similar to: "rpart" or "tree" function issue

I run the following code: library(rpart) data(kyphosis) fit <- rpart(Kyphosis ~ ., data=kyphosis) plot(fit) text(fit, use.n=TRUE) The text labels represent the count of each class at the leaf node. Unfortunately, the numbers are rounded and in scientific notation rather than the exact number of examples sorted by that node in each class. The plot is supposed to look like

I ran this code (several times) from the Quick-R web page ( http://www.statmethods.net/advstats/cart.html) but my cross-validation errors increase instead of decrease (same thing happens with an unrelated data set). Why does this happen? Am I doing something wrong? # Classification Tree with rpart library(rpart) # grow tree fit <- rpart(Kyphosis ~ Age + Number + Start,

Hi, I have a dataset that has 2 groups of samples. For each sample, then response measured is the number of success (no.success) obatined with the number of trials (no.trials). So a porportion of success (prpop.success) can be computed as no.success/no.trials. Now the objective is to test if there is a statistical significant difference in the proportion of success between the 2 groups of

Greetings R world, I know some version of the this question has been asked before, but i need to save the output of a loop into a data frame to eventually be written to a postgres data base with dbWriteTable. Some background. I have developed classifications models to help identify problem accounts. The logic is this, if the model classifies the record as including variable X and it turns out

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

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 wish to extract the terms from an rpart object. Specifically, I would like to be able to know what is the response variable (so I could do some manipulation on it). But in general, such a method for rpart will also need to handle a "." case (see fit2) Here are two simple examples: fit1 <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis) fit1$call fit2 <-

Hi, I am fairly new to GAM and started using package mgcv. I like the fact that optimal smoothing is automatically used (i.e. df are not determined a priori but calculated by the gam procedure). But the mgcv manual warns that p-level for the smooth can be underestimated when df are estimated by the model. Most of the time my p-levels are so small that even doubling them would not result

Greetings All, I am looking to write a parametrized Rscript that will accept a variable name(that also is the name of the flat file), transform the data into a data frame and preform various modeling on the structure and save the output and plot of the model. In this example i am using a rpart decision tree. The only problem i am having is integrating the parameter into the internal object name

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 <-

Hi R-helpers. I have the following problem: I would like to apply my function gain(df,X,A) to a list of arguments. df is a data frame X,A are the varibales od data frame. When I do > gain(kyphosis,"Kyphosis",c("Start","Number")) [1] "Start" "Number" I get the following error... Error in unique.default(x) : unique() applies only to vectors I

Dear R-users I am using RPART package to get regression trees. However having trouble getting the text function to put the full splitting rule number on the plot, instead to puts it in scientific notation. When a covariate has 1e4 or greater number of digits then the splitting rule number displayed on the plot is in scientific notation. But print.rpart displays the splitting rules in full.

I am using the rpart package and seem to have trouble with data sets that have columns with no data. I look at the column data in R and all values are NA. When this occurs, I get nothing back from the rpart function. Is there a way to get the rpart package to ignore these columns, without knowing what columns are empty? I have tried the na.action=na.omit and na.action=na.exclude, but neither one

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

I am working with a non-parametic smoothing operation using a Generalized Additive Model. It is a bivariate data set. I know how to do the smooth, and out comes a nice smooth curve. Now I want to find the value of the smoothed curve for several values of x (the abscissa). This can be done (please correct me if I am wrong) by using the predict.gam function. You feed the predict.gam function a

Hello, Due to exploration of the JIT capabilities offered through the {compiler} package, I came by the fact that using enableJIT(2) can *slow* the rpart function (from the {rpart} package) by a magnitude of about 10 times. Here is an example code to run: library(rpart) require(compiler) enableJIT(0) # just making sure that JIT is off # We could also use enableJIT(1) and it would be fine fo

I tried multiple imputation with aregImpute() and fit.mult.impute() in Hmisc 3.8-3 (June 2010) and R-2.12.1. The warning message below suggests that summary(f) of fit.mult.impute() would only use the last imputed data set. Thus, the whole imputation process is ignored. "Not using a Design fitting function; summary(fit) will use standard errors, t, P from last imputation only. Use

Hi, I am using rpart package to fit classification trees. library(rpart) fit <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis) plot(fit,uniform=T) text(fit, use.n=TRUE) But I am unable to label the branches (not the nodes) of the tree. Can somebody help me out in this? Thank you, Regards Utkarsh Singhal | Amba Research Ph +91 80 3980 8017 | Mob +91 99 0295 8815

Hi, is it possible to put a summary of an rpart-Object into a tktext-window? Here is what I'm trying to do: fit <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis) tt <- tktoplevel() tex <- tktext(tt) tkpack(tex) tkinsert(tex, "end", summary(fit)) But since the summary of an object is a list, I always get back the following error-message: cannot handle object of

Simple example: # Classification Tree with rpart library(rpart) # grow tree fit <- rpart(Kyphosis ~ Age + Number + Start, method="class", data=kyphosis) Now I would like to know how can I measure the "importance" of each of my three explanatory variables (Age, Number, Start) in the model? If this was a regression model, I could have looked at p values from the