Re-evaluating the tree in the random forest

Here's an example, using the iris data:

## Grow one tree, using all data, and try all

variables at all splits,

## using large nodesize to get smaller tree.
iris.rf <- randomForest(iris[-5], iris[[5]],

ntree=1, nodesize=20, mtry=4,
+                         sampsize=150,
replace=FALSE)

getTree(iris.rf, 1)

   left daughter right daughter split var split
point status prediction
1              2              3         3       
2.45      1          0
2              0              0         0       
0.00     -1          1
3              4              5         4       
1.75      1          0
4              6              7         3       
4.95      1          0
5              8              9         3       
4.85      1          0
6             10             11         4       
1.65      1          0
7              0              0         0       
0.00     -1          3
8              0              0         0       
0.00     -1          3
9              0              0         0       
0.00     -1          3
10             0              0         0       
0.00     -1          2
11             0              0         0       
0.00     -1          3

idx <- with(iris, Petal.Length > 2.45 &

Petal.Length < 3.5)

predict(iris.rf, iris[idx, -5])

[1] versicolor versicolor versicolor
Levels: setosa versicolor virginica

iris.rf$forest$xbestsplit[1,1] <- 3.5
predict(iris.rf, iris[newiris, -5])

[1] setosa setosa setosa
Levels: setosa versicolor virginica

Note how the predictions have changed.

HTH,
Andy

-----Original Message-----
From: Martin Lam [mailto:tmlammail at yahoo.com] 
Sent: Friday, September 09, 2005 9:04 AM
To: Liaw, Andy; r-help at stat.math.ethz.ch
Subject: RE: [R] Re-evaluating the tree in the

random forest

Hi,

Let me give a simple example, assume a dataset
containing 5 instances  with 1 variable and the

class

label:

[x1, y]:
[0.5, A]
[3.2, B]
[4.5, B]
[1.4, C]
[1.6, C]
[1.9, C]

Assume that the randomForest algorithm create this

(2

levels deep) tree:

Root node: question: x1 < 2.2?

Left terminal node:
[0.5, A]
[1.4, C]
[1.6, C]
[1.9, C]
Leaf classification: C

Right terminal node:
[3.2, B]
[4.5, B]
Leaf classification: B

If I change the question at the root node to "x1 <
1?", the instances in the left leaf node are not
correctly passed down the tree anymore.  
My original question was if there was a way to
re-evaluate the instances again into:

Root node: question: x1 < 1?

Left terminal node:
[0.5, A]
Leaf classification: A

Right terminal node:
[3.2, B]
[4.5, B]
[1.4, C]
[1.6, C]
[1.9, C]
Leaf classification: C

Cheers,

Martin

--- "Liaw, Andy" <andy_liaw at merck.com> wrote:

From: Martin Lam

Dear mailinglist members,

I was wondering if there was a way to

re-evaluate

the

instances of a tree (in the forest) again

after I

have

manually changed a splitpoint (or split

variable)

of a

decision node. Here's an illustration:

library("randomForest")

forest.rf <- randomForest(formula = Species ~

.,

data

= iris, do.trace = TRUE, ntree = 3, mtry = 2,
norm.votes = FALSE)

# I am going to change the splitpoint of the

root

node

of the first tree to 1
forest.rf$forest$xbestsplit[1,]
forest.rf$forest$xbestsplit[1,1] <- 1
forest.rf$forest$xbestsplit[1,]

Because I've changed the splitpoint, some

instances in

the leafs are not supposed where they should

be.

Is

there a way to reappoint them to the correct

leaf?

I'm not sure what you want to do exactly, but I
suspect you can use
predict().
 

I was also wondering how I should interpret

the

output

of do.trace:

ntree      OOB      1      2      3
    1:   3.70%  0.00%  6.25%  5.88%
    2:   3.49%  0.00%  3.85%  7.14%
    3:   3.57%  0.00%  5.56%  5.26%

What's OOB and what does the percentages mean?

OOB stands for `Out-of-bag'.  Read up on random
forests (e.g., the article
in R News) to learn about it.  Those numbers are
estimated error rates.  The
`OOB' column is across all data, while the

others

are for the classes.

Andy

 

Thanks in advance,

Martin


	
		

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