# Understanding decision trees

Im new to decision trees and am trying some decision tree modelling now with titanic data.

I have the following dataset.

 train <- read.csv(url("http://s3.amazonaws.com/assets.datacamp.com/course/Kaggle/train.csv"))


Now when I create a decision tree doing this:

my_tree_two <- rpart(Survived ~ Sex + Age, data=train, method="class")
my_tree_two


I get the following output.

n= 891 node), split, n, loss, yval, (yprob)
* denotes terminal node

1) root 891 342 0 (0.6161616 0.3838384)
2) Sex=male 577 109 0 (0.8110919 0.1889081)
4) Age>=6.5 553  93 0 (0.8318264 0.1681736) *
5) Age< 6.5 24   8 1 (0.3333333 0.6666667) *
3) Sex=female 314  81 1 (0.2579618 0.7420382) *


I understand most of it but what I do not understand is why age divided into age >= 6.5 and < 6.5? Could anybody elaborate on this?

• The algorithm always puts splits at the midpoint of two data values. Aug 17 '15 at 17:13
• A great tutorial of the Titanic data in R was created by Trevor Stephens. Decision tree page is here Aug 17 '15 at 17:15
• @MatthewDrury thanks for your input. So this is the top and the bottom of the intervals for Male en Female divided by two? Aug 17 '15 at 17:38

Apparently the goal is to split your observations in two classes, the class is given by the binary variable 'yval'

You 'predict' yval with a decision rule. Your decision rule is:

if ( sex == "male" ) {
# there are 577 males in your dataset
if ( Age >= 6.5)               # amongst the 577 males you have 553 with age >= 6.5
yval <- 0 # so the class is '0'
else   #Age < 6.5              # amongst the 577 males you have 24 with age < 6.5
yval<- 1
}
else  # sex is female
# there are 314 females in your dataset
yval <- 1.


Obviously, the class predicted by your decision rule can be erroneous and different from the class that was observed in your dataset.

In the leaf node 'sex == female' there are 25.8% of your cases that are class 0 and 74.2% that are 1, as, see the above decision rule, you decide to put them all in class 1 you make an error for 25.8% of all females.

• @MarcvanderPeet Is the confusion that 6.5 is not a true data value, i.e. not a valid value of Age? Aug 19 '15 at 14:40