Dear Donglei Hu,
If you have two correlation coefficients, you may try cordif {multilevel}
and cordif.dep {multilevel} for the independent correlations and for the
dependent correlations, respectively. However, they are both based on the
sampling distribution of correlation coeficient. A better ap
You could try fitting several cases:
- fit the reduced model with only one exp term, i.e. one of the alphas is 0
- fit the model with both alpha's constrained to be positive &
sufficiently away from 0
- fit the model with both alpha's negative & sufficiently away from 0
- fit the model with one po
On 18 Sep 2006 17:55:24 +0200,
Peter Dalgaard <[EMAIL PROTECTED]> wrote:
> "Gabor Grothendieck" <[EMAIL PROTECTED]> writes:
>> Here are some approaches:
>> - we only have 4 parameters so just use grid search to get starting
>> values as in:
>> https://stat.ethz.ch/pipermail/r-help/2005-September
Many thanks to Gabor and Deepayan. In case of conditioning, one can also
do this inside the panel function, using subscripts as necessary. The
segplot package is also a nice touch.
Ben
Gabor Grothendieck wrote:
> Try this:
>
> print(p)
> trellis.focus("panel", 1, 1)
> with(z, panel.segments(x,
I'm testing a FC5 machine for use in a student lab. R 2.3.1 is installed and
seems to work fine. There is one peculiarity - the logins are authenticating
to a server, and a "verbose" flag is set somewhere, leading to lots of
spurious messages like this
request done: ld 0xa227598 msgid 1
whic
On 9/16/06, Benjamin Tyner <[EMAIL PROTECTED]> wrote:
> For each level of the factor in dotplot, I have time points I'd like to
> connect with a line. In the example below, 'x' represents a starting
> time and 'd' a duration, and I wish to connect 'x' to 'x+d'. Ordinarily
> I would use Dotplot fro
*NOTICE OF CONSOLATION PRIZE WINNING*
This email confirms that you have been notified of by the DE GOLDSTAR E-GAMES
LOTTO BV DE Netherlands of your email lottery winning for 2006
GOLDSTAR LOTTO - Wheel E-game held on 18th September 2006.
We wish to congratulate you on the selection of yo
"David Barron" <[EMAIL PROTECTED]> writes:
> Is cor.test() in the stats packages what you mean?
No, he wants to compare two correlation coefficients, not test that
one is zero. That's usually a misguided question, but if need be, the
Fisher z transform atanh(r) can be used to convert r to an
appr
Ben Bolker <[EMAIL PROTECTED]> writes:
>
> Time for folks to re-read the posting guide?
>
> * When responding to a very simple question, use the following algorithm:
> 1. compose your response
> 2. type 4*runif(1) at the R prompt, and wait this many hours
> 3. c
Time for folks to re-read the posting guide?
* When responding to a very simple question, use the following algorithm:
1. compose your response
2. type 4*runif(1) at the R prompt, and wait this many hours
3. check for new posts to R-help; if no similar suggestion
Statistical Computing and Statistical Graphics Sections
American Statistical Association
Student Paper Competition 2007
The Statistical Computing and Statistical Graphics Sections of the ASA
are co-sponsoring a student paper competition on the topics of
Statistical Computing and Statistical Graphi
I have a few questions for R:
1. Other than using a built-in function such as mean(), how do I know if
it is installed in my current version of R?
2. To get help in R, I can use several ways:
?sort
help.search("sort")
help(sort)
apropos("sort")
the help menu
are there any other ways to get hel
yes, see ?points for the list and details.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained,
Is cor.test() in the stats packages what you mean?
On 18/09/06, Hu, Donglei <[EMAIL PROTECTED]> wrote:
> Hi,
>
>
>
> I calculated a few correlation coefficients. Now I want to know whether
> they are different from each other. Is there an R package that can do
> such a comparison? Thanks for an
On 18-Sep-06 Iñaki Murillo Arcos wrote:
> Hello,
>
> I don't know if the result of
>
> acos(0.5) == pi/3
>
> is a bug or not. It looks strange to me.
>
>Inaki Murillo
It is not a bug, but a feature, in that acos(0.5) and pi/3
are not computed in the same way, so (because of the small
i
Iñaki Murillo Arcos said the following on 9/18/2006 12:31 PM:
> Hello,
>
> I don't know if the result of
>
> acos(0.5) == pi/3
>
> is a bug or not. It looks strange to me.
>
>Inaki Murillo
>
> __
> R-help@stat.math.ethz.ch mailing list
> htt
I need to plot a prcomp object from package stats with custom symbols suitable
for B&W publication. My boss specifically wants filled and unfilled square,
triangle, circle, inverted triangle, diamond to represent 5 brain regions of 2
types of rat.
Can I specify these as a parameter?
Thanks,
On Mon, 18 Sep 2006, [iso-8859-1] I?aki Murillo Arcos wrote:
> Hello,
>
> I don't know if the result of
>
> acos(0.5) == pi/3
>
> is a bug or not. It looks strange to me.
Real numbers are strange. Would the result of:
acos(0.5) - pi/3
close enough to zero for you? Try help('Comparison
On 9/18/06, Iñaki Murillo Arcos <[EMAIL PROTECTED]> wrote:
> Hello,
>
> I don't know if the result of
>
> acos(0.5) == pi/3
>
> is a bug or not. It looks strange to me.
However,
> all.equal(acos(0.5), pi/3)
[1] TRUE
so you may want to check the FAQ entry on comparisons involving
floating poi
Iñaki Murillo Arcos <[EMAIL PROTECTED]> writes:
> Hello,
>
> I don't know if the result of
>
> acos(0.5) == pi/3
>
> is a bug or not. It looks strange to me.
Have a look in the R FAQ:
> acos(0.5) - pi/3
[1] 2.220446e-16
--
O__ Peter Dalgaard Øster Farimagsgade 5
How close do you think it should be, given finite resolution with digital
computing?
> acos(0.5) - pi/3
[1] 2.220446e-16
Charles Annis, P.E.
[EMAIL PROTECTED]
phone: 561-352-9699
eFax: 614-455-3265
http://www.StatisticalEngineering.com
-Original Message-
From: [EMAIL PROTECTED]
[m
If you had looked at help("==") you would have seen the following:
For numerical and complex values, remember == and != do not allow for
the finite representation of fractions, nor for rounding error. Using
all.equal with identical is almost always preferable.
Then if you had tried
> all.equal(a
Hi,
I calculated a few correlation coefficients. Now I want to know whether
they are different from each other. Is there an R package that can do
such a comparison? Thanks for any suggestion.
Best,
Donglei Hu
Department of Medicine
UCSF
[[alternative HTML version deleted]]
On Mon, 2006-09-18 at 19:31 +0200, Iñaki Murillo Arcos wrote:
> Hello,
>
> I don't know if the result of
>
> acos(0.5) == pi/3
>
> is a bug or not. It looks strange to me.
>
>Inaki Murillo
Seems reasonable to me:
> acos(0.5) == pi/3
[1] FALSE
> print(acos(0.5), 20)
[1] 1.047197551196
Hi,
are there any way to convert ISO8601 weeks to gregorian dates? Something
like
coverttodate(year=2006, week=38, day=1)
# Sept 18, 2006
Thanks in advance,
Ott
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PLE
Hello,
I don't know if the result of
acos(0.5) == pi/3
is a bug or not. It looks strange to me.
Inaki Murillo
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-p
On 9/18/2006 3:37 AM, Sean O'Riordain wrote:
> Good morning,
>
> I'm trying to concisely generate a single integer from 0 to n
> inclusive, where n might be of the order of hundreds of millions.
> This will however be used many times during the general procedure, so
> it must be "reasonably effici
There may be benefits to having a machine learning method that
explicitly targets covariate balance. We have experimented with
optimizing the weights directly to obtain the best covariate balance,
but got some strange solutions for simple cases that made us wary of
such methods.
Machine learning
Thanks very much, Greg. I will certainly look at glmpath.
My goal is to develop (nearly) automatic and flexible procedures for
estimating causal effects of risk factors in observational epidemiological
studies. A major part of this is the development of a propensity score
model (when the exposur
On Mon, 18 Sep 2006, Peter Dalgaard wrote:
> "Sarosh Jamal" <[EMAIL PROTECTED]> writes:
>
>> Hi there,
>>
>> I was updating the R-cmdr add-on (v.1.1-6 to the latest v.1.2) for R
>> (v.2.2.0) in a SunOS9 environment and came across some warnings during my
>> installation - it seems to download the
Check out Park & Hastie's glmpath package. They have a really clever
analysis and implementation of a generalized least angle regression.
Greg
>On Fri, 2006-09-15 at 18:49 -0400, Ravi Varadhan wrote:
> > Is there an R implementation of least angle regression for binary
response
> > modeling? I k
Seems to work also - thanks!
Tony Plate wrote:
> If I construct the matrix by list()ing together the expressions rather
> than c()ing, then it works OK:
>
> > x <- matrix(list( expression(x3-5*x+4), expression(log(x2-4*x
> > x[1,1]
> [[1]]
> expression(x3 - 5 * x + 4)
>
> > x[[1,1]]
> express
Hi Patrick,
thanks for pointing me to your work and Rmetrics.
I have a few questions on my mind right now. Do you have methods for
automatic download of price quote histories? I can use python to get
XML data on FOREX price quotes from the NYRB and other sites.
Together with Rpy and matplotlib,
On 9/12/06, Douglas Bates <[EMAIL PROTECTED]> wrote:
> On 9/11/06, Manuel Morales <[EMAIL PROTECTED]> wrote:
[snip]
> > Am I right that the MCMC sample can not be used, however, to evaluate
> > the significance of parameter groups. For example, to assess the
> > significance of a three-level factor
On 18 Sep 2006 19:53:59 +0200, Peter Dalgaard <[EMAIL PROTECTED]> wrote:
> "Sarosh Jamal" <[EMAIL PROTECTED]> writes:
>
> > Hi there,
> >
> > I was updating the R-cmdr add-on (v.1.1-6 to the latest v.1.2) for R
> > (v.2.2.0) in a SunOS9 environment and came across some warnings during my
> > instal
"Sarosh Jamal" <[EMAIL PROTECTED]> writes:
> Hi there,
>
> I was updating the R-cmdr add-on (v.1.1-6 to the latest v.1.2) for R
> (v.2.2.0) in a SunOS9 environment and came across some warnings during my
> installation - it seems to download the dependencies but runs into the
> following during i
If I construct the matrix by list()ing together the expressions rather
than c()ing, then it works OK:
> x <- matrix(list( expression(x3-5*x+4), expression(log(x2-4*x
> x[1,1]
[[1]]
expression(x3 - 5 * x + 4)
> x[[1,1]]
expression(x3 - 5 * x + 4)
> D(x[[1,1]], "x")
-5
>
The reason c() d
On 9/18/2006 1:16 PM, Evan Cooch wrote:
>
> Eik Vettorazzi wrote:
>> test=matrix(c( expression(x^3-5*x+4), expression(log(x^2-4*x
>> works.
> Well, not really (or I'm misunderstanding). Your code enters fine (no
> errors), but I can't access individual elements - e.g., test[1,1] gives
> me a
Hello,
are there potential drawbacks in the use of rank (ordinal) variables as
predictors in a "linear" mixed model for continuous dependents?
I am trying this approach to create mixed effects models independent of the
(monotonic) transform of the independent variables. The use of rank
transfor
Hi there,
I was updating the R-cmdr add-on (v.1.1-6 to the latest v.1.2) for R
(v.2.2.0) in a SunOS9 environment and came across some warnings during my
installation - it seems to download the dependencies but runs into the
following during install:
* Installing *source* package 'acepack' ...
**
Eik Vettorazzi wrote:
> test=matrix(c( expression(x^3-5*x+4), expression(log(x^2-4*x
> works.
Well, not really (or I'm misunderstanding). Your code enters fine (no
errors), but I can't access individual elements - e.g., test[1,1] gives
me an error:
> test=matrix(c( expression(x^3-5*x+4),
>From: =?ISO646-US?Q?Gunther_H=3Fning?= <[EMAIL PROTECTED]>
>Date: 2006/09/18 Mon AM 06:26:25 CDT
>To: 'Petr Pikal' <[EMAIL PROTECTED]>
>Cc: r-help@stat.math.ethz.ch
>Subject: Re: [R] Question on apply() with more information...
I think you want something like below but it
probably needs some fixi
Dear Bob,
It's not clear to me what you want to do. You provided
my example, which I already know, and you didn't
provide yours.
Marco
--- Bob Green <[EMAIL PROTECTED]> wrote:
> Marco,
>
> I saw your January e-mail to an enquiry about
> filling in missing values
> with a zero I tried adaptin
test=matrix(c( expression(x^3-5*x+4), expression(log(x^2-4*x
works.
btw. you recieved an error because D expects an expression and you offered
a list
> class(test[1])
[1] "list"
to get the error relating to the misuse of the tilde operator you have to
prompt the "correct" extractor "[["
f<
You may want to rethink your whole approach here:
1. Pie charts are usually a poor choice of graph, there are better
choices.
2. Adding percentages to a pie chart is a way of admitting that the pie
chart is not doing the job.
3. If you want people to compare percentages, then a table is what is
ne
Normally, I do symbolics in Maple, or Mathematica, but I'm trying to
write a simple script for students to handle some *very* simple
calculations (for other purposes) with matrix or vector elements, where
the elements are coded symbolically. What I've tried with *partial"
success is use of the
Peter et. al.:
>
> With those definitions (which are hardly universal), tolerance
> intervals are the same as prediction intervals with k == m == 1, which
> is what R provides.
>
>
I don't believe this is the case. See also:
http://www.itl.nist.gov/div898/handbook/prc/section2/prc263.htm
T
Sachin,
One of the references in the link is a EPA groundwater statistics training
manual from the early nineties. For a more complete related discussion see pp.
50-62 of the July 1992 EPA draft addendum on groundwater monitoring statistics
at
http://www.epa.gov/correctiveaction/resource/guidance
"Gabor Grothendieck" <[EMAIL PROTECTED]> writes:
> Here are some approaches:
>
> - we only have 4 parameters so just use grid search to get
> starting values as in:
>
> https://stat.ethz.ch/pipermail/r-help/2005-September/079617.html
>
> - there are singularities near beta_1 = beta_2 and near a
On Sat, 16 Sep 2006 21:17:21 -0500,
Bingshan Li <[EMAIL PROTECTED]> wrote:
> Hi there, I have a dataframe whose elements are numbers or characters. I
> want to extract the frequencies of each elements in the dataframe. For
> example,
> d = as.data.frame(matrix(c(1, 2, 3, 3), 2,2))
Don't use a da
I believe that your confusion is due to a typo in the formula in [3], it is
missing a sumation sign (and a subscript on x if you want to be picky). To get
the denominator you subtract the mean of your x variable from all the x-values,
square the differences, then sum them up (the missing sumat
Hi there,
I have a dataframe whose elements are numbers or characters. I want
to extract the frequencies of each elements in the dataframe. For
example,
d = as.data.frame(matrix(c(1, 2, 3, 3), 2,2))
What I want is first what are the elements in the data (1,2,3 here)
and second what are the
Here are some approaches:
- we only have 4 parameters so just use grid search to get
starting values as in:
https://stat.ethz.ch/pipermail/r-help/2005-September/079617.html
- there are singularities near beta_1 = beta_2 and near alpha_1 = 0
and near alpha_2 = 0 so reparameterize and use the uppe
Something like
pie(bull, labels=paste(names(bull),": ",bull,"%", sep=""))
On 18/09/06, ERICK YEGON <[EMAIL PROTECTED]> wrote:
>
> Hi Gurus, i have a small problem with working with graphs on R.
> Say i have data say bull-c(34,23,7,4) and i assign names to the elements
> in
> the brackets
> if i
ERICK YEGON said the following on 9/18/2006 8:22 AM:
> Hi Gurus, i have a small problem with working with graphs on R.
> Say i have data say bull-c(34,23,7,4) and i assign names to the elements in
> the brackets
> if i do
> Pie(bull) i get a pie chart of bull togtjer with the names.
> Question
Hi,
I'm trying to fit the following model to data using 'nls':
y = alpha_1 * beta_1 * exp(-beta_1 * x) +
alpha_2 * beta_2 * exp(-beta_2 * x)
and the call I've been using is:
nls(y ~ alpha_1 * beta_1 * exp(-beta_1 * x) +
alpha_2 * beta_2 * exp(-beta_2 * x),
start=list(alpha_1=
Sachin J <[EMAIL PROTECTED]> writes:
> Google search gave me this:
>
> http://ewr.cee.vt.edu/environmental/teach/smprimer/intervals/interval.html
>
> TIA
> Sachin
With those definitions (which are hardly universal), tolerance
intervals are the same as prediction intervals with k ==
Google search gave me this:
http://ewr.cee.vt.edu/environmental/teach/smprimer/intervals/interval.html
TIA
Sachin
Peter Dalgaard <[EMAIL PROTECTED]> wrote:
Sachin J writes:
> RUsers:
>
> Just confirming, does predict function with interval="prediction"
> option gives predict
Sachin J <[EMAIL PROTECTED]> writes:
> RUsers:
>
> Just confirming, does predict function with interval="prediction"
> option gives prediction interval or tolerance interval?. Sorry for
> reposting this question.
Is there any definition of tolerance interval that is different from
pre
[...]
>
>> cochranq.test(K)
>
>Cochran's Q Test for Dependent Samples
>
> data: K
> Cochran's Q = 23.9298, df = 11, p-value = 0.01303
>
Cochran's Q fits into the `coin' framework and thus:
> K <- as.table(matrix(c(1,1,0,0, 1,1,0,1, 1,1,1,1, 1,1,1,1, 1,0,0,0,
+ 1,1,1,1, 1,1,1,1, 0,0,0
Hi Gurus, i have a small problem with working with graphs on R.
Say i have data say bull-c(34,23,7,4) and i assign names to the elements in
the brackets
if i do
Pie(bull) i get a pie chart of bull togtjer with the names.
Question. How can i add values (percentages) in the graph
Thanks
On Mon, 2006-09-18 at 14:01 +1000, [EMAIL PROTECTED] wrote:
> Hi!
>
> I would like to conduct a Cochran`s Q Test in R, but have not found any
> suitable function.
>
> My first idea was: J <- as.table(matrix(c(6,16,2,4),ncol=2, dimnames =
> list("C" = c("Favorable","Unfavorable"),"Drug A Favorab
Ronaldo Reis-Jr. wrote:
> Hi,
>
> But, I have this warning in the enviromnent that have this problem:
>
> During startup - Warning message:
> using .GlobalEnv instead of 'package:lattice'
So, can you please be a bit more specific?
R version, your startup files such as Rprofile files etc?
Uwe L
RUsers:
Just confirming, does predict function with interval="prediction" option
gives prediction interval or tolerance interval?. Sorry for reposting this
question.
Thanks in advance
Sachin
David Barron <[EMAIL PROTECTED]> wrote:
Sorry, I think I may have mislead you; t
Dear list,
how can I write the full information of class summary.lm to a text, latex or
html file? xtable only extracts the table but not the rest of the information.
Best regards, Ulrich
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Hi,
But, I have this warning in the enviromnent that have this problem:
During startup - Warning message:
using .GlobalEnv instead of 'package:lattice'
maybe this is the problem?
How to fix it?
Thanks
ROnaldo
--
I've been on this lonely road so long,
Does anybody know where it goes,
I remembe
Hi,
I tried both ideas, but it isn't that what I'm looking for.
I want to avoid for loop, because the matrix is of big size(1200*1200
entries)
With a loop I would do:
for ( i in seq(along = SmoothList))
{
Xarry[i,] <- predict(SmoothList[[i]],Xarry[i,])$y
}
Actually I want to do more th
Hi,
This is a false problem.
This problem is only in a .RData file, not a global R problem.
Thanks
ROnaldo
--
I've been on this lonely road so long,
Does anybody know where it goes,
I remember last time the signs pointed home,
A month ago.
-- Carpenters, "Road Ode"
--
> Prof. Ro
I'm just getting started with R, having a lot of original work on
modeling and exploring the simulation by MCMC. I want to simulate the
prior and the posterior distribution of Dirichlet Process by MCMC.
Is there anyone in NYC that might be a good
tutor for me?
--
Dr. P. NGOM,
Faculté des Science
Hi Phil and Don,
Le 06-09-18 à 05:02, Phil Spector a écrit :
> Denis -
>As long as there is one data frame that has exactly 0 or 1
> observations
> for each level of the by variables, merge in R will operate
> identically
> to SAS without the need for any special options.
>The problem
Hi,
after a debian upgrade my dev.copy2eps dont work anymore. I have this message:
Error in matchFont(postscriptFonts(family)[[1]], old$encoding) :
unknown font
I try to change the font family in my .Rprofile:
setHook(packageEvent("graphics", "onLoad"),
function(...) {
You need to print() explicitly inside sourced-ed script: auto-printing is
not active.
On Mon, 18 Sep 2006, Ulrich Leopold wrote:
> Dear list,
>
> First my information:
> platform i386-pc-linux-gnu
> arch i386
> os linux-gnu
> system i386, linux-gnu
> status
>
"Ulrich Leopold" <[EMAIL PROTECTED]> writes:
> Now my question:
>
> How is it possible that a command in an R script is not executed completely
> whereas the same command is fully executed when I copy paste it to the R
> terminal?
It is fully executed, but you're not printing the results. Read
h
Hi
If I am correct apply do not choose from SmoothList as you expected.
Instead probably
lapply(SmoothList, predict,Xarray)
or
mapply(predict,SmoothList, Xarray)
can give you probably what you want.
HTH
Petr
On 18 Sep 2006 at 9:26, Gunther Höning wrote:
From: Gunther Hönin
Dear list,
First my information:
platform i386-pc-linux-gnu
arch i386
os linux-gnu
system i386, linux-gnu
status
major 2
minor 3.1
year 2006
month 06
day01
svn rev38247
language R
version.string Vers
check out 'cbind' to select columns.
regards
Isidora
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-c
Good morning,
I'm trying to concisely generate a single integer from 0 to n
inclusive, where n might be of the order of hundreds of millions.
This will however be used many times during the general procedure, so
it must be "reasonably efficient" in both memory and time... (at some
later stage in t
Ok.
I tried this too, but it still doesn't work.
Here some more information to try out, but just an excerpt of Xarray
x <- c(0.11,0.25,0.45,0.65,0.80,0.95,1)
Y <- matrix(c(15,83,57,111,150,168,175,37,207,142,277,375,420,437),nrow=2)
sm <- function(y,x){smooth.spline(x,y)}
SmoothList <- apply(Y,1
Anupam Tyagi yahoo.com> writes:
>
> New users may also want to look at SciViews R Graphical User Interface(GUI).
> It
> can be a good learning tool. Its text based editor is basic compared to WinEdt
> with the R editing plug-in, or ESS and (X)Emacs combination. But it has
> point-and-click menu
Hi!
I would like to conduct a Cochran`s Q Test in R, but have not found any
suitable function.
My first idea was: J <- as.table(matrix(c(6,16,2,4),ncol=2, dimnames =
list("C" = c("Favorable","Unfavorable"),"Drug A Favorable"=c("B
Favorable","B Unfavorable"
L <- as.table(matrix(c(2,4,6,6),n
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