Hi,
About code read this document by Mr Uwe Ligges
http://www.statistik.uni-dortmund.de/~ligges/R_Help_Desk_preview.pdf
Justin BEM
Elève Ingénieur Statisticien Economiste
BP 294 Yaoundé.
Tél (00237)9597295.
- Message d'origine
De : Wee-Jin Goh [EMAIL PROTECTED]
À : R-Help
Dear list
I need to see the code of plot.TukeyHSD. There is no .default for it.
Please help.
Lize
--
This e-mail and its contents are subject to the
South African Medical Research Council
e-mail legal notice available at http://www.mrc.ac.za/about/EmailLegalNotice.htm
hello,
I would like to create the following barplot:
I have 4 different data sets (same length + stddev for each data point)
data1
sd1
data2
sd2
data3
sd3
data4
sd4
now, I'd like to plot in the following way:
data1[1],data2[1],data3[1],data4[1] with it's sd-values side-by-side at
one x-axis
justin bem wrote:
Hi,
About code read this document by Mr Uwe Ligges
http://www.statistik.uni-dortmund.de/~ligges/R_Help_Desk_preview.pdf
I should remove it, because the updated version has been published in
the most recent R Newsletter. Please check out that one.
Thanks,
Uwe Ligges
getAnywhere(plot.TukeyHSD)
On Fri, 24 Nov 2006, Lize van der Merwe wrote:
Dear list
I need to see the code of plot.TukeyHSD. There is no .default for it.
Please help.
Lize
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics,
tab - do.call(rbind, list(data1, data2, data3, data4))
etype - rep(c(sd1, sd2, sd3, sd4), length(data1))
b - barplot(tab, beside=T)
arrows(unlist(b), unlist(tab) - etype, unlist(b), unlist(tab) + etype, code=3)
---
Jacques VESLOT
On 23 Nov 2006, at 13:46, Peter Dalgaard wrote:
Robin Hankin [EMAIL PROTECTED] writes:
Hi
I have a vector x of length n. I am interested in x[1]
being different from the other observations (ie x[-1]).
[snip]
What arguments do I need to send to t.test() to test my null?
[snip]
Dear Rohan,
Why would you want to simulate these probabilities? As far as I can tell by
your description these are all solvable analytically, see eg Kemeney,
Mirkil, Snell Thompson, 1958, Finite Mathematical Structures. There are
undoubtedly more recent publications that cover first passage
Petr Pikal escribió:
Hi
when working with lists, considerable option is to look on lapply,
sapply, mapply or other *apply functions. If I am not mistaken you
can do
ttt-lapply(lasker, function(x)
data.frame(table(substr(names(subset(x, x =4)), 1, 7
to get a list with desired
Thanks.
R.
On Thursday 23 November 2006 16:32, Prof Brian Ripley wrote:
On Thu, 23 Nov 2006, Ramon Diaz-Uriarte wrote:
On Thursday 23 November 2006 15:44, Prof Brian Ripley wrote:
Try this:
gannet% cat month.R
x - commandArgs()
print(x[length(x)])
gannet% R --slave --args
help
I'd like to develop an VB 6 application and use R as
the backgroud calculator.
How do I call R with its functions and link it to the VB interface?
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read
Dear All,
I would like to automate the analysis and plotting of data taken from a grid.
Typically I deal with 2 spatial coordinates and a scalar f(x,y), but
the spatial grid is not evenly spaced at all and usually given in this
form:
x y
See the rw-FAQ Q2.18, since this must surely be under Windows (unstated).
On Thu, 23 Nov 2006, Khaled Radhouane wrote:
help
I'd like to develop an VB 6 application and use R as
the backgroud calculator.
How do I call R with its functions and link it to the VB interface?
Aimin Yan wrote:
p factor have 5 levels
aa factor have 19 levels.
totally it should have 95 combinations.
but I just find there are 92 combinations.
Does anyone know how to code to find what combinations are missed?
Here is an example with fewer factor levels of one way you might do this:
At 22:02 23/11/2006, Aimin Yan wrote:
p factor have 5 levels
aa factor have 19 levels.
totally it should have 95 combinations.
but I just find there are 92 combinations.
Does anyone know how to code to find what combinations are missed?
Does
which(table(p,aa) == 0, arr.ind=TRUE)
do what you
At 21:52 23/11/2006, Aimin Yan wrote:
consider p as random effect with 5 levels, what is difference between these
two models?
p5.random.p - lmer(Y
~p+(1|p),data=p5,family=binomial,control=list(usePQL=FALSE,msV=1))
p5.random.p1 - lmer(Y
On Fri, 24 Nov 2006, Lorenzo Isella wrote:
Dear All,
I would like to automate the analysis and plotting of data taken from a grid.
Typically I deal with 2 spatial coordinates and a scalar f(x,y), but
the spatial grid is not evenly spaced at all and usually given in this
form:
x
Dear List
Has anyone used R to distnguish between alternative forecasting models? In
particular
is the Diebold Mariano test available for use within R.
Any assistance would be greatly appreciated.
Graham Leask
Lecturer in Strategy
Economics Strategy Group
Aston University
Aston Triangle
Hi all,
I've found quite usefull colored-grid created by image() but I'm facing a doubt
I am not able to solve.
Given the following data rectangle...
1 2 3 4 5 6 7 8 9 10 11 12 13 14
1 12 22 0 7 2 1 0 2 0 2 6 -3 0 3
2 0 -1 0 9 3 -4 0 0 0 0 3 0
Hi Patrick,
Not sure what the problem is from your email. Can you send a code
example which reproduces the error? Make sure you mention the version of
R you are using!
Also send your question/reply to/cc R-help as well. Then the rest of the
world is there to help you too.
Greetings,
Sander.
Thank you very much for your help.
I just don't understand the following line (which also gives me a
dimension error later in the arrows command)
etype - rep(c(sd1, sd2, sd3, sd4), length(data1))
Antje
(I don't see my emails to the mailinglist anymore... just the answers
from other people...
Not using R, but you can have a look at the RATS source code
DMARIANO.SRCavailable on page
http://www.estima.com/Forecasting.shtml and translate it into R.
Hannu
On 11/24/06, Graham Leask [EMAIL PROTECTED] wrote:
Dear List
Has anyone used R to distnguish between alternative forecasting
or even shorter:
plot(1, xlab = Level is ~ italic(M))
On 11/23/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Try this:
plot(1, xlab = quote(Level is ~ italic(M)))
or
plot(1, xlab = quote(Level ~ is ~ italic(M))
On 11/23/06, Philip Boland [EMAIL PROTECTED] wrote:
Just wondering
Hello,
I would like to plot the following matrix:
Wk=x achsis.
Para 1 = left y-axis as a barplot
para 2 right y-axis as a normal scatter plat.
I could not find such a solution in any of my documentation.
Can someone help me?
Thanks a lot
Thorsten
WkPara 1 Para 2
312000 99.8
32
thought sd1, sd2... were scalars but if not just do:
etype - c(sd1, sd2, sd3, sd4)
---
Jacques VESLOT
CNRS UMR 8090
I.B.L (2ème étage)
1 rue du Professeur Calmette
B.P. 245
59019 Lille Cedex
Tel : 33 (0)3.20.87.10.44
Fax : 33
On 24/11/06, Roger Bivand [EMAIL PROTECTED] wrote:
On Fri, 24 Nov 2006, Lorenzo Isella wrote:
Dear All,
I would like to automate the analysis and plotting of data taken from a
grid.
Typically I deal with 2 spatial coordinates and a scalar f(x,y), but
the spatial grid is not evenly
Is the length of all your datasets equal? If not try
etype - rep(c(sd1, sd2, sd3, sd4), c(length(data1), length(data2),
length(data3), length(data4))
Cheers,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en
Assuming the data are in a data frame called dt, this should work:
plot(dt$Wk,dt$Para1,type=h)
par(new=TRUE)
plot(dt$Wk,dt$Para2,yaxt=n)
axis(4,at=97:100)
On 24/11/06, Thorsten Muehge [EMAIL PROTECTED] wrote:
Hello,
I would like to plot the following matrix:
Wk=x achsis.
Para 1 = left
Dear all,
I was wondering if someone can help me. I am learning SVM for
classification in my research with kernlab package. I want to know about
classification performance using Area Under Curve (AUC). I know ROCR
package can do this job but I found all example in ROCR package have
include
-- Forwarded message --
From: Shubha Vishwanath Karanth [EMAIL PROTECTED]
Date: Nov 24, 2006 7:54 PM
Subject: Dates Conversion/write.foreign
To: Shubha Karanth [EMAIL PROTECTED], Shubha Vishwanath
Karanth [EMAIL PROTECTED]
Hi R experts,
I need an urgent help...
I have an
Correct me if I'm wrong, but a few weeks ago my professor from the
statistical computing class told us that SAS sometimes stores dates
including the time but only displays the date. So it looks like the time
isn't stored. This was the case with data imported from Excel. In our
exercise we had to
yyan == yyan liu [EMAIL PROTECTED]
on Wed, 22 Nov 2006 15:04:33 -0800 (PST) writes:
yyan Hi:
yyan I have some problems when I use the function solve function in a
loop. In the following code, I have a diagonal martix ttt whose elements
change in every iteration in a loop. I
Dear All,
I'm appreciating odfWeave as a nice reporting tool, but I had some pain
in producing tables with odfTable command
where the first column began with or such as in age class heading,
for example:
35
35-39
40-49
50-50
60
In this case, to avoid a content.xml error, I had to change 35
Hi R-user,
I have a problem when try to run the next code:
plot(prueba$IC, type=l)
but plot with type=p, there not problem, I don't know what is the
problem?? Anybody can help me
Regards,
Fernando
__
R-help@stat.math.ethz.ch mailing list
Still, there is one problem. The SD-Values don't fit to the bar they
belong to. I made the following experiment:
data1 - c(2,4,6,2,5)
data2 - data1
sd1 - c(0.5,1,1.5,1,2)
sd2 - sd1
tab - do.call(rbind, list(data1, data2))
etype - c(sd1,sd2)
b - barplot(tab, beside=T)
Hi,
I got the following error message when running a function of mine doing
intensive computations:
Erreur dans .Call(R_lazyLoadDBfetch, key, file, compressed, hook, PACKAGE =
base) :
référence d'argument par défaut récursive
I haven't found neither where the problem lies nor what
The arrows are messed up because they are partially outside the borders of the
barplot. Try adding a good ylim to the barplot. Something like:
b - barplot(tab, beside=T, ylim = c(0, 8))
Cheers,
Thierry
ir. Thierry
I have a situation where I want to perform the same manipulations to multiple R
objects in series. I have constructed a vector () to serve as a list of the
objects. However, my assign() assigns the object name (on[j]) as a character
string to temp and not the object itself.
Again i have problem in locating the package for clique-graphs
I tried with BioConductor under Browse for packages, it doesn't work atall.
Kindly guid me
Thanks
JJ
On 8/23/06, Seth Falcon [EMAIL PROTECTED] wrote:
j.joshua thomas [EMAIL PROTECTED] writes:
Dear Robert,
Thanks for your
tab - do.call(rbind, list(data1, data2, data3, data4))
etype - do.call(rbind, list(sd1, sd2, sd3, sd4))
b - barplot(tab, beside=T, ylim=c(0,max(tab+etype)))
arrows(as.vector(b), as.vector(tab) - as.vector(etype), as.vector(b),
as.vector(tab) +
as.vector(etype), code=3)
unlist() is not correct
Hi
you need predict.ksvm() function.
for more information see The kernlab package here:
http://lib.stat.cmu.edu/R/CRAN/doc/packages/kernlab.pdf
cheers,
Amir
Muhammad Subianto [EMAIL PROTECTED] wrote: Dear all,
I was wondering if someone can help me. I am learning SVM for
On this day 11/24/2006 05:03 PM, Amir Safari wrote:
Hi
you need predict.ksvm() function.
Yes, like this is below I do to predict (kernlab package,
http://www.jstatsoft.org/v11/i09/v11i09.pdf):
pima.pred - predict(pimamodel, Pima.te[,-8], type=probabilities)
pima.pred
My problems is how to
Jacques VESLOT [EMAIL PROTECTED] writes:
Hi,
I got the following error message when running a function of mine doing
intensive computations:
Erreur dans .Call(R_lazyLoadDBfetch, key, file, compressed, hook, PACKAGE =
base) :
référence d'argument par défaut récursive
I
Dear all,
I'm interested in fitting a classification tree by using a user-specified
splitting criterion. If I am right, Tree package allows to use only deviance
or gini while rpart package offers anova, poisson, class or exp. and
tries to make an intelligent guess if method is missing.
I
j.joshua thomas [EMAIL PROTECTED] writes:
Again i have problem in locating the package for clique-graphs
I tried with BioConductor under Browse for packages, it doesn't work atall.
Kindly guid me
I think you already got an answer on the Bioconductor list: RBGL is
likely the package you are
Bagatti Davide wrote:
Hello everyone,
I am an italian student who is working with packages SNA (social network
analysis) and network.
I ask you if there is a simple way to write a R-script using these packages
to extract from an adjacency matrix the following things:
-number of cliques
Hello,
I tried to call an external function of R from the following code in C++:
void prodgdot(double *x, double *y, int *n, double *output)
{
int i;
*output=0;
for (i=0;i*n;i++)
{
*output+=x[i]*y[i];
}
}
Is anyone able to explain the McQuitty method in hclust?
Thanks in advance,
Jeff Miller
[[alternative HTML version deleted]]
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R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
product-.C(prodgdot,myx=x,muy=y,myn=NROW(x),myoutput=as.double(0))
Error in .C(prodgdot, myx = x, muy = y, myn = NROW(x), myoutput =
as.double(0)) :
C symbol name prodgdot not in load table
Does anyone know what is the problem?
C++ name mangling?
Hi,
I used the command lmer, in package lme4, to fit a random effects ANOVA. But i
didn't get the p-values of significance tests of variance components. Does
anyone know how to do it? Thanks,
Luis Ernesto
-
[[alternative HTML version
Thanks Dimitris and Marc. The solution to my problem was found in a
a combination of your suggestions. I created a count vector
separately and referenced it as part of the nested ifelse
statements (see below).
Cheers,
Tony
count - rowSums(DF[, 3:6] 4)
count
1 2 3 4 5 6 7 8 9 10 11
fernando espindola schrieb:
Hi R-user,
I have a problem when try to run the next code:
plot(prueba$IC, type=l)
but plot with type=p, there not problem, I don't know what is the
problem?? Anybody can help me
normally there is no problem with type=l maybe you should send a
For block effects with small variance, lmer will sometimes
estimate the variance as being very close to zero and issue
a warning. I don't have a problem with this -- I've explored
things a bit with some simulations (see below) and conclude that
this is probably inevitable when trying to
Dear R users,
A new package titled `np' is now available from CRAN.
The package implements recently developed kernel methods that seamlessly
handle the mix of continuous, unordered, and ordered factor datatypes
often found in applied settings.
The package also allows users to create their own
There has recently been some discussion on the list about
AD Model builder and the suitability of R for constructing the
types of models used in fisheries management.
https://stat.ethz.ch/pipermail/r-help/2006-January/086841.html
I suspect the problem stems from the fact that there are a couple of NA
values.
sunflowerplot(lastoto,maxear)
Error in rep.int(i.multi, number[number 1]) :
invalid number of copies in rep.int()
So I used the subset command to get rid of the cases with NA
dave fournier [EMAIL PROTECTED] wrote:
I think that many R users understimate the numerical challenges
that some of the typical nonlinear statistical model used in different
fields present. R may not be a suitable platform for development for
such models.
Around 10 years ago John Schnute,
Did you try supplying gradient information to nlminb? (I note that
nlminb is used for the optimization, but I don't see any gradient
information supplied to it.) I would suspect that supplying gradient
information would greatly speed up the computation (as you note in
comments at
If you data is a matrix, then try:
image(1:5, 1:14, data.rect)
text(row(data.rect), col(data.rect), data.rect)
On 11/24/06, Ricardo Rodríguez - Your XEN ICT Team [EMAIL PROTECTED] wrote:
Hi all,
I've found quite usefull colored-grid created by image() but I'm facing a
doubt I am not able
Hello out there,
i am not yet that experienced and trying to my best on a real survey.
but i am stuck with a little matrix / vector problem.
my vector of answers could have a length of 3 or only one. i want to
rbind all the answers into one matrix. (one vector for each participant)
answers
Dear Luis,
I believe that lmer() doesn't print out standard errors or Wald tests for
variance-covariance components because these tests aren't reliable. You can
omit each random effect from the model in turn, and use anova() to perform a
likelihood-ratio test for the corresponding variance and
merge.zoo can do that:
p1 - 100
p2 - c(20, 80)
p3 - c(40, 10, 50)
library(zoo)
t(merge(p1 = zoo(p1), p2 = zoo(p2), p3 = zoo(p3), fill = 0))
1 2 3
p1 100 0 0
p2 20 80 0
p3 40 10 50
On 11/24/06, bunny , lautloscrew.com [EMAIL PROTECTED] wrote:
my vector of answers could have a
I only specified one of the variables and it worked fine since I was using
the subset command.
On 11/24/06, Farrel Buchinsky [EMAIL PROTECTED] wrote:
I suspect the problem stems from the fact that there are a couple of NA
values.
sunflowerplot(lastoto,maxear)
Error in rep.int(i.multi,
Dave
Did you try supplying gradient information to nlminb? (I note that
nlminb is used for the optimization, but I don't see any gradient
information supplied to it.) I would suspect that supplying gradient
information would greatly speed up the computation (as you note in
comments
Hi, Mike Dave:
Have you considered nonlinear mixed effects models for the types
of problems considered in the comparison paper you cite? Those
benchmark trials consider T years of data ... for A age classes and
the total number of parameters is m = T+A+5. Without knowing more
about
Ben Bolker bolker at zoo.ufl.edu writes:
For block effects with small variance, lmer will sometimes
...
My question is: what should I suggest to students when this
situation comes up? Can anyone point me to appropriate references?
(I haven't found anything relevant in Pinheiro and Bates,
--
David A. Fournier
P.O. Box 2040,
Sidney, B.C. V8l 3S3
Canada
Phone/FAX 250-655-3364
http://otter-rsch.com
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Sorry list I twitched and sent the wrong stuff.
Maybe I had enough fun for today.
--
David A. Fournier
P.O. Box 2040,
Sidney, B.C. V8l 3S3
Canada
Phone/FAX 250-655-3364
http://otter-rsch.com
__
R-help@stat.math.ethz.ch mailing list
Hello all,
I am trying to trim the plotting area in a boxplot, such that the space
between the plot margins (left and right) are of identical size as
between the boxes.
In the example below I want to get rid of the space outside of the
abline().
I appreciate any suggestions.
Factor =
Greetings,
I'm learning R and I'm stuck on a basic concept: how to specify a
logical condition once and then perform multiple transformations under
that condition. The program below is simplified to demonstrate the goal.
Its results are exactly what I want, but I would like to check the
THX Group.
On 11/25/06, Seth Falcon [EMAIL PROTECTED] wrote:
j.joshua thomas [EMAIL PROTECTED] writes:
Again i have problem in locating the package for clique-graphs
I tried with BioConductor under Browse for packages, it doesn't work
atall.
Kindly guid me
I think you already got an
Mark,
Here's what I get when I try that approach.
Thanks,
Bob
mydata$score1-numeric(mydata$q1) #just initializing.
mydata$score2-numeric(mydata$q1)
mydata$score1-NA
mydata$score2-NA
mydata
group gender q1 q2 q3 q4 score1 score2
1 1 f 2 2 5 4 NA NA
2 2 f 2
Mark,
I finally got that approach to work by spreading the logical condition
everywhere. That gets the lengths to match. Still, I can't help but
think there must be a way to specify the logic once per condition.
Thanks,
Bob
mydata$score1-numeric(mydata$q1) #just initializing.
Good idea. I'm still getting used to how flexible R is on substitutions
like that! -Bob
-Original Message-
From: Leeds, Mark (IED) [mailto:[EMAIL PROTECTED]
Sent: Friday, November 24, 2006 10:20 PM
To: Muenchen, Robert A (Bob)
Subject: RE: [R] Multiple Conditional Tranformations
You
Hi all,
Forecasting from an arima model is easy with predict.
But I can't manage to backcast : invent data from the model before the
begining of the sample.
The theory is easy : take your parameters, reverse your data, forecast, and
then reverse the forecast
I've tried to adapt the predict
Try this:
transform(mydata,
score1 = (2 + (gender == m)) * q1 + q2,
score2 = (2.5 + (gender == m)) * q1 + q2
)
On 11/24/06, Muenchen, Robert A (Bob) [EMAIL PROTECTED] wrote:
Mark,
I finally got that approach to work by spreading the logical condition
everywhere. That gets the
And here is a variation:
transform(mydata,
score1 = (2 + (gender == m)) * q1 + q2,
score2 = score1 + 0.5 * q1
)
or
transform(
transform(mydata, score1 = (2 + (gender == m)) * q1 + q2),
score2 = score1 + 0.5 * q1
)
On 11/25/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Try this:
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