similar to: log-minus-log plot

Displaying 20 results from an estimated 7000 matches similar to: "log-minus-log plot"

2007 Jun 17
1
error bars on survival curve
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
2009 Feb 05
4
See source code for survplot function in Design package
Dear R users, I know one way to see the code for a hidden function, say function_x, is using default.function_x (e.g. summary.default). But how can I see the code for imported packages that have no namespace (in this case Design)? Many Thanks Eleni
2006 Jan 20
3
command in survival package
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,]
2005 Dec 20
1
x axis
Hello, I write to know how can I modify the x axis : when I plot a survival object, R plots a graph with x values = 0, 10, 20, 30 while I want a graph with values 0, 6, 12, 18, 24 in the x axis. How can I do this? In R 2.1.1 version there was "time.inc" in survplot, but in version R 2.2.0 there isn't it! I am sorry for my english and I hope that you understand my problem. Thank you
2010 Dec 27
1
Problem using pkg "survival"
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:
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
2004 Apr 21
1
Boot package
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*
2008 Dec 06
1
Kaplan-Meier function from survfit
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 |
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
2006 May 05
2
How to access results of survival analysis
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 <-
2012 Dec 03
1
Confidence bands with function survplot
Dear all, I am trying to plot KM curves with confidence bands with function survplot under package rms. However, the following codes do not seem to work. The KM curves are produced, but the confidence bands are not there. Any insights? Thanks in advance. library(rms) ########data generation############ n <- 1000 set.seed(731) age <- 50 + 12*rnorm(n) label(age) <- "Age"
2012 Nov 06
3
Survplot, Y-axis in percent
Hi I am a new fan of R after getting mad with the graphical functional in SPSS. I have been able to create a nice looking Kaplan Meyer graph using Survplot function. However I have difficulties in turning the y axis to percent instead of the default 0-1 scale. Further I have tried the function yaxt="n" without any results. Any help in this matter will be appreciated. The code is
2009 Nov 06
1
Survival Plot in R 2.10.0
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
2004 Nov 23
6
Weibull survival regression
Dear R users, Please can you help me with a relatively straightforward problem that I am struggling with? I am simply trying to plot a baseline survivor and hazard function for a simple data set of lung cancer survival where `futime' is follow up time in months and status is 1=dead and 0=alive. Using the survival package: lung.wbs <- survreg( Surv(futime, status)~ 1, data=lung,
2009 Feb 02
1
survfit using quantiles to group age
I am using the package Design for survival analysis. I want to plot a simple Kaplan-Meier fit of survival vs. age, with age grouped as quantiles. I can do this: survplot(survfit(Surv(time,status) ~ cut(age,3), data=veteran) but I would like to do something like this: survplot(survfit(Surv(time,status) ~ quantile(age,3), data=veteran) #will not work ideally I would like to superimpose
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)
2006 Dec 09
2
Show number at risk on Kaplan Meier curve
Dear all, I'm using the "survival" package with R 2.4.0 on Mac OS X 10.4.8. I have two core statistics books (one of which is Altman's medical stats book) which suggest showing the number of individuals at risk at different time intervals on the Kaplan-Meier curve. My plot shows two curves that later cross, because of one significant outlier. I have two queries: Is there an
2008 Dec 16
2
"Dotted lines at the end of the KM-curve"
R-ers! Referees demand that the line in the KM-curve should be changed to dotted at the point where standarerror is <= 10 %. I don't think it's a good habit but I urgently need to implement such a thing in R with survfit, survplot or another program. They also want numbers at risk below the curve Some help, please.... Fredrik ######################## Fredrik Lundgren
2009 Jan 14
2
Kaplan-Meier Plot
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) <-
2006 May 30
1
position of number at risk in survplot() graphs
Dear R-help How can one get survplot() to place the number at risk just below the survival curve as opposed to the default which is just above the x-axis? I tried the code bellow but the result is not satisfactory as some numbers are repeated several times at different y coordinates and the position of the n.risk numbers corresponds to the x-axis tick marks not the survival curve time of