booop booop a écrit :
1...Kindly tell me is it possible to form
a matrix which contains a no of matrices.. for eg..
if a,b,c,d are matrices
and e is a matrix which contains a,b,c,d as rows and columns..
I don't think you can use matrix() to store other matrix() inside.
But
Paul Murrell a écrit :
You might also like to take a look at Figure 3.27
(and associated R code) on
http://www.stat.auckland.ac.nz/~paul/RGraphics/chapter3.html
Thank you very much for making this page available.
The power of R graphics, and the number of options,
makes such examples very
vincent == vincent [EMAIL PROTECTED]
on Thu, 29 Sep 2005 08:30:10 +0200 writes:
vincent Paul Murrell a écrit :
You might also like to take a look at Figure 3.27 (and
associated R code) on
http://www.stat.auckland.ac.nz/~paul/RGraphics/chapter3.html
vincent Thank you
Yves Magliulo said the following on 2005-09-28 17:05:
hi,
i'll try to help you, i send a mail about this subject last week... and
i did not have any response...
I'm using gam from package mgcv.
1)
How to interpret the significance of smooth terms is hard for me to
understand
Dear all,
Edmond Ng (http://multilevel.ioe.ac.uk/softrev/reviewsplus.pdf) provides
an example to fit the mixed effects meta-analysis in Splus 6.2. The
syntax is:
lme(fixed=d~wks, data=meta, random=~1|study, weights=varFixed(~Vofd),
control=lmeControl(sigma=1))
where d is the effect size, study
On Thu, 29 Sep 2005, Mike Cheung wrote:
Dear all,
Edmond Ng (http://multilevel.ioe.ac.uk/softrev/reviewsplus.pdf) provides
an example to fit the mixed effects meta-analysis in Splus 6.2. The
syntax is:
lme(fixed=d~wks, data=meta, random=~1|study, weights=varFixed(~Vofd),
On Thu, 29 Sep 2005 [EMAIL PROTECTED] wrote:
booop booop a écrit :
1...Kindly tell me is it possible to form
a matrix which contains a no of matrices.. for eg..
if a,b,c,d are matrices
and e is a matrix which contains a,b,c,d as rows and columns..
I don't think you can use
On Thu, 29 Sep 2005, Mulholland, Tom wrote:
Well I downloaded the data using the link in your message which suggests
that the code is right. I don't have its loaded (I assume it's from the
irregular time series package)
The package is called its, as the message correctly said.
so I can't
Hi
You can use e.g. plot - points construction
plot(Day, gene1)
points(Day, gene2, col=2)
see
?points
?lines
You also need to set appropriate y range by ylim argument to first
plot. And if you used search facility in CRAN and asked it
multiple plots on same x axis
you would have got many
Prof Brian Ripley a écrit :
On Thu, 29 Sep 2005 [EMAIL PROTECTED] wrote:
I don't think you can use matrix() to store other matrix() inside.
But array() is a solution to store matrix() inside.
(At least I have use it).
You _can_ do this with matrix() (although that was not quite what was
Mike
I do not believe this is availabe in either lme or lmer in R, only S-Plus.
-Original Message-
From: [EMAIL PROTECTED] on behalf of Mike Cheung
Sent: Thu 9/29/2005 4:32 AM
To: r-help@stat.math.ethz.ch
Cc:
Subject:[R] how to fix the level-1 variances in lme()?
Martin Maechler a écrit :
To really thank Paul, you should buy the book to which the above
web page refers :
http://www.stat.auckland.ac.nz/~paul/RGraphics/rgraphics.html
A very nice and useful book to have, indeed!
I just broked my piggy bank for Peter Dalgaard's one,
so I'll have to
hi,
thanks for reply.
1) sorry for the mistake, (GCV and UBRE estimate df in both case in
deed) i've made a confusion.
2) i use mgcv 1.1-8. (!) so in deed again, i need an update.
my questions :
3) i can see in the last version of summary.gam an estimated rank: how
do i interpret this?
4)
Hi Karin
I did not have seen any answer for your question yet so here is a
try.
I gues you want the horizotal layout or your boxplot.
boxplot(split(rnorm(30), rep(1:3, each=10)), horizontal =T,
names=letters[1:3])
boxplot(split(rnorm(30), rep(1:3, each=10)), horizontal =T,
names=c(NA,b,NA))
Sorry to revive and old topic, but writing to the clipboard seems to
have a problem for me: column names are ignored. Example:
# ~~~
# write.clipboard
# ~~~
write.clipboard = function(obj) {
write.table(obj, file(clipboard), sep=\t, row.names=F,
Hi list,
is there any direct way to obtain confidence intervals for the regression
slope from lm, predict.lm or the like?
(If not, is there any reason? This is also missing in some other statistics
softwares, and I thought this would be quite a standard application.)
I know that it's easy to
?confint
For example:
ctl - c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt - c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
group - gl(2,10,20, labels=c(Ctl,Trt))
weight - c(ctl, trt)
lm(weight ~ group)
Call:
lm(formula = weight ~ group)
Coefficients:
?confint
-Oprindelig meddelelse-
Fra: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] På vegne af Christian Hennig
Sendt: 29. september 2005 13:19
Til: r-help-request Mailing List
Emne: [R] Regression slope confidence interval
Hi list,
is there any direct way to obtain
On Thu, 29 Sep 2005, Christian Hennig wrote:
Hi list,
is there any direct way to obtain confidence intervals for the regression
slope from lm, predict.lm or the like?
There is a confint method: e.g.,
R fm - lm(dist ~ speed, data = cars)
R confint(fm, parm = speed)
2.5 % 97.5 %
?confint
Thank you to all of you.
As far as I see this is not mentioned on the lm help page (though I
presumably don't have the recent version), which I would
suggest...
Best,
Christian
On Thu, 29 Sep 2005, Chuck Cleland wrote:
?confint
For example:
ctl -
Dave,
qqmath(~val|ind,data=xx
,distribution=function(p) qt(p,df=19)
,ylab=Sample Quatinles
,xlab=Theoretical Quantiles
,aspect=1
,prepanel = prepanel.qqmathline
,panel=function(x,y)
{
panel.qqmathline(y, distribution=function(p)
Why not use vcov() and the normal approximation ?
ctl - c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt - c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
group - gl(2,10,20, labels=c(Ctl,Trt))
weight - c(ctl, trt)
lm.D9 - lm(weight ~ group)
cbind(estimate =
Sorry, I forgot confint and I made a mistake in my suggestion which
should be:
cbind(estimate = coef(lm.D9),
lower = coef(lm.D9) - 1.96 * sqrt(diag(vcov(lm.D9))),
upper = coef(lm.D9) + 1.96 * sqrt(diag(vcov(lm.D9
Best,
Renaud
Christian Hennig a écrit :
Hi list,
is there
memory.limit may not be the correct command. I use the command 'utils::
memory.size(3*1024)' to increase my memory size after using editbin to
modify the header of R to make it LARGEADDRESSAWARE as described in the
above FAQ. I am able to read about 2.7Gb into memory that way with 4Gb of
ram. Not
On Wed, 28 Sep 2005, Robert Bagchi wrote:
Hi Patrick
thanks for your advice. I have now tried glmmPQL, and it worked fine -
I'm getting consistent results between plots and models fitted by
glmmPQL. Plus it allows predict() and resid() which is another advantage
over lmer at present.
quick
See:
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/26922.html
On 9/29/05, Jose Quesada [EMAIL PROTECTED] wrote:
Sorry to revive and old topic, but writing to the clipboard seems to
have a problem for me: column names are ignored. Example:
# ~~~
# write.clipboard
#
Hello,
how do I obtain standard errors of variances and covariances of the
random effects with LME comparable to those of for example MlWin? I know
you shouldn't use them because the distribution of the estimator isn't
symmetric blablabla, but I need a measure of the variance of those
I have a file like this:
a 0.1
a 0.2
a 0.9
b 0.5
b 0.9
b 0.7
c 0.6
c 0.99
c 0.88
Which I would like to get to be the following matrix:
0.1 0.20.30.4 ...
a 12 0 0
b 00 0 0
You cannot. Also, it's not that the distribution of the random effects
is not symmetric, but that it *may* not be symmetric, and this is an
assumption that should be checked.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Roel de Jong
Sent: Thursday,
Petr Pikal [EMAIL PROTECTED] writes:
Hi Karin
I did not have seen any answer for your question yet so here is a
try.
I gues you want the horizotal layout or your boxplot.
boxplot(split(rnorm(30), rep(1:3, each=10)), horizontal =T,
names=letters[1:3])
boxplot(split(rnorm(30), rep(1:3,
you could use something like this (but maybe there are better
proposals):
dat - data.frame(g = rep(letters[1:3], each = 5), val = runif(15))
out - do.call(rbind, lapply(split(dat$val, dat$g), function(x){
f - factor(findInterval(x, vec = seq(0, 1, 0.1)), levels = 1:10)
table(f)
}))
On Thu, 29 Sep 2005, Karin Lagesen wrote:
I have a file like this:
a 0.1
a 0.2
a 0.9
b 0.5
b 0.9
b 0.7
c 0.6
c 0.99
c 0.88
Which I would like to get to be the following matrix:
0.1 0.20.30.4 ...
a 1
Hi!
I'm using R 2.0.1 on a Sun, with mgcv library version 1.3-1.
I would like to implement a new family for the function gam in mgcv
(truncated Poisson family as defined in Barry Welsh (2002), Ecological
Modelling).
I therefore defined a family function with all the necessary components
This can be shortened slightly using cut:
table(data.frame(g = dat$g, val = cut(dat$val, 0:10/10, lab = 1:10/10)))
On 9/29/05, Dimitris Rizopoulos [EMAIL PROTECTED] wrote:
you could use something like this (but maybe there are better
proposals):
dat - data.frame(g = rep(letters[1:3], each =
And slightly more to:
table(g = dat$g, val = cut(dat$val, 0:10/10, lab = 1:10/10))
On 9/29/05, Gabor Grothendieck [EMAIL PROTECTED] wrote:
This can be shortened slightly using cut:
table(data.frame(g = dat$g, val = cut(dat$val, 0:10/10, lab = 1:10/10)))
On 9/29/05, Dimitris Rizopoulos
On Thu, 2005-09-29 at 15:28 +0200, Karin Lagesen wrote:
Petr Pikal [EMAIL PROTECTED] writes:
Hi Karin
I did not have seen any answer for your question yet so here is a
try.
I gues you want the horizotal layout or your boxplot.
boxplot(split(rnorm(30), rep(1:3, each=10)),
Hi,
I need to install R-2.1.1 on a AIX machine. (i have tried several machine with
different os version : 5.1, 5.2 and 5.3)
I have tried several configuration which was advise on the R-install and admin
section :
but none goes to the end of compilation.
my configure command line is :
Dear all,
I want to know the probability of dying in a trap as a function of habitat
variables for a metapopulation of voles sampled in 44 stations.
My dataset is as follow:
Station tag number deadhabitat1habitat2
1 1 yes 20
Regarding non-parametric regression with interval censored survival data: Does
anyone know where to find an extension dealing with Cox regression for
(overlapping) interval censored data? If not, maybe someone has some ML
expressions for this interval scenario laying around that you are willing
See colAUC in caTools (there is a problem with 1.3 version, 1.4 is on the
way). See examples for other functions calculating AUC. An alternative
approach is:
x1 = x[y==1]; n1 = length(x1);
x2 = x[y==0]; n2 = length(x2);
r = rank(c(x1,x2))
auc =
Hi everyone,
I'm trying to solve a problem about how to get the
Fisher's discriminant functions of a lda (linear
discriminant analysis) object, I mean, the object
obtained from doing lda(formula, data) function of
the package MASS in R-project. This object gives me
the canonical linear
Dear R-Users,
I am trying to use mle() to optimize two (or more) parameters, but I want
to specify those parmeters in a data frame rather than having to spell
them out separately in the start variable of mle().
My call is
mle(negll, start=list(aps=init), fixed=list(measphot=newphot,
Hi
Works for me.
write.excel - function (tab, ...) write.table(tab, clipboard, sep
= \t, row.names = F)
write.excel(a)
from your example shows in Excel after ctrl-V as a table with names.
HTH
Petr
On 29 Sep 2005 at 12:12, Jose Quesada wrote:
Date sent: Thu, 29 Sep 2005
Hi
So you need x to be same in all subsequent plots e.g. from -5 to 5
like in this.
boxplot(split(rnorm(30), rep(1:3, each=10)), horizontal =T,
names=letters[1:3], ylim=c(-5,5))
You find it in a thorough reading bxp man page.
Currently, 'ylim' is used 'along the boxplot', i.e.,
Are you still interested in a reply to this post? I have not seen
any. If you are, it might help if you were more specific, e.g.,
following the posting guide www.R-project.org/posting-guide.html. I'm
not certain what you mean by a joint distribution defined by an
independent
I'm not certain what you are asking. Consider the following:
set.seed(1)
(irows - sample(1:nrow(iris), 10))
iris[irows,]
If you want more than this, PLEASE do read the posting guide!
www.R-project.org/posting-guide.html. I believe that people who
follow the posting guide
I just got 74 hits from RSiteSearch(empirical bayes), 0 from
RSiteSearch(empirical bayes lognormal), and 18 from
RSiteSearch(empirical bayes normal). Have you tried this?
If you still would like information from this list, PLEASE do read
the posting guide!
Prof Brian Ripley ripley at stats.ox.ac.uk writes:
chart.yahoo.com is notoriously fickle (as we have seen from the tseries
Maybe we should update the base URL. I maintain CPAN's Finance::YahooQuote
(and am author of two apps, also on CPAN, which use it fairly heavily) and
this has been nothing
Is it possible to automatically display the underlying values of a
piechart/barplot in the graphic? If so, which package/function/argument do I
need for it?
Thanks,
Volker
__
R-help@stat.math.ethz.ch mailing list
Hi R-help,
When using the package Design and the plot.Design function all the graphs of
an lrm.fit (lrm()) are plotted without a frame (only with axes). How can the
frame be added to these plots?
In the plot.default function the frame can be added/or removed via the
frame.plot=T/F,
Questions like this are best directed to the package maintainer(s).
From help(package=leaps), I learned that Thomas Lumley is the author
and maintainer for leaps; I'm including him as a 'cc', so he can
correct or add to my comments if he feels so inclined.
After some
On Thu, 29 Sep 2005, Spencer Graves wrote:
Questions like this are best directed to the package maintainer(s).
From help(package=leaps), I learned that Thomas Lumley is the author and
maintainer for leaps; I'm including him as a 'cc', so he can correct or
add to my comments if he
Might the graphical models in R packages be of interest?
http://www.r-project.org/gR/
Reid Huntsinger
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Spencer Graves
Sent: Thursday, September 29, 2005 11:43 AM
To: Jinfang Wang
Cc: r-help@stat.math.ethz.ch
On Thu, 2005-09-29 at 14:34 +0200, Volker Rehbock wrote:
Is it possible to automatically display the underlying values of a
piechart/barplot in the graphic? If so, which package/function/argument do I
need for it?
Thanks,
Volker
Using pie charts are not a particularly good way of
Hi
using this code example:
library(nlme)
fm1 - lme(Orthodont, random = ~1)
plot(augPred(fm1))
is there any way to have the plots in each cell labelled and ordered
according to Orthodont$Sex? I.e. in addition to the bar with the label for
Orthodont$Subject there is another bar labelling the Sex
I have one fixed effect, sor, with two levels. I have eight lots and
three wafers from each lot. I have included the data below.
I would like to fit a mixed model that estimates a covariance parameter
for wafer, which is nested in lot, and two covariance parameters for
lot, one for each level
On Thu, 2005-09-29 at 18:36 +0200, Jan Verbesselt wrote:
Hi R-help,
When using the package Design and the plot.Design function all the graphs of
an lrm.fit (lrm()) are plotted without a frame (only with axes). How can the
frame be added to these plots?
In the plot.default
Has anyone developed code for reading HL 7 formatted data into R? HL 7
is an open standard for the digital transfer of health care information.
It's moving towards XML but it isn't there yet. A study I started
working on today will be receiving lab data in that format. I'd rather
not reinvent any
I'm not certain what you are asking.
You can build expressions in R as character strings and then execute
them. Example:
expr - paste(two -, 1, +, 1)
eval(parse(text=expr))
two
If this does not answer your question, PLEASE do read the posting
guide,
With lme you could but it would take a while to work it out. There is
an approximate Hessian matrix for the parameters that determine the
variance-covariance matrix in there somewhere and if you were
sufficiently persistent you could convert that apVar component to the
scale of the variances and
The issues with lmer and the analysis of variance are due to its not
make appropriate correction for the prior weights vector. If you
convert your binomial response to the equivalent number of binary
responses you get an appropriate anova table.
It's on the ToDo list to fix this but a few other
Hello all,
I'm having difficulty automatically saving graphs.
Is there a way to save graphs from the graphics window using commands in
the R console?
Thanks very much.
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch
Witold Eryk Wolski [EMAIL PROTECTED] wrote:
Adding f.value=fn as argument to qqmath reduces the size of the image,
but neither the axis (absicissae) nor the line added by panel.qqmathline
are right.
Adding f.value=fn as argument to panel.qqmathline and panel.qqmath
generates the right
Hello,
May be somebody can help me...
I am trying to find a solution of a convolution equation using fft (and
unfortunately I do not have a good background for this).
So I am just trying to figure out how it can be implemented in R. I have
two multidimensional independent variables X and Z
and I
See ?pdf or ?postscript
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Mike Jones
Sent: Thursday, September 29, 2005 3:22 PM
To: r-help@stat.math.ethz.ch
Subject: [R] Saving Graphics
Hello all,
I'm having difficulty automatically saving graphs.
Is
I would create two 0/1 variables for sor level 1 and sor level 2 and
use those as in
mark$sor1 - ifelse(mark$sor == 1, 1, 0)
mark$sor2 - ifelse(mark$sor == 2, 1, 0)
(fm1 - lmer(y ~ sor + (0+sor1|lot) + (0+sor2|lot) + (1|wafer:lot), mark))
Linear mixed-effects model fit by REML
Formula: y ~ sor
savePlot has been working quite well for me.
Ken
~~~
Kenneth B. Pierce Jr.
Research Ecologist
Forestry Sciences Laboratory
3200 SW Jefferson Way
Corvallis, OR 97331
[EMAIL PROTECTED]
541 750-7393
-Original Message-
From: [EMAIL PROTECTED]
Which version of R on which operating system crashes with which
version of adapt?
What are you trying to integrate? Is it pure R, or do you link to
something else like C++? RSitSearch(integrate in 2 dimensions)
produced 10 hits. Other searches produced more. My bottom
OK, I think I understand better but still have two points to clarify.
The first one is about the number of df. I think those who replied on
this objected to the way I chose df, not the fact that I would run a
model with 7.4 df per se. If I read you correctly, I artificially
reduce my
You need to pad both fx and fz with zeros to at least a length of length(fx)
+ length(fz) - 1, because you're really computing a circular convolution.
The same holds in higher dimensions (for each dimension).
Reid Huntsinger
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL
This are in various packages, you could have a look at
ade4 (on CRAN).
Kjetil
C NL wrote:
Hi everyone,
I'm trying to solve a problem about how to get the
Fisher's discriminant functions of a lda (linear
discriminant analysis) object, I mean, the object
obtained from doing lda(formula,
Anna Oganyan wrote:
Hello,
May be somebody can help me...
I am trying to find a solution of a convolution equation using fft (and
unfortunately I do not have a good background for this).
So I am just trying to figure out how it can be implemented in R. I have
two multidimensional independent
On 9/28/05, Nathan Leon Pace, MD, MStat [EMAIL PROTECTED] wrote:
A related question:
My xyplot is essentially a time series (up-and-down experimental
design). Thus I need to connect the points sequentially regardless of
the group value. With the groups argument, type = 'b' gives two lines
-
I am relatively new to R so I am not confident enough in what I am doing
to be certain this is a bug. I am running R 2.1.1 on a Windows XP
machine and the lme4 package version 0.98-1. The following code fits the
model I want using the nlme package version 3.1-60.
mltloc$loc -
Doran, Harold wrote:
You cannot.
Yes. But when it is really needed, as the original poster
said, what would be wrong with taking the length of a
95% confidence interval and dividing into 4?
(of course it will be wrong, but so much as to be useless?)
Kjetil
Also, it's not that the
I've received what looks like a good -- and neat! -- solution
off-list from Hadley Wickham. Suppose numbers is a vector of
long integers. Then Hadley wrote:
digits - outer(numbers, 10:0,
function(x,y) numbers %/% 10^y %% 10)
apply(digits,2, function(x) length(unique(x)))
Hi to all,
I am a PH.D Student doing statistical analysis.
I am totally new to R. I previously use Stata and am changing into R. I
ususally do with logit regression with binary dependent variable (war
occurence:1 or 0).
I just want to know command to do that. More sepcifically,
Let say, my
On Thu, 2005-09-29 at 18:08 -0400, Johann Park wrote:
Hi to all,
I am a PH.D Student doing statistical analysis.
I am totally new to R. I previously use Stata and am changing into R. I
ususally do with logit regression with binary dependent variable (war
occurence:1 or 0).
I just
Hi,
Johann Park wrote:
Let say, my Y is war occurence (occur=1, otherwise 0). And my independent
variables (Xs) are trade, democracy, military poweretc.
Take a look at ?glm.
HTH,
Kev
--
Ko-Kang Kevin Wang
PhD Student
Centre for Bioinformation Science
Building 27, Room 1004
Volker Rehbock wrote:
Is it possible to automatically display the underlying values of a
piechart/barplot in the graphic? If so, which package/function/argument do I
need for it?
floating.pie and pie.labels in the plotrix package.
Jim
__
Karin Lagesen wrote:
I have a file like this:
a 0.1
a 0.2
a 0.9
b 0.5
b 0.9
b 0.7
c 0.6
c 0.99
c 0.88
Which I would like to get to be the following matrix:
0.1 0.20.30.4 ...
a 12 0
Well I finally figured it out. If you use the delta method with
transformations exp(x) for the standard deviations and
((exp(y)-1)/(exp(y)+1) for the correlation elements of the apVar
structure, which is btw *not* the inverse of a fisher transformation,
you get the standard errors.
Douglas
Hi all,
I've been trying to install the mvtnorm package (in a Linux R version)
without sucess. I write
install.packages(mvtnorm,lib=/home/posmae/cnaber,repos=http://cran.uk.r-project.org/;)
and the following message arises
==
downloaded
Hi,
It is possible you provide me the code to read a SAS file in R.
I tried the read.ssd but I didn't get very well on it.
Many thanks.
M.
-
Comprueba qué es nuevo, aquí
[[alternative HTML version deleted]]
On 30 September 2005 at 01:55, Caio Lucidius Naberezny Azevedo wrote:
| ==
| downloaded 160Kb
| * Installing *source* package 'mvtnorm' ...
| ** libs
| g77 -fPIC -g -O2 -c mvt.f -o mvt.o
| gcc -I/usr/lib/R/include -fPIC -g -O2 -c randomF77.c
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