This is about an Omegahat package.
Especially for packages not on CRAN, please ask the maintainer in the
first instance. There were Omegahat mailing lists: I do not know if they
are currently operational but this does indicate that R-help is not the
appropriate place to ask.
Your questions are
Dear colleagues,
I have the following code. This code is to 'filter' the data set.
It works on the data frame 'whole' with four numeric columns: a,b,d, and c.
Every row in the data frame is considered as a point in 3-D space.
Variables a,b, and d are the point's coordinates, and c is its
Dear All,
As far I as I have understood reading both your past posting and the
documentation, in order to have the command-line completion facility, I
have to run R within emacs.
However, as I try to start R within emacs as recommended:
C-u M-x R
emacs answers [no match]
the same if I provide the
On 03-Apr-05 Ashraf Chaudhary wrote:
Hi All:
I would like to generate a binomial random variable that
correlates with a normal random variables with a specified
correlation. Off course, the correlation coefficient would
not be same at each run because of randomness.
I greatly appreciate your
Hello
I am able to extract partial regression coefficients from a fitted model
object model, i.e.
model - lm(var.sel.gkm, weights = count.gkm, data = DATA)
summary(model)
write.table(model$coef, file = C:/coef_CO_gkm.txt, row.names = TRUE,
col.names = TRUE)
I was wondering if anyone could
Dear list,
perhpas this question is more suitable for R-dev but since I am not
really a developer I post it here first.
Apparently the following lines do not create any problem in R:
library(combinat)
r - 20; b - 2;
sum( sapply(0:r,function(x) nCm(r,x)^(2*b)) ) 2^64
while in C I obtain an
Any R output is an object you can manipulate basically using
exctractions functions $, [ and [[
To look at the content of the object, try:
str(model)
And (in this case)
str(summary(model))
Then you can extract what you need, such as:
summary(model)$adj.r.squared
[1] 0.02158191(an
one idea is to consider that the underlying (for ease normally
distributed) latent variables that produce the Bernoulli trials are
correlated with your original normal random variable.
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public
On Mon, 4 Apr 2005, Marco Chiarandini wrote:
Dear list,
perhpas this question is more suitable for R-dev but since I am not
really a developer I post it here first.
Apparently the following lines do not create any problem in R:
library(combinat)
r - 20; b - 2;
sum( sapply(0:r,function(x)
Hi,
One fruitful course for optimisation is to vectorise wherever
possible, and avoid for-loops.
Something like the code below might be a good place to start.
=
## Generate a thousand rows of data
cube.half.size - 2
mult.sigma - 2
n - 1000
whole - data.frame(a=runif(n), b=runif(n),
On Mon, 2005-04-04 at 10:11 +0200, Giorgio Corani wrote:
Dear All,
As far I as I have understood reading both your past posting and the
documentation, in order to have the command-line completion facility, I
have to run R within emacs.
However, as I try to start R within emacs as
Dear R-users,
I'm trying to add a title on a plot with both mathematical notation and
values substitution. I read the documentation and search the mailing list
but I was not able to solve my problem. Actually, there is a message by
Uwe Ligges on June 2003 which addresses a question very close to
Hi,
However, as I try to start R within emacs as recommended:
C-u M-x R
emacs answers [no match]
the same if I provide the whole path to the executable:
C-u M-x /usr/bin/R[no match]
given you have installed ESS (Emacs Speaks Statistics), you can start an
R session within Emacs
Dear Rich,
Thank you for reply. I think, optimization, you offered will satisfy
my needs.
I don't completely understand the following.
RF ## And wrap the original in a function for comparison:
RF ## This does not subset the way you want:
RF ## whole[-which(row.names(to.drop) %in%
Hi again,
The arguments to %in% are in the wrong order in your version. Because
of that, the statement
row.names(to.drop) %in% row.names(whole)
will be TRUE for the first nrow(to.drop) elements, and FALSE for the remainder.
To fix it, just switch the order around, or use the simpler version:
Dear R user,
I want to change each density and angle with symbol +,-,o,#
and *. How can I do that?
library(gplots)
barplot2(VADeaths,
density=c(5,7,11,15,17),
angle=c(65,-45,45,-45,90),
col = black,
legend = rownames(VADeaths))
title(main = list(Death Rates
Gidday,
See ?plotmath and demo(plotmath) for lots of information on plotting
with mathematical symbols.
This produces what you seem to be after (paste() being the missing
ingredient):
plot(1:10, main=substitute(paste(Monotonic Multigamma run ( *,
list(n==len, theta==t1), * )),
Luca Scrucca [EMAIL PROTECTED] writes:
Dear R-users,
I'm trying to add a title on a plot with both mathematical notation and
values substitution. I read the documentation and search the mailing list
but I was not able to solve my problem. Actually, there is a message by
Uwe Ligges on June
Dear Xin Meng,
This output presumably was produced by summarizing an object produced by aov().
The trick to figuring out what you want to do is to examine the structure of
the summary object (say, sumry), via str(sumry). In this case sumry[[Error:
Within]][[1]]$F value[1] should do what you
On Sat, 2 Apr 2005 23:00:43 -0500, Terry Mu [EMAIL PROTECTED] wrote :
thx, that's perfect. I thought of grep(), it also can do this.
I wonder if there is a document or book that explains things
categorically so it's easy to look up a function.
The HTML help does this: try help.start(), and
[EMAIL PROTECTED] writes:
Hello,
I am doing intensive tests on SVMs parameter selection. Once a while I got the
error:
Error in save.image(): image could not be renamed and is left in .RDataTmp1
I cannot use the information saves in .RDataTmp1.
Why? Anything wrong with load(.RDataTmp1)
Use the rownames() function to set the rownames equal to your first
column, and then drop the first column.
I don't know if there is a way to do it during retrieval from MySQL.
-Don
At 11:10 PM +0200 4/2/05, simone gabbriellini wrote:
Dear List,
I have this little problem:
I work with adiacency
hello R-Users,
I have this simple but not for me question:
I do:
res-dbSendQuery(con, SELECT * FROM tabellaProva)
myDataFrame-fetch(res)
myDataMatrix-as.matrix(myDataFrame[,-1])
namerows(myDataMatrix)-as.character(myDataFrame[,1])
and I have:
io tu
io 0 1
tu 1 0
my problem is that the
I have written a package, where a function definition includes a regexp
pattern including double backslashes, such as
myfunction - function (pattern = .*\\.txt$)
when I R CMD CHECK the corresponding .Rd file, I get warnings
(code/documentation mismatch), if I enforce two backslashes in the
Hi
I try to import html text and I need to split the fields at each td or
/td entry
How can I succeed? sep = 'td' doens't yield the right result
thanks for hints
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
Hello!
I am an 'R beginner'. I am trying to check if my data follow a negative
binomial function.
the command i've typed in is:
nbdo=rnegbin(58,mu=27.82759,theta=0.7349851)
ks.test(do$DO,nbdo)
Each time i do that, p given is different and i get this warning message:
'Warning message:
cannot
simone gabbriellini [EMAIL PROTECTED] writes:
hello R-Users,
I have this simple but not for me question:
I do:
res-dbSendQuery(con, SELECT * FROM tabellaProva)
myDataFrame-fetch(res)
myDataMatrix-as.matrix(myDataFrame[,-1])
namerows(myDataMatrix)-as.character(myDataFrame[,1])
joerg van den hoff wrote:
I have written a package, where a function definition includes a regexp
pattern including double backslashes, such as
myfunction - function (pattern = .*\\.txt$)
when I R CMD CHECK the corresponding .Rd file, I get warnings
(code/documentation mismatch), if I enforce
Luca Scrucca wrote on 4/4/2005 1:50 PM:
Dear R-users,
I'm trying to add a title on a plot with both mathematical notation and
values substitution. I read the documentation and search the mailing list
but I was not able to solve my problem. Actually, there is a message by
Uwe Ligges on June 2003
Christoph Lehmann wrote:
Hi
I try to import html text and I need to split the fields at each td or
/td entry
How can I succeed? sep = 'td' doens't yield the right result
If it fits pairwise together, use
sep=c(td, /td)
if not, you can read the whole lot with readLines and strsplit for both
What does 'with ties in' mean?
with some identical elements (par ex., au moins une paire ex-equo)
HTH
Ken Knoblauch
Inserm U371, Cerveau et Vision
Department of Cognitive Neurosciences
18 avenue du Doyen Lepine
69675 Bron cedex
France
tel: +33 (0)4 72 91 34 77
fax: +33
Agnes Gault wrote:
Hello!
I am an 'R beginner'. I am trying to check if my data follow a negative
binomial function.
the command i've typed in is:
nbdo=rnegbin(58,mu=27.82759,theta=0.7349851)
ks.test(do$DO,nbdo)
Each time i do that, p given is different and i get this warning message:
simone gabbriellini wrote:
Dear List,
I have this little problem:
I work with adiacency matrix like:
data me you
me0 1
you 1 0
I store those matrix in a mysql database
actually I use RMySQL with:
res-dbSendQuery(connection, SELECT * FROM table)
You can import the whole thing and use on it strsplit
?strsplit
Eric
Eric Lecoutre
UCL / Institut de Statistique
Voie du Roman Pays, 20
1348 Louvain-la-Neuve
Belgium
tel: (+32)(0)10473050
[EMAIL PROTECTED]
http://www.stat.ucl.ac.be/ISpersonnel/lecoutre
If the statistics are boring, then
Christoph Lehmann wrote:
entry from html:
tr bgcolor=#9090f0td align=rightbBM/b/tdtd
0.952/tdtd 0.136/tdtd 6.984/tdtd0.00/td/tr
tr bgcolor=#9090f0td align=rightbBH/b/tdtd
1.338/tdtd 0.136/tdtd 9.821/tdtd0.00/td/tr
using
left.data- scan(paste(path, left.file, sep = ), what =
simone gabbriellini wrote:
hello R-Users,
I have this simple but not for me question:
I do:
res-dbSendQuery(con, SELECT * FROM tabellaProva)
myDataFrame-fetch(res)
myDataMatrix-as.matrix(myDataFrame[,-1])
namerows(myDataMatrix)-as.character(myDataFrame[,1])
and I have:
Agnes Gault wrote:
Hello!
I am an 'R beginner'. I am trying to check if my data follow a negative
binomial function.
the command i've typed in is:
nbdo=rnegbin(58,mu=27.82759,theta=0.7349851)
ks.test(do$DO,nbdo)
Each time i do that, p given is different
The p-values are different each time
You need another OS. Standard/32-bit Windows (XP, 2000 etc) can't use
more than 4 GB of RAM. Anyway, if you try to buy a box with 16 GB of
RAM, the seller will probably warn you about Windows and recommend a
suitable OS.
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL
Dear users
I need to perform a loglinear analysis of a real data set for a course
project. I need a real data set with contingency tables in at least 3
dimensional, each with
more than 2 levels.
Thanks
Joe Warfield
[[alternative HTML version deleted]]
I have two data sets that I converted to its objects to get:
ts1
date settle
1 2000-09-29 107.830
2 2000-10-02 108.210
3 2000-10-03 108.800
4 2000-10-04 108.800
5 2000-10-05 109.155
ts2
date settle
1 2000-09-25 107.610
2 2000-09-26 107.585
3 2000-09-27 107.385
4 2000-09-28
Hi,
I have a set of x,y data points and each data point lies between (0,0)
and (1,1). Of this set I have selected all those that lie in the lower
triangle (of the plot of these points).
What I would like to do is to divide the region (0,0) to (1,1) into
cells of say, side = 0.01 and then count
On Mon, 4 Apr 2005 13:46:46 -0400 Omar Lakkis wrote:
I have two data sets that I converted to its objects to get:
ts1
date settle
1 2000-09-29 107.830
2 2000-10-02 108.210
3 2000-10-03 108.800
4 2000-10-04 108.800
5 2000-10-05 109.155
ts2
date settle
1 2000-09-25
Dear useRs,
I have just uploaded to CRAN first version of my package outliers for
testing data for outlying observations. It contains all types of Dixon
and Grubbs test and the Cochran test for outlying variance.
Until placing in package collection, the files are also availalble at my
Warfield Jr., Joseph D. wrote:
Dear users
I need to perform a loglinear analysis of a real data set for a course
project. I need a real data set with contingency tables in at least 3
dimensional, each with
more than 2 levels.
Thanks
Joe Warfield
[[alternative HTML version deleted]]
On Monday 04 April 2005 13:22, Rajarshi Guha wrote:
Hi,
I have a set of x,y data points and each data point lies between
(0,0) and (1,1). Of this set I have selected all those that lie in
the lower triangle (of the plot of these points).
What I would like to do is to divide the region
Dnia 2005-04-04 20:59, Uytkownik Ben Fairbank napisa:
Forbidden
You don't have permission to access /outliers/ on this server.
Bad Options in httpd.conf, just corrected. Thanks.
__
Hi all;
I'm trying to fit a reparameterization of the
assymptotic regression model as that shown in
Ratkowsky (1990) page 96.
Y~y1+(((y2-y1)*(1-((y2-y3)/(y3-y1))^(2*(X-x1)/(x2-x1/(1-((y2-y3)/(y3-y1))^2))
where y1,y2,y3 are expected-values for X=x1, X=x2, and
X=average(x1,x2), respectively.
From: Deepayan Sarkar [EMAIL PROTECTED] Mon, 4 Apr 2005 13:52:48 -0500
On Monday 04 April 2005 13:22, Rajarshi Guha wrote:
Hi,
I have a set of x,y data points and each data point lies between
(0,0) and (1,1). Of this set I have selected all those that lie in
the lower triangle (of
Christian Mora wrote:
Hi all;
I'm trying to fit a reparameterization of the
assymptotic regression model as that shown in
Ratkowsky (1990) page 96.
Y~y1+(((y2-y1)*(1-((y2-y3)/(y3-y1))^(2*(X-x1)/(x2-x1/(1-((y2-y3)/(y3-y1))^2))
where y1,y2,y3 are expected-values for X=x1, X=x2, and
On Mon, 2005-04-04 at 14:22 -0400, Rajarshi Guha wrote:
Hi,
I have a set of x,y data points and each data point lies between (0,0)
and (1,1). Of this set I have selected all those that lie in the lower
triangle (of the plot of these points).
What I would like to do is to divide the region
I said:
myfun - function(x, y, ints) {
fx - x %/% (1/ints)
fy - y %/% (1/ints)
txy - hist(fx + ints*fy+ 1, breaks=0:(ints*ints), plot=FALSE)$counts
dim(fxy) - c(ints, ints)
^^^
return(txy)
}
Of course it should be:
dim(txy) - c(ints, ints)
^^^
Sorry about that,
Hello
I am getting memory allocation errors when running a function that uses
locfit within a for loop. After 25 or so loops, it gives this error.
Error: cannot allocate vector of size 281250 Kb
Running on linux cluster with a Gb of RAM. Problem never happens on my
OS X (less memory).
Perhaps the following, substituting your vectors of x and y for
runif(1)
x-trunc(100*runif(1))
y-trunc(100*runif(1))/100
length(unique(x+y))
[1] 6390
Ben Fairbank
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Rajarshi Guha
Sent: Monday,
I need capabilities, for my data analysis, like the Pinheiro Bates
S-Plus/R package nlme() but with binomial family and logit link.
I need multiple crossed, possibly interacting fixed effects (age cohort of
twin when entered study, sex of twin, sampling method used to acquire twin
pair, and
The code sniplet you provided is nowhere near correct or sufficient for
anyone to help. Please (re-)read the posting guide and try again.
Andy
From: Mike Hickerson
Hello
I am getting memory allocation errors when running a function
that uses
locfit within a for loop. After 25 or so
The questioner clearly wants generalized linear mixed models. lmer in
package lme4 may be more appropriate. (Prof. Bates is a co-author.).
glmmPQL should do the same job, though, but with less accuracy.
Simon.
check glm()
On Apr 4, 2005 6:46 PM, William M. Grove [EMAIL PROTECTED] wrote:
I need
Hi
Jan Sabee wrote:
Dear R user,
I want to change each density and angle with symbol +,-,o,#
and *. How can I do that?
library(gplots)
barplot2(VADeaths,
density=c(5,7,11,15,17),
angle=c(65,-45,45,-45,90),
col = black,
legend = rownames(VADeaths))
title(main =
Hi
Thanks Martin and Adai.
This gives me a good starting point.
Paul
Adaikalavan Ramasamy wrote:
Thank you to Paul Murrell and Martin Maechler for their help.
pushHexport() and the rest of the codes have done the trick.
I spent the afternoon trying to code up something that might be used as
Dear R-list,
i have 6 different sets of samples. Each sample has about 5000 observations,
with each observation comprised of 150 baseline covariates (X), 125 of which
are dichotomous. Roughly 20% of the observations in each sample are treatment
and the rest are control units.
i am doing
Dear R
Should I be concerned if the loadings to a Principle Component Analysis are
as follows:
Loadings:
Comp.1 Comp.2 Comp.3 Comp.4
X100m -0.500 0.558 0.661
X200m -0.508 0.379 0.362 -0.683
X400m -0.505 -0.274 -0.794 -0.197
X800m -0.486 -0.686 0.486 0.239
60 matches
Mail list logo
