R help - Dec 2003 - 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 the time.
Someone gave me a reference to a paper in JASA by Gary
(1992) for this type of plot, but I didn't have the
access to the paper. So I am wondering if anyone knows
how to do this in R or S-Plus?

In addition, can anyone explain to me what are the
following "type" options in predict.coxph()
predicting?

predict(fit,type='lp',se.fit=T)
predict(fit,type='risk',se.fit=T)
predict(fit,type='expected',se.fit=T)
predict(fit,type='terms',se.fit=T)


Thank you very much

On Thu, 11 Dec 2003 15:23:37 -0800 (PST)
array chip <arrayprofile at yahoo.com> wrote:
> 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 the time.
> Someone gave me a reference to a paper in JASA by Gary
> (1992) for this type of plot, but I didn't have the
> access to the paper. So I am wondering if anyone knows
> how to do this in R or S-Plus?
This will get you started.  Look into the documentation for confidence
limits.

library(Design)  # if S-Plus do  library(Hmisc,T);library(Design,T)
dd <- datadist(yourdataframe)
options(datadist='dd')
f <- cph(Surv( ) ~ x1+x2, surv=T, x=T, y=T)
plot(f, x1=NA, time=2)  # plots 2-year survival

Frank Harrell
> 
> In addition, can anyone explain to me what are the
> following "type" options in predict.coxph()
> predicting?
> 
> predict(fit,type='lp',se.fit=T)
> predict(fit,type='risk',se.fit=T)
> predict(fit,type='expected',se.fit=T)
> predict(fit,type='terms',se.fit=T)
> 
> 
> Thank you very much
> 
> 
> 
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---
Frank E Harrell Jr   Professor and Chair           School of Medicine
                     Department of Biostatistics   Vanderbilt University