similar to: Bug(?) in predict.tree() --- Evaluation order of Boolean operators

? data=read.table("http://statcourse.com/research/boston.csv", , sep=",", header = TRUE) ? library(rpart) ? fit=rpart (MV~ CRIM+ZN+INDUS+CHAS+NOX+RM+AGE+DIS+RAD+TAX+ PT+B+LSTAT) predict(fit,data[4,]) plot only reveals part of the tree in contrast to the results on obtains with CART or C5 -------- Original Message -------- Subject: Re: [R] help with rpart From: Sarah

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

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

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)

Hi, I'm trying to use read.fwf temp = read.fwf ("Raw data.txt", widths = c (11, 21, 10, rep (16, 6)) ,skip = 2, n = 2, stringsAsFactors = FALSE, strip.white = TRUE) but no matter what I do the strings are turned into factors. I believe it's the "n=2" parameter that causes the problem as it seems to work without this. Am I missing something? Thanks in advance,

Hi I'm having a problem using the zoo library and I can't see what I'm doing wrong. For example setting up the data > t1 = zoo (matrix (1:12, nrow = 3), order.by = as.Date (c("2008-08-01","2008-08-02","2008-08-03"))) > colnames (t1) = c ("A", "B", "C", "D") > t2 = zoo (matrix (1:12, nrow = 3), order.by =

Hello, and sorry if this is already explained somewhere. I couldn't find anything. R (2.3.1, Windows) seems to perform some kind of lazy evaluation when evaluating defaults in function calls that, at least for me, leads to unexpected results. Consider the following, seemingly equivalent functions: > foo1 <- function(x, y=x) { + x <- 0 + y + } > foo1(1) [1] 0 > foo2

Hi All, I am running quantile regression in a "for loop" starting with 1 variable and adding a variable at a time reaching a maximum of 20 variables. I get the following warning messages after my "for" loop runs. Should I be concerned about these messages? I am building predictive models and am not interested in inference. Warning messages: 1: In

I am having difficulty with evaluation/environment construction for a formula to be evaluated by model.matrix(). Basically, I want to construct a model matrix that first looks in "newdata" for the values of the model parameters, then in "object at data". Here's what I've tried: 1. model.matrix(~f,data=c(newdata,object at data)) -- fails because something (terms()?)

Hi, Using the Rconsole file I can specify the size and location of the Rgui windows on NT. e.g. # Dimensions (in characters) of the console. rows = 51 columns = 100 How can I specify the size of the help windows that popups when I ask for help? e.g. '?help' I would like the popup window to have say rows = 51 and columns = 100, just like the main window but a different location on the

Gabor As someone trying to the rest of my team using Subversion (which I have used for a while, but more as a backup / record of changes), have you a neat / automated way of building a package from a subversion repository? Thanks David Jessop -------------------------- David Jessop Global Head of Quantitative Research UBS Investment Research +44 20 7567 9882 ----- Original Message -----

Hi, I want to call R from within Java, using jri as per http://www.rosuda.org/software/jri/ So I am following the instructions in the README file for JRI 0.2-4. I have run 'sh configure.win' and 'make' and they seemed to be successful. (See below for the output from make, for example.) But when I try 'run.bat rtest' (with and without R command line arguments) the output

Hi, I have a basic question about aligning date labels for the x-axis in an xyplot so that they align with the grid lines from the panel.grid argument. For example, with x <- data.frame( date = seq(as.Date("2005/01/01"), as.Date("2006/06/01"), length.out = 20), value = runif(20)) xyplot(value ~ date, data = x, panel = function(x, y, subscripts, ...) {

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 probably making a simple mistake but I can't see it > X Error: Object "X" not found > exists("X") [1] FALSE > lapply("X", exists) [[1]] [1] TRUE Why is lapply producing true? Is it something to do with the first argument of lapply also being called 'X'? > version _ platform i386-pc-mingw32 arch i386

Hi, Given a function fun <- function(a, b) a + b how do I generate the function 'function(x, y) x + y'? Working from the help files and Bill Venables' R-news article (June 2002), I have tried various permutations with substitute without success. e.g. do.call("substitute", list(fun, list(a = as.name("x"), b = as.name("y")))) Regards, John.

Here's one not-so-straightforward way: > f <- function(a, b) a + b > flist <- as.list(f) > names(flist)[1:2] <- c("x", "y") > flist[[3]] <- do.call("substitute", list(body(f), list(a=as.name("x"), b=as.name("y")))) > g <- as.function(flist) > g function (x, y) x + y HTH, Andy > From: john.gavin at ubs.com

Duncan, I agree completely with "don't use attach"; if I could get all the users of the survival package to agree as well the problem in question would go away :-) I'm thinking about ways to add more effective error surveillance. Your suggestion was not horribly complex and I'll look into it further. My first foray failed because what I want (I think) is the

I've been tracking down a survival problem from R-help today. A short version of the primary issue is reconstructed by the following simple example: library(survival) attach(lung) fit <- coxph(Surv(time, status) ~ log(age)) predict(fit, newdata=data.frame(abe=45)) Note the typo in the last line of "abe" instead of "age". Instead of an error message, this returns