similar to: rpart help please

Displaying 20 results from an estimated 10000 matches similar to: "rpart help please"

2002 Oct 17
1
manova with Error?
Let's say I have a within-subject experiment with 2 observables, obs1 and ob2 and 2 independent factors, fac1 and fac2. I can do summary( aov( obs1~fac1*fac2 + Error(Subject/(fac1*fac2)) ) ) summary( aov( obs2~fac1*fac2 + Error(Subject/(fac1*fac2)) ) ) to test the 2 observables separately. > summary( fit<-manova( cbind(obs1,obs2)~fac1*fac2 + Error(Subject/(fac1*fac2)) ) ) gives
2004 Mar 18
1
help with aov
Hi all, Suppose the following data and the simple model y<-1:12+rnorm(12) fac1<-c(rep("A",4),rep("B",4),rep("C",4)) fac2<-rep(c("D","C"),6) dat<-data.frame(y,fac1,fac2) tmp<-aov(y~fac1+fac2,dat) the command tmp$coeff gives the fllowing results : (Intercept) fac1B fac1C fac2D 3.307888 2.898187 7.409010
2001 Jul 02
1
text.rpart: Unwanted NA labels on terminal nodes (PR#1009)
Brian The following (which is new to rw1030) occurs with both Windows 98 & Windows ME. I have not tested behaviour under Unix or Linux, but I expect it is no different. text.rpart() prints unwanted NAs (presumably in the splitting criterion position) on terminal nodes. Criterion <- factor(paste("Leaf", 1:5)) Node <- factor(1:5)
2008 Dec 20
1
How test contrasts/coefficients of Repeated-Measures ANOVA?
Hi all, I'm doing a Repeated-Measures ANOVA, but I don't know how to test its contrasts or where to find the p-values of its coefficients. I know how to find the coefficient estimates of a contrast, but not how to test these estimates. First I do something like: y.aov <- aov(y ~ fac1 * fac2 + Error(subj/(fac1 * fac2)), data=data) Then, with coef(y.aov) I get the coefficients
2008 Sep 09
1
How do I compute interactions with anova.mlm ?
Hi, I wish to compute multivariate test statistics for a within-subjects repeated measures design with anova.mlm. This works great if I only have two factors, but I don't know how to compute interactions with more than two factors. I suspect, I have to create a new "grouping" factor and then test with this factor to get these interactions (as it is hinted in R News 2007/2), but
2008 Jul 24
2
factor question
Hi! Suppose I have a factor: > fac1 <- factor(rep(c("dog","cat","tree"),c(2,3,4))) > fac1 [1] dog dog cat cat cat tree tree tree tree Levels: cat dog tree > length(fac1) [1] 9 Now I want to get rid of the dogs: > fac1 <- fac1[3:9] > fac1 [1] cat cat cat tree tree tree tree Levels: cat dog tree > How should I remove the dog from the
2018 Aug 14
2
Xenial rpart package on CRAN built with wrong R version?
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
2008 Jul 22
2
rpart$where and predict.rpart
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
2002 Jun 04
2
Scaling on a data.frame
Hey, hopefully there is an easy way to solve my problem. All that i think off is lengthy and clumsy. Given a data.frame d with columns VALUE, FAC1, FAC2, FAC3. Let FAC1 be something like experiment number, so that there are exactly the same number of rows for each level of FAC1 in the data.frame. Now i would like to scale all values according to the center of its experiment. So i can apply s
2009 Jun 09
3
rpart - the xval argument in rpart.control and in xpred.rpart
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
2014 Aug 13
1
Request to review a patch for rpart
Dear list For my work, it would be helpful if rpart worked seamlessly with an empty model: library(rpart); rpart(formula=y~0, data=data.frame(y=factor(1:10))) Currently, an unrelated error (originating from na.rpart) is thrown. At some point in the near future, I'd like to release a package to CRAN which uses rpart and relies on that functionality. I have prepared a patch (minor
2001 Aug 12
2
rpart 3.1.0 bug?
I just updated rpart to the latest version (3.1.0). There are a number of changes between this and previous versions, and some of the code I've been using with earlier versions (e.g. 3.0.2) no longer work. Here is a simple illustration of a problem I'm having with xpred.rpart. iris.test.rpart<-rpart(iris$Species~., data=iris[,1:4], parms=list(prior=c(0.5,0.25, 0.25))) + ) >
2003 Jun 02
1
Ploting rpart objects / namespace problems
I've written a small adaptation of the text.rpart function of the rpart package to better suite my tree presentation needs (I basically left the code untouched, only changing some numbers relating to label positioning). When I changed to version 1.7.0 this function stopped working printing the error : couldn't find function "rpartco" As far as I understand this has to do
2006 Aug 09
2
How to draw the decision boundaries for LDA and Rpart object
Hello useR, Could you please tell me how to draw the decision boundaries in a scatterplot of the original data for a LDA or Rpart object. For example: > library(rpart) >fit.rpart <- rpart(as.factor(group.id)~., data=data.frame(Data) ) How can I draw the cutting lines on the orignial Data? Or is there any built in functions that can read the rpart object 'fit.rpart' to do
2007 Oct 30
2
flexible processing
Hello, unfortunately, I don't know a better subject. I would like to be very flexible in how to process my data. Assume the following dataset: par1 <- seq(0,1,length.out = 100) par2 <- seq(1,100) fac1 <- factor(rep(c("group1", "group2"), each = 50)) fac2 <- factor(rep(c("group3", "group4", "group5", "group6"), each =
2005 Sep 24
1
rpart Error in yval[, 1] : incorrect number of dimensions
I tried using rpart, as below, and got this error message "rpart Error in yval[, 1] : incorrect number of dimensions". Thinking it might somehow be related to the large number of missing values, I tried using complete data, but with the same result. Does anyone know what may be going on, and how to fix it? I have traced two similar error messages in the Archive, but following the
2007 Oct 17
1
passing arguments to functions within functions
Dear R Users, I am trying to write a wrapper around summarize and xYplot from Hmisc and am having trouble understanding how to pass arguments from the function I am writing to the nested functions. There must be a way, but I have not been able to figure it out. An example is below. Any advice would be greatly appreciated. Thanks, Dan # some example data df=expand.grid(rep=1:4,
2004 Sep 06
1
rpart problem
Dear all, I am having some trouble with getting the rpart function to work as expected. I am trying to use rpart to combine levels of a factor to reduce the number of levels of that factor. In exploring the code I have noticed that it is possible for chisq.test to return a statistically significant result whilst the rpart method returns only the root node (i.e. no split is made). The following
2009 May 22
1
regrouping factor levels
Hi all, I had some trouble in?regrouping factor levels for a variable. After some experiments, I have figured out how I can recode to modify the factor levels. I would now like some help to understand why some methods work and others don't. Here's my code : rm(list=ls()) ###some trials in recoding factor levels char<-letters[1:10] fac<-factor(char) levels(fac) print(fac) ##first
2018 Aug 14
2
Xenial rpart package on CRAN built with wrong R version?
I?m not using the PPA, but the CRAN mirror from https://cran.r-project.org/bin/linux/ubuntu/#installation: deb https://cloud.r-project.org/bin/linux/ubuntu xenial-cran35/ On 14 Aug 2018, at 12:25, George N. White III <gnwiii at gmail.com<mailto:gnwiii at gmail.com>> wrote: qOn Tue, 14 Aug 2018 at 05:04, Ulrich KELLER <ulrich.keller at uni.lu<mailto:ulrich.keller at