[R] Nomogram with stratified cph in Design package

[Frank E Harrell Jr]

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
> 
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-- 
Frank E Harrell Jr   Professor and Chair           School of Medicine
                      Department of Biostatistics   Vanderbilt University