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BART (version 2.9.9)

predict.recurbart: Predicting new observations with a previously fitted BART model

Description

BART is a Bayesian “sum-of-trees” model.
For a numeric response \(y\), we have \(y = f(x) + \epsilon\), where \(\epsilon \sim N(0,\sigma^2)\).

\(f\) is the sum of many tree models. The goal is to have very flexible inference for the uknown function \(f\).

In the spirit of “ensemble models”, each tree is constrained by a prior to be a weak learner so that it contributes a small amount to the overall fit.

Usage

# S3 method for recurbart
predict(object, newdata, mc.cores=1, openmp=(mc.cores.openmp()>0), ...)

Value

Returns an object of type recurbart with predictions corresponding to newdata.

Arguments

object

object returned from previous BART fit with recur.bart or mc.recur.bart.

newdata

Matrix of covariates to predict the distribution of \(t\).

mc.cores

Number of threads to utilize.

openmp

Logical value dictating whether OpenMP is utilized for parallel processing. Of course, this depends on whether OpenMP is available on your system which, by default, is verified with mc.cores.openmp.

...

Other arguments which will be passed on to pwbart.

Details

BART is an Bayesian MCMC method. At each MCMC interation, we produce a draw from the joint posterior \((f,\sigma) | (x,y)\) in the numeric \(y\) case and just \(f\) in the binary \(y\) case.

Thus, unlike a lot of other modelling methods in R, we do not produce a single model object from which fits and summaries may be extracted. The output consists of values \(f^*(x)\) (and \(\sigma^*\) in the numeric case) where * denotes a particular draw. The \(x\) is either a row from the training data (x.train) or the test data (x.test).

See Also

recur.bart, mc.recur.bart, recur.pwbart, mc.recur.pwbart, mc.cores.openmp

Examples

Run this code
## load 20 percent random sample
data(xdm20.train)
data(xdm20.test)
data(ydm20.train)

##test BART with token run to ensure installation works
## with current technology even a token run will violate CRAN policy
## set.seed(99)
## post <- recur.bart(x.train=xdm20.train, y.train=ydm20.train,
##                    nskip=1, ndpost=1, keepevery=1)

if (FALSE) {
set.seed(99)
post <- recur.bart(x.train=xdm20.train, y.train=ydm20.train)
## larger data sets can take some time so, if parallel processing
## is available, submit this statement instead
## post <- mc.recur.bart(x.train=xdm20.train, y.train=ydm20.train,
##                      mc.cores=8, seed=99)

require(rpart)
require(rpart.plot)

dss <- rpart(post$yhat.train.mean~xdm20.train)

rpart.plot(dss)
## for the 20 percent sample, notice that the top splits
## involve cci_pvd and n
## for the full data set, notice that all splits
## involve ca, cci_pud, cci_pvd, ins270 and n
## (except one at the bottom involving a small group)

## compare patients treated with insulin (ins270=1) vs
## not treated with insulin (ins270=0)
N.train <- 50
N.test <- 50
K <- post$K ## 798 unique time points

## only testing set, i.e., remove training set
xdm20.test. <- xdm20.test[N.train*K+(1:(N.test*K)), ]
xdm20.test. <- rbind(xdm20.test., xdm20.test.)
xdm20.test.[ , 'ins270'] <- rep(0:1, each=N.test*K)

## multiple threads will be utilized if available
pred <- predict(post, xdm20.test., mc.cores=8)

## create Friedman's partial dependence function for the
## intensity/hazard by time and ins270
NK.test <- N.test*K
M <- nrow(pred$haz.test) ## number of MCMC samples, typically 1000

RI <- matrix(0, M, K)

for(i in 1:N.test)
    RI <- RI+(pred$haz.test[ , (N.test+i-1)*K+1:K]/
              pred$haz.test[ , (i-1)*K+1:K])/N.test

RI.lo <- apply(RI, 2, quantile, probs=0.025)
RI.mu <- apply(RI, 2, mean)
RI.hi <- apply(RI, 2, quantile, probs=0.975)

plot(post$times, RI.hi, type='l', lty=2, log='y',
     ylim=c(min(RI.lo, 1/RI.hi), max(1/RI.lo, RI.hi)),
     xlab='t', ylab='RI(t, x)',
     sub='insulin(ins270=1) vs. no insulin(ins270=0)',
     main='Relative intensity of hospital admissions for diabetics')
lines(post$times, RI.mu)
lines(post$times, RI.lo, lty=2)
lines(post$times, rep(1, K), col='darkgray')

## RI for insulin therapy seems fairly constant with time
mean(RI.mu)

}

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