Nomogram with stratified cph in Design package
David Winsemius wrote:
On Apr 25, 2009, at 6:57 PM, reneepark wrote:
Hello,
I am using Dr. Harrell's design package to make a nomogram. I was able to
make a beautiful one without stratifying, however, I will need to
stratify
to meet PH assumptions. This is where I go wrong, but I'm not sure where.
Non-Stratified Nomogram:
f<-cph(S~A+B+C+D+E+F+H,x=T,y=T,surv=T,time.inc=10*12,method="breslow")
srv=Survival(f)
srv120=function(lp) srv(10*12,lp)
quant=Quantile(f)
med=function(lp) quant(.5,lp)
at.surv=c(0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9)
at.med=c(120,80,60,40,30,20,15,10,8,6,4,2,0)
nomogram(f,lp=F, fun=list(srv120, med),funlabel=c("120-mo
Survival","Median
Survival"),fun.at=list(at.surv, at.med))
I get a the following warning:
Warning message:
In approx(fu[s], xseq[s], fat) : collapsing to unique 'x' values
However, a great nomogram is constructed.
But then I try to stratify...
Stratified Nomogram:
f<-cph(S~A+B+C+D+E+F+strat(H),x=T,y=T,surv=T,time.inc=10*12,method="breslow")
srv=Survival(f)
surv.p <- function(lp) srv(10*12, lp, stratum="Hist=P")
surv.f <- function(lp) srv(10*12, lp, stratum="Hist=F")
surv.o <- function(lp) srv(10*12, lp, stratum="Hist=O")
quant=Quantile(f)
med.p <- function(lp) quant(.5, lp, stratum="Hist=P")
med.f <- function(lp) quant(.5, lp, stratum="Hist=F")
med.o <- function(lp) quant(.5, lp, stratum="Hist=O")
nomogram(f, fun=list(surv.p, surv.f, surv.o, med.p, med.f, med.o),
+ funlabel=c("S(120|P)","S(120|F)","S(120|O)",
+ "med(P)","med(F)","med(O)"),
+ fun.at=list(c(0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9),
+ c(120,80,60,40,30,20,15,10,8,6,4,2,0)))
the final nomogram only gives me a survival probability line for one
of the
3 Hist categories "S(120|P)". It does show the letters "S(120|F)" but
there
is no survival probability line; there is nothing for the last
category O,
and no median risk at all.
Those outputs seem consistent with the fact that stratification is not computing separate models, but rather a pooled model. See Section 19.1.7 of RMS.
But you can think of stratification as using a different transformation for each stratum, and as long as you create a separate function for each level of the stratification variable, as Rene did, all should be well.
I considered the idea that I was exceeding some sort of space limitation, and tried to set total.sep.page=T, but it didn't change the output.
Does a "median risk' exist when you stratify? You are allowing 3 separate survival functions to be created so that you estimate the remaining parameters. It's possible that you can extract information about them, but you may be on your own about how to recombine them.
Yes it exists, using the separate function approach. Rene if you can duplicate the problem with a simple simulated or real dataset and send that to me I can try to go through this step by step. It's probably a scaling, units of measurement, or extrapolation problem where the median is not defined. You can evaluate the created functions yourself a several settings to see if the results are reasonable and to learn where extrapolation is not possible because of truncated follow-up. Frank
I get the following error message: Error in axis(sides[jj], at = scaled[jj], label = fat[jj], pos = y, cex.axis = cex.axis, : no locations are finite I would very much appreciate any assistance in this matter. Thank you very much. ~Renee Park
David Winsemius, MD Heritage Laboratories West Hartford, CT
______________________________________________ R-help at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Frank E Harrell Jr Professor and Chair School of Medicine
Department of Biostatistics Vanderbilt University