similar to: predict risk (% of death) at a certain time point

Displaying 20 results from an estimated 7000 matches similar to: "predict risk (% of death) at a certain time point"

2012 Nov 26
1
Plotting an adjusted kaplan-meier curve
Dear R-users I am trying to make an adjusted Kaplan-Meier curve (using the Survival package) but I am having difficulty with plotting it so that the plot only shows the curves for the adjusted results. My data come from a randomised controlled trial, and I would like the adjusted Kaplan-Meier curve to only show two curves for the adjusted survival: one for those on treatment (Treatment==1)
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 =
2010 Apr 19
2
Kaplan-Meier survfit problem
When I try to the code from library(survival) of library(ISwR), the following code survfit(Surv(days,status==1)) that could produce Kaplan-Meier estimates shows the following error "Error in survfit(Surv(days, status == 1)) : Survfit requires a formula or a coxph fit as the first argument" How it can be done in R.2.10 -- View this message in context:
2001 Mar 11
2
Kaplan-Meier for left-truncated, right-censored data
Is it possible to calculate Kaplan-Meier for left-truncated, right-censored data using survival5? -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at
2012 Nov 26
1
Plotting an adjusted survival curve
First a statistical issue: The survfit routine will produce predicted survival curves for any requested combination of the covariates in the original model. This is not the same thing as an "adjusted" survival curve. Confusion on this is prevalent, however. True adjustment requires a population average over the confounding factors and is closely related to the standardized
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
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) <-
2004 May 04
2
Epidemiology Tools
Hi all, Please help on this. We will be teaching epidemiology using opensource software. What are R built-in functions or functions in available packages that are capable of doing these: a) Logistic regression (glm?) b) Conditional logistic regression c) Logistic regression with random effects d) Beta-binomial regression e) Poisson regression f) Weibull regression (eha?) g) Exponential
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
2003 Dec 11
1
plot of survival probability vs. covariate
Hi everyone, I am fitting a cox proportional hazard model with a continuous variable "x" as the covariate: fit<-coxph(Surv(time, status)~x) Now I wanted to make a plot of survival probability vs. the covariate, and the 95% confidence interval for the survival probability. It's just like a Kaplan-Meier Survival curve, except now the x axis represents the value of covariate, not
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
2010 Nov 18
3
Plotting number of patients at risk below survival curve
Dear list, does anyone know of a R-package that has implemented the increasingly popular inclusion of the number of patients at risk below Kaplan-Meier curves like in http://bloodjournal.hematologylibrary.org/content/vol116/issue19/images/large/zh89991058760001.jpeg any hint (or negative answer) is much appreciated. Thanks Thorsten -- Thorsten Raff 2nd Medical Department, University
2011 Mar 14
1
pairwise p-values in KM plots?
How can I compute pairwise p-values in Kaplan-meier plots for three or more groups? bin.1<-cut(score,c(-1000,-1,1,1000),c("low","intermediate","high")) I use "km.coxph.plot" currently which reports one p-value. Thanks very much. [[alternative HTML version deleted]]
2008 Sep 29
1
Hazard plot
Hi All, This sounds a relatively simple query, and I hope it is! I am looking at a continuous variable, age. I am looking at time to 12-month remission and can calculate the HR and 95% confidence interval are follows: coxfita = coxph(Surv(rem.Remtime,rem.Rcens)~nearma$all.age,data=nearma) exp(coxfita$coefficients) exp(confint(coxfita)) However, because I am looking at age as a continuous
2004 Jan 07
2
Survival, Kaplan-Meier, left truncation
Dear all, I have data from 1970 to 1990 for people above age 50. Now I want to calculate survival curves by age starting at age 50 using the Kaplan Meier Estimator. The problem I have is that there are already people in 1970 who are older than 50 years. I guess this is called delayed entry or left truncation (?). I thought the code would be: roland <- survfit(Surv(time=age.enter,
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
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 Jul 13
2
life table and Kaplan-Meier
Hello, I have a question about the function lifetab in package KMsurv. The description of the output value surv says "the estimated survival function at the start of the intervals". Are these estimates the ones calculated via Kaplan-Meier probability of survival ? Thanks in advance! -- View this message in context:
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): ----------------
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)