similar to: text.rpart: Unwanted NA labels on terminal nodes (PR#1009)

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 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. I uses rpart to build a regression tree. Y is continuous. Now, I try to predict on a new set of data. In the new set of data, one of my x (call Incoterm, a factor) has a new level. I wonder why the error below appears as the guide says "For factor predictors, if an observation contains a level not used to grow the tree, it is left at the deepest possible node and

Hi, I am trying to write my own split function for rpart. The aim is to do, instead of anova, a linear regression to determine the split (minimize some criterion like sum of rss left and right of the split). The regression (lm) should simply use the dependent and independent variables passed to rpart. I am aware of the example provided in the rpart source code, but stumbled on similar problems

Hi all, I am trying to get to grips with rpart, and find it not very easy given the information that comes with the package. Contrary to e.g. the ctest package docs, it doesn't say when "an rpart" could be used, and/or how to interpret the results. Here are a few of the open questions I have: 1) Read in ?rpart: ...method: one of.... If y is a survival object... A similar

Dear r-list: I am using rpart to build a tree on a dataset. First I obtain a perhaps too large tree: > arbol.bsvg.02 <- rpart(formula, data = bsvg, subset=grp.entr, control=rpart.control(cp=0.001)) > arbol.bsvg.02 n= 100000 node), split, n, loss, yval, (yprob) * denotes terminal node 1) root 100000 6657 0 (0.93343000 0.06657000) 2) meses_antiguedad_svg>=10.5 73899 3658

Hi, all, I am wondering if it is possible to set parameters of 'rpart' and 'tree' such that they will produce the exact same tree? Thanks. Auston Wei Statistical Analyst Department of Biostatistics and Applied Mathematics The University of Texas MD Anderson Cancer Center Tel: 713-563-4281 Email: wwei@mdanderson.org [[alternative HTML version deleted]]

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Wondered about the best way to control for input variables that have a large number of levels in 'rpart' models. I understand the algorithm searches through all possible splits (2^(k-1) for k levels) and so variables with more levels are more prone to be good spliters... so I'm looking for ways to compensate and adjust for this complexity. For example, if two variables produce

Dear R friends- I'm attempting to generate a regression tree with one gradient predictor and multiple responses, trying to test if change in size (turtle.data$Clength) acts as a single predictor of ten multiple diet taxa abundances (prey.data) Neither rpart or mvpart seem to allow me to do multiple responses. (Or if they can, I'm not using the functions properly.) > library(rpart)

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

Hi all, I apologies in advance if I am missing something very simple here, but since I failed at resolving this myself, I'm sending this question to the list. I would appreciate any help in understanding how the rpart function is (exactly) computing the "improve" (which is given in fit$split), and how it differs when using the split='information' vs split='gini'

HI, Dear R community, I am writing the following function to create one data set(*tree.pred*) and one vector(*valid.out*) from loops. Later, I want to use the data set from this loop to plot curves. I have tried return, list, but I can not use the *tree.pred* data and *valid.out* vector. auc.tree<- function(msplit,mbucket) { * tree.pred<-data.frame()

Greetings, I am using rpart for classification with "class" method. The test data is the Indian diabetes data from package mlbench. I fitted a classification tree firstly using the original data, and then exchanged the order of Body mass and Plasma glucose which are the strongest/important variables in the growing phase. The second tree is a little different from the first one. The

Hi R-helpers, I am using rpart package for decision tree using R.We are invoking R environment through JRI from our java application.Hence, the result of R command is returned in REXP and we use geterrMessage() to retrieve the error. When we execute the following command, cnr_model<-rpart(as.factor(Species)~Sepal Length+Sepal Width+Petal Length, method="class",

Hi, I am using rpart to do leave one out cross validation, but met some problem, Data is a data frame, the first column is the subject id, the second column is the group id, and the rest columns are numerical variables, > Data[1:5,1:10] sub.id group.id X3262.345 X3277.402 X3369.036 X3439.895 X3886.935 X3939.054 X3953.777 X3970.352 1 32613 HAM_TSP 417.7082 430.4895 619.4776 720.8246

Hi, I am currently studying Decision Trees by using rpart package in R. I artificially created a data set which includes the dependant variable (y) and a few independent variables (x1, x2...). The dependant variable y only comprises 0 and 1. 90% of y are 1 and 10% of y are 0. When I apply rpart to it, there is no splitting at all. I am wondering whether this is because of the

Hi, Being a novice this is my first usage of R. I am trying to use rpart for building a decision tree in R. And I have the following dataframe Outlook Temp Humidity Windy Class Sunny 75 70 Yes Play Sunny 80 90 Yes Don't Play Sunny 85 85 No Don't Play Sunny 72 95 No Don't Play Sunny 69 70 No Play Overcast 72 90 Yes Play Overcast 83 78 No Play Overcast 64 65 Yes Play Overcast 81 75

Dear useRs: Is there a way I could predict the terminal node associated with a new data entry in an rpart environment? In the example below, if I had a new data entry with an AM of 5, I would like to link it to the terminal node 2. My searches led to http://tolstoy.newcastle.edu.au/R/e4/help/08/07/17702.html but I do not seem to be able to operationalize Professor Ripley's suggestions. Many

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