similar to: Class probabilities in 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 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! Could you please explain the difference between "prior" and "weight" in rpart? It seems to be the same. But in this case why including a weight option in the latest versions? For an unbalanced sampling what is the best to use : weight, prior or the both together? Thanks a lot. Aur?lie Davranche.

Hi, When I look at the summary of an rpart object run on my data, I get 7 nodes but when I plot the rpart object, I get only 3 nodes. Should the number of nodes not match in the results of the 2 functions (summary and plot) or it is not always the same? Look forward to your reply, Carol -------------------------------------------- ?summary(rpart.res) Call: rpart(formula = mydata$class ~ ., data

Hello All, Am I misreading the documentation? The text.rpart documentation says: "label a column name of x$frame; values of this will label the nodes. For the "class" method, label="yval" results in the factor levels being used, "yprob" results in the probability of the winning factor level being used, and ?specific yval level? results in the probability of

I am performing a binary classification using a classification tree. Ironically, the data themselves are 2483 tree (real biological ones) locations as described by a suite of environmental variables (slope, soil moisture, radiation load, etc). I want to separate them from an equal number of random points. Doing eda on the data shows that there is substantial difference between the tree and random

Hi, I've searched online, in a few books, and in the archives, but haven't seen this. I believe that xerror is scaled to rel error on the first split. After fitting an rpart object, is it possible with a little math to determine the percentage of true classifications represented by a xerror value? -seth -- View this message in context:

Hi Everyone, This example code results in R 'crashing'; that is the R application closes with no warnings or error messages. #----------------------- myD <- read.table(stdin(), header=TRUE, nrows=20) Broth Salt pH Temp N Y Growth 1 310 9.0 2.92 10 90.0 NA 0 2 615 6.0 7.82 30 1.0 2 1 3 217 2.0 7.34 10 7.0 8

Hello, I have attempted to email the author of this package without success, just wondering if anybody else has experienced this. I am having an using rpart on 4000 rows of data with 13 attributes. I can run the same test on 300 rows of the same data with no issue. When I run on 4000 rows, Rgui.exe runs consistently at 50% cpu and the UI hangs.... it will stay like this for at least 4-5hours if

Dear r-help-list: If I use the rpart method like cfit<-rpart(y~.,data=data,...), what kind of tree is stored in cfit? Is it right that this tree is not pruned at all, that it is the full tree? If so, it's up to me to choose a subtree by using the printcp method. In the technical report from Atkinson and Therneau "An Introduction to recursive partitioning using the rpart

--- included message ---- Thus, my question is: *What common measures exists for ranking/measuring variable importance of participating variables in a CART model? And how can this be computed using R (for example, when using the rpart package)* ---end ---- Consider the following printout from rpart summary(rpart(time ~ age + ph.ecog + pat.karno, data=lung)) Node number 1: 228 observations,

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

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

I am using rpart to build a model for later predictions. To save the prediction across restarts and share the data across nodes I have been using "save" to persist the result of rpart to a file and "load" it later. But the saved size was becoming unusually large (even with binary, compressed mode). The size was also proportional to the amount of data that was used to create the

Hi, I am using rpart decision trees to analyze customer churn. I am finding that the decision trees created are not effective because they are not able to recognize factors that influence churn. I have created an example situation below. What do I need to do to for rpart to build a tree with the variable experience? My guess is that this would happen if rpart used the loss matrix while creating

I've been using the package rpart with R 1.3.0 for Windows to produce simple classification trees for some measurement data from paleontological specimens. Both the rpart documentation and the output confirm that the program produces splits on continuous data that leave "holes" in the data. It is probably of little practical importance, but is there a reason why the binary

Hi Everyone, Is there a way to get predict.rpart() to return the nodes reached by the new examples in addition to the predicted probabilities it already returns? In other words, I would like to know the leaf node in the tree object that each new example data drops down to. Thanks in advance for your help. Osei

Hi there. I was wondering if somebody knows how to perform a bagging procedure on a classification tree without running the classifier with weights. Let me first explain why I need this and then give some details of what I have found out so far. I am thinking about implementing the bagging procedure in Matlab. Matlab has a simple classification tree function (in their Statistics toolbox) but

Hello, I use rpart to predict survival time and have a problem in interpreting the output of ?estimated rate?. Here is an example of what I do: > stagec <- > read.table("http://www.stanford.edu/class/stats202/DATA/stagec.data", > col.names=c("pgtime", "pgstat", "age","eet", "g2", "grade", "gleason", >

Dear R-list, I am using the rpart/mvpart-package for selecting a right-sized regression tree by 10-fold cross-validation. My question: Is there a possibility to find out for every observation in which of the ten folds it is lying? I want to use the same folds for validating another regression method (moving averages) in order to choose the better one. Thanks a lot, Pedro