Bioinformatics Asked by Nienke on December 17, 2020
I have fitted a cox model on a pooled dataset of multiple studies, say study A, B and C. As these studies have different baseline hazard functions, I stratified for ‘study’ in the model.
Like this:
df$SurvObj <- with(df, Surv(event_rd, event == 1))
fit <- coxph(SurvObj ~ cov1 + cov2 + cov3 + cov4 + strata(study),data=df)
Now, I want to assess whether I can use the beta coefficients from the above model to predict the event probability in a new study, study D.
The baseline hazard in study D is different from those in study A, B and C.
When I have
predict(fit,type="lp",newdata=studyD,reference="strata")
I get this error:
Error in model.frame.default(data = studyD, formula = ~hba1c + sbp +
: factor strata(study) has new levels study=4
Why does R require that I match the strata of study D with those in study A,B and/or C when I only want to output the linear predictor (type="lp"
)? As the linear predictor is independent from the baseline hazard function in this case.
I suppose it should be possible to extract the linear predictor for individuals in study D and then manually calculate the event probability using the baseline hazard of study D.
Does anybody know how to do this?
I could not find any question like this so maybe I am thinking the wrong way.
Unlike wikipedia, the linear predictor (LP) in coxph
is made dependent from the baseline hazard function $lambda$. For example, if a model has two predictors x1
, x2
stratified by sex
, the linear predictor would be like:
$$LP = log ( lambda _{sex}) + beta _{1}x _{1}+beta _{2}x _{2}$$
Where the $log(lambda _{sex})$ can be obtained by predict()
with x1=x2=0
.
We could calculate the LP without strata information (sex) by hand using the $beta _{1} , beta _{2}$ provided by coef(fit)
and a customised $lambda _{unkown , sex}$.
Below is a simulation comparing the results from predict()
and by hand coef(fit)
.
## Toy data
df <- list(time=c(4,3,1,1,2,2,3),
status=c(1,1,1,0,1,1,0),
x1=c(0,2,1,1,1,0,0),
x2=1:7,
sex=c(0,0,0,0,1,1,1))
## Fit a stratified model
fit <- coxph(Surv(time, status) ~ x1 + x2 + strata(sex), df)
coef(fit)
# x1 x2
# 0.79451881 -0.01917633
## Baseline linear predictor
lambda_sex1 <- predict(fit, newdata=list(x1=0, x2=0, sex=0))
# -0.746578
lambda_sex2 <- predict(fit, newdata=list(x1=0, x2=0, sex=1))
# -0.1497816
## by predict function
predict(fit, newdata=list(x1=1, x2=2, sex=0))
# 0.009588167
predict(fit, newdata=list(x1=1, x2=2, sex=1))
# 0.6063845
## by hand
lambda_sex1 + coef(fit)[1]*1 + coef(fit)[2]*2
# 0.009588167
lambda_sex2 + coef(fit)[1]*1 + coef(fit)[2]*2
# 0.6063845
Finally, to calculate "a" linear predictor for an unknown sex
:
# prevented by predict function
predict(fit, newdata=list(x1=1, x2=2, sex=2))
#Error in model.frame.default(data = list(x1 = 1, x2 = 2, sex = 2), formula = ~x1 + :
# factor strata(sex) has new level sex=2
# by hand, assuming lambda for sex 2 is 0
0 + coef(fit)[1]*1 + coef(fit)[2]*2
# 0.7561661
Answered by Ryan SY Kwan on December 17, 2020
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