Hello,
I have a simple question on rug().
Currently there is only one color possible for the rug.
Is it possible to plot a the rug with different colors, for each rug item ?
Thx.
Bjoern
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mail
Hello,
I encountered the following problem with the parameter scaled in ksvm()
from package kernlab:
[Package kernlab version 0.9-5]
library(kernlab)
> svp =ksvm(x=mydata,y=y,scaled=T)
Using automatic sigma estimation (sigest) for RBF or laplace kernel
> svp
Hello,
I would like to know, whether for the support vector classification function
ksvm()
the response values stored in [EMAIL PROTECTED] are cross validated
outputs/predictions:
Example code from package kernlab, function ksvm:
library(kernlab)
## train a support vector machine
filter <-
ksv
Hello,
I have two questions concerning the RWeka package:
1.) First question:
How can one perform a cross validation, -say 10fold- for a given data
set and given model ?
2.) Second question
What is the correct syntax for the parametrization of e.g. Kernel
Hello,
I remarked that the function
## Default S3 method:
text (x, y = NULL, labels = seq(along = x), adj = NULL,pos = NULL, offset =
0.5, vfont = NULL,cex = 1, col = NULL, font = NULL, ...)
accepts vectors of arguments (of the same length) except for the parameter adj.
When passing a vector of
Hi,
many thanks for the answer.
It ist true, that for example
m1 <- SMO(Species ~ ., data = iris, control = Weka_control( K =
"weka.classifiers.functions.supportVector.PolyKernel"))
m2 <- SMO(Species ~ ., data = iris, control = Weka_control( K =
"weka.classifiers.functions.supportVector
Hello,
I have some trouble in achieving the desired parametrisation
for the weka classifier functions, using the package RWeka.
The problem is, that the functions
result=classifier(formula, data, subset, na.action, control =
Weka_control(mycontrol))
do not seem to be manipulated by the
Hello,
package "arules" is very well suited for mining association rules,
interfacing implementations of "apriori" and "eclat".
What is not realised is an implementation of the Frequent Pattern Tree (Han,
J., Pei, J. and Yiwen, Y. (2000).
There are several free implementations avail
Hello,
I wonder if
image(t(x)[ncol(x):1, ])
can do the job correct!
perhaps this does the job better:
image(t(x)[,nrow(x):1])
Björn
From: Prof Brian Ripley
Date: Fri 27 Aug 2004 - 06:43:50 EST
On Thu, 26 Aug 2004, Glynn, Earl wrote:
> Start with:
>
> > x <- c(1:7,1)
> > dim(x) <- c(2,4)