similar to: Trying to run simple survival program in R but does not work

Hello, I now receive an error message when obtaining a survival plot, which was not previously received with the same code. I recently updated all my packages. It does not seem to be a peculiarity of my data as I receive the error using data available in R. A plot is produced but I am uncertain regarding the error message. > library(surv2sample) > data(gastric) > fit <-

Hello, it seems that the main results of survival analysis with package survival are shown only as side effects of the print method. If I compute e.g. a Kaplan-Meier estimate by > km.survdur<-survfit(s.survdur) then I can simply print the results by > km.survdur Call: survfit(formula = s.survdur) n events median 0.95LCL 0.95UCL 100.0 58.0 46.8 41.0 79.3 Is

Hi, I am wondering if anyone can explain to me if cumulative incidence (CI) is just "1 minus kaplan-Meier survival"? Under what circumstance, you should use cumulative incidence vs KM survival? If the relationship is just CI = 1-survival, then what difference it makes to use one vs. the other? And in R how I can draw a cumulative incidence plot. I know I can make a Kaplan-Meier

Hi, Sorry to trouble the list - I would like to ask a question - I can't find the answer in the r-help archives. I am trying to plot 2 survival curves in different colors. What I have tried is this plot(survfit(sim.surv ~ sim.km.smpl),col="red") but both the survival curves turn red... Your advice would be most appreciated! Thank you. Min-Han

Hello everybody, does anybody know how the function plot.survfit exactly works? I'd like to plot the log of the cummulative hazard against the log time by using plot.survfit(...fun="cloglog") which does not work correctly. The scales are wrong and there is an error message about infinit numbers. It must have something to do with the censored data, doesn't it? #Example:

First a statistical issue: The survfit routine will produce predicted survival curves for any requested combination of the covariates in the original model. This is not the same thing as an "adjusted" survival curve. Confusion on this is prevalent, however. True adjustment requires a population average over the confounding factors and is closely related to the standardized

Hi, I am attempting to graph a Kaplan Meier estimate for some claims using the survfit function. However, I was wondering if it is possible to plot a cdf of the kaplan meier rather than the survival function. Here is some of my code: library(survival) Surv(claimj,censorj==0) survfit(Surv(claimj,censorj==0)~1) surv.all<-survfit(Surv(claimj,censorj==0)~1) summary(surv.all) plot(surv.all)

Dear R-help list, I have a problem with the tick marks of a Kaplan-Meier survival plot. Here is a sample: follow.up<-c(10,20,30,40,50,60,70,80,90,100) #months dead<-c(1,1,1,0,1,1,0,0,0,0) KM <-survfit(Surv(follow.up, dead)) plot(KM) The result is a nice plot. However, our research group thinks it may be a better idea to place the ticks to the years on the time scale, i.e. 0, 12, 24, 36

Buenas, Como pudeo calcular el tiempo de vida? Os cuento, tengo una serie de cuchillas y quiero ver el consumo de las mismas y he pensado en hacer un estudio por tiempo de vida. No se como hacerlo con R Gracias Jesús [[alternative HTML version deleted]]

Dear Sirs, I want to estimate the survival mean of a few specific teams. I'm trying to calculate it through a Kaplan Meier estimator. For doing so, I load the "survival" package and run the following instructions: "options(survfit.print.mean=TRUE)" allows showing the mean and mean standard error "KM=survfit(Surv(Dias,Censura))"

Los datos no son de desgaste de cuchilla, sino de consumo de las mismas. Por ello tengo los datos de la siguiente forma: Unidades cambiadas Fecha En unidades cambiadas, suele ser una y en fecha el dia que se hizo el cmabio. Con eso no se muy bien como estructurar los datos para hacer el análisis. Gracias Jesús > Date: Mon, 7 Dec 2015 16:27:18 +0100 > From: griera en yandex.com

I would like to produce a complimentary log-log survival plot with only the points appearing on the graph. I am using the code below, taken from the plot.survfit page of help for the the survival package (version 2.35-7). I am running in R 2.10.0 on Windows XP, and the list of packages following the error is loaded. Is there some specific 'type= ' syntax, or an additional parameter that

Hi all, I am trying to draw a Kaplan-Meier curve and I found online that Kaplan - Meier estimates are computed with a function called km in the event package. Is there an update for that because when I choose to download packages in R,. there is no package called event, even though I have selected all the repositories. Thanks in advance, Eleni [[alternative HTML version deleted]]

Dear R-helpers, I am working with 'cph' models from 'rms' library. When I build simple survival models, based on 'Surv(time, event)', everything is fine and I can make nice plots using plot(Predict(f, time=3)). However, recently I tried to be more specific and used 'Surv(start, stop, event)' type model. Using this model 'plot(Predict(f))' works OK, but

Hi, I am taking a survival class. Recently I need to do the Nelson-Aalen estimtor in R. I searched through the R help manual and internet, but could not find such a R function. I tried another way by calculating the Kaplan-Meier estimator and take -log(S). However, the function only provides the summary of KM estimator but no estimated values. Could you please help me with this? I would

Hi all, I have installed in R 2.6 the survival package and I am trying to analyze some data and present plots. The problem occurs when I try to plot lines on the same plot. E.g. take simulated data for time (t), event (e) and groups (group1 and group2) and do: ### this is how I create the plots plot1<-survfit(Surv(t,e)~ as.factor(group1)) plot2<-survfit(Surv(t,e)~ as.factor(group2))

Hi, there is a function to plot survival curves: library(survival) plot.KM <- function(survival, x, x_cut.off, main='', label='') { plot(survfit(survival ~ I(x >= x_cut.off)), main=main) legend('bottomleft', c(expression(label >= x_cut.off),expression(label < x_cut.off))) } Now, I need to determine as the argument what appears in the legend. I want

I have some data which is censored and I want to determine the median. Its actually cost data for a cohort of patients, many of whom are still on treatment and so are censored. I can do the same sort of analysis for a survival curve and get the median survival... ...but can I just use the survival curve functions to plot an X axis that is $ rather than date? If not is there some other way to

Pero como haría el data frame?? Porque las cuchillas son de la misma referencia. En realidad es para ver cada cuanto se gstan las cuchillas y ver que pedidos hay que hacer de las mismas. La tabla que tengo es: 25 enero-> 1 cuchilla gastada 30 enero -> 1 cuchilla gastada 3 de febrero -> 2 cuchillas gastadas 5 de febrero -> 1 cuchilla gastada Y así.... No tiene necesariamente que ser

Hi everyone, I am not new to R, but new to running survival models in R. I am trying to create some basic KM curves, using the following code: library(survival) library(KMsurv) (import data etc - basic right censored, with continuously observed time of death) sleepfit <- survfit(Surv(timeb, death), data = sleep) Here timeb is measured is survival in years, death is a 1/0 indicator (1 =