You are mixing up two of the steps in rpart. 1: how to find the best candidate split and
2: evaluation of that split.
With the "class" method we use the information or Gini criteria for step 1. The code
finds a worthwhile candidate split at 0.5 using exactly the calculations you outline. For
step 2 the criteria is the "decision theory" loss. In your data the estimated rate is 0
for the left node and 15/45 = .333 for the right node. As a decision rule both predict
y=0 (since both are < 1/2). The split predicts 0 on the left and 0 on the right, so does
nothing.
The CART book (Brieman, Freidman, Olshen and Stone) on which rpart is based highlights the
difference between odds-regression (for which the final prediction is a percent, and error
is Gini) and classification. For the former treat y as continuous.
Terry T.
On 05/15/2017 05:00 AM, r-help-requ...@r-project.org wrote:
The following code produces a tree with only a root. However, clearly the tree
with a split at x=0.5 is better. rpart doesn't seem to want to produce it.
Running the following produces a tree with only root.
y <- c(rep(0,65),rep(1,15),rep(0,20))
x <- c(rep(0,70),rep(1,30))
f <- rpart(y ~ x, method='class', minsplit=1, cp=0.0001,
parms=list(split='gini'))
Computing the improvement for a split at x=0.5 manually:
obs_L <- y[x<.5]
obs_R <- y[x>.5]
n_L <- sum(x<.5)
n_R <- sum(x>.5)
gini <- function(p) {sum(p*(1-p))}
impurity_root <- gini(prop.table(table(y)))
impurity_L <- gini(prop.table(table(obs_L)))
impurity_R <- gini(prop.table(table(obs_R)))
impurity <- impurity_root * n - (n_L*impurity_L + n_R*impurity_R) # 2.880952
Thus, an improvement of 2.88 should result in a split. It does not.
Why?
Jonathan
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