\r with cat performs a carriage return and line feed in R-GUI.
for (i in 1:10) { cat(\rFoo,i) }
Foo 1
Foo 2
Foo 3
Foo 4
Foo 5
Foo 6
Foo 7
Foo 8
Foo 9
Foo 10
_
Performs a carriage return only on the console.
for (i in 1:10) { cat(\rFoo,i) }
Foo 10 _
Cheers,
Martin
On 18/03/2006, at
G'day Rolf,
RT == Rolf Turner [EMAIL PROTECTED] writes:
RT Is there a way of calculating the derivative of a function
RT returned by splinefun()?
Yes, of course. :)
As somebody else once said: this is R, anything can be done, the
question is just how easy it is to do so.
It may depend
Hello, I would like to call a function that can take infinite time to
be executed in some circumstances (which can not be easily detected).
So, I would like that once the function is being executed for more
than two seconds it is stopped. I have found documentation for timers
but i did not found
Hello!
I am testing if my data are distributed under Laplace's distribution,
which I managed to do with:
library(rmutil)
ks.test(jitter(x), plaplace, mean(x), sd(x))
Nevertheless, I am trying to bootstrap ks.test without any success.
Something is wrong in my commands:
data =
Gabor Grothendieck wrote:
If you are just looking for something simple that may be good enough
then assign the largest one to group 1, the second largest to group 2,
..., the 8th largest to group 8 and then start over again with group 1
and so on.
# test data
set.seed(1)
x - sample(100,
On 3/18/06, Dan Bolser [EMAIL PROTECTED] wrote:
Gabor Grothendieck wrote:
If you are just looking for something simple that may be good enough
then assign the largest one to group 1, the second largest to group 2,
..., the 8th largest to group 8 and then start over again with group 1
and
BeT == Berwin A Turlach [EMAIL PROTECTED]
on Sat, 18 Mar 2006 17:08:47 +0800 writes:
BeT G'day Rolf,
RT == Rolf Turner [EMAIL PROTECTED] writes:
RT Is there a way of calculating the derivative of a function
RT returned by splinefun()?
BeT Yes, of course. :)
BeT As
Hi,
I have a function of the second grade, with 2 parameters:
y~A^2 + A + B^2 + B
The response y is a measurement for the precision of the analytical method,
where A en B are method parameters. As its neccesary to keep the precision of
the analytical methad as good as possible, its usefull
Mathias Ehrich wrote:
Hi all,
this might be a very stupid question, but I tried for several hours now
and I'm helpless. I'm running windows and R 2-2-1. Recently, every time
when I have an error in my code a message box pops up telling me the
error message. The code won't continue before I
pau carre wrote:
Hello, I would like to call a function that can take infinite time to
be executed in some circumstances (which can not be easily detected).
So, I would like that once the function is being executed for more
than two seconds it is stopped. I have found documentation for timers
Where did you find the function forecast? It is NOT a standard R
function.
I just tried 'methods(class=ar)', which produced the following:
[1] predict.ar* print.ar*
Non-visible functions are asterisked
The documentation for 'predict.ar' includes the
Have you considered the Bioconductor project (www.bioconductor.org)?
If you are not already familiar with their capabilities, I suggest you
review their capabilities and consider posting a question to their
listserve if you don't find the answer without that.
hope this
Hi everybody,
let us assume i have the following matrixX and vectorY
matrixX - runif(100)
dim(matrixX) - c(10,10)
vectorY - as.matrix(as.character(seq(1,10)))
if I define:
subsample-c(2)
i can extract the rows from matriX based on the elements in vectorY which
are listed in subsample
On 3/18/06, Uwe Ligges [EMAIL PROTECTED] wrote:
pau carre wrote:
Hello, I would like to call a function that can take infinite time to
be executed in some circumstances (which can not be easily detected).
So, I would like that once the function is being executed for more
than two seconds
matrixX[vectorY %in% subsample,]
?%in%
Roberto Furlan wrote:
Hi everybody,
let us assume i have the following matrixX and vectorY
matrixX - runif(100)
dim(matrixX) - c(10,10)
vectorY - as.matrix(as.character(seq(1,10)))
if I define:
subsample-c(2)
i can extract the rows from matriX
daer list,
i want to save a result as mentioned above.
for (i in 1:100) {
a[i] - write.table(a[i], C:/.../resulti)
}
how is the correct way to save the results?
thank you
stefan
__
R-help@stat.math.ethz.ch mailing list
On 3/17/2006 9:44 AM, Jeffrey Racine wrote:
Hi, and thanks in advance for your time.
Background - I am working on a package and wish to have a routine's
progress reported. The routine can take some time, and I would like to
inform the user about the routine's progress. I have scoured the
Is there any possibility to divide too long text in a plot to two or more
lines, when using labels-parameter in the text()-command?
Here is an example picture:
http://users.utu.fi/attenka/253.jpeg
My example script is something like this:
text(1,0.7,labels=Chordnames[fnid(pcs%%12)]) #
See ?sprintf and/or ?paste.
/Henrik
On 3/18/06, Stefan Semmeling [EMAIL PROTECTED] wrote:
daer list,
i want to save a result as mentioned above.
for (i in 1:100) {
a[i] - write.table(a[i], C:/.../resulti)
}
how is the correct way to save the results?
thank you
stefan
Hi,
I have read about the use of symbols() to draw circles of different
sizes, but I have not been able to find out how to add a legend to
such a graph, legend that would display some specific sizes and their
meaning.
Before finding the symbols function in Paul Murrell's book, I had
Hi All,
I have the following adjacency matrix for a directed graph:
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]00000000
[2,]00000000
[3,]10000000
[4,]00100000
Denis Chabot wrote:
Hi,
I have read about the use of symbols() to draw circles of different
sizes, but I have not been able to find out how to add a legend to
such a graph, legend that would display some specific sizes and their
meaning.
Before finding the symbols function in Paul
On Sat, 18 Mar 2006, Denis Chabot wrote:
Hi,
I have read about the use of symbols() to draw circles of different
sizes, but I have not been able to find out how to add a legend to
such a graph, legend that would display some specific sizes and their
meaning.
Before finding the
On Sat, 18 Mar 2006, Atte Tenkanen wrote:
Is there any possibility to divide too long text in a plot to two or more
lines, when using labels-parameter in the text()-command?
Here is an example picture:
http://users.utu.fi/attenka/253.jpeg
My example script is something like this:
On 3/18/06, Atte Tenkanen [EMAIL PROTECTED] wrote:
Is there any possibility to divide too long text in a plot to two or more
lines, when using labels-parameter in the text()-command?
Here is an example picture:
http://users.utu.fi/attenka/253.jpeg
My example script is something like this:
Thank you for answering, both Roger and Henrik
There are only 5 cases which need divisions, so I inserted \n to them and
now everything works.
Atte
On 3/18/06, Atte Tenkanen [EMAIL PROTECTED] wrote:
Is there any possibility to divide too long text in a plot to two or
more
lines, when using
You might find the 2nd part of the following response useful
https://stat.ethz.ch/pipermail/r-help/2006-March/090611.html
And if you want to RTFM, I guess sections 2.5, 2.7, 5.1, 5.2 of
http://cran.r-project.org/doc/manuals/R-intro.html might be useful.
PS:
1) R-help is designed for and by
Do you by any chance want to sample from each group equally to get an
equal representation matrix ? Here is an example of the input :
mydf - data.frame( value=1:100, value2=rnorm(100),
grp=rep( LETTERS[1:4], c(35, 15, 30, 20) ) )
which has 35 observations from A, 15 from B,
Berton,
What you say makes sense and helps. I could do a parametric
bootstrapping. However, I cannot make predict.nlme work at all, even
without the se's. It would save me a lot of time to be able to use
the predict as the statistic in boot.
Does predict.nlme work at all? It is a listed
29 matches
Mail list logo
