Hi,
Can anyone suggest some short term statistical courses using R that one
could take? Name or Links will be much appreciated.
many thanks,
j
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On Thu, 16 Oct 2008, Michael Just wrote:
Hello,
I have never used source and I am a R beginner.
Not a 'beginner' R-help poster, though. (32 posts this month so far under
this name.)
If I have text file that contains 1,000's of lines of code. Can I use
source to bring this code into R a
On Sun, 12 Oct 2008, Peter Dalgaard wrote:
Prof Brian Ripley wrote:
Please do RTFM, for the help says
df: degrees of freedom (> 0, maybe non-integer). 'df = Inf' is
allowed. For 'qt' only values of at least one are currently
supported.
On Sun, 12 Oct 2008, Enrico R
How about this:
> df1=data.frame(v1=c(1,1,2,3,2,4,1))
> df1$v2 <- ave(df1$v1,df1$v1,FUN=length)
> df1
v1 v2
1 1 3
2 1 3
3 2 2
4 3 1
5 2 2
6 4 1
7 1 3
On Thu, Oct 16, 2008 at 1:27 PM, Lijiang Guo <[EMAIL PROTECTED]> wrote:
> Dear R-helpers,
>
> I have a data frame with 3 variables
Hello,
I have never used source and I am a R beginner. If I have text file that
contains 1,000's of lines of code. Can I use source to bring this code into
R and execute the code? Does it run the code one line at a time? Is there a
best way to setup source() for maximum effieciency? After reading ?
On Oct 16, 2008, at 1:27 AM, Lijiang Guo wrote:
Dear R-helpers,
I have a data frame with 3 variables, each record is a unique
combination of
the three variables. I would like to count the number of unique
values of v3
in each v1, and save it as a new variable v4 in the same data frame.
e.
My first thought was that this is R < 2.7.2 and you have encountered
(from the CHANGES file)
o Rscript -e (and Rterm -e) failed on Vista because the MSVCRT
function 'tmpfile' is broken on that platform.
But that couldn't possibly be the case, as the posting guide required you
to upd
Dear R-helpers,
I have a data frame with 3 variables, each record is a unique combination of
the three variables. I would like to count the number of unique values of v3
in each v1, and save it as a new variable v4 in the same data frame.
e.g.
df1
[v1] [v2] [v3]
[1,] "a" "C" "1"
[2,] "b" "
Hello R-Help
I have a question about the "behind the scenes" behaviour of the Rscript -e
command and particularly its interaction with Sweave and tempdir().
We are trying to deploy R as a web service to do water quality analyses and
have been writing software to call Sweave via Rscript eg:
cm
Hi Jörg,
our package distrTeach has functions to visualize the central limit
theorem and the law of large numbers for "arbitrary" univariate
distributions; e.g.,
library(distrTeach)
D <- sin(Norm()) + Pois()
plot(D)
illustrateCLT(D, len = 10)
illustrateLLN(D, m = 10)
Best
Matthias
Jörg Groß
Thanks, Roger, your demo is interesting. I'm thinking about improving it later.
I've also made a demo for the CLT in my package 'animation', in which
there's also normality testing for the sample means, because I don't
think "bell-shaped" alone means normality - so I performed the
Shapiro-Wilk tes
Thanks Gabor,
To be clear, would something like testframe$est[[i]] <- fp$estimate be
valid within my loop, as in (assuming I created testframe before the
loop):
for (i in 1:length(V4) ) {
x = read.csv(as.character(V4[[i]]), header = FALSE, na.strings="");
y = x[,1];
fp = fitdistr(y,"expo
On Wed, 15 Oct 2008, Mathew Rosales wrote:
Hi All,
Is there a package in R that does double bootstrap?
Yes, package 'boot'. It's support software for a book, and you will need
to look in the book to see how to do it.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of App
I'd like to suggest you "combine" them in R instead of using other
software. For example, you can use low-level plot commands to add the
points to the image plot.
If you insist on combining them together, you may use the utility
"composite" in ImageMagick, but I don't recommend you to combine a pd
Thank you Prof. Ripley.
I appreciate this.
Have a good day.
Ted
On Thu, Oct 16, 2008 at 12:20 AM, Prof Brian Ripley
<[EMAIL PROTECTED]> wrote:
> On Wed, 15 Oct 2008, Ted Byers wrote:
>
>> Thanks Jim,
>>
>> I hadn't seen the distinction between the commandline in RGui and what
>> happens within
On Wed, 15 Oct 2008, Minwook Kim wrote:
Hello,
If package check showed the Error, How can I use that?
I want to use caMassClass in Mac.
By compiling from the sources, but please ask about Mac packages on
R-sig-mac.
You need to click on ERROR to see what the problem was:
Packages r
On Wed, 15 Oct 2008, cls59 wrote:
On Unix/Linux platforms, you can use the included Rscript utility by adding
the following shebang at the top of your program. Command line arguments can
then be retrieved using the commandArgs function:
#!/usr/bin/Rscript
args <- commandArgs(trailingOnly = TR
Hi All,
Is there a package in R that does double bootstrap?
Thanks,
Matt
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commente
On Wed, 15 Oct 2008, Ted Byers wrote:
Thanks Jim,
I hadn't seen the distinction between the commandline in RGui and what
happens within my code.
I have, however seen other differences I don't understand. For
example, looking at the documentation for RScript, I see:
Rscript [options] [-e expr
testframe$newvar <- ...whatever...
(or see ?transform for another way)
adds a new column to the data frame. The table does not
have to pre-exist in your MySQL database and you don't need
a create statement; however, if the table does pre-exist the columns
of your data frame and those of the datab
Thanks Gabor,
I get how to make a frame using existing vectors. In my example, the
following puts my first three columns into a frame (and displays it:
> testframe <- data.frame(mid=V1,year=V2,week=V3)
> testframe
mid year week
1 251 2008 18
2 251 2008 19
3 251 2008 20
4 251 2008
Ajay,
Bayesm deals with this very issue in choice modelling (a form of econometric
modelling as outlined in the article). I think those guys (the developers of
Bayesm) and the apprach they recommend for navigating the likelihood function
through a bayesian approachs makes a lot of sense to me,
Hello,
If package check showed the Error, How can I use that?
I want to use caMassClass in Mac.
CRAN Package Check Results for Package caMassClass
Last updated on 2008-10-15 23:56:28.
Flavor Version Tinstall Tcheck Ttotal Status Flags
r-devel-linux-ix86 1.6 13.53 439.10 452.63 N
On 16/10/2008, at 2:27 PM, Moshe Olshansky wrote:
Hi Rolf,
Thank you for making me aware of the existence of PolynomF package.
By the way, your solution needs a small modification:
a <- polynom(c(-3,1)) and not polynom(-3,1) and similar for b.
Indeed; I typed the code into the email separat
On 16/10/2008, at 2:29 PM, Guillaume Filteau wrote:
Hello all,
When I create a matrix, is there a way to make it start at [0,0],
instead of [1,1]?
That way, a 2x2 matrix would go from [0,0] to [1,1], instead of [1,1]
to [2,2].
First, see fortune(36).
Then, if you ***MUST***, install packag
Put the data in an R data frame and use dbWriteTable() to
write it to your MySQL database directly.
On Wed, Oct 15, 2008 at 9:34 PM, Ted Byers <[EMAIL PROTECTED]> wrote:
>
> Here is my little scriptlet:
>
> optdata =
> read.csv("K:\\MerchantData\\RiskModel\\AutomatedRiskModel\\soptions.dat",
> hea
If you are willing to use character instead of numeric you can:
> xx <- matrix(1:4, 2, 2, dimnames = list(as.character(0:1), as.character(0:1)))
> xx
0 1
0 1 3
1 2 4
> xx["0", "1"]
[1] 3
On Wed, Oct 15, 2008 at 9:29 PM, Guillaume Filteau <[EMAIL PROTECTED]> wrote:
> Hello all,
>
> When I creat
On Unix/Linux platforms, you can use the included Rscript utility by adding
the following shebang at the top of your program. Command line arguments can
then be retrieved using the commandArgs function:
#!/usr/bin/Rscript
args <- commandArgs(trailingOnly = TRUE)
args is now a character vector c
Hello all,
When I create a matrix, is there a way to make it start at [0,0],
instead of [1,1]?
That way, a 2x2 matrix would go from [0,0] to [1,1], instead of [1,1]
to [2,2].
Best,
Guillaume
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R-help@r-project.org mailing list
https://stat.ethz.
Here is my little scriptlet:
optdata =
read.csv("K:\\MerchantData\\RiskModel\\AutomatedRiskModel\\soptions.dat",
header = FALSE, na.strings="")
attach(optdata)
library(MASS)
setwd("K:\\MerchantData\\RiskModel\\AutomatedRiskModel")
for (i in 1:length(V4) ) {
x = read.csv(as.character(V4[[i]]),
Thanks Jim,
I hadn't seen the distinction between the commandline in RGui and what
happens within my code.
I have, however seen other differences I don't understand. For
example, looking at the documentation for RScript, I see:
Rscript [options] [-e expression] file [args]
And the example:
Rs
You have to explicitly 'print' the value of x in the loop:print(x)
'x' by itself is just it value. At the command line, typing an
objects name is equivalent to printing that object, but it only
happens at the command line. If you want a value printed, the 'print'
it. Also works at the comma
I am not sure what you mean by "p is not updated". I took the basic
outline you have and as you can see, p has the correct value after the
loop.
> p<-0.1
> n<-10
> result<-list()
> while (p<=1) {
+ for (i in 1:n) {
+ result[[i]]<-p
+ p<-p+0.1
+ print(p)
+ }
+ }
[1]
Frank E Harrell Jr wrote:
Gad Abraham wrote:
This approach leaves much to be desired. I hope that its
practitioners start gauging it by the mean squared error of predicted
probabilities.
Is the logic here is that low MSE of predicted probabilities equals a
better calibrated model? What abou
Create an empty matrix first and then fill it in. That
will avoid the overhead in repeatedly expanding it. If
you don't know how many rows then make it 1000 rows
and remove the unused ones once finished.
On Wed, Oct 15, 2008 at 4:05 PM, culpritNr1 <[EMAIL PROTECTED]> wrote:
>
> Hello fellow R s
Ajay Shah mayin.org> writes:
>
> I wondered was people on this list felt about this article:
> http://www.voxeu.org/index.php?q=node/2363
> which talks about the problems of obtaining sound answers in numerical
> optimisation in settings such as MLE or NLS.
It seems perfectly reasonable (it
Dennis Lee Bieber wrote:
On Tue, 14 Oct 2008 09:10:14 -0400, "Daniel Malter" <[EMAIL PROTECTED]>
declaimed the following in gmane.comp.lang.r.general:
Other than that I agree with Duncan, your teacher is being paid for helping
you with these very basic problems.
"Me three"...
Dear J.,
Perhaps the following will clarify what you got:
(1) For those unfamiliar with it, contr.Sum() is like contr.sum(), except
for (in my opinion) printing more informative coefficient labels. There is
no contr.Poly(), but since you didn't use it, there was no error.
(2) Here's what I got w
Hi All,
Thanks for all these great efficient solutions.
Another one in the mix that Phil S.
sent on to me:
sapply(strsplit(tst,":"),function(x)x[2])
Matt
>-Original Message-
>From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
>Sent: Thursday, 16 October 2008 8:38 AM
>To: Redding, Matt
Hi Dylan,
You might want to have a look at the plyr package which is designed to
make these sorts of tasks easier - http://had.co.nz/plyr. The site
includes a ~20 page introductory pdf.
Hadley
On Wed, Oct 15, 2008 at 3:45 PM, dylan boyd <[EMAIL PROTECTED]> wrote:
> Another request for help imp
On Wed, 15 Oct 2008, Werner Wernersen wrote:
Hi,
I was wondering why the results from lm and dynlm are not the same for
what I think is the same model.
...because it's not the same model :-)
I haven't looked at this in detail, but:
set.seed(123456)
e1 <- rnorm(100)
e2 <- rnorm(100)
y1 <-
On 16/10/2008, at 11:07 AM, Prof Brian Ripley wrote:
[[ ]] works on vectors!
So it does. My bad.
letters[[3]]
[1] "c"
See help("[["). But strictly V4 is a factor and hence not a
vector: [[ ]]
also works on factors.
Yes. I had forgotten that not everyone sets options
(stringsAsF
Here are several solutions:
#1
# replace first three and last 3 characters with nothing
x <- c("23:10:34", "01:02:03")
gsub("^...|...$", "", x)
#2
# backref = -1 says extract the portion in parens
# it would have returned a list so we use simplify = c
library(gsubfn)
strapply(x, ":(..):", backref
Try this:
format(strptime(tst, "%H:%M:%S"), "%M")
On Wed, Oct 15, 2008 at 6:54 PM, Redding, Matthew <
[EMAIL PROTECTED]> wrote:
> Hi All,
>
> Is there a means to extract the "10" from "23:10:34" in one pass using
> strsplit (or something else)?
> tst <- "23:10:34"
>
> For example my attempt
> s
On 16/10/2008, at 10:44 AM, Erin Hodgess wrote:
Dear R people:
Is there a way to perform simple polynomial multiplication; that is,
something like
(x - 3) * (x + 3) = x^2 - 9, please?
I looked in poly and polyroot and expression. There used to be a
package that had this, maybe?
The Polynom
It's no harder: use sub() which is vectorized. You've not actually
defined the pattern, but here is my guess at your intentions:
x <- c("23:10:34", "23:20:30")
sub("([0-9][0-9]:)([0-9][0-9])(:[0-9][0-9])", "\\2", x)
Another interpetation would be
sub("([^:]*:)([^:]*)(:[^:]*)", "\\2", x)
for
One way is to use convolution (?convolve):
If A(x) = a_p*x^p + ... + a_1*x + a_0
and B(x) = b_q*x^q + ... + b_1*x + b_0
and if C(x) = A(x)*B(x) = c_(p+q)*x^(p+q) + ... + c_0
then c = convolve(a,rev(b),type="open")
where c is the vector (c_(p+q),...,c_0), a is (a_p,...,a_0) and b is
(b_q,...,b_0)
Matthew -
Redding, Matthew wrote:
Hi All,
Is there a means to extract the "10" from "23:10:34" in one pass using
strsplit (or something else)?
tst <- "23:10:34"
For example my attempt
strsplit(as.character(tst),"^[0-9]*:")
gives
[[1]]
[1] "" "" "34"
Why not simply,
strsplit(tst, ":")
Hi All,
Just to make that question a bit harder - how
do I apply that string extraction to vector of these time strings?
Thanks,
Matt Redding
>-Original Message-
>From: [EMAIL PROTECTED]
>[mailto:[EMAIL PROTECTED] On Behalf Of Redding, Matthew
>Sent: Thursday, 16 October 2008 7:54
Hi,
I was wondering why the results from lm and dynlm are not the same for what I
think is the same model.
I have just modified example 4.2 from the Pfaff book, please see below for the
code and results.
Can anyone tell my what I am doing wrongly?
Many thanks,
Werner
set.seed(123456)
e1 <-
This was also posted on R-sig-mac, and I've answered it there.
Please don't cross-post.
On Wed, 15 Oct 2008, Gang Chen wrote:
When invoking dev.new() on my Mac OS X 10.4.11, I get an X11 window
instead of quartz which I feel more desirable. So I'd like to set
the default device to quartz. Howe
Joe Kaser wrote:
Thanks for the help. ifelse does the job.
Could you elaborate, or give an example of the "awkward" things ifelse
might do to classed objects?
It strips the class. One thing that usually gets me is this:
> dd <- as.Date(c("2008-1-2","2007-3-21"))
> ifelse(dd>as.Date("2008-1-
[[ ]] works on vectors!
letters[[3]]
[1] "c"
See help("[["). But strictly V4 is a factor and hence not a vector: [[ ]]
also works on factors.
On Thu, 16 Oct 2008, Rolf Turner wrote:
On 16/10/2008, at 10:03 AM, jim holtman wrote:
try putting as.character in the call:
x = read.csv(as.c
On 16/10/2008, at 10:47 AM, Wacek Kusnierczyk wrote:
Rolf Turner wrote:
On 16/10/2008, at 12:28 AM, Henrique Dallazuanna wrote:
Try this:
lapply(1:n, rnorm)
That has nothing to do with what the inquirer *asked*.
probably because it's not what the inquirer knew he *should* have
asked.
Neat? It mihgt be very usefull, but it's rather a dirty hack in a
dirty language than a 'neat' solution.
\misiek
> That is neat Gabor. Thanks, Ted
> Gabor Grothendieck wrote:
>
> The gsubfn package can do quasi perl-style interpolation by
> prefacing any function call with fn$.
>
> library(gs
Dear Erin,
Yes. Take a look at page 13 in [1].
HTH,
Jorge
[1] http://cran.r-project.org/web/packages/Ryacas/vignettes/Ryacas.pdf
On Wed, Oct 15, 2008 at 5:44 PM, Erin Hodgess <[EMAIL PROTECTED]>wrote:
> Dear R people:
>
> Is there a way to perform simple polynomial multiplication; that is,
Ah! Thanks.
ifelse() appears to do the job.
note: subtracting 12 from chron rather than noon doesn't seem to work. I
think (but am not totally sure) chron interprets 12 as fraction of a day -
i.e. 12/1 days...
On Wed, Oct 15, 2008 at 1:33 PM, Greg Snow <[EMAIL PROTECTED]> wrote:
> The if state
Hi All,
Is there a means to extract the "10" from "23:10:34" in one pass using
strsplit (or something else)?
tst <- "23:10:34"
For example my attempt
strsplit(as.character(tst),"^[0-9]*:")
gives
[[1]]
[1] "" "" "34"
Obviously it is matching the first two instances of [0-9]. Note that
there
Thanks for the help. ifelse does the job.
Could you elaborate, or give an example of the "awkward" things ifelse might
do to classed objects?
On Wed, Oct 15, 2008 at 1:38 PM, Peter Dalgaard <[EMAIL PROTECTED]>wrote:
> Joe Kaser wrote:
>
>> Hello,
>>
>> I've been learning R functions recently and
Dear R people:
Is there a way to perform simple polynomial multiplication; that is,
something like
(x - 3) * (x + 3) = x^2 - 9, please?
I looked in poly and polyroot and expression. There used to be a
package that had this, maybe?
thanks,
Erin
--
Erin Hodgess
Associate Professor
Department o
Rolf Turner wrote:
>
> On 16/10/2008, at 12:28 AM, Henrique Dallazuanna wrote:
>
>> Try this:
>>
>> lapply(1:n, rnorm)
>
> That has nothing to do with what the inquirer *asked*.
>
probably because it's not what the inquirer knew he *should* have asked.
it does solve the task he used as an example i
Hi all,
This is probably going to come off as unnecessary (and show my ignorance)
but I am trying to understand the parameter estimates I am getting from R
when doing an ANCOVA. Basically, I am accustomed to the estimate for the
categorical variable being equivalent to the respective cell means
I'm using stableFit from the package fBasics to estimate the parameters of a
truncated normal distribution (I'm interested in the parameters of the
underlying normal distribution). It is correct to generalize this truncated
normal distribution as a stable distribution ?
Thanks
David
--
View thi
Hello fellow R sufferers,
Is there a way to perform an appending operation in place?
Currently, the way my pseudo-code goes is like this
for (i in 1:1000) {
if (some condition) {
newRow <- myFunction(myArguments)
X <- rbind(X, newRow) # <- this is the bottleneck!!
}
Actually, I'd tried single brackets first. Here is what I got:
> for (i in 1:length(V4) ) { x = read.csv(V4[i], header = FALSE,
> na.strings="");x }
Error in read.table(file = file, header = header, sep = sep, quote = quote, :
'file' must be a character string or connection
>
the advice to
When invoking dev.new() on my Mac OS X 10.4.11, I get an X11 window
instead of quartz which I feel more desirable. So I'd like to set
the default device to quartz. However I'm confused because of the
following:
> Sys.getenv("R_DEFAULT_DEVICE")
R_DEFAULT_DEVICE
"quartz"
> getOption("devi
On 16/10/2008, at 12:28 AM, Henrique Dallazuanna wrote:
Try this:
lapply(1:n, rnorm)
That has nothing to do with what the inquirer *asked*.
On Wed, Oct 15, 2008 at 8:19 AM, Megh Dal <[EMAIL PROTECTED]>
wrote:
Can anyone please tell me how to define a "list". Suppose I want
to define
Hello,
I have read the R memory pages.
I realized after my post that I would not have enough memory to
accomplish this task.
The command I'm using to convert the list into a data-frame is as such:
som <- do.call("rbind", somlist)
Where som is the dataframe resulting from combining all the data
On 16/10/2008, at 10:03 AM, jim holtman wrote:
try putting as.character in the call:
x = read.csv(as.character(V4[[i]]), header = FALSE
No. This won't help. V4 is a column of the data frame optdata,
and hence is a vector. Not a list! Use single brackets --- V4[i] ---
and all will be well
try putting as.character in the call:
x = read.csv(as.character(V4[[i]]), header = FALSE
On Wed, Oct 15, 2008 at 4:46 PM, Ted Byers <[EMAIL PROTECTED]> wrote:
>
> Here is what I tried:
>
> optdata =
> read.csv("K:\\MerchantData\\RiskModel\\AutomatedRiskModel\\soptions.dat",
> header = FALSE, na.s
Here is what I tried:
optdata =
read.csv("K:\\MerchantData\\RiskModel\\AutomatedRiskModel\\soptions.dat",
header = FALSE, na.strings="")
optdata
attach(optdata)
for (i in 1:length(V4) ) { x = read.csv(V4[[i]], header = FALSE,
na.strings="");x }
And here is the outcome (just a few of the 60 reco
Hi Siang Li. It would help if you explained what packages you are
using. auto.arima() is in the forecast package and arimax appears to
be from the TSA package.
Using auto.arima() to select the orders is inappropriate because you
are ignoring the regressors. auto.arima() does not currently handle
Another request for help implementing the 'apply' functions to avoid a
loop structure...
I am working with a data set that includes lab measurements taken at
different dates for the subjects, with some subjects having more
results than others. I would like to average lab results for each
subject
Joe Kaser wrote:
Hello,
I've been learning R functions recently and I've come across a problem that
perhaps someone could help me with.
# I have a a chron() object of times
hours=chron(time=c("01:00:00","18:00:00","13:00:00","10:00:00"))
# I would like to subtract 12 hours from each time el
The if statement is for program flow control (do this bunch of code if this
condition is true, else do this), the vectorized version is the ifelse
function, that is the one that you want to use.
Also note (this is for times in general, not sure about chron specifically)
subtracting noon from th
Hello,
I've been learning R functions recently and I've come across a problem that
perhaps someone could help me with.
# I have a a chron() object of times
> hours=chron(time=c("01:00:00","18:00:00","13:00:00","10:00:00"))
# I would like to subtract 12 hours from each time element, so I created
Galton's 19th century mechanical version of this is the quincunx. I
have a
(very primitive) version of this for R at:
http://www.econ.uiuc.edu/~roger/courses/476/routines/quincunx.R
url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Depart
Jörg Groß wrote:
Hi,
Is there a way to simulate a population with R and pull out m samples,
each with n values
for calculating m means?
I need that kind of data to plot a graphic, demonstrating the central
limit theorem
and I don't know how to begin.
So, perhaps someone can give me some t
To remove all numbers and periods at the end do:
> vars <- sub("[0-9.]+$", "", pick )
If there are other periods that this does not remove (not at the end) then you
can do a second pass:
> vars <- sub("\\.", "", vars)
Part of the reason that your code for the period does not work is that an
u
I can't help with your inquiry generally, but I can address the
issue of ``optional braces''.
On 16/10/2008, at 8:34 AM, Ted Byers wrote:
I take it FUN = function(x, grp) quantile(x$DATA, names=FALSE) is the
function definition for a function called FUN. I would guess,
then, tha
Try this;
gsub("\\.|[0-9]", "", pick)
On Wed, Oct 15, 2008 at 4:42 PM, John Poulsen <[EMAIL PROTECTED]> wrote:
> Hello R-users,
>
> I have code that gives me the important variables from an analysis. I need
> to input these variables into a different analysis. To do this, I need to
> modify th
A case for [g]sub:
gsub(".", "", sub("[0-9]*$", "", pick), fixed = TRUE)
On Wed, 15 Oct 2008, John Poulsen wrote:
Hello R-users,
I have code that gives me the important variables from an analysis. I need
to input these variables into a different analysis. To do this, I need to
modify them
Hi Joerg,
Is there a way to simulate a population with R and pull out m samples,
each with n values
for calculating m means?
I need that kind of data to plot a graphic, demonstrating the central
limit theorem
and I don't know how to begin.
So, perhaps someone can give me some tips and hints
Hi all,
I am running OLS models with a power parameter: y=a+bx^p, and I want to iterate
through p with a set increment. The function I constructed is something like
below:
test<-function(data, model,...){
f<-formula(model) #model="y~I(x^p)"
p<-0.1
n<-10
result<-list()
while (p<=1) {
for (i in 1
On 10/15/2008 3:48 PM, Jörg Groß wrote:
Hi,
Is there a way to simulate a population with R and pull out m samples,
each with n values
for calculating m means?
I need that kind of data to plot a graphic, demonstrating the central
limit theorem
and I don't know how to begin.
The easiest
Hello R-users,
I have code that gives me the important variables from an analysis. I
need to input these variables into a different analysis. To do this, I
need to modify them slightly... 1) remove all numbers at the end of the
variables, 2) remove all periods.
I tried to do it with the aw
Look at the clt.examp function in the TeachingDemos package, it generates
samples from normal, uniform, gamma (default exponential), and beta (default
U-shaped) distributions and plots histograms of the means along with a
reference line of a normal distribution with the same mean and sd. The de
Hi,
Is there a way to simulate a population with R and pull out m samples,
each with n values
for calculating m means?
I need that kind of data to plot a graphic, demonstrating the central
limit theorem
and I don't know how to begin.
So, perhaps someone can give me some tips and hints ho
I have examined the documentation for batch mode use of R:
R CMD BATCH [options] infile [outfile]
The documentation for this seems rather spartan.
Running "R CMD BATCH --help" gives me info on only two options: one for
getting help and the other to get the version. I see, further on, that
th
In the example in the documentation, I see:
rs <- dbSendQuery(con,
"select Agent, ip_addr, DATA from pseudo_data order by Agent")
out <- dbApply(rs, INDEX = "Agent",
FUN = function(x, grp) quantile(x$DATA, names=FALSE))
Maybe I am a bit thick, but it took me a while, and a
Or ?"Memory-limits" (and the posting guide of course).
On Wed, 15 Oct 2008, Prof Brian Ripley wrote:
See ?"Memory-size"
On Wed, 15 Oct 2008, B. Bogart wrote:
Hello all,
I'm working with a large data-set, and upgraded my RAM to 4GB to help
with the mem use.
I've got a 32bit kernel with 64GB
Hi,
have a look to Dirks tutorial at the UseR2008. This should be a good
starting point:
http://www.statistik.uni-dortmund.de/useR-2008/tutorials/eddelbuettel.html
Markus
Rajasekaramya wrote:
> Hi there,
>
> I am looking for R/parallel package or some other package that would speed
> up the an
i'm in the process of switch from Maple to R and am trying to code the
following function:
w(d)=\int -Inf_Inf A(x) |int 0_(t(d)-x) Fy dydx
can some one point me in the right direction? i don't seem to be able to
figure it out on my own.
Dr. Wade Winterhalter
University of Central Florida
D
Hi there,
I am looking for R/parallel package or some other package that would speed
up the analysis.I am working on computatioanly intensive data so any
suggestions would be really helpful.
Kindly let me know if any
--
View this message in context:
http://www.nabble.com/R-Parallel-tp1425p1
Surely you jest! You need to consult your local statistician.
-- Bert Gunter
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of mirre simons
Sent: Wednesday, October 15, 2008 2:08 AM
To: r-help@r-project.org
Subject: [R] LME gives estimates with a random fac
Thanks for the reply,
that's exactly what I wanted. I also figured that the function rowSds() does
exist. I just needed to load the generfilter package.
Thanks again.
Nutter, Benjamin wrote:
>
> ?apply
>
> e.g. apply(matrix,1,sd)
>
> -Original Message-
> From: [EMAIL PROTECTED] [ma
> apply( mymatrix, 1, sd )
Will give the row standard deviations of a matrix or matrix like data frame.
There are also some functions in add-on packages that do row sd's or var's.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
801.408.8111
>
Doen't work.
\misiek
Prof Brian Ripley wrote:
> See ?"Memory-size"
> On Wed, 15 Oct 2008, B. Bogart wrote:
[...]
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do
On 10/10/2008 8:55 AM, Oliver Bandel wrote:
Zitat von Duncan Murdoch <[EMAIL PROTECTED]>:
On 10/10/2008 8:13 AM, Oliver Bandel wrote:
> Hello,
>
>
> I tried to use rgl.snapshot and it failed.
>
> The error message was not very verbose:
>
>
>
> ==
>>
>> plot3d( motion[[idx+2]
Thank you very much Greg.
Apparently, my learning curve is still well below it's horizontal
asymptote.
And no wonder in the vast universe of the R-project, eh?
Thanx, DaveT.
>-Original Message-
>From: Greg Snow [mailto:[EMAIL PROTECTED]
>Sent: October 15, 2008 01:52 PM
>To: Thompson, Davi
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