Contrasts
Multiple at options
If you have more than one at option then you can perform contrasts.
For example, if you predict survival curves for more than one covariate pattern
can take differences or ratios of these predictions.
This is best shown through example. I first load the Rotterdam 2 breast cancer data.
. use https://www.pclambert.net/data/rott2b, clear
(Rotterdam breast cancer data (augmented with cause of death))
. stset os, f(osi==1) scale(12) exit(time 60)
Survival-time data settings
Failure event: osi==1
Observed time interval: (0, os]
Exit on or before: time 60
Time for analysis: time/12
--------------------------------------------------------------------------
2,982 total observations
0 exclusions
--------------------------------------------------------------------------
2,982 observations remaining, representing
753 failures in single-record/single-failure data
13,038.968 total analysis time at risk and under observation
At risk from t = 0
Earliest observed entry t = 0
Last observed exit t = 5
I will fit a simple model modelling the effect of hormon and use a
natural spline for age.
. stpm3 i.hormon @ns(age, df(3)), scale(lncumhazard) df(5)
Iteration 0: log likelihood = -2120.3313
Iteration 1: log likelihood = -2120.0705
Iteration 2: log likelihood = -2120.0694
Iteration 3: log likelihood = -2120.0694
Number of obs = 2,982
Wald chi2(4) = 69.03
Log likelihood = -2120.0694 Prob > chi2 = 0.0000
------------------------------------------------------------------------------
| Coefficient Std. err. z P>|z| [95% conf. interval]
-------------+----------------------------------------------------------------
xb |
hormon |
yes | .3966145 .1041656 3.81 0.000 .1924536 .6007754
_ns_f1_age1 | -1.769774 1.089136 -1.62 0.104 -3.904441 .3648924
_ns_f1_age2 | -1.233782 .4602279 -2.68 0.007 -2.135812 -.3317519
_ns_f1_age3 | -1.709216 .4899396 -3.49 0.000 -2.66948 -.7489521
-------------+----------------------------------------------------------------
time |
_ns1 | -22.10362 2.190797 -10.09 0.000 -26.39751 -17.80974
_ns2 | 6.252487 1.18838 5.26 0.000 3.923304 8.58167
_ns3 | -1.093169 .0583025 -18.75 0.000 -1.207439 -.9788977
_ns4 | -.5425182 .0376756 -14.40 0.000 -.6163609 -.4686754
_ns5 | -.376672 .0361107 -10.43 0.000 -.4474477 -.3058964
_cons | -.1364105 .2354592 -0.58 0.562 -.597902 .325081
------------------------------------------------------------------------------
Extended functions
(1) @ns(age, df(3))
Warning: This is a test version of stpm3
. estimates store stpm3_ns
I will now predict the survival function for a 70 year woman with and without hormonal treatment.
. predict S70h0 S70h1, survival ci ///
> at1(age 70 hormon 0) ///
> at2(age 70 hormon 1) ///
> timevar(0 5, step(0.1)) ///
> frame(f1)
Predictions are stored in frame - f1
The predicted survival functions are plotted below.
. frame f1 {
. twoway (line S70h0 S70h1 tt) ///
> , xtitle("Years since surgery") ///
> ytitle("S(t)") ///
> legend(order(1 "hormon=0" 2 "hormon=1") ring(0) pos(1) cols(1))
. }
Visually we can see the difference in the lines which is over
10 percentage points at 5 years. We can calculate the difference using
the contrast(difference) option. We can name the contrast variable
using the contrastvar() option rather than rely on the default name.
. predict S70h0 S70h1, survival ci ///
> at1(age 70 hormon 0) ///
> at2(age 70 hormon 1) ///
> contrast(difference) ///
> contrastvar(Sdiff) ///
> timevar(0 5, step(0.1)) ///
> frame(f1, replace)
Predictions are stored in frame - f1
Once the contrast is stored, it can be plotted together with a 95% confidence interval.
. frame f1 {
. twoway (rarea Sdiff_lci Sdiff_uci tt, color(red%30)) ///
> (line Sdiff tt, color(red)) ///
> , xtitle("Years since surgery") ///
> ytitle("Difference in S(t)") ///
> ylabel(,format(%3.2f)) ///
> legend(off)
. }
By default at1() is the reference, so the difference is at2() - at1().
The reference level can be changed using the atreference() option.
. predict S70h0 S70h1, survival ci ///
> at1(age 70 hormon 0) ///
> at2(age 70 hormon 1) ///
> atreference(2) ///
> contrast(difference) ///
> contrastvar(Sdiff) ///
> timevar(0 5, step(0.1)) ///
> frame(f1, replace)
Predictions are stored in frame - f1
Once the contrast is stored, it can be plotted together with a 95% confidence interval.
. frame f1 {
. twoway (rarea Sdiff_lci Sdiff_uci tt, color(red%30)) ///
> (line Sdiff tt, color(red)) ///
> , xtitle("Years since surgery") ///
> ytitle("Difference in S(t)") ///
> ylabel(,format(%3.2f)) ///
> legend(off)
. }
More than 2 at options
You can have as many at options as you want. The following predicts survival
at 10 year intervals between 40 and 90 years at diagnosis.
. local j 1
. foreach a of numlist 40(10)90 {
2. local atlist `atlist' at`j'(age `a' hormon 0)
3. local ++j
4. }
. predict S*, survival ci ///
> `atlist' ///
> timevar(0 5, step(0.1)) ///
> frame(f2, replace)
Predictions are stored in frame - f2
. frame f2 {
. twoway (line S? tt, ) ///
> , xtitle("Years since surgery") ///
> ytitle("S(t)") ///
> ylabel(,format(%3.2f)) ///
> legend(order(1 "40" 2 "50" 3 "60" 4 "70" 5 "80" 6 "90") ///
> ring(0) pos(7) cols(1))
. }
If you want to make contrasts between the different ages
then we need to make one of the at options the reference.
Below I make the 2nd at option the reference, which is for 50 year olds (they have the best survival).
. local atlist
. local j 1
. foreach a of numlist 40(10)90 {
2. local atlist `atlist' at`j'(age `a' hormon 0)
3. local ++j
4. }
. predict S*, survival ci ///
> `atlist' ///
> contrast(difference) ///
> contrastvar(Sdiff*) ///
> atreference(2) ///
> timevar(0 5, step(0.1)) ///
> frame(f2, replace)
Predictions are stored in frame - f2
. frame f2 {
. twoway (line Sdiff? tt, ) ///
> , xtitle("Years since surgery") ///
> ytitle("S(t)") ///
> ylabel(,format(%3.2f)) ///
> title("Difference in survival compared to 50 year old") ///
> legend(order(1 "40" 2 "60" 3 "70" 4 "80" 5 "90") ///
> ring(0) pos(7) cols(1))
. }
