Hybrid_R_splitCompare2.R

Hybrid_R_splitCompare2.R

### compare with other ways of building a depth 3 tree
library(policytree)
rm(list = ls())
# set.seed(20)
n = 1000
p = 2
d = 3
X = round(matrix(rnorm(n*p),n,p), 2)
Y = matrix(rnorm(n*d), n, d)

htree = hybrid_policy_tree(X,Y,depth = 3, search.depth = 2) # suggested approach

# 2): Make a depth 2 tree to form the first split. Make a depth 2 tree in the two immediate leaf nodes,
# then make a depth 2 tree in each 4 terminal nodes, then "prune" that back up to depth 3.

root = policy_tree(X, Y, depth = 2)
svar=root$nodes[[1]]$split_variable
sval=root$nodes[[1]]$split_value
leaf = ifelse(X[,svar] <= svar, 1, 2)
sample.by.leaf = split(seq_len(n), leaf)
pp2 = rep(NA, n)

for (S in sample.by.leaf) {
  subroot = policy_tree(X[S,,drop=F], Y[S,,drop=F],depth=2)
  var=subroot$nodes[[1]]$split_variable
  val=subroot$nodes[[1]]$split_value
  subleft = S[X[S, var] <= val]
  subright = S[X[S, var] > val]
  subs = list(left=subleft, right=subright)
  for (s in subs) {
    pt = policy_tree(X[s,,drop=F], Y[s,,drop=F],depth=2)
    svar=pt$nodes[[1]]$split_variable
    sval=pt$nodes[[1]]$split_value
    left = s [X[s, svar] <= sval]
    right = s [X[s, svar] > sval]
    left.action = which.max(colSums(Y[left,]))
    right.action = which.max(colSums(Y[right,]))
    for (i in s) {
      if (i %in% left) {
        pp2[i] = left.action
      } else {
        pp2[i] = right.action
      }
    }
  }
}

mean(Y[cbind(1:n, pp2)]) # 2)

mean(Y[cbind(1:n, predict(htree, X))]) # This is better

# the ratio mean(Y[cbind(1:n, predict(htree, X))]) / mean(Y[cbind(1:n, pp2)]) from 500 repetitions:
# Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
# 1.098   1.530   1.787   1.951   2.151  17.655