similar to: ROC curve in randomForest

After ctree builds a tree, how would I determine the direction missing values follow by examining the BinaryTree-class object? For instance in the example below Bare.nuclei has 16 missing values and is used for the first split, but the missing values are not listed in either set of factors. (I have the same question for missing values among numeric [non-factor] values, but I assume the answer

Dear All, I'm getting a error while trying to apply the BreastCancer dataset (package=mlbench) to kknn (package=kknn) that I don't understand as I'm new to R. The codes are as follow: rm = (list = ls()) library(mlbench) data(BreastCancer) library(kknn) BCancer = na.omit(BreastCancer) d = dim(BCancer)[1] i1 = seq(1, d, 2) i2 = seq(2, d, 2) t1 = BCancer[i1, ] t2 =

Hi All, I'm using the party package to grow conditional inference trees for survival analysis. When I used party version party_0.9-9991 everything worked well, but when I update to party_1.0-2 (due to using 64bit R), I get an error. For simplicity I will show the error I get for the example in the party documentation: ### survival analysis if (require("ipred")) {

R-help, I am trying to perform a basic anlaysis of the BreastCancer data from "mlbench" using the svm() function in "e1071". I use the following code library("e1071") library("mlbench") data(BreastCancer) BC <- subset(BreastCancer, select=-Id) pairs(BC) model <- svm(Class ~ ., data=BC, cross=10) ## plot(model, BC, ) tobj <- tune.svm(Class ~ .,

Hello everybody, this is my first request about R so I am sorry if I send it to a bad mail or if I am not very clear. So my problem is about the use of rfImpute from randomForest package. I am interested in imputations of missing values and I read that randomForest can make it. So i write the following code : set.seed(100); library(mlbench) library(randomForest) data(BreastCancer)

Apologies - I was not trying to correct you Brian, but to explore how the situation could arise. I'm sure you had a good idea why the namespace (or a reference to it) had been saved, but this was not clear to me and I thought, possibly not to others either. Thanks for putting me right over parent environments vs. enclosures - again I was not trying to correct you with the point I made there,

Last week Giovanni Parrinello posted a message asking why various packages were loaded when he loaded an .Rdata file. Brian Ripley replied saying he thought it was because the saved workspace contained a reference to the namespace of ipred. (Correspondence copied below). This begs the question: how did the reference to the namespace of ipred come to be in the .Rdata file? Brian did say it is

Hi Everybody, i again have the problem in using tune.knn(), its giving an error saying missing values are not allowed.... again here is the script for BreastCancer Data, library(e1071) library(mda) trdata<-data.frame(train,row.names=NULL) attach(trdata) xtr <- subset(trdata, select = -Class) ytr <- Class bestpara <-tune.knn(xtr,ytr, k = 1:25, tunecontrol = tune.control(sampling

Last week Giovanni Parrinello posted a message asking why various packages were loaded when he loaded an .Rdata file. Brian Ripley replied saying he thought it was because the saved workspace contained a reference to the namespace of ipred. (Correspondence copied below). This begs the question: how did the reference to the namespace of ipred come to be in the .Rdata file? Brian did say it is

Hello Everybody, I am reffering David Meyer's Benchmarking Support Vector Machines , Report No.78 (Nov.2002), i am newly working with R but i am not sure how it is handling missing values in the benchmark datasets, I would be very thankful to you if you could let me know how to handle those missing numerical & categorical variables in the data (e.g. BreastCancer). because, i am

Hello, This is a re-submittal of question I submitted last week, but haven't rec'd any responses. I need to extract the probabilities used to construct the barplots displayed as part of the graph produced by plot("ctree"). For example, library(party) iris.ct <- ctree(Species ~ . , data = iris) plot(iris.ct) Instead of a simple example with only 4 terminal nodes, my

Hi, I have three original curves as follows, n<-seq(20,200,by=10) t<-c(0.1138, 0.1639, 0.2051, 0.2473, 0.2890, 0.3304, 0.3827, 0.4075, 0.4618, 0.4944, 0.5209, 0.5562, 0.5935, 0.6197, 0.6523, 0.6771, 0.6984, 0.7209, 0.7453) es<-c(0.3682, 0.4268, 0.5585, 0.6095, 0.7023, 0.7534, 0.8225, 0.8471, 0.8964, 0.9098, 0.9371, 0.9514, 0.9685, 0.9747, 0.9812, 0.9859, 0.9905, 0.9923, 0.9940)

Dear all, I was intrigued by the ctree command and wanted to check it out. I first ran the demo with example(ctree) and did get the survival graphs in the end. Upon doing this with my own data and yielding a "Invalid operation on a survival time" I tried to rerun example(ctree) and now I also get "Invalid operation on a survival time" after the example runs plot(GBSG2ct)...

Hi, I am working on a regression tree in Rpart that uses a continuous response variable that is ordered. I read a previous response by Pfr. Ripley to a inquiry regarding the ability of rpart to handle ordinal responses in 2003. At that time rpart was unable to implement an algorithm to handle ordinal responses. Has there been any effort to rectify this in recent years? Thanks! Stacey On

Hi, I am having difficulty understanding some lme output -- I haven't found too many examples to help explain to me how to interpret the coefficients and would appreciate any help. I am fitting a model: fit <- lme(y ~ pre + group + time + group:time, random=~1|subject, na.action=na.omit, data=mydata) ...for a dataset where there are two groups being followed over time. pre is

Why are the results not reliable? ________________________________ From: ESCHEN Rene [mailto:rene.eschen@unifr.ch] Sent: Wednesday, August 23, 2006 3:48 AM To: Spencer Graves; r-help@stat.math.ethz.ch Cc: Doran, Harold Subject: RE: [R] Random structure of nested design in lme The output of the suggested lmer model looks very similar to the output of aov, also when I ran the model

Hi All, i've got a problem using the ctree function of the party package. I've searched around a lot but couldn't manage it. I've got an ordered factor as response. As far as i know i have to use scores to be able to use this ordered factor. But if i do so i get a tree which predicts all observations as the first level of my ordered factor. In order to test if i did anything

Hi, I am new to R and am using the ctree() function to do customer segmentation. I am using the following code to generate the tree: treedata$Response<-factor(treedata$Conversion) fit<-ctree(Response ~ .,controls=ctree_control(mincriterion=0.99,maxdepth=4),data=treedata) plot(fit) print(fit) The variable "Response" above equals 1 if the customer responded to an offering and

Hello, My name is Rodrigo, I am using R program and I have a trouble. I am trying to do a dendrogram with genetics information. Let me explain... The Similarity Matrix was already did, and with this matrix I want to construct a dendrogram. So, the distance is done. I need to transform this matrix (that I have) in a dendrogram, I woud be very grateful if someone could help me. PS: I am sending

Dear all How does print.htest display the p-value in scientific notation? > (x <- cor.test(iris[[1]], iris[[3]])) Pearson's product-moment correlation data: iris[[1]] and iris[[3]] t = 21.65, df = 148, p-value < 2.2e-16 alternative hypothesis: true correlation is not equal to 0 95 percent confidence interval: 0.8270 0.9055 sample estimates: cor 0.8718 Above the p-value comes