similar to: Rpart and case weights: working with functions

Hello, I want to add text annotation about correlation on "pairs" plots. I found that I could pass a function to the "panel" argument of pairs : panel.annot <- function(x, y, ...) { points(x, y, ...) c <- cor.test(x, y) legend("topleft", legend=substitute(rho == r, list(r=sprintf("%.2f", c$estimate))), bty="n") } And then :

Hello, I've got a function that takes a numeric vector (x), computes a transformation value (myAttr) for x, transforms x according to myAttr and then sets myAttr as an attribute of x before returning x, so I can easily know what myAttr was used (basically it's a power transformation and myAttr is the lambda). myFunction.numeric <- function(x, ...) { myAttr <- calcMyAttr(x, ...) x

Hi there. I was wondering if somebody knows how to perform a bagging procedure on a classification tree without running the classifier with weights. Let me first explain why I need this and then give some details of what I have found out so far. I am thinking about implementing the bagging procedure in Matlab. Matlab has a simple classification tree function (in their Statistics toolbox) but

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

Hello all R-help list subscribers, I'd like to create a regression tree of a data set with binary response variable. Only 5% of observations are a success, so the regression tree will not find really any variable value combinations that will yield more than 50% of probability of success. I am however interested in areas where the probability of success is noticeably higher than 5%, for

Hi! Could you please explain the difference between "prior" and "weight" in rpart? It seems to be the same. But in this case why including a weight option in the latest versions? For an unbalanced sampling what is the best to use : weight, prior or the both together? Thanks a lot. Aur?lie Davranche.

Dear r-help mailing list, Is there a way to incorporate weights into the minsplit criteria in rpart, when the weights are uneven? I could not find a way for the minsplit threshold to take the weights into account, and when the weights are uneven it becomes an issue, as the following example shows. My current workaround is to expand the data into one in which each row is an observation, but that

Dear R community, I am trying to write my own user defined split function for rpart. I read the example in the tests directory and I understand the general idea of the how to implement user defined splitting functions. However, I am having troubles with addressing the data frame used in calling rpart in my split functions. For example, in the evaluation function that is called once per node,

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

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)

Hello all, I am currently working with rpart to classify vegetation types by spectral characteristics, and am comming up with poor classifications based on the fact that I have some vegetation types that have only 15 observations, while others have over 100. I have attempted to supply prior weights to the dataset, though this does not improve the classification greatly. Could anyone supply some

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

Not sure if anyone has posted on this problem ... I want to use rpart to build a binary tree on a relatively large dataset with ~1400 data points and 15 predictors. But I've noticed that rpart fails almost immediately in the call to C_s_to_rp, as that code returns nonsense. Looking at the code itself isn't terribly helpful, and there don't seem to be any hard limits coded anywhere.

Dear all, I'm trying to manage with user defined split function in rpart (file rpart\tests\usersplits.R in http://cran.r-project.org/src/contrib/rpart_3.1-34.tar.gz - see bottom of the email). Suppose to have the following data.frame (note that x's values are already sorted) > D y x 1 7 0.428 2 3 0.876 3 1 1.467 4 6 1.492 5 3 1.703 6 4 2.406 7 8 2.628 8 6 2.879 9 5 3.025 10 3 3.494

Philipp Benner reported a Debian bug report against r-cran-rpart aka rpart. In short, the issue has to do with how rpart evaluates a formula and supporting arguments, in particular 'weights'. A simple contrived example is ----------------------------------------------------------------------------- library(rpart) ## using data from help(rpart), set up simple example myformula <-

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

I have been trying to write my own user defined function in Rpart.I imitated the anova splitting rule which is given as an example.In the work I am doing ,I am calculating the concentration index(ci) ,which is in between -1 and +1.So my deviance is given by abs(ci)*(1-abs(ci)).Now when I run rpart incorporating this user defined function i get the following error message: Error in

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