function but it seems i cant get it how do l instal it if its not there thanks --------------------------------- [[alternative HTML version deleted]]

Dear all, I am doing library(survival) fit <- coxph(Surv(futime,fustat) ~ rx, ovarian) plot(survfit(fit,newdata=ovarian),col=c(1,2)) legend("bottomleft", legend=c("rx = 0", "rx = 1"), lty=c(1,2),col=c(1,2)) Is this correct to compare these two groups? Is the 0.31 the p-value that the median f two groups are equal Why lty does not work here? Many thanks

Hi Dear R-users I have a database with 18000 observations and 20 variables. I am running cox regression on five variables and trying to use survfit to plot the survival based on a specific variable without success. Lets say I have the following coxph: >library(survival) >fit <- coxph(Surv(futime, fustat) ~ age + rx, data = ovarian) >fit what I am trying to do is plot a survival

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