similar to: rpart vs. tree and deviance calculations

Hi, I am doing a regression tree using the package 'rpart' but could not able to calculate the residual mean deviance. Please help. Narayan [[alternative HTML version deleted]]

I have used packages rpart, mvpart and tree for classification and regression trees. I want to calculate fraction of null deviance explained by each node and variable in the tree. For instance, at the first split, this would be (1 - (sum of residual deviance in each of the two leaves)/deviance at the root). In the subsequent splits, this formula is slightly different. There probably is a function

Hi, I have a problem with library (rpart) (and/or library(tree)). I use a data.frame with variables "pnV22" (observation: 1, 0 or yes, no) "JTemp" (mean temperature) "SNied" (summer rain) I used function "rpart" to build a model: library(rpart) attach(data.frame) result <- rpart(pnV22 ~ JTemp + SNied) I got the following tree: n=55518 (50

Hi All, I am fitting a tree to censored survival data using the rpart package and wanted to better understand the results. I am trying to interpret the output from the tree. I am interested in understanding what "yval" is for a survival tree. I see in the output of summary, the phrase "estimated rate". The estimated rate is 1 for the entire tree, and more of less for each

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

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 all, I apologies in advance if I am missing something very simple here, but since I failed at resolving this myself, I'm sending this question to the list. I would appreciate any help in understanding how the rpart function is (exactly) computing the "improve" (which is given in fit$split), and how it differs when using the split='information' vs split='gini'

The following code tr <- rpart(Y ~ ., dat, method="class") dev <- residuals(tr, "deviance") produces the following error Error in log(x) : Non-numeric argument to mathematical function > .Traceback [[1]] [1] "log(yhat)" # line 588 of rpart [[2]] [1] "switch(type, usual = as.integer(y != yhat), pearson = (1 - yhat)/yhat, " [2] "

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

Hi, I think I'm missing something very obvious, but I am missing it, so I would be very grateful for help. I'm using rpart to analyse data on skull base morphology, essentially predicting sex from one or several skull base measurements. The sex of the people whose skulls are being studied is known, and lives as a factor (M,F) in the data. I want to get back predictions of gender, and

All the documentation that I have on survival splitting is found in the technical report you mention. However, there is both a short form and a long form of this on our web site, did you get the larger one (52 pages)? I admit it is not a lot. There are no other split algorithms implimented for survival data. It would be possible to add your own. Attached is a slightly updated version of the

Full_Name: Bill Wheeler Version: 2.0.1 OS: Windows 2000 Submission from: (NULL) (67.130.36.229) The program to reproduce the error is below. I am calling rpart with a user-defined split function for a binary response variable and one continuous independent variable. The split function works for some datasets but not others. The error is: Error in "$<-.data.frame"(`*tmp*`,

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 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

partition.tree plots in 2d the partition of a classification tree produced by the function tree (assuming the data frame from which it was computed has two continuous predictors). I get an error when I feed a tree produced by rpart to partition.tree (since trees produced by rpart are superclasses of those produced by tree). Is there an equivalent of partition.tree for objects of class rpart?

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 Experts, I am new to R, using decision tree model for getting segmentation rules. A) Using behavioural data (attributes defining customer behaviour, ( example balances, number of accounts etc.) 1. Clustering: Cluster behavioural data to suitable number of clusters 2. Decision Tree: Using rpart classification tree for generating rules for segmentation using cluster number(cluster id) as target

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

Hi Prof. Williams, thanks for your suggestion. The updated code is below. It turns out it was a matter of displaying the second column in yval to get the number of N and subtracting it from the n column in the frame to get the number of Y remaining in a binary example. once this is added now the function returns the rules along with Y and N count affected by the resulting rule. I am ccing

Dear all, I would like to model the relationship between y and x. y is binary variable, and x is a count variable which may be possion-distribution. I think it is better to divide x into intervals and change it to a factor before calling glm(y~x,data=dat,family=binomail). I try to use rpart. As y is binary, I use "class" method and get the following result. >