similar to: How to personalize the rpart function: t.default(x)

Hallo! I am a student of the Politecnico di Milano (Milan, italy) and I'm working on CARTs. I'm trying to use the R rpart function with a personalized splitfunction... but I'm not able to do it! More precisely, I would like to know what is the meaning of the function 'init', 'split' and 'eval' named in the help page.I can't find any answer in

I'd like to create a bidimensional presentation of a classification tree built using the rpart() function. I've seen that a partition.tree() function exists for the tree() function. Does a similar function exist for the rpart() function? Thanks a lot Simone Vantini

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

Hallo R-users, I would like to know if there is a way to get the effective degrees of freedom in local polinomial regression. At the moment, to carry out the local polinomial regression, I am using the function locpoly() in library KernSmooth (I need to estimate both the regression function and its derivatives, up to the 3rd one). Thanks, Simone -- Simone Vantini MOX - Modeling and

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

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

Hello, I just upgraded my Ubuntu Xenial system to R 3.5.1 (from 3.4.?) by changing the sources.list entry and doing an "apt-get dist-upgrade". Everything works except loading the rpart package in R: > library(rpart) Error: package or namespace load failed for ?rpart?: package ?rpart? was installed by an R version with different internals; it needs to be reinstalled for use with

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

I am trying to create a classification tree using either tree or rpart but when it comes to plotting the results the formatting I get is different than what I see in all the tutorials. What I would like to see is the XX/XX format but all I get is a weird decimal value. I was also wondering how you know which is yes and which is no in each leaf of the tree? Is yes always on the left?

While I'm very pleased with the results I get with rpart and rpart.plot, I would like to change the scientific notation of the dependent variable in the plots into integers. Right now all my 5 or more digit numbers are displayed using scientific notation. I managed to find this: http://tolstoy.newcastle.edu.au/R/e8/help/09/12/8423.html but I do not fully understand what to change, and to

Dear list For my work, it would be helpful if rpart worked seamlessly with an empty model: library(rpart); rpart(formula=y~0, data=data.frame(y=factor(1:10))) Currently, an unrelated error (originating from na.rpart) is thrown. At some point in the near future, I'd like to release a package to CRAN which uses rpart and relies on that functionality. I have prepared a patch (minor

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

Hello useR, Could you please tell me how to draw the decision boundaries in a scatterplot of the original data for a LDA or Rpart object. For example: > library(rpart) >fit.rpart <- rpart(as.factor(group.id)~., data=data.frame(Data) ) How can I draw the cutting lines on the orignial Data? Or is there any built in functions that can read the rpart object 'fit.rpart' to do

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))) + ) >

After entering ?library(rpart)?, I tried to plot an existing rpart tree, and got this error message: Error: couldn't find function "plot.rpart". However, ??plot.rpart? does bring up the help for the function. The same things occur for text.rpart, although print(my.tree) does work. So, I tried to re-install rpart using Packages | Install from CRAN, but then I get this

Dear useRs: I try to plot an rpart object but cannot get a nice tree structure plot. I am using plot.rpart and text.rpart (please see below) but the branches that connect the nodes overlap the text in the ellipses and rectangles. Is there a way to get a clean nice tree plot (as in the Rpart Mayo report)? I work under Windows and use R2.11.1 with rpart version 3.1-46. Thank you. Tudor ...

I have a class with 732 members, so using rpart.plot is giving me a tiny plot in the middle of the window. Is there a good way to modify the plot, or replace the long list with something like "group1"? -- View this message in context: http://r.789695.n4.nabble.com/Plotting-rpart-trees-with-long-list-of-class-members-tp4635671.html Sent from the R help mailing list archive at

Hi, Suppose i have generated an object using the following : fit <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis) And when i print fit, i get the following : n= 81 node), split, n, loss, yval, (yprob) * denotes terminal node 1) root 81 17 absent (0.7901235 0.2098765) 2) Start>=8.5 62 6 absent (0.9032258 0.0967742) 4) Start>=14.5 29 0 absent (1.0000000

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

I've written a small adaptation of the text.rpart function of the rpart package to better suite my tree presentation needs (I basically left the code untouched, only changing some numbers relating to label positioning). When I changed to version 1.7.0 this function stopped working printing the error : couldn't find function "rpartco" As far as I understand this has to do