Please note that I am on the membership list and inadvertently sent this
request from an incorrect mail account. Please excuse me for any confusion that
I may have caused. The following is the same request that I sent earlier.
I have installed Sweave as recommended.
http://lifeasclay.w
Try adding bty='n'
Michael
On Aug 18, 2012, at 9:21 AM, ARI BEN wrote:
>
>
> Envoyé de mon iPhone
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/po
I have installed Sweave as recommended.
http://lifeasclay.wordpress.com/tag/sweave/.Placing a test.Rnw file, for
instance, in
"/Library/Frameworks/R.framework/Versions/2.15/Resources/library/utils/Sweave"
generates test.tex files that generate pdf files with LaTeX. To be honest,
p
Hello,
Try the following.
# This needs several other packages
# install.packages("XLConnect")
require(XLConnect)
fpattern <- "Book.*.xls?" # pattern for filenames
output.file <- "Test.xls"
lfiles <- list.files(pattern = fpattern)
# Read first worksheet from each file
data.lst <-lapply(lfiles
Envoyé de mon iPhone
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Hey
my name is daniel, i am writing my bachelor thesis and wondering if you can
help me.
i am trying to generate a regression tree via rpart. to
reduce the error of the model i use cross validation, but instead
of reducing the cross validation error (xerror) is increasing the more splits
there ar
Hi to people in the forum,
I'm doing some circular statistics in the package circular. My data is in
hours and in months, and hours are easily recognised in the functions.
However, it is not the case of months having only 12 potential values. I'm
thinking in transform the data by dividing the cir
Karl:
My ignorance of optimization makes any further comments hazardous.
Indeed, my initial reply may already have gone too far, as I my not
either understand you or NM. So I'm just going to shut up.
-- Bert
On Sat, Aug 18, 2012 at 9:33 AM, Karl Ove Hufthammer wrote:
> You’re right that the ste
You’re right that the step size should be effectively adjusted using
alpha, beta and gamma in later iterations, but the problem is that the
values used for the first simplex generated depends on the differences
between the initial values, which makes no sense, as this make
optimisation problem not
Well, I'm no optimization guru, but a quick reading of Wikipedia said
tha step size depends on the initial value configuration and is then
"adjusted" by the algorithm using alpha, beta and gamma scaling
parameters thru the optimization. So it seems that it is supposed to
work exactly as you describ
On 17.08.2012 14:02, Michael Meyer wrote:
Do
splits <- strsplit(filePath,"file://%22,fixed=true)[[1/]]
fileName <- splits[length(splits)])
And what is your question now?
If this is an answer, and we really appreciate people who try to help,
please
a) cite the question and
b) provide answ
On 17.08.2012 19:32, Fg Nu wrote:
Uwe,
I did give you the output from R CMD INSTALL (or at least, what I thought was
the relevant part).
Here is the output in toto:
* installing to library 'C:/programming/r/revolutions/R-2.14.2/library'
* installing *source* package 'KernGPLM' ...
** libs
Hello all! Thank you for your advice.
The solution finally was like this:
first install filehash package
and then library(tikzDevice)
Just in case any one needs it.
Regards
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing
HI,
Try this:
m[m==0]<-NA
apply(m,1,mean,na.rm=T)
#[1] 2.5 NaN 1.0
A.K.
- Original Message -
From: Yingwei Lee
To: r-help@r-project.org
Cc:
Sent: Friday, August 17, 2012 11:28 PM
Subject: [R] Row means of matrix.
Hi all,
I'm just having trouble getting row means of a matrix
Dear all,
I’m having some problems getting optim with method="Nelder-Mead" to work
properly. It seems like there is no way of controlling the step size,
and the step size seems to depend on the *difference* between the
initial values, which makes no sense. Example:
f=function(xy, mu1, mu2) {
On 08/18/2012 03:32 AM, Bert Gunter wrote:
Folks:
...
So contrary opinions
cheerily welcomed. But perhaps these comments might be helpful to
those who have been "bitten" by factors or just wonder what all the
fuss is about.
I tend to use stringsAsFactors=FALSE quite a bit, as I am often
manipul
On Fri, Aug 17, 2012 at 07:34:35PM +0100, Rui Barradas wrote:
> Hello,
>
> No, factors may use less memory. System dependent?
>
> > x <-sample(c("small","medium","large"),1e4,rep=TRUE)
> > y <- factor(x)
> > object.size(x)
> 80184 bytes
> > object.size(y)
> 40576 bytes
> >
> > sessionInfo()
> R v
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