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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]]

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)

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

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

What is the best way to report the standard error when publishing Kaplan-Meier plots? In my field (Vascular Surgery), practitioners loosely refer to the "10% error" cutoff as the point at which to stop drawing the KM curve. I am interpreting this as the *standard error of the cumulative hazard*, although I'm having a difficult time finding some guidelines about this (perhaps I am

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

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

Helo: After changing "involuntarily" some of the graphics parameters with the command par() (I did not know that changes with this command are permanent), now when I made a plot of the survival Kaplan-Meier function, the Y axis does not start at 1, and the X axis does starts at 0. The commands that I use are: library(survival) BROWN.SPV = Surv(BROWN$TEMPS, BROWN$DEF)

Hi, I'm trying to print the p-values from the output of a CPH test onto a Kaplan Meier plot. Can this be done? I only really want the p-values from the CPH test to appear but if this can't be done I am willing to have the entire CPH output. This is what I am currently trying: (it doesn't print the CPH output) plot_KM <- function(field) { library(survival) y =

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]]

Hi All, I'm trying to figure out the cumulative incidence curve in R in some limited time. I found in package "cmprsk", the command "plot.cuminc" can get this curve. But I noticed that there is no mark for the censored time there, comparing with the KM curve by "plot.survfit". Here are my codes (attached is the data): ----------------

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

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

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:

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

Hello R users I have a question with Kaplan-Meier Curve with respect to my research. We have done a retrospective study on fillings in the tooth and their survival in relation to the many influencing factors. We had a long follow-up time (upto 8yrs for some variables). However, we decided to stop the analysis at the 6year follow up time, so that we can have uniform follow-up time for all the

When I try to the code from library(survival) of library(ISwR), the following code survfit(Surv(days,status==1)) that could produce Kaplan-Meier estimates shows the following error "Error in survfit(Surv(days, status == 1)) : Survfit requires a formula or a coxph fit as the first argument" How it can be done in R.2.10 -- View this message in context:

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

Hola, te vale esto? Es forma estandar de representar graficos supervivencia Basado en esto: https://rviews.rstudio.com/2017/09/25/survival-analysis-with-r/ set.seed(20) DATOS <- data.frame ( ID = c (1:10) , TIEMPO = sample(1:40, 10, replace=F) , DEF = sample(0:1, 10, replace=T) ) DATOS library(survival) DATOS$DEF <- as.numeric(DATOS$DEF) DATOS$TIEMPO <-