Displaying 20 results from an estimated 4000 matches similar to: "How to show non user defined data set such as cu.summary (from rpart)?"
2007 Feb 18
3
User defined split function in rpart
Dear R community,
I am trying to write my own user defined split function for rpart. I read
the example in the tests directory and I understand the general idea of the
how to implement user defined splitting functions. However, I am having
troubles with addressing the data frame used in calling rpart in my split
functions.
For example, in the evaluation function that is called once per node,
2012 Oct 23
1
Compiling samba4 hangs at [1815/3978] Compiling librpc/ndr/ndr_basic.c
Hi,
I have tried both RC4 and from the repository but I can't seem to get
samba4 to compile. I have rebuilt the OS (Centos 6.3) from scratch and I am
still having the same issue. I get:
WAF_MAKE=1 ./buildtools/bin/waf build
Waf: Entering directory `/opt/samba-master/bin'
Selected embedded Heimdal build
[ 133/3978] Generating VERSION
[ 168/3978] Generating smbd/build_options.c
[1815/3978]
2003 Jun 17
1
User-defined functions in rpart
This question concerns rpart's facility for user-defined functions that
accomplish splitting.
I was interested in modifying the code so that in each terminal node,
a linear regression is fit to the data.
It seems that from the allowable inputs in the user-defined functions,
that this may not be possible, since they have the form:
function(y, wt, parms) (in the case of the
2007 Jan 03
1
User defined split function in Rpart
Dear all,
I'm trying to manage with user defined split function in rpart
(file rpart\tests\usersplits.R in
http://cran.r-project.org/src/contrib/rpart_3.1-34.tar.gz - see bottom of
the email).
Suppose to have the following data.frame (note that x's values are already
sorted)
> D
y x
1 7 0.428
2 3 0.876
3 1 1.467
4 6 1.492
5 3 1.703
6 4 2.406
7 8 2.628
8 6 2.879
9 5 3.025
10 3 3.494
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
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
2005 Aug 26
1
Help in Compliling user -defined functions in Rpart
I have been trying to write my own user defined function in Rpart.I
imitated the anova splitting rule which is given as an example.In the
work I am doing ,I am calculating the concentration index(ci) ,which
is in between -1 and +1.So my deviance is given by
abs(ci)*(1-abs(ci)).Now when I run rpart incorporating this user
defined function i get the following error message:
Error in
2011 Aug 06
4
compiling buoh comic reader for C-6
Hi all!
I'm trying to build the buoh comic reader for centos 6 and my head is
getting tired of repeated forceful contact with the wall.
I built it on 5.6 without undue problems, but somehow 6.0 is beating me.
(and no, the 5.6 binary doesn't work, at least not without a lot of
futzing around, on 6.0. tried that first.)
before I give gory details, has anyone already done this, from whom
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
2004 May 13
2
R 1.9.0 and pred.rpart
I have just upgraded from R 1.7.3 to R 1.9.0 and have found that the
predict function no longer works for rpart:
> predict(hmmm,sim3[1:10,])
Error in predict.rpart(hmmm, sim3[1:10, ]) :
couldn't find function "pred.rpart"
I have re-installed the rpart package to no avail. Any ideas?
Giles Hooker
2011 Sep 07
2
rpart/tree issue
I am trying to create a classification tree using either tree or rpart
but when it comes to plotting the results the formatting I get is
different than what I see in all the tutorials. What I would like to
see is the XX/XX format but all I get is a weird decimal value. I was
also wondering how you know which is yes and which is no in each leaf of
the tree? Is yes always on the left?
2011 Aug 25
2
rpart: plot without scientific notation
While I'm very pleased with the results I get with rpart and
rpart.plot, I would like to change the scientific notation of the
dependent variable in the plots into integers. Right now all my 5 or
more digit numbers are displayed using scientific notation.
I managed to find this:
http://tolstoy.newcastle.edu.au/R/e8/help/09/12/8423.html
but I do not fully understand what to change, and to
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
2007 Feb 15
2
Does rpart package have some requirements on the original data set?
Hi,
I am currently studying Decision Trees by using rpart package in R. I
artificially created a data set which includes the dependant variable
(y) and a few independent variables (x1, x2...). The dependant variable
y only comprises 0 and 1. 90% of y are 1 and 10% of y are 0. When I
apply rpart to it, there is no splitting at all.
I am wondering whether this is because of the
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
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 Aug 15
3
How to reinstall rpart?
After entering ?library(rpart)?, I tried to plot an existing rpart tree, and
got this error message: Error: couldn't find function "plot.rpart".
However, ??plot.rpart? does bring up the help for the function. The same
things occur for text.rpart, although print(my.tree) does work.
So, I tried to re-install rpart using Packages | Install from CRAN, but
then I get this
2011 Apr 08
4
Rpart decision tree
Dear useRs:
I try to plot an rpart object but cannot get a nice tree structure plot. I
am using plot.rpart and text.rpart (please see below) but the branches that
connect the nodes overlap the text in the ellipses and rectangles. Is there
a way to get a clean nice tree plot (as in the Rpart Mayo report)? I work
under Windows and use R2.11.1 with rpart version 3.1-46.
Thank you.
Tudor
...
2012 Jul 06
2
Plotting rpart trees with long list of class members
I have a class with 732 members, so using rpart.plot is giving me a tiny plot
in the middle of the window. Is there a good way to modify the plot, or
replace the long list with something like "group1"?
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2010 Dec 13
2
rpart.object help
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