similar to: How do I specify a partially completed survival analysis model?

--- begin inclusion -- After I simulate Time and Censor data vectors denoting the censoring time and status respectively, I can call the following function to fit the data into the Cox model (a is a data.frame containing 4 columns X1, X2, Time and Censor): b = coxph (Surv (Time, Censor) ~ X1 + X2, data = a, method = "breslow"); Now the purpose of me doing simulation is that I have

Dear UseRs, Suppose I have data regarding smoking habits of a prospective cohort and wish to determine the risk ratio of colorectal cancer in the smokers compared to the non-smokers. What do I do at the end of the study with people who die of heart disease? Can I just censor them exactly the same as people who become uncontactable or who die in a plane crash? If not, why not? I'm thinking

Hello folks, I am wondering how to do survival analysis with time-related IVs in R. For example, > > If we have time-related variables, such as the Overall Condition of 1990, 1991 etc., how can we include these variables in coxph model? > > > > If we can not use coxph model, do we need to rearrange the dataset to make it something like: > > ID time age

I was wondering if someone familiar with survival analysis can help me with the following. I would like to fit a Weibull curve, that may be dependent on a covariate, my dataframe "labdata" that has the fields "cov", "time", and "censor". Do I do the following? wieb<-survreg(Surv(labdata$time, labadata$censor)~labdata$cov,

Hi all, I have a little problem. I make an weibull survival analysis using the survival package. It,s OK, them I have the functions. I plot this funcions with curve(). I want to make a plot with the real survival points (proportion of alive x time) and them add the curves to points. I have the time to dead, the censor data and my trataments. To analysis the model is: model1 <-

For adjusted survival curves I took the sample code from here: https://rpubs.com/daspringate/survival and adapted for my date, but ... have a QUESTION. library(survival) library(survminer) df<-read.csv("base.csv", header = TRUE, sep = ";") head(df) ID start stop censor sex age stage treatment 1 1 0 66 0 2 1 3 1 2 2 0 18 0 1 2 4 2 3 3 0 43 1 2 3 3 1 4 4 0 47 1 2 3 NA 2 5 5

Dear r-helpers, This is my first time to run survival analysis. Currently, I have a data set which contains two variables, the variable of time to event (or time to censoring) and the variable of censor indicator. For the indicator variable, it was coded as 0 and 1. 0 represents right censor, 1 means event of interest. Now I try to use "survfit" in the package of "survival". I

For adjusted survival curves I took the sample code from here: https://rpubs.com/daspringate/survival and adapted for my date, but got error. I would like to understand what is my mistake. Thanks! #ADAPTATION FOR MY DATA library(survival) library(survminer) df<-read.csv("F:/R/data/base.csv", header = TRUE, sep = ";") head(df) ID start stop censor sex age stage treatment 1

List, Consider the following data. gender mygroup id 1 F A 1 2 F B 2 3 F B 2 4 F B 2 5 F C 2 6 F C 2 7 F C 2 8 F D 2 9 F D 2 10 F D 2 11 F D 2 12 F D 2 13 F D 2 14 M A 3 15 M A 3 16 M A 3 17

Hi experts, http://r.789695.n4.nabble.com/file/n4034318/Parametric_survival_analysis_2nd-order_efffect.JPG Parametric_survival_analysis_2nd-order_efffect.JPG As we know a normal survival regression is the equation (1) Well, I'ld like to modify it to be 2nd-order interaction model as shown in equation(2) Question: Assume a and z is two covariates. x = dummy variable (1 or 0) z = factors

Dear all, I would have some questions on the coxph function for survival analysis, which I use with frailty terms. My model is: mdcox<-coxph(Surv(time,censor)~ gender + age + frailty(area, dist='gauss'), data) I have a very large proportion of censored observations. - If I understand correctly, the function mdcox$frail will return the random effect estimated for each group on the

Hi, I was wondering if someone can help me interpret the results of running weibreg. I run the following and get the following R output. > weibreg(Surv(time, censor)~covar) fit$fail = 0 Call: weibreg(formula = Surv(time, censor)~covar) Covariate Mean Coef Rel.Risk L-R p Wald p covar 319.880 -0.002 0.998 0.000 log(scale) 0.000 8.239

Dear R-devel, I'm working on Prof. Loader's new version of locfit to try to get it pass R CMD check. I'm almost there, but I have a problem with some Rd files that I hope some one can help me resolve. Here's an example: In the package there's a function called locfit.censor(). This function can be used in a few different ways: locfit.censor(x, y, cens, ...)

I have the complete data like id time censor 1 10 0 1 20 0 1 30 0 2 10 0 2 20 1 2 30 0 2 40 0 3 10 0 3 20 0 3 30 1 .... for id 1, i want to select the last row since all censor indicator is 0; for id 2, i want to select the row where censor ==1; for id 3, i also want to select the row where censor==1. So if there is a 1 for censor, then I want to select such a row, otherwise I want to select the

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 was trying to fit a parametric survival model with Weibull distribution on counting process type of data (NOT interval censor data), but the survreg(Surv(T1,T2,event)~x,data,dist="weibull") did not seem to work. Anyone can help me with that? Thanks, Rachel Memorial Sloan-Kettering Cancer Center -- View this message in context:

Hey all, I am trying to use the survreg function in R to estimate the mean and standard deviation to come up with the MLE of alpha and lambda for the weibull distribution. I am doing the following: times<-c(10,13,18,19,23,30,36,38,54,56,59,75,93,97,104,107,107,107) censor<-c(1,0,0,1,0,1,1,0,0,0,1,1,1,1,0,1,0,0) survreg(Surv(times,censor),dist='weibull') and I get the following

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 all, The survivals functions can be tested by the Log-rank test and others, for example the Gehan-Breslow. The graham breslow work with the alpha values. But I don't know how is the Gehan-Breslow test with R. Somebody know a type function?.. or other suggestions? Any help will be really appreciated Jos? Bustos Marine Biologist Master Apllied Stat Program University of Concepci?n