similar to: RV: help

Displaying 20 results from an estimated 6000 matches similar to: "RV: help"

2011 Apr 05
6
simple save question
Hi, When I run the survfit function, I want to get the restricted mean value and the standard error also. I found out using the "print" function to do so, as shown below, print(km.fit,print.rmean=TRUE) Call: survfit(formula = Surv(diff, status) ~ 1, type = "kaplan-meier") records n.max n.start events *rmean *se(rmean) median 200.000
2015 Dec 07
2
Tiempo de vida
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
2015 Dec 08
2
Tiempo de vida
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
2013 Mar 04
2
survfit plot question
Hello, I create a plot from a coxph object called fit.ads4: plot(survfit(fit.ads4)) plot is located at: https://www.dropbox.com/s/9jswrzid7mp1u62/survfit%20plot.png I also create the following survfit statistics: > print(survfit(fit.ads4),print.rmean=T) Call: survfit(formula = fit.ads4) records n.max n.start events *rmean *se(rmean) median 0.95LCL 0.95UCL 203.0
2013 Apr 29
1
R help - bootstrap with survival analysis
Hi, I'm not sure if this is the proper way to ask questions, sorry if not. But here's my problem: I'm trying to do a bootstrap estimate of the mean for some survival data. Is there a way to specifically call upon the rmean value, in order to store it in an object? I've used print(...,print.rmean=T) to print the summary of survfit, but I'm not sure how to access only rmean
2003 Oct 06
3
tick marks: 0, 12, 24, 36 ...
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
2002 Aug 02
1
survival analysis: plot.survfit
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:
2005 Feb 04
5
How to access results of survival analysis
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
2012 Feb 20
1
Reporting Kaplan-Meier / Cox-Proportional Hazard Standard Error, km.coxph.plot, survfit.object
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
2008 Feb 14
4
Kaplan Meier function
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]]
2011 Jun 27
7
cumulative incidence plot vs survival plot
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
2015 Aug 02
3
ayuda con análisis de supervivencia
Hola a todos, -Estoy estudiando el efecto de dos genotipos (~tratamientos) en la aparición de síndrome metabólico (MetS) con datos longitudinales recogidos a tiempo 0,7,10,15,20 y 25 años. -He hecho un dataframe con las siguientes variables MetS: Síndrome Metabólico (Si=1,No=0) bmi: Indice de masa corporal (IMC) cuando se produce la conversión a MetS+ . Para los que permancen MetS-, esta variable
2004 Oct 05
2
Nelson-Aalen estimator in R
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
2012 Jan 05
2
Problem with axes in a plot of Kaplan-Meier
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)
2004 Sep 03
1
Printing output on Plot
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 =
2006 Aug 17
1
putting the mark for censored time on 1-KM curve or competing risk curve
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): ----------------
2004 Oct 05
1
save print survfit object to data frame
Hello, I have estimated a survival model with six strata: >model.b <- survfit(Surv(time=start.tijd,time2=eind.tijd2,event=va)~strata(product.code) , data=wu.wide) I would like to save the output of >print(model.b,print.n="records",show.rmean=FALSE) in a dataframe so that I can export it later. How do I do this? Note that summary(model.b) gives an error: Error in
2015 Dec 10
2
Tiempo de vida
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
2010 Sep 10
2
survfit question
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)
2012 Mar 07
4
Difference in Kaplan-Meier estimates plus CI
I thought this would be trivial, but I can't find a package or function that does this. I'm hoping someone can guide me to one. Imagine a simple case with two survival curves (e.g. treatment & control). I just want to calculate the difference in KM estimates at a specific time point (e.g. 1 year) plus the estimate's 95% CI. The former is straightforward, but the estimates not