similar to: Problem using pkg "survival"

Hi there, I have a question about one command sentence when I follow the example in the book of "Survival analysis in S": > aml1<-aml[aml$group==1] but I got the error warning: NULL data frame with 23 rows Thus, I couldn't keep going on the next command: esf.fit<-survfit(Surv(aml1,status)~1). and also when I try > aml1<-aml[aml$group==1,]

Hi List, A friend of mine recently asked the same question as Heinz T?chler. Since I've already written the code I'd like to share with the list. # x is an object returned by "survfit"; # "smed" returns a matrix of 5 columns of # n, events, median, 0.95LCL, 0.95UCL. # The matrix returned has rownames as the # group labels (eg., treatment arms) if any. smed <-

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 am using plot(survfit(Surv(time,status) ~...) and would like to add error bars rather than the confidence intervals. Am I able to do this at specified times? e.g. when time = 20 & 40. leukemia.surv <- survfit(Surv(time, status) ~ x, data = aml) plot(leukemia.surv, lty = 2:3,xlim = c(0,50)) #can i add error bars at times 20 & 40? legend(100, .9, c("Maintenance", "No

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

Dear mailing list, I tried to run the example for the conditional bootstap written in the help file of censboot. I got the following result: STRATIFIED CONDITIONAL BOOTSTRAP FOR CENSORED DATA Call: censboot(data = aml, statistic = aml.fun, R = 499, F.surv = aml.s1, G.surv = aml.s2, strata = aml$group, sim = "cond") Bootstrap Statistics : original bias std. error t1*

Hi! I came across R just a few days ago since I was looking for a toolbox for cox-regression. I?ve read "Cox Proportional-Hazards Regression for Survival Data Appendix to An R and S-PLUS Companion to Applied Regression" from John Fox. As described therein plotting survival-functions works well (plot(survfit(model))). But I?d like to do some manipulation with the survival-functions

Hi All, Please pardon me if I am missing something obvious here. How do I get the Kaplan-Meier estimate function that is created by survfit and plotted by the code. fit <- survfit(Surv(time, status) , data=aml) plot(fit) That is, I need a function that will give me the survival estimate at a given time: \hat{S}(t). Thanks in advance. Ritwik Sinha ritwik.sinha at gmail.com | +12033042111 |

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

Dear experts, I would like to know how to plot the log-minus-log plot for survival analysis (to check the proportional assumption) in R. Using the AML example. fit <- survfit(Surv(time, status) ~ x, data=aml) length(fit$surv) #20 as the length of fit$surv is shorter than aml$x and aml$time. I don't know how to plot. Thank you. Regards, CH -- CH Chan Research Assistant - KWH

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

Hello: Using the built-in dataset aml as an example: data(aml) If I use instead dummy variables: aml$x1 = (aml$x=="maintained")aml$x2 = (aml$x=="unmaintained") and I want to plot the survival curve using x1, x2, and I just want the 2 levels, rather than 4 curves from: fit <- survfit(Surv(time, status) ~ x1+x2, data=aml) plot(fit) I guess because there are 2 levels

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 all, I want to plot a kaplan Meier plot with the following functions, but I fail to produce the plot I want: library(survival) tim <- (1:50)/6 ind <- runif(50) ind[ind > 0.5] <- 1; ind[ind < 0.5] <- 0; MS <- runif(50) pred <- vector() pred[MS < 0.3] <- 0; pred[MS >= 0.3] <- 1 df <- as.data.frame(cbind(MS, tim, pred, ind)) names(df) <-

Hola Jesús, La respuesta, desde mi punto de vista, es un poco off-topic de lo que se trata en esta lista, pero comento como lo veo yo. Con el nivel de detalle que tienes, puedes hacer varias cosas: - Simplemente mantén en tu almacén un número de cuchillas mayor que la última vez que tuviste que pedirlas con urgencia. En los entornos de Producción, efectivamente el que rompas el stock

Dear all, I am trying to produce survfit plots in a trellis environment and I would like the plots to be logarithmic. I am trying this: print(Ecdf(~time | size*type, groups=alg,data=B,subscripts=TRUE, panel=function(x,groups,subscripts) { t <- survfit(Surv(time[subscripts],event[subscripts])~groups[subscripts],data=B)

Hello friends and fellow R users, I have successfully tabulated and entered my survival data into R and have generated survival curves. But I would like to be able to determine what the survival rates are now at one month, three months, six months and one year. I have a data set, via.wall, which I have entered into R, and which generates the following Surv object: Surv(Days,Status==1) [1]

Dear help news reader, I'm trying to draw a Kaplan-Meier curve and would like to ask the news group for some help Supposing I have study comapring two drugs, "A", and "B" and I recorde the time to get to the clinical endpoint (Time), in my case becommming virus free. I have setup the following frame: Time c Drug 1 5 1 A 2 7 1 B 3 2 1 A 4 10 1

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

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))"