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\title{Spline terms with interactions in a Cox model}
\author{Terry Therneau}

\begin{document}
\maketitle
This is a topic that comes up just often enough in my work that I end up
re-discovering how to do it correctly about once a year.
A note showing how may be useful to others, it is certainly a useful
reference for me.

As an example we will use the effect of age on survival in the 
\texttt{flchain} data set, a population based sample of subjects from
Olmsted County, Minnesota.
If we look at a simple model using age and sex we see that both 
are very significant.
<<fit1, fig=TRUE>>=
require(survival)
options(show.signif.stars=FALSE) # display intelligence
fit1 <- coxph(Surv(futime, death) ~ sex + pspline(age, 3), data=flchain)
fit1
termplot(fit1, term=2, se=TRUE, col.term=1, col.se=1,
         ylab="log hazard")
@ 

We used a smoothing spline because the printout then nicely
segregates the linear and non-linear effects.
The non-linearity is not very large, as compared to the linear portion,
but still may be important.

We would like to go forward and fit separate age curves for the males and
the females, since the above fit makes an unwarranted 
assumption that the male/female
ratio of death rates will be the same at all ages. 
The primary problem is that a formula of 
\texttt{sex * pspline(age)} does not work;
the coxph routine is not clever enough to do the right thing automatically.
(Perhaps some future version will be sufficiently bright, but don't hold
your breath).
If we were using regression splines instead, e.g. \texttt{ns(age, df=4)},
the coxph routine would succeed but the plotting would fail; the 
solution below works for both cases.

We need to create our own dummy variables to handle the interaction.

<<fit2>>=
agem <- with(flchain, ifelse(sex=="M", age, 60))
agef <- with(flchain, ifelse(sex=="F", age, 60))
fit2 <- coxph(Surv(futime, death) ~ sex + pspline(agef, df=3)
              + pspline(agem, df=3), data=flchain)
anova(fit2, fit1)
@ 
 
The gain in this particular problem is not great, but we will forge ahead.
You might well ask why we used 60 as a dummy value of \texttt{agem} for
the females instead of 0?
There is nothing special about the choice, and any value within
the range of ages would do as well, though I try to pick one where the
standard errors of the curves are not outrageous.  
If a value of 0 is used it forces the pspline function to create a basis
set that includes all the empty space between 0 and 50, and do predictions
at 0; these last can become numerically unstable leading to errors or
incorrect values.

The Cox model deals with relative hazards, when doing a plot we will usually
want to specify who our reference is.  
By default the termplot function uses an average reference, that is, any
plot will be centered to have an average log hazard of 0.
In this case, we decided to use 65 year old females as our reference, with
all of the hazards relative to them. 

<<plot2, fig=TRUE>>=
# predictions
pterm <- termplot(fit2, term=2:3, se=TRUE, plot=FALSE)
# reference
refdata <- data.frame(sex=c('F', 'M'), agef=c(65, 60), agem=c(60,65))
pred.ref <- predict(fit2, newdata=refdata, type="lp")
# females
tempf <- pterm$agef$y + outer(pterm$agef$se, c(0, -1.96, 1.96))
frow <- which(pterm$agef$x == 65)
tempf <- tempf  - tempf[frow,1]  # shift curves
# males
tempm <- pterm$agem$y + outer(pterm$agem$se, c(0, -1.96, 1.96))
mrow  <- which(pterm$agem$x == 65)
tempm <- tempm + diff(pred.ref) - tempm[mrow,1]
# plot
matplot(pterm$agef$x, exp(tempf), log='y', col=1, 
        lty=c(1,2,2), type='l', lwd=c(2,1,1),
        xlab="Age", ylab="Relative risk of death")
matlines(pterm$agem$x, exp(tempm), log='y', 
         col=2, lwd=c(2,1,1), lty=c(1,2,2))
legend(80, 1, c("Female", "Male"), lty=1, lwd=2, col=1:2, bty='n')
@ 

\begin{enumerate}
  \item The termplot routine is used to get the data points for the plot,
    without executing a plot, by use of the \texttt{plot=FALSE} argument.
    The result is a list with one element per term; each element of the list
    contains x, y, and se components.
  \item We had decided to center the female curve at age 65, risk =1.  The
    relative offset for the male curve can be derived directly from
    \texttt{fit2} by adding up the right coefficients, and I used to do it that
    way but would get it wrong one time out of two.  
    So instead use the \texttt{predict} routine to get predicted log hazards
    for males and females at a particular age.  This tells me how far apart
    the curves should be at that point.  We force the females to go through
    0, which is exp(0) =1 on the hazard scale.
  \item Get the predicted curve and confidence bands for the females
    as a matrix \texttt{tempf}, and then shift them by subtracting the value
    for a 65 year old female.  Do the same for males, plus adding in
    the curve separation at age 65 from \texttt{pred.ref}.
  \item The male and female portions don't have quite the same set of age
    values, there are no 95 year old males in the data set for example,
    so the plot needs to be done in two steps.
\end{enumerate}

The final curves for males and female are not quite parallel.
One thing the plot does not display is that the spacing between the
male and female points also has a standard error.
This moves the entire bundle of three red curves up and down.  It is not
clear how best to add this information into the plot.
For questions of parallelism and shape, as here, it seemed best to
ignore it, which is what the termplot function also does.
If someone were reading individual male/female differences off the plot
a different choice would be appropriate.
\end{document}

