similar to: Possible update to survival

Thanks for the reproducable example. I can confirm that it fails on my machine using survival 2-37.5, the next soon-to-be-released version, The issue is with NextMethod, and my assumption that the called routine inherited everything from the parent, including the environment chain. A simple test this AM showed me that the assumption is false. It might have been true for Splus. Working this

ExpeRts, I'm trying to compare three survival curves using the function "survdiff" in the survival package. Following is my code and corresponding error message. > survdiff(Surv(st_months, status) ~ strata(BOR), data=mydata) Error in survdiff(Surv(st_months, status) ~ strata(BOR), data = mydata) : No groups to test When I check the "strata" of the variable. I get .

Dear useRs, What is the syntax to obtain survival curves for single strata on many subjects? I have a model based on Surv(time,response) object, so there is a single row per subject and no start,stop and no switching of strata. The newdata has many subjects and each subject has a strata and the survival based on the subject risk and the subject strata is needed. If I do newpred <-

Dear R users: I am not entirely convinced that clogit gives me the correct result when I use pspline() and maybe you could help correct me here. When I add a constant to my covariate I expect only the intercept to change, but not the coefficients. This is true (in clogit) when I assume a linear in the logit model, but the same does not happen when I use pspline(). If I did something similar

[code]>require("survival") > coxph(Surv(futime,fustat)~age + strata(rx),ovarian) Call: coxph(formula = Surv(futime, fustat) ~ age + strata(rx), data = ovarian) coef exp(coef) se(coef) z p age 0.137 1.15 0.0474 2.9 0.0038 Likelihood ratio test=12.7 on 1 df, p=0.000368 n= 26, number of events= 12 > coxph(Surv(futime,fustat)~age, ovarian, subset=rx==1)

The latest version of the survival package has two important additions. In prior code the call coxph(Surv(time, status) ~ age + strata(inst), data=lung) could fail if a version of either Surv() or strata() existed elsewhere on the search path; the wrong function could be picked up. Second, a model with survival::strata(inst) in the formula would not do what users expect. These

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:

I am wondering how to estimate the survival curve for a particular case(s) given a coxph model using this example code: #fit a cox proportional hazards model and plot the #predicted survival curve fit <- coxph( Surv(futime,fustat)~resid.ds+strata(rx)+ecog.ps+age,data=ovarian[1:23,]) z <- survfit(fit,newdata=ovarian[24:26,],individual=F) zs <- z$surv zt <-

The combination of survfit, coxph, and factors is getting confused. It is not smart enough to match a new data frame that contains a numeric for sitenew to a fit that contained that variable as a factor. (Perhaps it should be smart enough to at least die gracefully -- but it's not). The simple solution is to not use factors. site1 <- 1*(coxsnps$sitenew==1) site2 <-

Hi, I am working on transforming a SAS code to R code. It's about the survival analysis and the SAS code is as below: -------------------------------------- proc lifetest data=surdata plot=(s); time surv*censht(1); strata educ; title 'Day 1 homework'; run; ---------------------------------------- here is the data: subject surv censht educ 1 78 1 1 2

Hello, I'm using plot.survfit to plot cumulative incidence of an event. Essentially, my code boils down to: cox <-coxph(Surv(EVINF,STATUS) ~ strata(TREAT) + covariates, data=dat) surv <- survfit(cox) plot(surv,mark.time=F,fun="event") Follow-up time extends to 54 weeks, but the last event occurs at week 30, and no more people are censored in between. Is there a

Hi, I'm trying to use the R Survival analysis on a windows 7 system. The input data format is described at the end of this mail. 1/ I tried to perform a survival analysis including stratified variables using the following formula. cox.xtab_miR=coxph(Surv(time, status) ~ miR + strata(sex,nbligne, age), data=matrix) and obtain the following error message Warning message: In fitter(X, Y,

Well, I would suggest using the code already in place in the survival package. Here is my code for your problem. I'm using a copy of the larynx data as found from the web resources for the Klein and Moeschberger book. larynx <- read.table("larynx.dat", skip=12, col.names=c("stage", "time", "age", "year",

Hello R users,  I am trying to obtain a direct adjusted survival curve. I am sending my whole code (see below). It's basically the larynx cancer data with Stage 1-4. I am using the cox model using coxph option, see the fit3 coxph. When I use the avg.surv option on fit3, I get the following error: "fits<-avg.surv(fit3, var.name="stage.fac", var.values=c(1,2,3,4), data=larynx)

Hi, I'm going crazy trying to plot a quite simple graph. i need to plot estimated hazard rate from a cox model. supposing the model i like this: coxPhMod=coxph(Surv(TIME, EV) ~ AGE+A+B+strata(C) data=data) with 4 level for C. how can i obtain a graph with 4 estimated (better smoothed) hazard curve (base-line hazard + 3 proportional) to highlight the effect of C. thanks!! laudan [[alternative

windows xp R 2.8.1 I am trying to plot the -log(log(survival)) to visually test the proportional hazards assumption of a Cox regression. The plot, which should give two lines (one for each treatment) gives only one line and a warning message. I would appreciate help getting two lines, and an explanation of the warning message. My problem may the that I have very few events in one of my strata,

I am not sure if this is the right place for that kind of questions, but I wondered that the recommended package survival did not pass R's check procedure without warnings: 1) unbalanced braces: * Rd files with unbalanced braces: * man/Surv.Rd * man/cluster.Rd * man/cox.zph.Rd * man/coxph.Rd * man/coxph.detail.Rd * man/date.ddmmmyy.Rd * man/lines.survfit.Rd *

Hi, Im an honours student at Monash University. I'm trying to analyse some data for my project, which involved 2 treatments. My subjects were exposed to both treatments, and i gave them 60 minutes to perform a certain behaviour. 3 of my subjects performed the behaviour in one treatment but not the other. Therefore, i need to do a survival analysis using paired data. Im little confused about

Dear R-help, I would like to compute the variance for the proportion of treatment effect by a surrogate in a survival model (Lin, Fleming, and De Gruttola 1997 in Statistics in Medicine). The paper mentioned that the covariance matrix matches that of the covariance matrix estimator for the marginal hazard modelling of multiple events data (Wei, Lin, and Weissfeld 1989 JASA), and is implemented

Calum had a long question about drawing survival curves after fitting a Weibull model, using pweibull, which I have not reproduced. It is easier to get survival curves using the predict function. Here is a simple example: > library(survival) > tfit <- survreg(Surv(time, status) ~ factor(ph.ecog), data=lung) > table(lung$ph.ecog) 0 1 2 3 <NA> 63 113 50 1