similar to: Offset - usersplits function package RPART

Hello list, I'm having a problem with custom functions in rpart, and before I tear my hair out trying to fix it, I want to make sure it's actually a problem. It seems that, when you write custom functions for rpart (init, split and eval) then rpart no longer cross-validates the resulting tree to return errors. A simple test is to use the usersplits.R function to get a simple, custom

Dear all, I'm trying to manage with user defined split function in rpart (file rpart\tests\usersplits.R in http://cran.r-project.org/src/contrib/rpart_3.1-34.tar.gz - see bottom of the email). Suppose to have the following data.frame (note that x's values are already sorted) > D y x 1 7 0.428 2 3 0.876 3 1 1.467 4 6 1.492 5 3 1.703 6 4 2.406 7 8 2.628 8 6 2.879 9 5 3.025 10 3 3.494

Dear all, I'm writing my own method to be used in Rpart by defining the list of functions named init, split and eval. I'm following the example given in the file 'tests/usersplits.R' in the sources. By now I'm able to define the split function (and it works correctly in the tree construction) while I have some problems with the init and the eval function. The task I'm

Hi I have installed rpart in my Linux (PLD) but I don't know how I may find help conect this package? Here is my instalaction: > install.packages("rpart",dependencies=TRUE) --- Please select a CRAN mirror for use in this session --- trying URL 'http://r.meteo.uni.wroc.pl/src/contrib/rpart_3.1-46.tar.gz' Content type 'application/x-gzip' length 136572 bytes (133

Dear R community, I am trying to write my own user defined split function for rpart. I read the example in the tests directory and I understand the general idea of the how to implement user defined splitting functions. However, I am having troubles with addressing the data frame used in calling rpart in my split functions. For example, in the evaluation function that is called once per node,

Dear R community, as stated in Breiman et.al. (1984) and De'Ath & Fabricius (2000) using sums of absolute deviations about the median as an impurity measure gives robust trees. I would like to use this method in rpart. Has somebody already tried this method in rpart? Is there maybe already a script available somewhere? I am aware of the possibility to define usersplits myself with

Hallo! I am a student of the Politecnico di Milano (Milan, italy) and I'm working on CARTs. I'm trying to use the R rpart function with a personalized splitfunction... but I'm not able to do it! More precisely, I would like to know what is the meaning of the function 'init', 'split' and 'eval' named in the help page.I can't find any answer in

Dear R users, I know cross-validation does not work in rpart with user defined split functions. As Terry Therneau suggested, one can use the xpred.rpart function and then summarize the matrix of the predicted values into a single "goodness" value. I need only a confirmation: set for example xval=10, if I correctly understood a single column of the matrix obatined by xpred.rpart gives

If I understand correctly, rpart() will pick predictor at each node automatically. I am wondering if there is a way to force rpart() including a specific predictor. The reason I am asking is that I'd like to use rpart() to detect interaction terms for some variables. Thanks.

Is there an optimal / minimum sample size for attempting to construct a classification tree using /rpart/? I have 27 seagrass disturbance sites (boat groundings) that have been monitored for a number of years. The monitoring protocol for each site is identical. From the monitoring data, I am able to determine the level of recovery that each site has experienced. Recovery is our

After fitting and pruning an rpart model, it is often the case that one or more of the original predictors is not used by any of the splits of the final tree. It seems logical, therefore, that values for these "unused" predictors would not be needed for prediction. But when predict() is called on such models, all predictors seem to be required. Why is that, and can it be easily

I've been groping my way through a classification/discrimination problem, from a consulting client. There are 26 observations, with 4 possible categories and 24 (!!!) potential predictor variables. I tried using lda() on the first 7 predictor variables and got 24 of the 26 observations correctly classified. (Training and testing both on the complete data set --- just to get started.) I

Dear R users, I'm working with the rpart package and want to evaluate the performance of user defined split functions. I have some problems in understanding the meaning of the xval argument in the two functions rpart.control and xpred.rpart. In the former it is defined as the number of cross-validations while in the latter it is defined as the number of cross-validation groups. If I am

Hi, I am using the rpart package to build a classification tree. I did manage to build a tree with data on a previous project. However, when attampting to build a tree on a project I am working on, I seem to be getting the error shown below: > nhg3.rp <- rpart(profitresp ~., nhg3, method="class") > plot(nhg3.rp, branch=0.4, uniform=T); text(nhg3.rp, digits=3) Error in

Hi everyone, I have a problem using rpart (R 2.0.1 under Unix) Indeed, I have a large matrix (9271x7), my response variable is numeric and all my predictor variables are categorical (from 3 to 8 levels). Here is an example : > mydata[1:5,] distance group3 group4 group5 group6 group7 group8 pos_1 0.141836040224967 a c e a g g pos_501

I'd like to create a bidimensional presentation of a classification tree built using the rpart() function. I've seen that a partition.tree() function exists for the tree() function. Does a similar function exist for the rpart() function? Thanks a lot Simone Vantini

Hi dear R-help memebers, When building a CART model (specifically classification tree) using rpart, it is sometimes obvious that there are variables (X's) that are meaningful for predicting some of the outcome (y) variables - while other predictors are relevant for other outcome variables (y's only). *How can it be estimated, which explanatory variable is "used" for which of

Dear All, I have a question regarding predict.rpart. I use rpart to build classification and regression trees and I deal with data with relatively large number of input variables (predictors). For example, I build an rpart model like this rpartModel <- rpart(Y ~ X, method="class", minsplit =1, minbucket=nMinBucket,cp=nCp); and get predictors used in building the model like

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

This question concerns rpart's facility for user-defined functions that accomplish splitting. I was interested in modifying the code so that in each terminal node, a linear regression is fit to the data. It seems that from the allowable inputs in the user-defined functions, that this may not be possible, since they have the form: function(y, wt, parms) (in the case of the