11
$\begingroup$

I have used rpart.control for minsplit=2, and got the following results from rpart() function. In order to avoid overfitting the data, do I need to use splits 3 or splits 7? Shouldn't I use splits 7? Please let me know.

Variables actually used in tree construction:

[1] ct_a ct_b usr_a

Root node error: 23205/60 = 386.75

n= 60        

    CP nsplit rel error  xerror     xstd
1 0.615208      0  1.000000 1.05013 0.189409
2 0.181446      1  0.384792 0.54650 0.084423
3 0.044878      2  0.203346 0.31439 0.063681
4 0.027653      3  0.158468 0.27281 0.060605
5 0.025035      4  0.130815 0.30120 0.058992
6 0.022685      5  0.105780 0.29649 0.059138
7 0.013603      6  0.083095 0.21761 0.045295
8 0.010607      7  0.069492 0.21076 0.042196
9 0.010000      8  0.058885 0.21076 0.042196
$\endgroup$
2
  • 1
    $\begingroup$ I answered this in the follow-up you posted to previous Q. Given that, there was no need for this. I mentioned that you shouldn't edit Q's to follow-up for future reference! $\endgroup$ Commented Jul 25, 2011 at 17:02
  • 1
    $\begingroup$ To avoid searching for the related question in the future, here is the link to the previous Q: stats.stackexchange.com/questions/13446/…. $\endgroup$
    – chl
    Commented Jul 25, 2011 at 21:49

1 Answer 1

10
$\begingroup$

The convention is to use the best tree (lowest cross-validate relative error) or the smallest (simplest) tree within one standard error of the best tree. The best tree is in row 8 (7 splits), but the tree in row 7 (6 splits) does effectively the same job (xerror for tree in row 7 = 0.21761, which is within (smaller than) the xerror of best tree plus one standard error, xstd, (0.21076 + 0.042196) = 0.252956) and is simpler, hence the 1 standard error rule would select it.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.