similar to: User defined split function in Rpart

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Dear R-users, I'm working with the Rpart package and trying to understand how the procedure select the best split in the case the method "class" and the splitting index "Gini" are used. In particular I'd like to have look to the source code that works out the best split for un unordered predictor. Does anyone can suggest me which functions in the sources I should

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()

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)

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 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

Hello. I have used the "party" package to generate a regression tree as follows: >origdata<-read.csv("origdata.csv") >ctrl<-ctree_control(mincriterion=0.99,maxdepth=10,minbucket=10) >test.ct<-ctree(Y~X1+X2+X3,data=origdata,control=ctrl) The above works fine. Orig data was my training data. I now have a test data file (testdata), and

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 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

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

Dear List, Does anybody no how to compare mean survival times for two (more) groups in R? What test statistics should I use? Thank you very much! Joe [[alternative HTML version deleted]]

Dear List, quantile(x) function allows to obtain specified quantiles of a vector of observations x. Is there an analogous function to compute quantiles in the case one have the vector of the observations x and the correspondent vector f of relative frequencies ? Thank you Paolo Radaelli [[alternative HTML version deleted]]

Brian The following (which is new to rw1030) occurs with both Windows 98 & Windows ME. I have not tested behaviour under Unix or Linux, but I expect it is no different. text.rpart() prints unwanted NAs (presumably in the splitting criterion position) on terminal nodes. Criterion <- factor(paste("Leaf", 1:5)) Node <- factor(1:5)

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

Should one expect minor numerical differences between 64bit and 32bit R on Windows? Hunting around the lists I've not been able to find a definitive answer yet. Seems plausible using different precision arithmetic, but waned to confirm from those who might know for sure. BACKGROUND A colleague was trying to replicate some modelling results (from a soon to be published book) using rpart, ada,

Dear R users, I still feel new to R so please apologize if I am doing something stupid here. My use of the polygon() function produces a result that I cannot comprehend: In a plot, I would like to shade the area below a normal distribution. However, I do not want the entire area to be shaded, but just the area on the right side of a vertical line that I draw through the distribution (in

I've just installed R 1.0.8 (for Windows) and I tried to install the package Quantreg directly from Cran but it's not in the list of downlodable packages. I tried also downloading the zip file and then install it but there is an error. How can I do it? Thank you [[alternative HTML version deleted]]

How does one use the vcut from Vorbis-tools package? I tried to enter samples, miliseconds, seconds, h:mm:ss as cut point, but nothing worked (I even tried to enter 1 as cutpoint, but I always got "Cutpoint not within stream." message)... I'm doing this on Windows, if it matters... -- Jernej Simoncic, jernej.simoncic@guest.arnes.si http://www2.arnes.si/~sopjsimo/ ICQ: 26266467

Dear Mrs Rachel Pearce, I am looking for a function "cutpt-coxph" in R - like you did some years ago. How have you solved the problem? Have you found it or a similar function? thank you, Sincerely, Friederike "The title says it all really; I am looking for a function along the lines of cutpt.coxph as described in "Survival Analysis Using S" (Tableman and Kim), Chapter

I am new to R and I am trying to do a monte carlo simulation where I generate data and interject error then test various cut points; however, my output was garbage (at x equal zero, I did not get .50) I am basically testing the performance of classifiers. Here is the code: n <- 1000; # Sample size fitglm <- function(sigma,tau){ x <- rnorm(n,0,sigma) intercept <- 0 beta