> 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 ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
