Displaying 20 results from an estimated 1100 matches similar to: "memory error with rpart()"
2012 Apr 03
1
rpart error message
Hi R-helpers,
I am using rpart package for decision tree using R.We are invoking R
environment through JRI from our java application.Hence, the result of R
command is returned in REXP and we use geterrMessage() to retrieve the
error.
When we execute the following command,
cnr_model<-rpart(as.factor(Species)~Sepal Length+Sepal Width+Petal Length,
method="class",
2005 Jan 27
0
how to evaluate the significance of attributes in tree gr owing
FWIW, I wrote a little function to extract variable importance as defined in
the CART book a while ago. It's rather limited: Only works for regression
problem, and you need to set maxsurrogate=0 and maxcompete=0. It may (or
may not) help you:
varimp.rpart <- function(x) {
dev <- x$frame[, c("var", "dev")]
dev <- dev[dev$var != "<leaf>",
2002 Nov 25
3
Full enumeration, how can this be vectorized
Hi all,
I am currently using the code snippet below to minimize a function which I
wasn't able to minimize properly with optim (cliffy surface, constraints,
poles). Obviously this iteration is probably the poorest way of implementing
such a function and it is expectedly slow. I've already written some
vectorized functions, but I'm afraid that I'm missing the "big
2007 Dec 10
1
Multiple Reponse CART Analysis
Dear R friends-
I'm attempting to generate a regression tree with one gradient predictor and multiple responses, trying to test if change in size (turtle.data$Clength) acts as a single predictor of ten multiple diet taxa abundances (prey.data) Neither rpart or mvpart seem to allow me to do multiple responses. (Or if they can, I'm not using the functions properly.)
> library(rpart)
2002 Mar 14
1
Row-binding factor variables looses the ordered attribute.
Hi everyone,
I've just notice a problem with ordered factor variables. It appears
that row-binding two ordered factors together looses the ordered
attribute. The following example happens both in R1.3.1 and R1.4.1 (on
RedHat 7.2):
> y <- ordered(gl(3,6),labels=1:3)
> y
[1] 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3
Levels: 1 < 2 < 3
> data <-
2002 Apr 29
3
Organizing the help files in a package
Dear all!!
I am using R1.4.1 on windows 98.
I had been trying to organize the package and has already been able to
document some of the functions in to .Rd (R documentation) files. From these
.Rd files I generated plain text files as well as html files.
I have also given the 00Index file in each of the directories:
html/
help/
data/
man/
Problem: I don't get the help using comand
2002 Apr 01
3
svd, La.svd (PR#1427)
(I tried to send this earlier, but it doesnt seem to have come through,
due to
problems on my system)
Hola:
Both cannot be correct:
> m <- matrix(1:4, 2)
> svd(m)
$d
[1] 5.4649857 0.3659662
$u
[,1] [,2]
[1,] -0.5760484 -0.8174156
[2,] -0.8174156 0.5760484
$v
[,1] [,2]
[1,] -0.4045536 0.9145143
[2,] -0.9145143 -0.4045536
> La.svd(m)
$d
[1]
2002 Apr 16
2
multiple plot devices
Hello,
sorry but i found no way or help to work
with multiple graph devices (Rdocs,SearchIndex).
When is use the function only the graphic device of the last variable is open.
How is it possible to let the several plot-device open or save this in a file with different names ?
(win 2000 - R1.4.1)
thanks for advance
& regards,Christian
normal <- function(x) {
par(mfrow=c(2,2))
2012 Jan 06
1
Please help!! How do I set graphical parameters for ploting ctree()
I'm trying to understand how to set graphical parameters for trees created with the party package. For example take the following code:
library(party)
data(airquality)
airq <- subset(airquality, !is.na(Ozone))
airct <- ctree(Ozone ~ ., data = airq,
controls = ctree_control(maxsurrogate = 3))
plot(airct)
My problem is, I've got a ctree that has
2002 Apr 01
1
An introduction to R (PR#1426)
(I sent this earlier, but it seems not to have come through, due to
problems witkh my system)
The following command from appendix A, "a sample session", isnt correct:
contour(x, y, fa, nint=15)
when used R protests:
Warning message:
parameter "nint" couldn't be set in high-level plot() function
it should probably be nlevels, as used a few lines before.
This is
2002 Apr 25
1
Rdbi package and PgSQL
I can use the Rdbi package to connect to a PostGreSQL server fine but
when I use the dbDisconnect(), I get a segmentation error and it throws
me out of R. I'm using RH7.2, R1.4.1, Rdbi 0.1-2, and Rdbi.PgSQL 0.1-2.
Anyone else seen anything like this and have an possible answer?
Andrew Schuh
-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-
r-help mailing
2009 May 08
1
Get (feature, threshold) from Output of rpart() for Stump Tree
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
2004 Oct 30
2
How to add values to an array at any position.
Hi,
How to add values to an array at any position.
Asking because of the following:
e.g.
y<-c(0.1,NaN,0.2,NaN) #or data frame
x<-na.omit(y)
take some columns from x and
do some computation with functions which do not allow NaN 's.
After the computing add NaN's at positions stored in
attr(x,"na.action")
of the result vector.
/E
--
Dipl. bio-chem. Witold Eryk
2013 Jul 14
2
creating dummy variables based on conditions
Hello everyone,
I have a dataset which includes the first three variables from the demo
data below (year, id and var). I need to create the new variable ans as
follows
If var=1, then for each year (where var=1), i need to create a new dummy
ans which takes the value of 1 for all corresponding id's where an instance
of one was recorded. Sample data with the output is shown below.
year
2003 Feb 06
1
rdbi segmentation fault (fwd)
one more bit of information about this problem. If I start R as the
user "postgres" i dont have the segmentation fault.
---------- Forwarded message ----------
Date: Wed, 5 Feb 2003 19:19:39 -0500 (EST)
From: Rafael A. Irizarry <ririzarr at jhsph.edu>
Reply-To: rafa at jhu.edu
To: "R-Help (E-mail)" <r-help at r-project.org>
Subject: rdbi segmentation fault
hi! i
2009 Jul 26
3
Question about rpart decision trees (being used to predict customer churn)
Hi,
I am using rpart decision trees to analyze customer churn. I am finding that
the decision trees created are not effective because they are not able to
recognize factors that influence churn. I have created an example situation
below. What do I need to do to for rpart to build a tree with the variable
experience? My guess is that this would happen if rpart used the loss matrix
while creating
2007 Dec 10
1
What is happening here - rsync can't copy where cp can, Input/Output errors
I am using rsync version 2.6.9 protocol version 29 on a Fedora 7
system to backup files to a network drive.
rsync is getting I/O errors when copying maildir files (I'm not sure
if the error is happening with other files but these are the only ones
I have found so far).
I have done some tests on one particular directory, "cp -R" succeeds
but "rsync -r" fails:-
home#
2011 Jun 21
0
How does rpart computes "improve" for split="information"?? (which seems to be different then the "gini" case)
Hello dear R-help members,
I would appreciate any help in understanding how the rpart function computes
the "improve" (which is given in fit$split) when using the
split='information' parameter.
Thanks to Professor Atkinson help, I was able to find how this is done in
the case that split='gini'. By following the explanation here:
2007 Dec 12
1
Problem with filenames with commas in them
This is a continuation of my previous problem where cp could copy
files whereas rsync couldn't. It turns out that the problem is with
files which have commas in their names, rsync can write the initial
version of the file but it can't check/rewrite them.
Here is the error I get from rsync when trying to overwrite the files
(using --inplace) :-
home# rsync -r --inplace .in
2011 Jun 13
1
In rpart, how is "improve" calculated? (in the "class" case)
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'