similar to: Age as time-scale in a cox model

Dear R users, My question is more methodology related rather than specific to R usage. Using time on study as time in a cox model, eg: library(Design) stanf.cph1=cph(Surv(time, status) ~ t5+id+age, data=stanford2, surv=T) #In this case the 1000-day survival probability would be: stanf.surv1=survest(stanf.cph1, times=1000) #Age in this case is a covariate. #I now want to compare the above

I am having trouble formatting some survival data to use in a time dependent cox model. My time dep. variable is habitat and I have it recorded for every day (with some NAs). I think it is working properly except for calculating the death.time. This column should be 1s or 0s and as I have it only produces 0s. Any help will be greatly appreciated.

Hi I am doing a survival analysis and have run into a couple of things that I was hoping I could get some advice on. The first thing is that when I run an ordinary Cox regression in R I get the same results that I do using Stata (provided that I specify the Efron method for handling ties) but I also get the following warning message: "Warning message: In coxph(Surv1 ~ Out + site + haem) :

Hello R users Would somebody know how to estimate survival from a frailty cox model, using the function survfit and the argument newdata ? (or from any other way that could provide individual expected survival with standard error); Is the problem related to how the random term is included in newdata ? kfitm1 <- coxph(Surv(time,status) ~ age + sex + disease + frailty(id,

Dear all, I am new to R and may make beginner mistakes. Sorry. I am learning using R to do survival analysis. As a start I used the example script code provided in the documentation of predict.survreg of the survival package: # Draw figure 1 from Escobar and Meeker fit <- survreg(Surv(time,status) ~ age + age^2, data=stanford2, dist='lognormal') plot(stanford2$age, stanford2$time,

Hola a todos, -Estoy estudiando el efecto de dos genotipos (~tratamientos) en la aparición de síndrome metabólico (MetS) con datos longitudinales recogidos a tiempo 0,7,10,15,20 y 25 años. -He hecho un dataframe con las siguientes variables MetS: Síndrome Metabólico (Si=1,No=0) bmi: Indice de masa corporal (IMC) cuando se produce la conversión a MetS+ . Para los que permancen MetS-, esta variable

Dear useRs, There is a new version 0.1-2 of the package surv2sample available on CRAN. Users of the previous versions should update because a bug in the function cif2.ks has been fixed. General information about the package: surv2sample provides various two-sample tests for right-censored survival data. Three main areas and corresponding methods are: * comparison of two survival

Dear useRs, There is a new version 0.1-2 of the package surv2sample available on CRAN. Users of the previous versions should update because a bug in the function cif2.ks has been fixed. General information about the package: surv2sample provides various two-sample tests for right-censored survival data. Three main areas and corresponding methods are: * comparison of two survival

Terry, I recently noticed the censor argument of survfit. For some analyses it greatly reduces the size of the resulting object, which is a nice feature. However, when combined with the id argument, only 1 prediction is made. Predictions can be made individually but I'd prefer to do them all at once if that change can be made. Chris ##################################### # CODE # create

Hi, I am just wondering if there is a test available for testing if a linear fit of an independent variable in a Cox regression is enough? Thanks for any suggestions. John Zhang ____________________________________________________________________________________ [[elided Yahoo spam]]

Dear all, I was wondering if anyone can help? I am an R user but recently I have resorted to SAS to calculate the probability of the event (and the associated confidence interval) for the Cox model with combinations of risk factors. For example, suppose I have a Cox model with two binary variables, one for gender and one for treatment, I wish to calculate the probability of survival for the

I've a data set with 60000 rows of data representing 6000+ distinct loans. I did a coxph() regression on it (see call below), but a subsequent survfit() call on the coxph object is almost certainly wrong. It gives n=6 when it should be more like 6000+ (I think) > survfit(resultag) Call: survfit.coxph(object = resultag) n events median 0.95LCL 0.95UCL 6 489 Inf

Dear Thomas, The head of my dataset > head(wsuv) parcel sp time censo treatment species 1 S8 Poecilanthe effusa ( Hub. ) Ducke. 1 1 1 1 2 S8 Poecilanthe effusa ( Hub. ) Ducke. 1 1 1 1 3 S8 Poecilanthe effusa ( Hub. ) Ducke. 1 1 1 1 4 S8 Poecilanthe effusa ( Hub. ) Ducke. 1 1 1

Dear all, I'm having difficulty getting access to data generated by survfit and print.survfit when they are using with a Cox model (survfit.coxph). I would like to programmatically access the median survival time for each strata together with the 95% confidence interval. I can get it on screen, but can't get to it algorithmically. I found myself examining the source of print.survfit to

Hello, I'm trying to estimate a cox model with a frailty variable and time-dependent covariate (below there is the statement I use and the error message). It's seems to be impossible, because every time I add the time-dependent covariate the model doesn't converge. Instead, if I estimate the same model without the time-dependent covariate it's converge. I'd like knowing if

Hi all, Is there an equivalent to predict(...,type="linear") of a Proportional hazard model for a Cox model instead? For example, the Figure 13.12 in MASS (p384) is produced by: (aids.ps <- survreg(Surv(survtime + 0.9, status) ~ state + T.categ + pspline(age, df=6), data = Aidsp)) zz <- predict(aids.ps, data.frame(state = factor(rep("NSW", 83), levels =

OS X R 4.3.3 Colleagues I have created objects using the Surv function in the survival package: > FIT.1 Call: survfit(formula = FORMULA1) n events median 0.95LCL 0.95UCL SUBDATA$ARM=1, SUBDATA[, EXP.STRAT]=0 18 13 345 156 NA SUBDATA$ARM=2, SUBDATA[, EXP.STRAT]=1 13 5 NA 186 NA SUBDATA$ARM=2, SUBDATA[, EXP.STRAT]=2 5

Hi Dennis, look at the help page for summary.survfit, the Value n.event. G?ran On 2024-05-15 22:41, Dennis Fisher wrote: > OS X > R 4.3.3 > > Colleagues > > I have created objects using the Surv function in the survival package: >> FIT.1 > Call: survfit(formula = FORMULA1) > > n events median 0.95LCL 0.95UCL >

Hi, I have the binned data (observed and generated from model)  that I would like to test using the anderson-darling goodness of fit test.  But I'm not sure which package in R to use. I tried ad.test(...) but it does not recognise the test by Vito Ricci in FITTING DISTRIBUTIONS WITH R   > ad.test(hist_hume_beec[,1],hist_hume_beec[,2]) Error: could not find function "ad.test"

Hello, I am trying to compute the mdeian of the survival time from the function survfit: > fit <- survfit(Surv(time, status) ~ 1) > fit Call: survfit(formula = Surv(time, status) ~ 1) records n.max n.start events median 0.95LCL 0.95UCL 111 111 111 20 NA NA NA The results is NA? the fit$surv gives values between 1 and 0.749! Am I doing this correct?