I have downloaded the latest version of R and continue to have the same problem
I did with the previuos version which is that I am unable to run a two-sample
t-test in R-commander. I can run a paired t-test and a one sample t-test but
not a two-sample t-test or one factor ANOVA. The data have be
On Thu, 28 Dec 2006, Weiwei Shi wrote:
> n-fold cv is easy to be implemented in R by yourself. you can create
> 10 folds' samples by yourself and disable cv in rpart. Then you can
> use the same sample sets to compare different methods.
BUT, rpart uses CV to select the cost-complexity parameter,
Try this:
lda(formula(paste(names(iris)[5],"~.")),iris)
You have to create *formula* object from string and pass it to lda().
On 12/29/06, Feng Qiu <[EMAIL PROTECTED]> wrote:
> Hi Gabor:
> Thank you! But it didn't work. Since lda() takes the variable
> name as the input parameter. S
Try this:
idx <- match("Species", names(iris))
lda(iris[-idx], iris[,idx])
On 12/29/06, Feng Qiu <[EMAIL PROTECTED]> wrote:
> Hi Gabor:
> Thank you! But it didn't work.
> Since lda() takes the variable
> name as the input parameter.
That is not correct. It takes the variable itself,
Generally, you can create formula from string:
lda(formula(paste(names(iris)[5],"~.")),iris)
On 12/29/06, Feng Qiu <[EMAIL PROTECTED]> wrote:
> Hi Gabor:
> Thank you! But it didn't work. Since lda() takes the variable
> name as the input parameter. So what I was trying to do is "make
You have to create *formula* object from string and pass it to lda().
See my previous post.
On 12/29/06, Feng Qiu <[EMAIL PROTECTED]> wrote:
> Hi Gabor:
> Thank you! But it didn't work. Since lda() takes the variable
> name as the input parameter. So what I was trying to do is "make th
Hi Gabor:
Thank you! But it didn't work. Since lda() takes the variable
name as the input parameter. So what I was trying to do is "make the name
dynamically". I used sprintf() to generate a variable name, such as "V16".
But it seems that the function doesn't recognize the generated n
You could try something like:
> test <- GAGurine[ c(TRUE,TRUE,TRUE,FALSE), ]
or if you want a random sequential sample
> test <- GAGurine[ sample(c(TRUE,TRUE,TRUE,FALSE)), ]
hope this helps,
-Original Message-
From: [EMAIL PROTECTED] on behalf of Pedro Mardones
Sent: Mon 12/25/2006 10
Farrel Buchinsky gmail.com> writes:
> Clearly g1 and g2 are not exactly the same even though they are the same
> class.
> I can make g2 behave as g1 if i type
> g3<-genotype(g2)
> g3 then behaves as it should.
>
> The issue may have arrisen since g2 is just a subset of a much bigger object
> that
I have been merrily using the genetics package and more specifically have
been using the makeGenotypes and genotypes function. I check my
accomplishments by going
> class(g2)
[1] "genotype" "factor"
and likewise
> class(g1)
[1] "genotype" "factor"
Yet when I execute a command such as allele cou
I'm trying to fit a nested mixed model using lmer and have some
questions about the output and my model formulations.
I have replicate measures on Lines which are strictly nested within
Populations.
(a) So if I want to fit a model where Line is a random effect and
Populations are fixed and the
Dear list,
I'm trying to compare two sets of variables using canonical analysis. My X
variables include 3 climate indices, all continuous. My Y variables describe
a set of 3 environmental moisture measurements. However, I would also like
to include a class variable for habitat type.
My colleau
One can use sapply to get a consistent set of idioms despite the underlying
inconsistency of the functions themselves (using the builtin dataset anscombe):
sapply(anscombe, median)
sapply(anscombe, min)
sapply(anscombe, mean)
On 12/28/06, ivo welch <[EMAIL PROTECTED]> wrote:
> dear r experts:
>
>
dear r experts:
I know its almost surely documented. if d is a data frame
median(d) <-- fails (message is "need numeric data")
min(d) <-- succeeds, but gives a scalar.
mean(d) <-- gives a vector
I am wondering whether this could be changed into something more
consistent. most naturally, the
n-fold cv is easy to be implemented in R by yourself. you can create
10 folds' samples by yourself and disable cv in rpart. Then you can
use the same sample sets to compare different methods.
HTH
weiwei
On 12/28/06, Pedro Ramirez <[EMAIL PROTECTED]> wrote:
> Dear R-list,
>
> I am using the rpart
Hi all,
I'm running a vector-time series model with the vars package. When I
test the univariate and multivariate normality of the residuals using
normality(), I get the results, but also this warning
Warning messages:
1: longer object length
is not a multiple of shorter object length
Dear R-list,
I am using the rpart/mvpart-package for selecting a right-sized regression tree
by
10-fold cross-validation. My question: Is there a possibility to find out for
every
observation in which of the ten folds it is lying? I want to use the same folds
for
validating another regression m
Dear R-list,
I am using the rpart/mvpart-package for selecting a right-sized regression tree
by 10-fold cross-validation. My question: Is there a possibility to find out
for every observation in which of the ten folds it is lying? I want to use the
same folds for validating another regression m
use colClasses to define the columns that you want to read is as date as
'character' to keep the leading zeros.
On 12/28/06, Brian Edward <[EMAIL PROTECTED]> wrote:
>
> Thanks for the help in addressing my query regarding the formatting of
> dates. All the suggestions were very helpful and I seem
What is this error message about and how do I troubleshoot it?
> sqlTables(channel)
Error in .Call("RODBCFetchRows", attr(channel, "handle_ptr"), max, buffsize, :
negative length vectors are not allowed
The channel was created as such
channel <- odbcConnect("Labdata")
I think this is telli
Following the posting guide suggests
RSiteSearch("bitmap")
which leads to
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/28384.html
and
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/65284.html
On Thu, 28 Dec 2006, Michael Wolosin wrote:
> All -
>
> I'm creating som
Thanks for the help in addressing my query regarding the formatting of
dates. All the suggestions were very helpful and I seem to be on track with
solving the problem. Don, you are exactly right in your posting -- that
is, the issue of the leading zero. And that is where my problem is. When I
On Thu, 28 Dec 2006, yangguoyi.ou wrote:
> There is a bug in Matrix package, please check it, thanks!
You didn't say what R code you ran to get that output or why you think it
is wrong.
Let us experiment to see if we can guess what the problem is from the
limited information provided
> x<-t(Ma
Note that you don't really need list or llist in this case:
summarize(x, by = mydata[2:3], FUN = mean, na.rm = TRUE)
Also you could use aggregate:
aggregate(mydata[1], mydata[2:3], mean, na.rm = TRUE)
On 12/28/06, Muenchen, Robert A (Bob) <[EMAIL PROTECTED]> wrote:
> Hi All,
>
>
>
> I'm using
Muenchen, Robert A (Bob) wrote:
> Hi All,
>
>
>
> I'm using the Hmisc summarize function and used list instead of llist to
> provide the by variables. It generated an error message. Is this a bug,
> or do I misunderstand how Hmisc works with lists? The program below
> demonstrates the error mes
I am appreciated that you still remember my question.
I try this one, it works now.
thank you very, very much.
Aimin
At 08:19 PM 12/27/2006, Henrik Bengtsson wrote:
>Hi,
>
>a follow up after realizing that you might not have started the Matlab
>application to listen on port 9998. Try:
>
>Matlab$
Hi All,
I'm using the Hmisc summarize function and used list instead of llist to
provide the by variables. It generated an error message. Is this a bug,
or do I misunderstand how Hmisc works with lists? The program below
demonstrates the error message.
Thanks,
Bob
x<-1:8
group <- c(1,
Dear R-enthusiasts,
I am trying to do a Cochran-Armitage test for trend in R. After consulting
google I found Torsten Hothorn's remark that the 'coin' library could be
used.
lungtumor <- data.frame(dose = rep(c(0, 1, 2), c(40, 50, 48)),
tumor = c(rep(c(0, 1), c(38, 2)),
Bram Kuijper wrote:
> Hi everyone,
>
> I am using the lattice package to plot some simulation results, by using
> the function xyplot(). However, I cannot find a way to plot multiple
> lines within the same xyplot and to have each of the lines be drawn in a
> different color.
>
> This is what
Try this:
library(lattice)
x <- 1:10
xyplot(x/max(x) ~ x, type = "n", col = "blue", ylim = 0:1)
for(i in 1:3) {
trellis.focus("panel", 1, 1)
panel.lines(x, x^i/max(x^i), col = i)
trellis.unfocus()
}
On 12/28/06, Bram Kuijper <[EMAIL PROTECTED]> wrote:
> Hi everyone,
>
Hi everyone,
I am using the lattice package to plot some simulation results, by using
the function xyplot(). However, I cannot find a way to plot multiple
lines within the same xyplot and to have each of the lines be drawn in a
different color.
This is what I am currently doing:
xyplot(a + b
Uwe Ligges schrieb:
>
> Yes: Use the postscript() device explicitly and define the width and
> height in the function call.
>
>
I tried it before but I think I did a mistake:
Does the width and height command not work with
If paper="letter" ?
The plot was cutted at the upper boarder of the "
On 12/28/06, Vladimir Eremeev <[EMAIL PROTECTED]> wrote:
> > Hi everyone:
>
> Please, refer to the chapter 9 of the "R language definition" ($R_HOME/doc/
> manual/r-lang.pdf)
>
> The simplest way is using print() functions.
Note that on Mac and Windows you may need to use flush.console() as well
i
> Hi everyone:
Please, refer to the chapter 9 of the "R language definition" ($R_HOME/doc/
manual/r-lang.pdf)
The simplest way is using print() functions.
Besides the debugging functions described in the manual, you can use the very
powerful package debug, developed by Mark Bravington.
>
[Forwarding to r-help for completeness. /Henrik]
-- Forwarded message --
From: Henrik Bengtsson <[EMAIL PROTECTED]>
Date: Dec 28, 2006 9:45 PM
Subject: Re: [R] Query regarding linking R with Matlab
To: [EMAIL PROTECTED]
Hi.
On 12/28/06, Bhanu Kalyan.K <[EMAIL PROTECTED]> wrote:
Tord Snäll wrote:
> Hi,
> I would like to plot a map of US counties using different colors. map()
> seems to be the function to use, e.g.
> library(maps); map('usa'); map('county', 'colorado', add=T,fill = T,
> col=c(1:5))
> plots Colorado counties using colours 1 to 5.
>
> However, I want each
Knut Krueger wrote:
> R 2.4.0 for Windows
> The following plot appears as a squared window (as all r-plots)
> Not all subtitles are visible, but all subtitle will appear, when
> changing the aspect ratio of the plot window with the mouse to a wide
> format.
> But does not work when using the
See package pixmap and Paul Murrel's talk from useR!2006:
http://www.r-project.org/useR-2006/Slides/Murrell.pdf
Uwe Ligges
Michael Wolosin wrote:
> All -
>
> I'm creating some plots in R that I would like to overlay on images that
> are created outside of R.
>
> I've used "image" before to pl
On 12/28/06, Bhanu Kalyan.K <[EMAIL PROTECTED]> wrote:
> Respected Sir,
>
> It worked.
>
> > open(matlab)
> [1] TRUE
Good.
>
> But however, the 'evaluate' function is not responding.
> When i give the command as:
>
> > res <- evaluate(matlab, "A=1+2;", "B=ones(2,20);")
>
> The R interface is not
Dear R user,
I am new with split-plot designs and I have problems with multiple comparisons.
This data correspond to an split-plot experiment with two replications
(bloque).(Hoshmand, 2006 pp 138). Briefly, the whole-plot factor is
Nitrogen concentration ("nitrogeno") and the subplot factor is
There is a bug in Matrix package, please check it, thanks!
Matlab result:
x =
1 2 3 4 5
6 7 8 910
1112131415
1617181920
2122232425
>> lu(x)
ans =
21. 22.
R 2.4.0 for Windows
The following plot appears as a squared window (as all r-plots)
Not all subtitles are visible, but all subtitle will appear, when
changing the aspect ratio of the plot window with the mouse to a wide
format.
But does not work when using the save as postscript menu item from
Thank you for your answer
> Actually I think there may be a bug here since the axes do not intersect.
>
Should I report this or did you report it already as a possible bug?
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