Dear R users
(B
(BI have a data frame containing character and numeric variables, whose
(Bname is seishin. When I tried to assign NA to "" in the data frame, R, 2.00
(Bshowed an error message, such as
(B
(B> seishin[seishin==""]<-NA
(BError: NAs are not allowed in subscripted assignments
(
Here's my report on the issue:
The command 'system' combined with either 'scan'
or 'read.table' (depending on what I want to do with the
output in each step) worked beautifully.
Cheers,
Tim Liao
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Hi, Uwe.
Thank you for help.
Yesterday another R-user Mr. Ramasamy told me to do a simple program
such as (changing too the directories, now C:/minhadll/):
/*file conv.c*/
#include
void printhello (){
Rprintf("%s", "hello world\n");
}
When I use C:\R\rw2000\bin>RCMD SHLIB -o c:/
On Sat, 13 Nov 2004, Rob Steele wrote:
Does R do its own swapping out to disk? I disabled Linux swapping and the
No.
system still gets stuck in Purgatory where there's little CPU activity but
the disk goes like crazy. That's with R having almost the whole machine to
itself and running a memory
Anne Piotet wrote:
I have a problem with plot for summary
s<-summary(response~ x1+x2+x3+x4+x5+x6+x7+
+ x8+ x9+x10)
plot(s)
Error in plot.new() : Figure margins too large
I tried to set the margins to null with par(mai=c(0,0,0,0))
but keep getting the same error message
What is wrong?
Anne,
Emily Baldock wrote:
I am trying to get a subscript into the title.
My expression is quote(bold(apoA[bold("1")]))
I have managed to get something near to what I want with
sd2 <-1.882
measure <- "g/L"
direction <- "increase"
chem <- quote(bold(apoA[bold("1")]))
plot(0,0)
titletxt <- substitute(pas
O. Neto wrote:
Hi R-Users
I wrote 1 week ago asking about a message that appears when I try run
dyn.load.
I'm trying to do an example in C code from "Writing R Extension" to
learn how to do it.
I have R 2.0.0, Rtools, Perl and MinGW as describe in
http://www.murdoch-sutherland.com/Rto
Does R do its own swapping out to disk? I disabled Linux swapping and
the system still gets stuck in Purgatory where there's little CPU
activity but the disk goes like crazy. That's with R having almost the
whole machine to itself and running a memory hungry compute only function.
I've seen t
bogdan romocea wrote:
Dear R users,
This is a KDE beginner's question.
I have this distribution:
length(cap)
[1] 200
summary(cap)
Min. 1st Qu. MedianMean 3rd Qu.Max.
459.9 802.3 991.6 1066.0 1242.0 2382.0
I need to compute the sum of the values times their probability of
oc
Alexandre Galvão Patriota wrote:
Hi Douglas, I need to find the matrix V=ZDZ'+R, for
example:
require(nlme)
data(Orthodont)
attach(Orthodont)
fm1 <- lme(distance ~ age + Sex, data =
Orthodont,random=~age)
X<-model.matrix(distance ~ age + Sex)
Z<-model.matrix(distance ~ age + Subject -1)
D<-diag(nco
Hi Douglas, I need to find the matrix V=ZDZ'+R, for
example:
require(nlme)
data(Orthodont)
attach(Orthodont)
fm1 <- lme(distance ~ age + Sex, data =
Orthodont,random=~age)
X<-model.matrix(distance ~ age + Sex)
Z<-model.matrix(distance ~ age + Subject -1)
D<-diag(ncol(Z))
cova<-VarCorr(fm1,rdig=7)
Prof Brian Ripley <[EMAIL PROTECTED]> writes:
> > [EMAIL PROTECTED]:~/r-devel> echo
> > 'set.seed(1);M<-matrix(rnorm(9e6),3e3);system.time(solve(M))' |
> > BUILD-GOTO/bin/R -q --vanilla
> >> set.seed(1);M<-matrix(rnorm(9e6),3e3);system.time(solve(M))
> > [1] 29.12 1.39 32.21 0.00 0.00
> >>
>
On Sat, 13 Nov 2004, Ann Huxtable wrote:
Hello,
I have only recently started using R. I have two data samples that I want to
carry out some initial explorative data analysis to:
i). Determine the distribution of the data
ii). Determine whether both datasets are from the same distribution.
I have
Alexandre Galvão Patriota wrote:
The model is Y = XB + Zg + e
where
g~N(0, D)
e~N(0, R)
How to extract the VAR(g)= D, VAR(e)=R and V=ZDZ'+R?
thanks
The VarCorr function can provide D.
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On Sat, 13 Nov 2004, Peter Dalgaard wrote:
Peter Dalgaard <[EMAIL PROTECTED]> writes:
Prof Brian Ripley <[EMAIL PROTECTED]> writes:
ATLAS, you need to build shared ATLAS libraries (rather than
static). This requires some modifications to the configuation files
for ATLAS. But my experience shows t
Andy,
Thanks a lot for the clarifications. I was running a simulation a
number of times and trying to come up with a number to summarize the
results. And, I failed to realize from the beginning that what I was
trying to compute was just the mean.
Regards,
b.
--- "Liaw, Andy" <[EMAIL PROTECT
Peter Dalgaard <[EMAIL PROTECTED]> writes:
> Prof Brian Ripley <[EMAIL PROTECTED]> writes:
>
> > > ATLAS, you need to build shared ATLAS libraries (rather than
> > > static). This requires some modifications to the configuation files
> > > for ATLAS. But my experience shows that R itself builds
Hello,
I have only recently started using R. I have two data samples that I want to
carry out some initial explorative data analysis to:
i). Determine the distribution of the data
ii). Determine whether both datasets are from the same distribution.
I have managed to create unit probability histog
First thing you probably should realize is that density is _not_
probability. A probability density function _integrates_ to one, not _sum_
to one. If X is an absolutely continuous RV with density f, then Pr(X=x)=0
for all x, and Pr(a < X < b) = \int_a^b f(x) dx.
sum x*Pr(X=x) (over all possible
Dear wizaRds, Kevin,
I came up with a solution to perform a r doc search in the Firefox
search bar, you'll have to create a file called R.src in the search
plugins directory : (for me it is : C:/Program Files/Mozilla
Firefox/searchplugins) and put a png image of the R logo in the same
directo
Maybe a litle simpler solution:
x<-rnorm(50)
den<-density(x) #see ?density for more details
den$x[which(den$y==max(den$y))]
Hope it helps,
Ales Ziberna
- Original Message -
From: "Gregor GORJANC" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Friday, November 12, 2004 6:38 PM
Subject
Ah, that is now a recursive linear filter. In fact filter() would do both
your examples.
On Sat, 13 Nov 2004, James Muller wrote:
Take 3:
# p is a vector
myfunc <- function (p) {
x <- rep(0,length(p))
x[1] <- p[1]
for (i in c(2:length(p))) {
x[i] <- 0.8*p[i] + 0.2*x[i-1] # note the x
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