Displaying 20 results from an estimated 404 matches for "survfits".
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survfit
2012 Jul 06
1
How to compute hazard function using coxph.object
My question is, how to compute hazard function(H(t)) after building the
coxph model. I even aware of the terminology that differs from hazard
function(H(t)) and the hazard rate(h(t)). Here onward I wish to calculate
both.
Here what I have done in two different methods;
##########################################################################################
2010 Jul 15
1
Standard Error for individual patient survival with survfit and summary.survfit
I am using the coxph, survfit and summary.survfit functions to calculate an estimate of predicted survival with confidence interval for future patients based on the survival distribution of an existing cohort of subjects. I am trying to understand the calculation and interpretation of the std.err and confidence intervals printed by the summary.survfit function.
Using the default confidence
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
2007 Dec 09
2
Getting estimates from survfit.coxph
Dear all,
I'm having difficulty getting access to data generated by survfit and
print.survfit when they are using with a Cox model (survfit.coxph).
I would like to programmatically access the median survival time for
each strata together with the 95% confidence interval. I can get it on
screen, but can't get to it algorithmically. I found myself examining
the source of print.survfit to
2013 Jun 25
1
censor=FALSE and id options in survfit.coxph
Terry,
I recently noticed the censor argument of survfit. For some analyses it greatly reduces the size of the resulting object, which is a nice feature.
However, when combined with the id argument, only 1 prediction is made. Predictions can be made individually but I'd prefer to do them all at once if that change can be made.
Chris
#####################################
# CODE
# create
2009 Feb 25
3
survival::survfit,plot.survfit
I am confused when trying the function survfit.
my question is: what does the survival curve given by plot.survfit mean?
is it the survival curve with different covariates at different points?
or just the baseline survival curve?
for example, I run the following code and get the survival curve
####
library(survival)
fit<-coxph(Surv(futime,fustat)~resid.ds+rx+ecog.ps,data=ovarian)
2007 Apr 29
2
how to code the censor variable for "survfit"
Dear r-helpers,
This is my first time to run survival analysis. Currently, I have a
data set which contains two variables, the variable of time to event
(or time to censoring) and the variable of censor indicator. For the
indicator variable, it was coded as 0 and 1. 0 represents right
censor, 1 means event of interest. Now I try to use "survfit" in the
package of "survival". I
2009 Jul 15
2
Substituting a user-defined function for a standard function in graphics
Dear R-help readers,
I'm trying to substitute my own version of plot.default, in order to
capture low-level graphics input for further manipulation. I seem to be
having difficulties getting the versions of functions that I intend to use
called at the right places.
Specifically, I have an object produced by survfit from the survival
package, and I want to save the values that
2006 Oct 25
1
Incorrect 'n' returned by survfit()
I've a data set with 60000 rows of data representing 6000+ distinct loans. I did a coxph() regression on it (see call below), but a subsequent survfit() call on the coxph object is almost certainly wrong. It gives n=6 when it should be
more like 6000+ (I think)
> survfit(resultag)
Call: survfit.coxph(object = resultag)
n events median 0.95LCL 0.95UCL
6 489 Inf
2014 Mar 06
1
Survfit Error
Hi everyone,
I am not new to R, but new to running survival models in R.
I am trying to create some basic KM curves, using the following code:
library(survival)
library(KMsurv)
(import data etc - basic right censored, with continuously observed time of death)
sleepfit <- survfit(Surv(timeb, death), data = sleep)
Here timeb is measured is survival in years, death is a 1/0 indicator (1 =
2011 Oct 01
4
Is the output of survfit.coxph survival or baseline survival?
Dear all,
I am confused with the output of survfit.coxph.
Someone said that the survival given by summary(survfit.coxph) is the
baseline survival S_0, but some said that is the survival S=S_0^exp{beta*x}.
Which one is correct?
By the way, if I use "newdata=" in the survfit, does that mean the survival
is estimated by the value of covariates in the new data frame?
Thank you very much!
2009 Feb 17
3
Survival-Analysis: How to get numerical values from survfit (and not just a plot)?
Hi!
I came across R just a few days ago since I was looking for a toolbox
for cox-regression.
I?ve read
"Cox Proportional-Hazards Regression for Survival Data
Appendix to An R and S-PLUS Companion to Applied Regression" from John Fox.
As described therein plotting survival-functions works well
(plot(survfit(model))). But I?d like to do some manipulation with the
survival-functions
2010 Aug 31
1
Speeding up prediction of survival estimates when using `survifit'
Hi,
I fit a Cox PH model to estimate the cause-specific hazards (in a competing risks setting). Then , I compute the survival estimates for all the individuals in my data set using the `survfit' function. I am currently playing with a data set that has about 6000 observations and 12 covariates. I am finding that the survfit function is very slow.
Here is a simple simulation example
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 Mar 18
1
median survival time from survfit
Hello,
I am trying to compute the mdeian of the survival time from the function
survfit:
> fit <- survfit(Surv(time, status) ~ 1)
> fit
Call: survfit(formula = Surv(time, status) ~ 1)
records n.max n.start events median 0.95LCL 0.95UCL
111 111 111 20 NA NA NA
The results is NA? the fit$surv gives values between 1 and 0.749! Am I doing
this correct?
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)
2005 Sep 19
2
Problem with tick marks in lines.survfit (package survival)
I have attempted to follow posting guidelines but I have failed to find out
what I am doing wrong here.
I am trying to use lines.survfit to plot a second curve onto a survival
curve produced by plot.survfit. In my case this is to be a progression free
survival curve superimposed upon an overall survival curve, but I will
illustrate my problem using the example given in the help for
2013 Mar 14
1
cnfidence intervals for survfit()
Hi, I am wondering how the confidence interval for Kaplan-Meier estimator is calculated by survfit(). For example,
> summary(survfit(Surv(time,status)~1,data),times=10)
Call: survfit(formula = Surv(rtime10, rstat10) ~ 1, data = mgi)
time n.risk n.event survival std.err lower 95% CI upper 95% CI
10 168 55 0.761 0.0282 0.707 0.818
I am trying to reproduce the
2009 Sep 08
1
Obtaining value of median survival for survfit function to use in calculation
Hi,
I'm sure this should be simple but I can't figure it out! I want to get the median survival calculated by the survfit function and use the value rather than just be able to print it. Something like this:
library(survival)
data(lung)
lung.byPS = survfit(Surv (time, status) ~ ph.ecog, data=lung)
# lung.byPS
Call: survfit(formula = Surv(time, status) ~ ph.ecog, data = lung)
1
2010 Sep 23
2
extending survival curves past the last event using plot.survfit
Hello,
I'm using plot.survfit to plot cumulative incidence of an event.
Essentially, my code boils down to:
cox <-coxph(Surv(EVINF,STATUS) ~ strata(TREAT) + covariates, data=dat)
surv <- survfit(cox)
plot(surv,mark.time=F,fun="event")
Follow-up time extends to 54 weeks, but the last event occurs at week
30, and no more people are censored in between. Is there a