R help - Dec 2010 - rpart.object help

Hi,

Suppose i have generated an object using the following :
fit <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis) 

And when i print fit, i get the following :

n= 81 

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

 1) root 81 17 absent (0.7901235 0.2098765)  
   2) Start>=8.5 62  6 absent (0.9032258 0.0967742)  
     4) Start>=14.5 29  0 absent (1.0000000 0.0000000) *
     5) Start< 14.5 33  6 absent (0.8181818 0.1818182)  
      10) Age< 55 12  0 absent (1.0000000 0.0000000) *
      11) Age>=55 21  6 absent (0.7142857 0.2857143)  
        22) Age>=111 14  2 absent (0.8571429 0.1428571) *
        23) Age< 111 7  3 present (0.4285714 0.5714286) *
   3) Start< 8.5 19  8 present (0.4210526 0.5789474) *

Is it possible to extract the splits alone as a matrix using rpart.object?
If so, how?

Regards,
Jagdeesh
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On Sun, 12 Dec 2010, jagdeesh_mn wrote:
> Hi,
>
> Suppose i have generated an object using the following :
> fit <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis)
>
> And when i print fit, i get the following :
>
> n= 81
>
> node), split, n, loss, yval, (yprob)
>      * denotes terminal node
>
> 1) root 81 17 absent (0.7901235 0.2098765)
>   2) Start>=8.5 62  6 absent (0.9032258 0.0967742)
>     4) Start>=14.5 29  0 absent (1.0000000 0.0000000) *
>     5) Start< 14.5 33  6 absent (0.8181818 0.1818182)
>      10) Age< 55 12  0 absent (1.0000000 0.0000000) *
>      11) Age>=55 21  6 absent (0.7142857 0.2857143)
>        22) Age>=111 14  2 absent (0.8571429 0.1428571) *
>        23) Age< 111 7  3 present (0.4285714 0.5714286) *
>   3) Start< 8.5 19  8 present (0.4210526 0.5789474) *
>
> Is it possible to extract the splits alone as a matrix using rpart.object?
> If so, how?
What do you think 'rpart.object' is?  There is no such function in R.
If you read help("rpart.object") it describes the returned object. 
You are probably looking for fit$frame, but if you want something 
else, study

rpart:::print.rpart

to see how that output is computed.
> Regards,
> Jagdeesh
-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595

> On Sun, 12 Dec 2010, jagdeesh_mn wrote:
> 
>> Hi,
>>
>> Suppose i have generated an object using the following :
>> fit <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis)
>>
>> And when i print fit, i get the following :
>>
>> n= 81
>>
>> node), split, n, loss, yval, (yprob)
>>      * denotes terminal node
>>
>> 1) root 81 17 absent (0.7901235 0.2098765)
>>   2) Start>=8.5 62  6 absent (0.9032258 0.0967742)
>>     4) Start>=14.5 29  0 absent (1.0000000 0.0000000) *
>>     5) Start< 14.5 33  6 absent (0.8181818 0.1818182)
>>      10) Age< 55 12  0 absent (1.0000000 0.0000000) *
>>      11) Age>=55 21  6 absent (0.7142857 0.2857143)
>>        22) Age>=111 14  2 absent (0.8571429 0.1428571) *
>>        23) Age< 111 7  3 present (0.4285714 0.5714286) *
>>   3) Start< 8.5 19  8 present (0.4210526 0.5789474) *
>>
>> Is it possible to extract the splits alone as a matrix using
>> rpart.object?
>> If so, how?
> 
> 
>> Regards,
>> Jagdeesh
The best description of the rpart object is obtained with
help(rpart.object).  Each row of $frame describes one primary split.
More detailed descriptions of the (1 + ncompete + nprimary) split
variables for the node are found in the $splits and $csplits component.
You would need to look at summary.rpart to see how that is all indexed.
I would suggest grabbing a copy of the source code, since that contains
comments, which are stripped out when you print the R internal version.

Terry Therneau