surv.bart {BART}  R Documentation 
Here we have implemented a simple and direct approach to utilize BART in survival analysis that is very flexible, and is akin to discretetime survival analysis. Following the capabilities of BART, we allow for maximum flexibility in modeling the dependence of survival times on covariates. In particular, we do not impose proportional hazards.
To elaborate, consider data in the usual form:
(t_i, \delta_i, {x}_i)
where t_i
is the event time,
\delta_i
is an indicator distinguishing events
(\delta=1
) from rightcensoring
(\delta=0
), {x}_i
is a vector of covariates, and
i=1, ..., N
indexes subjects.
We denote the K
distinct event/censoring times by
0<t_{(1)}<...<t_{(K)}<\infty
thus
taking t_{(j)}
to be the j^{th}
order
statistic among distinct observation times and, for convenience,
t_{(0)}=0
. Now consider event indicators y_{ij}
for each subject i
at each distinct time t_{(j)}
up to and including the subject's observation time
t_i=t_{(n_i)}
with
n_i=\sum_j I[t_{(j)}\leq t_i]
.
This means y_{ij}=0
if j<n_i
and
y_{in_i}=\delta_i
.
We then denote by p_{ij}
the probability
of an event at time t_{(j)}
conditional on no previous event. We
now write the model for y_{ij}
as a nonparametric probit
regression of y_{ij}
on the time t_{(j)}
and the covariates
{x}_i
, and then utilize BART for binary responses. Specifically,
y_{ij}\ =\ \delta_i I[t_i=t_{(j)}],\ j=1, ..., n_i
; we have
p_{ij} = F(\mu_{ij}),\ \mu_{ij} = \mu_0+f(t_{(j)}, {x}_i)
where F
denotes the standard normal cdf (probit link).
As in the binary
response case, f
is the sum of many tree models.
surv.bart( x.train=matrix(0,0,0),
y.train=NULL, times=NULL, delta=NULL,
x.test=matrix(0,0,0),
K=NULL, events=NULL, ztimes=NULL, zdelta=NULL,
sparse=FALSE, theta=0, omega=1,
a=0.5, b=1, augment=FALSE, rho=NULL,
xinfo=matrix(0,0,0), usequants=FALSE,
rm.const=TRUE, type='pbart',
ntype=as.integer(
factor(type, levels=c('wbart', 'pbart', 'lbart'))),
k=2, power=2, base=.95,
offset=NULL, tau.num=c(NA, 3, 6)[ntype],
ntree=50, numcut=100, ndpost=1000, nskip=250,
keepevery = 10L,
printevery=100L,
id=NULL, ## surv.bart only
seed=99, ## mc.surv.bart only
mc.cores=2, ## mc.surv.bart only
nice=19L ## mc.surv.bart only
)
mc.surv.bart( x.train=matrix(0,0,0),
y.train=NULL, times=NULL, delta=NULL,
x.test=matrix(0,0,0),
K=NULL, events=NULL, ztimes=NULL, zdelta=NULL,
sparse=FALSE, theta=0, omega=1,
a=0.5, b=1, augment=FALSE, rho=NULL,
xinfo=matrix(0,0,0), usequants=FALSE,
rm.const=TRUE, type='pbart',
ntype=as.integer(
factor(type, levels=c('wbart', 'pbart', 'lbart'))),
k=2, power=2, base=.95,
offset=NULL, tau.num=c(NA, 3, 6)[ntype],
ntree=50, numcut=100, ndpost=1000, nskip=250,
keepevery = 10L,
printevery=100L,
id=NULL, ## surv.bart only
seed=99, ## mc.surv.bart only
mc.cores=2, ## mc.surv.bart only
nice=19L ## mc.surv.bart only
)
x.train 
Explanatory variables for training (in sample)
data. 
y.train 
Binary response dependent variable for training (in sample) data. 
times 
The time of event or rightcensoring. 
delta 
The event indicator: 1 is an event while 0 is censored. 
x.test 
Explanatory variables for test (out of sample) data. 
K 
If provided, then coarsen 
events 
If provided, then use for the grid of time points. 
ztimes 
If provided, then these columns of 
zdelta 
If provided, then these columns of 
sparse 
Whether to perform variable selection based on a sparse Dirichlet prior rather than simply uniform; see Linero 2016. 
theta 
Set 
omega 
Set 
a 
Sparse parameter for 
b 
Sparse parameter for 
rho 
Sparse parameter: typically 
augment 
Whether data augmentation is to be performed in sparse variable selection. 
xinfo 
You can provide the cutpoints to BART or let BART
choose them for you. To provide them, use the 
usequants 
If 
rm.const 
Whether or not to remove constant variables. 
type 
Whether to employ AlbertChib, 
ntype 
The integer equivalent of 
k 
k is the number of prior standard deviations 
power 
Power parameter for tree prior. 
base 
Base parameter for tree prior. 
offset 
With binary
BART, the centering is 
tau.num 
The numerator in the 
ntree 
The number of trees in the sum. 
ndpost 
The number of posterior draws returned. 
nskip 
Number of MCMC iterations to be treated as burn in. 
printevery 
As the MCMC runs, a message is printed every printevery draws. 
keepevery 
Every keepevery draw is kept to be returned to the user. 
numcut 
The number of possible values of c (see usequants).
If a single number if given, this is used for all variables.
Otherwise a vector with length equal to ncol(x.train) is required,
where the 
id 

seed 

mc.cores 

nice 

surv.bart
returns an object of type survbart
which is
essentially a list. Besides the items listed
below, the list has a binaryOffset
component giving the value
used, a times
component giving the unique times, K
which is the number of unique times, tx.train
and
tx.test
, if any.
yhat.train 
A matrix with ndpost rows and nrow(x.train) columns.
Each row corresponds to a draw 
yhat.test 
Same as yhat.train but now the x's are the rows of the test data. 
surv.test 
The survival function, 
yhat.train.mean 
train data fits = mean of yhat.train columns. 
yhat.test.mean 
test data fits = mean of yhat.test columns. 
surv.test.mean 
mean of surv.test columns. 
varcount 
a matrix with ndpost rows and nrow(x.train) columns. Each row is for a draw. For each variable (corresponding to the columns), the total count of the number of times that variable is used in a tree decision rule (over all trees) is given. 
Note that yhat.train and yhat.test are
f(t, x)
+ binaryOffset
. If you want draws of the probability
P(Y=1  t, x)
you need to apply the normal cdf (pnorm
)
to these values.
## load survival package for the advanced lung cancer example
data(lung)
N < length(lung$status)
table(lung$ph.karno, lung$pat.karno)
## if physician's KPS unavailable, then use the patient's
h < which(is.na(lung$ph.karno))
lung$ph.karno[h] < lung$pat.karno[h]
times < lung$time
delta < lung$status1 ##lung$status: 1=censored, 2=dead
##delta: 0=censored, 1=dead
## this study reports time in days rather than weeks or months
## coarsening from days to weeks or months will reduce the computational burden
##times < ceiling(times/30)
times < ceiling(times/7) ## weeks
table(times)
table(delta)
## matrix of observed covariates
x.train < cbind(lung$sex, lung$age, lung$ph.karno)
## lung$sex: Male=1 Female=2
## lung$age: Age in years
## lung$ph.karno: Karnofsky performance score (dead=0:normal=100:by=10)
## rated by physician
dimnames(x.train)[[2]] < c('M(1):F(2)', 'age(39:82)', 'ph.karno(50:100:10)')
table(x.train[ , 1])
summary(x.train[ , 2])
table(x.train[ , 3])
##test BART with token run to ensure installation works
set.seed(99)
post < surv.bart(x.train=x.train, times=times, delta=delta,
nskip=1, ndpost=1, keepevery=1)
## Not run:
## run one long MCMC chain in one process
## set.seed(99)
## post < surv.bart(x.train=x.train, times=times, delta=delta, x.test=x.test)
## in the interest of time, consider speeding it up by parallel processing
## run "mc.cores" number of shorter MCMC chains in parallel processes
post < mc.surv.bart(x.train=x.train, times=times, delta=delta,
mc.cores=8, seed=99)
pre < surv.pre.bart(times=times, delta=delta, x.train=x.train,
x.test=x.train)
K < pre$K
M < nrow(post$yhat.train)
pre$tx.test < rbind(pre$tx.test, pre$tx.test)
pre$tx.test[ , 2] < c(rep(1, N*K), rep(2, N*K))
## sex pushed to col 2, since time is always in col 1
pred < predict(post, newdata=pre$tx.test, mc.cores=8)
pd < matrix(nrow=M, ncol=2*K)
for(j in 1:K) {
h < seq(j, N*K, by=K)
pd[ , j] < apply(pred$surv.test[ , h], 1, mean)
pd[ , j+K] < apply(pred$surv.test[ , h+N*K], 1, mean)
}
pd.mu < apply(pd, 2, mean)
pd.025 < apply(pd, 2, quantile, probs=0.025)
pd.975 < apply(pd, 2, quantile, probs=0.975)
males < 1:K
females < males+K
plot(c(0, pre$times), c(1, pd.mu[males]), type='s', col='blue',
ylim=0:1, ylab='S(t, x)', xlab='t (weeks)',
main=paste('Advanced Lung Cancer ex. (BART::lung)',
"Friedman's partial dependence function",
'Male (blue) vs. Female (red)', sep='\n'))
lines(c(0, pre$times), c(1, pd.025[males]), col='blue', type='s', lty=2)
lines(c(0, pre$times), c(1, pd.975[males]), col='blue', type='s', lty=2)
lines(c(0, pre$times), c(1, pd.mu[females]), col='red', type='s')
lines(c(0, pre$times), c(1, pd.025[females]), col='red', type='s', lty=2)
lines(c(0, pre$times), c(1, pd.975[females]), col='red', type='s', lty=2)
## End(Not run)