Displaying 20 results from an estimated 10000 matches similar to: "Survival questions"
2007 Jun 10
0
Question on weighted Kaplan-Meier analysis of case-cohort design
I have a study best described as a retrospective case-cohort design:
the cases were all the events in a given time span surveyed, and the
controls (event-free during the follow-up period) were selected in
2:1 ratio (2 controls per case). The sampling frequency for the
controls was about 0.27, so I used a weight vector consisting of 1
for cases and 1/0.27 for controls for coxph to adjust
2012 Apr 30
0
need help with avg.surv (Direct Adjusted Survival Curve), Message-ID:
Well, I would suggest using the code already in place in the survival
package. Here is my code for your problem.
I'm using a copy of the larynx data as found from the web resources for
the Klein and Moeschberger book.
larynx <- read.table("larynx.dat", skip=12,
col.names=c("stage", "time", "age", "year",
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,
2012 Apr 29
0
need help with avg.surv (Direct Adjusted Survival Curve)
Hello R users,
I am trying to obtain a direct adjusted survival curve. I am sending my whole code (see below). It's basically the larynx cancer data with Stage 1-4. I am using the cox model using coxph option, see the fit3 coxph. When I use the avg.surv option on fit3, I get the following error: "fits<-avg.surv(fit3, var.name="stage.fac", var.values=c(1,2,3,4), data=larynx)
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
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:
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
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)
2017 Aug 27
0
How to get CI from surfit object in survival
Did you not notice the conf.type = "none" argument to your survfit
call and the associated documentation in the survfit help?
-- Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Sat, Aug 26, 2017 at 5:18 PM, Adrian Johnson
2013 Nov 04
0
Fwd: Re: How to obtain nonparametric baseline hazard estimates in the gamma frailty model?
-------- Original Message --------
Subject: Re: How to obtain nonparametric baseline hazard estimates in the gamma frailty model?
Date: Mon, 04 Nov 2013 17:27:04 -0600
From: Terry Therneau <therneau.terry at mayo.edu>
To: Y <yuhanusa at gmail.com>
The cumulative hazard is just -log(sfit$surv).
The hazard is essentially a density estimate, and that is much harder. You'll notice
2010 Dec 10
2
survival package - calculating probability to survive a given time
Dear R users,
i try to calculate the probabilty to survive a given time by using the
estimated survival curve by kaplan meier.
What is the right way to do that? as far as is see i cannot use the
predict-methods from the survival package?
library(survival)
set.seed(1)
time <- cumsum(rexp(1000)/10)
status <- rbinom(1000, 1, 0.5)
## kaplan meier estimates
fit <- survfit(Surv(time,
2011 Jun 27
7
cumulative incidence plot vs survival plot
Hi, I am wondering if anyone can explain to me if cumulative incidence (CI) is
just "1 minus kaplan-Meier survival"? Under what circumstance, you should use
cumulative incidence vs KM survival? If the relationship is just CI =
1-survival, then what difference it makes to use one vs. the other?
And in R how I can draw a cumulative incidence plot. I know I can make a
Kaplan-Meier
2017 Aug 27
2
How to get CI from surfit object in survival
Dear Bert,
thank you for suggestion. I am aware of R-help function. I must
apologize, my earlier question could lead to assumptions otherwise.
As you can see below, I only get Std. error but not lower and Upper CIs.
I was wondering if there is another argument or method, could give CIs
which I cannot find anywhere.
Thanks
Adrian
> Nsurv <-
2008 Sep 30
0
Hazard curves
-- begin included message -----
I am looking at a continuous variable, age. I am looking at time to
12-month remission and can calculate the HR and 95% confidence interval are
follows:
coxfita = coxph(Surv(rem.Remtime,rem.Rcens)~nearma$all.age,data=nearma)
exp(coxfita$coefficients)
exp(confint(coxfita))
However, because I am looking at age as a continuous variable I cannot draw
a Kaplan-Meier
2009 Jun 20
1
Plotting Cumulative Hazard Functions with Strata
Hello:
So i've fit a hazard function to a set of data using
kmfit<-survfit(Surv(int, event)~factor(cohort))
this factor variable, "cohort" has four levels so naturally the strata
variable has 4 values.
I can use this data to estimate the hazard rate
haz<-n.event/n.risk
and calculate the cumulative hazard function by
H<--log(haz)
Now, I would like to plot this
2013 Mar 15
0
confidence interval for survfit
The first thing you are missing is the documentation -- try ?survfit.object.
fit <- survfit(Surv(time,status)~1,data)
fit$std.err will contain the standard error of the cumulative hazard or -log(survival)
The standard error of the survival curve is approximately S(t) * std(hazard), by the delta
method. This is what is printed by the summary function, because it is what user's
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
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2011 Jul 15
1
Plotting survival curves from a Cox model with time dependent covariates
Dear all,
Let's assume I have a clinical trial with two treatments and a time to
event outcome. I am trying to fit a Cox model with a time dependent
treatment effect and then plot the predicted survival curve for one
treatment (or both).
library(survival)
test <-
list(time=runif(100,0,10),event=sample(0:1,100,replace=T),trmt=sample(0:1,100,replace=T))
model1 <- coxph(Surv(time,
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)
2008 Mar 02
1
Problem plotting curve on survival curve (something silly?)
OK this is bound to be something silly as I'm completely new to R -
having started using it yesterday. However I am already warming to its
lack of 'proper' GUI... I like being able to rerun a command by editing
one parameter easily... try and do that in a Excel Chart Wizzard!
I eventually want to use it to analyse some chemotherapy response /
survival data. That data will not be