similar to: Question about Rpart function

Hi, I am using rpart with method="anova".Can we compute 95% confidence intervals and prediction interval for the predicted mean at each node? Thanks, Puja [[alternative HTML version deleted]]

Hello, I am trying to predict the classes of a test data set after training an rpart tree. When I run: predict(rpart_object_based_on_training_data, newdata = "testdata", type = "class", na.action = na.pass) I get an error message saying that a variable that is present in both training and test data sets has new levels in the test set. This is true that there are new levels

Hi all, What parameter do I normally change in the rpart function? How do I set the "cp" option? Is there a way to read off error rate directly from the "rpart" function for training data; I imagine for testing data I have to apply a "predict", but for training data I guess the error count would be somewhere existing once the "rpart" function is

I've been groping my way through a classification/discrimination problem, from a consulting client. There are 26 observations, with 4 possible categories and 24 (!!!) potential predictor variables. I tried using lda() on the first 7 predictor variables and got 24 of the 26 observations correctly classified. (Training and testing both on the complete data set --- just to get started.) I

Hi all, I am very new to R (only two days of studies). I know a little bit of statistical learning and looking for an implementation of CART and random forest and therefore I am now studying R. I tested with rpart and randomForest package, they are quite good. However, I need a classification tree and for each training data, there is different loss. (i.e. the loss function is not purely 0-1,

Hello, I am working with rpart function but geting some error in prediction. the same code works fine with iris dataset. but applying other dataset it doesn't work. sample code is given for reference. > acc_model<-rpart(V1~V2+V3+V4+V5+V6+V7+V8, data=accEx.train) > plotcp(acc_model) >

Dear R users, I know cross-validation does not work in rpart with user defined split functions. As Terry Therneau suggested, one can use the xpred.rpart function and then summarize the matrix of the predicted values into a single "goodness" value. I need only a confirmation: set for example xval=10, if I correctly understood a single column of the matrix obatined by xpred.rpart gives

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, I have a technical question about rpart: according to Breiman et al. 1984, different costs for misclassification in CART can be modelled either by means of modifying the loss matrix or by means of using different prior probabilities for the classes, which again should have the same effect as using different weights for the response classes. What I tried was this: library(rpart)

I make classification tree like this code p.t2.90 <- rpart(y~aa_three+bas+bcu+aa_ss, data=training,method="class",control=rpart.control(cp=0.0001)) Here I want to set weight for 4 predictors(aa_three,bas,bcu,aa_ss). I know that there is a weight set-up in rpart. Can this set-up satisfy my need? If so, could someone give me an example? Thanks, Aimin Yan

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

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

RPART doesn't seem to handle the degenerate case when all training samples are drawn from a single class: > TrainType [1] 0 0 0 0 > TrainDat V1 V2 V3 V4 V5 1 0.6434392 0.5105860 0.3048803 0.3161728 0.5449632 2 0.1710005 0.5973921 0.1267061 0.6146834 0.7299928 3 0.6919125 0.8880789 0.9123243 0.9061885 0.9553663 4 0.3094843 0.6475508

So there are a couple parts to this question. I am trying to implement the rpart/random forest algorithms on a transaction lists. That is to say i am trying to train models in order to deduce what are the most predictive transactions within a customers history in order apply this model to future test data and identify accounting irregularities(ie. this account had x and y so they should have also

Hello there. I have fitted a rpart model. > rpartModel <- rpart(y~., data=data.frame(y=y,x=x),method="class", ....) and can use rpart$where to find out the terminal nodes that each observations belongs. Now, I have a set of new data and used predict.rpart which seems to give only the predicted value with no information similar to rpart$where. May I know how

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

There have been various messages about packages tree and rpart whilst I have been travelling, and I have now prepared updates. tree ==== Tree is one of the oldest packages on CRAN (Feb 2000 apart from adding the maintainer field), and I had been hoping that it would fade away in favour of rpart. 1) sys.parent needed to be replaced by parent.frame in all but the most recent R (post 1.3.0).

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 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 am doing > library(rpart) > m <- rpart("y ~ x", D[insample,]) > D[outsample,] y x 8 0.78391922 0.579025591 9 0.06629211 NA 10 NA 0.001593063 > p <- predict(m, newdata=D[9,]) Error in model.frame(formula, rownames, variables, varnames, extras, extranames, : invalid result from na.action How do I persuade him to give me NA