Displaying 20 results from an estimated 6000 matches similar to: "Cox proportional hazards model"
2011 Jan 16
1
Help in Coxme
I am a relative newbie to survival analysis and R in general, but
would like to use the coxme package to analyse some data I currently
have.
The data is relative to survival times of drosophila melanogaster
populations to infection with pathogens, and has the variables:
Time,
Status,
Treatment (4 treatments + 2 controls)
Population
Replicate
?and I'm currently using the following call
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"
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
2024 May 15
2
Extracting values from Surv function in survival package
OS X
R 4.3.3
Colleagues
I have created objects using the Surv function in the survival package:
> FIT.1
Call: survfit(formula = FORMULA1)
n events median 0.95LCL 0.95UCL
SUBDATA$ARM=1, SUBDATA[, EXP.STRAT]=0 18 13 345 156 NA
SUBDATA$ARM=2, SUBDATA[, EXP.STRAT]=1 13 5 NA 186 NA
SUBDATA$ARM=2, SUBDATA[, EXP.STRAT]=2 5
2024 May 16
1
Extracting values from Surv function in survival package
Hi Dennis,
look at the help page for summary.survfit, the Value n.event.
G?ran
On 2024-05-15 22:41, Dennis Fisher wrote:
> OS X
> R 4.3.3
>
> Colleagues
>
> I have created objects using the Surv function in the survival package:
>> FIT.1
> Call: survfit(formula = FORMULA1)
>
> n events median 0.95LCL 0.95UCL
>
2009 Apr 14
1
Function call error in cph/survest (package Design)
Dear UseR,
I do not know if this a problem with me, my data or cph/survest in package
design. The example below works with a standard data set, but not with my
data, but I cannot locate the problem.
Note that I am using an older package of survival to avoid a problem with
the newly renamed function in survival meeting Design.
Dieter
# First, check standard example to make sure
library(Design)
2015 Feb 04
2
Interpretación de coeficientes en un cox proportional hazards con variable strata
Buenas.
Abajo pongo la salida de un modelo de cox , dónde he estratificado por
una variable de país (Countryb) y por otra (Q6). Además hay interacción
entre la variable mobilityPDurG2 (es una variable 0,1, y 0 es la
categoría de referencia) país.
La categoría de referencia para país es "united kingdom".
Mi duda surge si quiero calcular el hazard ratio para los que tienen un
1
2007 May 07
1
Predicted Cox survival curves - factor coding problems..
The combination of survfit, coxph, and factors is getting confused. It is
not smart enough to match a new data frame that contains a numeric for sitenew
to a fit that contained that variable as a factor. (Perhaps it should be smart
enough to at least die gracefully -- but it's not).
The simple solution is to not use factors.
site1 <- 1*(coxsnps$sitenew==1)
site2 <-
2013 Jan 31
1
obtainl survival curves for single strata
Dear useRs,
What is the syntax to obtain survival curves for single strata on many subjects?
I have a model based on Surv(time,response) object, so there is a single row per subject and no start,stop and no switching of strata.
The newdata has many subjects and each subject has a strata and the survival based on the subject risk and the subject strata is needed.
If I do
newpred <-
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
2007 Dec 04
2
weighted Cox proportional hazards regression
I'm getting unexpected results from the coxph function when using
weights from counter-matching. For example, the following code
produces a parameter estimate of -1.59 where I expect 0.63:
d2 = structure(list(x = c(1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1,
1, 0, 0, 1, 0, 1, 0, 1, 0, 1), wt = c(5, 42, 40, 4, 43, 4, 42,
4, 44, 5, 38, 4, 39, 4, 4, 37, 40, 4, 44, 5, 45, 5, 44, 5), riskset =
2009 Mar 14
1
obtaining the values for the hazard function in a cox regression
Hello ,
I am hoping for some advice regarding obtaining the values for the
hazard function in a cox regression that I have undertaken. I have a
model in the following form, analysed with the package survival (v.
2.34-1) and a log-log plot obtained using Design (v. 2.1-2).
For two variables, the lines in the survival curves crossed. The
statistician I been obtaining advice from (who does not
2011 May 06
2
coxph and survfit issue - strata
Dear users,
In a study with recurrent events:
My objective is to get estimates of survival (obtained through a Cox model) by rank of recurrence and by treatment group.
With the following code (corresponding to a model with a global effect of the treatment=rx), I get no error and manage to obtain what I want :
data<-(bladder)
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
2012 Nov 17
4
survfit & number of variables != number of variable names
This works ok:
> cox = coxph(surv ~ bucket*(today + accor + both) + activity, data = data)
> fit = survfit(cox, newdata=data[1:100,])
but using strata leads to problems:
> cox.s = coxph(surv ~ bucket*(today + accor + both) + strata(activity),
> data = data)
> fit.s = survfit(cox.s, newdata=data[1:100,])
Error in model.frame.default(data = data[1:100, ], formula = ~bucket + :
2012 Sep 05
1
showing ticks for censored data in survfit() in the rms package
The answer to this may be obvious, but I was wondering in the rms
package and the survfit(), how you can plot the censored time points
as ticks.
Take for example,
library(survival)
library(rms)
foo <- data.frame(Time=c(1,2,3,4,5,6,10), Status=c(1,1,0,0,1,1,1))
answer <- survfit(Surv(foo$Time, foo$Status==1) ~1)
# this shows the censored time points as ticks at Time = 3 and 4
plot(answer)
2009 May 11
1
Warning trying to plot -log(log(survival))
windows xp
R 2.8.1
I am trying to plot the -log(log(survival)) to visually test the proportional hazards assumption of a Cox regression. The plot, which should give two lines (one for each treatment) gives only one line and a warning message. I would appreciate help getting two lines, and an explanation of the warning message. My problem may the that I have very few events in one of my strata,
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
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
2007 Oct 29
3
using survfit
hie
when i use plot.survfit to plot more than one graph why I only see the last graph how do i see the other graphs.for example
n=20
n1=n/2
n2=n/4
a11=4;a12=4 ;a21=4 ;a22=4
t1<-array(1,c(n1))
t2<-array(2,c(n1))
treatgrp=matrix(c(t1,t2))