similar to: Greenwood's Variance

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

Hello all. I've been attempting to utilize the "survival" pkg ( http://cran.r-project.org/web/packages/survival/index.html), while reading through this guide (http://www.ms.uky.edu/~mai/Rsurv.pdf). I figured working through the guide would be the best way to go, before attempting my own data. I tried to utilize the Kaplain-Meier estimator as shown in the guide:

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*

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

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 It seems that survfit() doesn't accept the argumnet 'error' as below >survfit(fit, error='greenwood') Error in survfit.coxph(fit, error = "greenwood") : unused argument(s) (error ...) Isn't is allowed to do that for a coxph object? Regards Soren Windows XP, SP2 R 2.3.0

Dear all, I run the example of "censboot" contained in "boot" package. But, I can't find the confidence interval of the resulted "censboot" object. Any idea ? > aml.fun <- function(data) { + surv <- survfit(Surv(time, cens)~group, data=data) + out <- NULL + st <- 1 + for (s in 1:length(surv$strata)) { +

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

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 <-

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 |

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 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,]

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

I would like to join repeated measures for patients across two visits using a line. The program below uses symbols to represent each patient. Basically, I would like to join each pair of symbols. library(lattice) patient <- c(1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9) var <- c(826,119,168,90,572,323,122,10,42,900,250,180,120,650,400,130,12,33) visit <- c(1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2)

The latest version of the survival package has two important additions. In prior code the call coxph(Surv(time, status) ~ age + strata(inst), data=lung) could fail if a version of either Surv() or strata() existed elsewhere on the search path; the wrong function could be picked up. Second, a model with survival::strata(inst) in the formula would not do what users expect. These

I have an outlier that I would still like to display, but would prefer to shorten the axis. For example, display 0% - 40%, and 90% - 100%. Is this possible? I am using an xyplot. Thanks Murray -- Murray Pung Statistician, Datapharm Australia Pty Ltd 0404 273 283 [[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)

Is there any way of adding text to a density plot? I have had a go using the text() function but I think the error is because this function doesn't work with densityplot(). Alternatively, I understand I can achieve pretty much the same result if I plot a density kernel estimate using plot() (which allows text()), but I do prefer densityplot(). Also, is it possible to specify the dimensions

I would like to set the y axis limit of an xyplot using the object 'ylimit', but receive this error: [1] 990 Error in extend.limits(limitlist[[i]], axs = axs) : improper length of lim I get the same error if I use ylim. library(lattice) trellis.device(col = FALSE, theme = lattice.getOption("col.whitebg")) name <- "Variable name" symbols <-

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) <-