[R] timing predict.tree()

[Martin Maechler]

>>>>> "Saket" == Saket Joshi <joshi at engr.orst.edu> writes:

    Saket> Hi all,
    Saket> I am running R1.5.0 under Unix.
    Saket> I am repeating my earlier question with a few details added.

    Saket> I have the following tree fitted as the tree object 'my.tree':

    Saket> node), split, n, deviance, yval
    Saket> * denotes terminal node

    Saket> 1) root 5807 0.9998 0.0001722
    Saket> 2) V604 < 0.5 5798 0.0000 0.0000000 *
    Saket> 3) V604 > 0.5 9 0.8889 0.1111000 *


    Saket> And I have a data.frame called 'new.temp' for which I
    Saket> want to make predictions with this tree. new.temp is
    Saket> a data.frame with a single row and the required
    Saket> number of columns.

    Saket> When I run:
    Saket> predict.tree(my.tree, new.temp)

    Saket> it takes 60 to 70 seconds to get the output. I had
    Saket> expected it to take far less because the tree is very
    Saket> small and the new data contains just 1 row.

    Saket> My question is: Does this function 'predict.tree'
    Saket> provided in the library package 'tree' really consume
    Saket> so much time even for minor calculations or is
    Saket> something wrong?

there is something wrong with your situation and you still gave not enough
details for us to find out.

For me, the three statements

  library(tree)
  example(predict.tree)
  predict(shuttle.tr, shuttle[254,,drop = FALSE])

all give output very quickly.

BTW : Note that the use of the `tree' package is not really
      recommended, but rather use the recommended package
      `rpart'

Martin Maechler <maechler at stat.math.ethz.ch>	http://stat.ethz.ch/~maechler/
Seminar fuer Statistik, ETH-Zentrum  LEO C16	Leonhardstr. 27
ETH (Federal Inst. Technology)	8092 Zurich	SWITZERLAND
phone: x-41-1-632-3408		fax: ...-1228			<><
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