similar to: rpart, resolution

? data=read.table("http://statcourse.com/research/boston.csv", , sep=",", header = TRUE) ? library(rpart) ? fit=rpart (MV~ CRIM+ZN+INDUS+CHAS+NOX+RM+AGE+DIS+RAD+TAX+ PT+B+LSTAT) predict(fit,data[4,]) plot only reveals part of the tree in contrast to the results on obtains with CART or C5 -------- Original Message -------- Subject: Re: [R] help with rpart From: Sarah

1. How can I plot the entire tree produced by rpart? 2. How can I submit a vector of values to a tree produced by rpart and have it make an assignment? Mark

? data=read.table("http://statcourse.com/research/boston.csv", , sep=",", header = TRUE) ? library(rpart) ? fit=rpart (MV~ CRIM+ZN+INDUS+CHAS+NOX+RM+AGE+DIS+RAD+TAX+ PT+B+LSTAT) Please: Show me the tree. Mark -------- Original Message -------- Subject: Re: [R] help with rpart From: "Stephen Milborrow" <[1]milbo at sonic.net>

Hi, I am using the packages tree and rpart to build a classification tree to predict a 0/1 outcome. The package rpart has the advantage that the function plotcp gives a visual representation of the cross-validation results with a horizontal line indicating the 1 standard error rule, i.e. the recommendation to select the most parsimonious model (the smallest tree) whose error is not more than one

Greetings, I checked the Indian diabetes data again and get one tree for the data with reordered columns and another tree for the original data. I compared these two trees, the split points for these two trees are exactly the same but the fitted classes are not the same for some cases. And the misclassification errors are different too. I know how CART deal with ties --- even we are using the

Hi all, I'm a newbee using R. I need to do a classification tree using the rpart package. Basically I have a set of birds of known sex and several morphological measurements and we want to predict the sex using the morphology. I read my csv file and it shows up in R no problem, looks fine but when I execute the following rpart command

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

Jonathan, Try making a list instead of an array. See ?list. Also, did you look into random forests? I'm not sure what you want to do, but there might be methods there to do some of the work for you. Sean On 3/19/04 1:12 PM, "Jonathan Williams" <jonathan.williams at pharmacology.oxford.ac.uk> wrote: > I would like to collect the trees grown by rpart fits in an array,

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).

Greetings, I am using rpart for classification with "class" method. The test data is the Indian diabetes data from package mlbench. I fitted a classification tree firstly using the original data, and then exchanged the order of Body mass and Plasma glucose which are the strongest/important variables in the growing phase. The second tree is a little different from the first one. The

Dear helpers, I am a Dutch student from the Erasmus University. For my Bachelor thesis I have written a script in R using boosting by means of classification and regression trees. This script uses the function the predefined function rpart. My input file consists of about 4000 vectors each having 2210 dimensions. In the third iteration R complains of a lack of memory, although in each iteration

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

Hi, I have a question regarding how to get some partial information from the output of rpart, which could be used as the first argument to predict. For example, in my code, I try to learn a stump tree (decision tree of depth 2):    "fit        <- rpart(y~bx, weights = w/mean(w), control = cntrl)     print(fit)     btest[1,]  <- predict(fit, newdata = data.frame(bx)) " I found

Hello list, I'm trying to generate classifiers for a certain task using several methods, one of them being decision trees. The doubts come when I want to estimate the cross-validation error of the generated tree: tree <- rpart(y~., data=data.frame(xsel, y), cp=0.00001) ptree <- prune(tree, cp=tree$cptable[which.min(tree$cptable[,"xerror"]),"CP"]) ptree$cptable

Hi folks, I'm getting the following error message when using predict.rpart on a large dataset (say 100,000 records) within Splus: Terminating S Session: Signal: bad address signal Has anyone run into this error, discovered a work-around, etc.? More generally, I'm trying to use predict.rpart to generate predictions from very large, geographic datasets (we're talking all of

I'm in a situation where I say: > predict(m.rpart, newdata=D[N1+t,]) 0 1 173 0.8 0.2 which I interpret as meaning: an 80% chance of "0" and a 20% chance of "1". Okay. This is consistent with: > predict(m.rpart, newdata=D[N1+t,], type="class") [1] 0 Levels: 0 1 But I'm puzzled at the following. If I say: > predict(m.rpart,

Hi, I am currently studying Decision Trees by using rpart package in R. I artificially created a data set which includes the dependant variable (y) and a few independent variables (x1, x2...). The dependant variable y only comprises 0 and 1. 90% of y are 1 and 10% of y are 0. When I apply rpart to it, there is no splitting at all. I am wondering whether this is because of the

I have just upgraded from R 1.7.3 to R 1.9.0 and have found that the predict function no longer works for rpart: > predict(hmmm,sim3[1:10,]) Error in predict.rpart(hmmm, sim3[1:10, ]) : couldn't find function "pred.rpart" I have re-installed the rpart package to no avail. Any ideas? Giles Hooker

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 using an algorigm to split my data set into two random sections repeatedly and constuct a model using rpart() on one, test on the other and average out the results. One of my variables is a factor(crop) where each crop type has a code. Some crop types occur infrequently or singly. when the data set is randomly split, it may be that the first data set has a crop type which is not present in