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)

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) <-

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