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Rachana Jain wrote:
> Hello,
>
> I am trying to use the "tree" package in R with RSperl.
>
> I have generated a regression tree using the "tree" package in R and saved it
> in a file say "RegTree_2.Rdata"
>
> If I use the predict function from R command line, it works fine. However, I
> want to use it from within perl using RSperl. To do that, here is what I do,
>
> my $regTree = "RegTree_2.Rdata";
> my $tree = &R::load($regTree);
> my $ans = &R::call("predict.tree", $tree, $_[0]);
>
> On running my code, I get the following error:
> Error in predict.tree("rtree_2", list("5031601", "GS0098", 0, "1", "1", :
> not legitimate tree
>
One of the things that comes to mind immediately is that the return
value of the R function load is a character vector giving the names of
the variables that were assigned during the restoration of the objects
from the file. Thus, the value of $tree in Perl will be a simple
string (or an array of strings).
The call to predict.tree bypasses the generic predict method.
So if you called predict.tree() with a string rather than a tree
you will see the "not legitimate tree" error.
Why you see list("5031601",...) is not clear to me unless
that is currently what is in $tree and so you saved not just
the tree but a lot more besides.
Also, predict will want a data frame as the second argument.
How does $_[0] expand to such an R object when passed from Perl
to the R function?
D.
> Any help would be appreciated.
>
> Thanks
> -Rachana
>
>
> Rachana Jain
> [EMAIL PROTECTED]
> Ph: 408-499-9802 (Cell)
> 513-556-7509 (Res.)
> 2931 Scioto Lane Apt. No. 1002
> Cincinnati OH - 45219, U.S.A.
>
>
>
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