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Hello, I'm using mvpart option xv="1se" to compute a regression tree of good size with the 1-SE rule. To better understand 1-SE rule, I took a look on its coding in mvpart, which is : Let z be a rpart object , xerror <- z$cptable[, 4] xstd <- z$cptable[, 5] splt <- min(seq(along = xerror)[xerror <= min(xerror) + xvse * xstd]) I interprete this as following: the

Hi Listers I need to calculate and then plot a frequency histogram of the best tree calculated using the 1-se rule. I have included some code that has worked well for me in the past but it was only for selecting the minimum cross-validation error. I include the code for my model, some relevant output and the code for selecting and plotting the frequency histogram of minimum xerror. Here is the

I am running some models (for the first time) using rpart and am getting results I don't know how to interpret. I'm using cross-validation to prune the tree and the results look like: Root node error: 172.71/292 = 0.59148 n= 292 CP nsplit rel error xerror xstd 1 0.124662 0 1.00000 1.00731 0.093701 2 0.064634 1 0.87534 1.08076 0.092337 3 0.057300 2

Hi, I've searched online, in a few books, and in the archives, but haven't seen this. I believe that xerror is scaled to rel error on the first split. After fitting an rpart object, is it possible with a little math to determine the percentage of true classifications represented by a xerror value? -seth -- View this message in context:

Full_Name: Gregory R. Warnes Version: R 1.3.1 OS: Solaris 2.8 Submission from: (NULL) (192.77.198.200) rpart library version 3.1-2 Error message: > plotcp(fit.thirds.1,ylim=c(0.7,1.5)); Error in plot.default(ns, xerror, axes = FALSE, xlab = "cp", ylab = "X-val Relative Error", : formal argument "ylim" matched by multiple actual arguments > This can be

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 All, I have some questions on using library rpart. Given my data below, the plotcp gives me increasing 'xerrors' across different cp's with huge xstd (plot attached). What causes the problem or it's not a problem at all? I am thinking 'xerror's should be decreasing when 'cp' gets smaller. Also what the 'xstd' really tells us? If the error bars for

Dear R-list members, I am doing a CART analysis in R using the rpart function in the rpart package: Phrag.rpart=rpart(PhragDiff~., data = Phrag, method="anova", xval=10). I used the xerror values in the CP table to prune the tree to 4 nsplits: CP nsplit rel error xerror xstd 1 0.098172 0 1.00000 1.02867 0.12768 2 0.055991 3 0.70548 1.00823 0.12911 3

Dear r-help-list: If I use the rpart method like cfit<-rpart(y~.,data=data,...), what kind of tree is stored in cfit? Is it right that this tree is not pruned at all, that it is the full tree? If so, it's up to me to choose a subtree by using the printcp method. In the technical report from Atkinson and Therneau "An Introduction to recursive partitioning using the rpart

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 All, I have to say upfront that I am a complete neophyte when it comes to programming. Nevertheless I enjoy the challenge of using R because of its incredible statistical resources. My problem is this .........I am running a regression tree analysis using "rpart" and I need to run the calculation repeatedly (say n=50 times) to obtain a distribution of results from which I will pick

Hello: I am using rpart and would like more background on how the splits are made and how to interpret results - also how to properly use text(.rpart). I have looked through Venables and Ripley and through the rpart help and still have some questions. If there is a source (say, Breiman et al) on decision trees that would clear this all up, please let me know. The questions below pertain to a

HI, Dear R community, How to extract the variables actually used in tree construction? I want to extract these variables and combine other variable as my features in next step model building. > printcp(fit.dimer) Classification tree: rpart(formula = outcome ~ ., data = p_df, method = "class") Variables actually used in tree construction: [1] CT DP DY FC NE NW QT SK TA WC WD WG WW

Hello list, I'm having a problem with custom functions in rpart, and before I tear my hair out trying to fix it, I want to make sure it's actually a problem. It seems that, when you write custom functions for rpart (init, split and eval) then rpart no longer cross-validates the resulting tree to return errors. A simple test is to use the usersplits.R function to get a simple, custom

Dear all, I'm getting an error in one of the stock examples in the 'mvpart' package. I tried: require(mvpart) data(spider) fit3 <- rpart(gdist(spider[,1:12],meth="bray",full=TRUE,sq=TRUE)~water+twigs+reft+herbs+moss+sand,spider,method="dist") #directly from ?rpart summary(fit3) ...which returned the following: Error in apply(formatg(yval, digits - 3), 1,

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

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

Il 21/10/2016 17:20, m.roth at 5-cent.us ha scritto: > John R Pierce wrote: >> On 10/21/2016 2:03 AM, Alessandro Baggi wrote: >>> >>> My ssds are failing? >> >> SSD's wear out based on writes per block. they distribute those >> writes, but once each block has been written X number of times, they are >> no longer reliable. >> >>

Dear useRs, I may be temporarily (I hope :-)) confused, and I hope that someone can answer this question that bugs me at the moment: In the CP table of rpart, I thought the following equation should hold: rel error = rel error(before) - (nsplit - nsplit(before)) * CP(before), where (before) always denotes the entry in the row above. While this equation holds for many rows of the CP tables

Hello, When trying to run mvpart either specifying my own parameters or using the defaults, I get the following error: Error in all(is.leaf) : unused argument(s) (c(FALSE, TRUE, FALSE, FALSE, TRUE, TRUE, TRUE)) As far as I can tell, is.leaf is part of the dendrogam package, so I'm assuming there's some problem with the graphical parameters. However running same formula and data