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A few weeks back I posted that the subset feature of rpart was not working when predicting a categorical variable. I was able to figure out a simple solution to the problem that I hope can be included in future editions of rpart. I also include a fix for another related problem. The basic problem is that when predicting a categorical using a subset, the subset may not have all the categories

(Apparently, I posted this to the wrong place. I am hopefully posting this is the correct place now. If not, please advise.) A few weeks back I posted that the subset feature of rpart was not working when predicting a categorical variable. I was able to figure out a simple solution to the problem that I hope can be included in future editions of rpart. I also include a fix for another related

Hi all, I am using R1.5.0 under Unix, I have a couple of questions here. 1. My program is running out of memory. I am writing a program to grow a list of trees using rpart() on a subset of a large dataset(5807x693) with a different response for every tree. I saw that after each tree was constucted, 116 MB of data was being added to the Vcells. I have no idea what this data is. My dataset is

Dear all, I'm getting an error in one of the stock examples in the 'mvpart' package. I tried: require(mvpart) data(spider) fit3 <- rpart(gdist(spider[,1:12],meth="bray",full=TRUE,sq=TRUE)~water+twigs+reft+herbs+moss+sand,spider,method="dist") #directly from ?rpart summary(fit3) ...which returned the following: Error in apply(formatg(yval, digits - 3), 1,

Hi, I am reading the source code of rpart. I have problems understand the following code and would appreciate for any helps. In rpart.s, there is a line: rpfit &lt;- .C(C_s_to_rp, &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; n = as.integer(nobs), &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; nvarx

I just updated rpart to the latest version (3.1.0). There are a number of changes between this and previous versions, and some of the code I've been using with earlier versions (e.g. 3.0.2) no longer work. Here is a simple illustration of a problem I'm having with xpred.rpart. iris.test.rpart<-rpart(iris$Species~., data=iris[,1:4], parms=list(prior=c(0.5,0.25, 0.25))) + ) >

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

Hi, I just noticed that rpart behaves unexpectecly, when performing classification learning and specifying a loss matrix. if the response variable y is a factor and if not all levels of the factor occur in the observations, rpart exits with an error: > df=data.frame(attr=1:5,class=factor(c(2,3,1,5,3),levels=1:6)) > rpart(class~attr,df,parms=list(loss=matrix(0,6,6))) Error in

Hi, there: I am working on a classification problem by using rpart. when my response variable y is binary, the trees grow very fast, but if I add one more case to y, that is making y has 3 cases, the tree growing cannot be finished. the command looks like: x<-rpart(r0$V142~.,data=r0[,1:141], parms=list(split='gini'), cp=0.01) changing cp or removing parms does not help.

Hi All, I have some questions on using library rpart. Given my data below, the plotcp gives me increasing 'xerrors' across different cp's with huge xstd (plot attached). What causes the problem or it's not a problem at all? I am thinking 'xerror's should be decreasing when 'cp' gets smaller. Also what the 'xstd' really tells us? If the error bars for

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, I am trying to use the rpart library with my own set of functions on a survival object. I get an immeadiate segmentation fault when i try calling rpart with my list of functions. I get the same problem with the logrank example from Therneau,s S-rpart library though their anova example works. Should I report this as a bug, as even if my functions are structured improperly, that should lead to

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

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 all, I have a 100 iteration loop. Within each loop, there are some calls to rpart() like: ctl <- rpart.control(maxcompete=0, maxsurrogate=0, maxdepth=10) temp <- rpart(y~., x, w=wt, method="class", parms=list(split="gini"), control=ctl) res <- log(predict.rpart(temp, type="prob")) newres <- log(predict.rpart(temp, newdata=newx,

I have been trying to write my own user defined function in Rpart.I imitated the anova splitting rule which is given as an example.In the work I am doing ,I am calculating the concentration index(ci) ,which is in between -1 and +1.So my deviance is given by abs(ci)*(1-abs(ci)).Now when I run rpart incorporating this user defined function i get the following error message: Error in

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",

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

Full_Name: Bill Wheeler Version: 2.0.1 OS: Windows 2000 Submission from: (NULL) (67.130.36.229) The program to reproduce the error is below. I am calling rpart with a user-defined split function for a binary response variable and one continuous independent variable. The split function works for some datasets but not others. The error is: Error in "$<-.data.frame"(`*tmp*`,

Dear users, I'm using rpart for classification trees, but my code isn't working when I try to use all the variables in my data frame. This data frame was created from a data frame with 1775 variables, but I choose only 13. arv13<-rpart(iv~.,data=gn,method="class",parms=list(split="information")) #Error: Error in `[.data.frame`(frame, predictors) : undefined