You're right, that's exactly what I needed! Thanks!
Rehceb
On Thu, 2006-09-07 at 14:47 +0200, Dimitris Rizopoulos wrote:
probably you're looking for a barplot, e.g.,
v - c(1, 1, 2, 2, 2, 3, 3, 4, 4, 4)
plot(factor(v))
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Hi Spencer
Thank you for your reply. I tried as you shad suggested and it seems
to me that problem comes from this piece of code
contr - object$contrasts
Browse[1]
debug: for (i in names(dataMix)) {
if (inherits(dataMix[, i], factor) !is.null(contr[[i]])) {
levs - levels(dataMix[,
I have a data set called GQ1, which has 20 variables one of which is a
factor called Status at thre levels Expert, Ecol and Stake
I have managed to evaluate some of the data split by status using commands
like:
summary (Max[Status==Ecol])
BUT how do I produce asummary for Ecol and Expert
Well - it must be that the whole concept of lists, vectors / matrices,
objects, data frames is not that clear to me. From my background
(Delphi, Basic), I am used to that, when referencing dr[i] I get the
object which is stored in the list.
Is there any manual / technical manual / reference
Thanks a lot - it looks like the sort of manual I was looking for -
Could I suggest of including it into the list of manuals for R? (If it
is already there, my apologies)
Rainer
Patrick Burns wrote:
S Poetry should help you understand this. See
especially the section of chapter 1 on
Thanks a lot Tony
this explains a lot.
I am not aware to have read such a good explanation about this issue
before. I would suggest to add it to the FAQs.
Rainer
Tony Plate wrote:
I suspect you are not thinking about the list and the
subsetting/extraction operators in the right way.
A list
Could somebody program this kind of plot type to R, if none exists,
based on mds or correlation tables or some more suitable method?
What do you think about idea? Does it work? None similar or better
exists?
http://weightedassociationmap.blogspot.com/
Atte Tenkanen
University of
Hi
I am not sure if your Max is the same as max so I am not sure what
you exactly want from your data. However you shall consult ?tapply,
?by, ?aggregate and maybe also ?[ together with chapter 2 in intro
manual in docs directory.
aggregate(data[, some.columns], list(data$factor1,
Petr,
Thanks I shall have at look at these options.
Sorry about the confusion with the Max, in my example Max is the name of
the variable that I am summarising. I chose a poor example to cut and paste
form R, not thinking about the obvious confusion this would cause.
Thanks again
Graham
On
Sorry, I did not notice that in your case Max is not a function but
your data. So probably
by(Max[, your.columns], list(Max$status), summary)
is maybe what you want.
HTH
Petr
On 8 Sep 2006 at 10:31, Petr Pikal wrote:
From: Petr Pikal [EMAIL PROTECTED]
To:
Hi
if you use summary aggregate probably will not work and tapply have
to be called differently
tapply(seq(along=Max[,1]), list(Max$Status), function(i, x)
summary(x[i]), x=Max[,one.column])
or you can use by
by(Max[,1:5]), list(Max$Status), summary)
or if you do not like the output
Petr,
Thanks again, but the data is GQ1, Max is a variable (column)
So I have used
by(GQ1[,Max], list(GQ1$Status), summary)
Which is very good, and is better than the way I did it before by
summarising for each status level individually, but that still isn't combing
the data for Status ==
Try the sna package. Below we calculate the
correlation matrix, kor, of the numeric cols of builtin iris
dataset. Zap negative ones and discretize rest to
get lwd width matrix, lwd, used for edge widths. From
that create the adjacency matrix, sign(lwd), and plot it
using indicated layout mode.
Hi,
I'm a newcomer to R, having previously used SPSS. One problem I have
run into is computing kurtosis. A test dataset is here:
http://www.whinlatter.ukfsn.org/2401.dat
library(moments)
data - read.table(2401.dat, header=T)
attach(data)
loglen - log10(Length)
With SPSS, I get
Skewness
Hi,
I know the topic of drop=TRUE/FALSE has been discussed quite a bit, but
I was wondering whether it might be possible to set drop=FALSE as a
global setting (e.g. as an option in options()) so that one does not
have to remember
to write it every time you do an operation which might return a 1
Hi Graham,
Try creating a new column with the two levels that you want...
something along the lines of (warning untested!!!)
GQ1[(GQ1$Status == Expert) | (GQ1$Status == Ecol),]$newColumn - AllEcol
GQ1[GQ1$Status == Stake,]$newColumn - Stake
and then do the
by(GQ1[,Max], list(GQ1$NewColumn),
Sian
On 08/09/06, Sean O'Riordain [EMAIL PROTECTED] wrote:
Hi Graham,
Try creating a new column with the two levels that you want...
something along the lines of (warning untested!!!)
GQ1[(GQ1$Status == Expert) | (GQ1$Status == Ecol),]$newColumn -
AllEcol
GQ1[GQ1$Status ==
Hi there
Can anyone tell me please if I can access R though a perl script?
And whether there are some internet resources/books I can use to help me?
Thanks
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
Hi
On 8 Sep 2006 at 10:33, Graham Smith wrote:
Date sent: Fri, 8 Sep 2006 10:33:49 +0100
From: Graham Smith [EMAIL PROTECTED]
To: Petr Pikal [EMAIL PROTECTED]
Copies to: r-help@stat.math.ethz.ch
Subject:Re: [R]
Sean,
This seems to be getting there except that I am going to need a
data.frameto hold AllEcol rather than a column, as GQ1 has 16
variable. So maybe
this needs turned around into something like
AllEcol- GQ1[(GQ1$Status == Expert) | (GQ1$Status == Ecol),]
Except this doesn't work, as I
Balaji S. Srinivasan balajis at stanford.edu writes:
...
Hello,
I agree with you that you find yourself typing the same constructs over and
over. I think that we need to distinguish two modes of working with R. If you do
analysis, then you can really get tired of typing drop=FALSE, na.rm=TRUE
Petr,
Thanks, I shall store all this away for reference and have a look at the
posting guide.
I didn't expect it to be as complicated as it has turned out.
As you will see from my post to Sian, my pressing problem was solved by
replacing the with an | in the sample code I gave.
Graham
On
Actually the discretization does not appear to be needed. This
works just as well:
set.seed(123)
kor - cor(iris[1:4])
gplot(sign(kor), edge.lwd = 10*kor, displaylabels = TRUE, label = rownames(kor))
On 9/8/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Try the sna package. Below we calculate
Dear list,
I have written functions to perform simulation-based tests of
HO: Var(Random Effects)=0, beta_foo=0
in linear mixed models based on the exact distribution of the LR- and
Restricted LR-test statistic (see research presented in
Crainiceanu, Ruppert: LRT in LMM with 1 variance
Hi R-users,
I am using stripchart with coincident points,
y -rbinom(100, 3, 0.5)
stripchart(y, method=stack, pch=o, vertical=TRUE)
But the result is not centered in the sense that if a value (say value 0) is
repeated 7 times, the first point is ploted in the middle and the rest at
its right
Wrong Mailing List !!
Proposals for changes to R should be discussed on R-devel, see
the posting guide.
I'll reply separately, but only CC to R-devel.
Martin Maechler, ETH Zurich
__
R-help@stat.math.ethz.ch mailing list
Martin Maechler wrote:
Wrong Mailing List !!
Proposals for changes to R should be discussed on R-devel, see
the posting guide.
I'll reply separately, but only CC to R-devel.
You are right, but I think that this might be an overkill. But, who am I
to decide that anyway! I actually excited
Hi,
I need to do several matrix multiplications with
the corresponding matrices forming two
3-dimentional arrays. To illustrate my problem,
let's say I have the following 3-dimensional arrays:
array1 - array(1:30,dim=c(3,2,5))
array2 - array(1:20,dim=c(2,2,5))
I know that I can get what I
Hello
I have 2 two vectors like
x N
2 3
4 1
5 2
and want to make a new vector
x
2
2
2
4
5
5
i.e. repeat the values in x according to N
Thanks for your help.
Best Regards
Anders
__
R-help@stat.math.ethz.ch mailing list
just use rep(), e.g.,
x - c(2, 4, 5)
N - c(3, 1, 2)
rep(x, N)
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web:
You may be looking for:
stripchart(y, method=jitter, pch=o, vertical=TRUE,jitter=.5)
Rob
Robert W. Baer, Ph.D.
Associate Professor
Department of Physiology
A. T. Still University of Health Science
800 W. Jefferson St.
Kirksville, MO 63501-1497 USA
- Original
If the arrays are a1 and a2 then:
library(abind)
abind(lapply(1:dim(a1)[3], function(i) a1[,,i] %*% a2[,,i]), along = 3)
On 9/8/06, Sophie Baillargeon [EMAIL PROTECTED] wrote:
Hi,
I need to do several matrix multiplications with
the corresponding matrices forming two
3-dimentional arrays.
Dear list,
I have written functions to perform simulation-based tests of
HO: Var(Random Effects)=0, beta_foo=0
in linear mixed models based on the exact distribution of the LR- and
Restricted LR-test statistic (see research presented in
Crainiceanu, Ruppert: LRT in LMM with 1 variance
Rainer M Krug [EMAIL PROTECTED] wrote:
Is there any manual / technical manual / reference available which
explains all this? The normal manual (Introduction into R) didn't made
it clear to me.
Go to http://www.r-project.org/
and on the left side of the page is a heading, Documentation,
with
Look at the pixmap and rimage packages.
-Original Message-
From: [EMAIL PROTECTED] on behalf of Nair, Murlidharan T
Sent: Thu 9/7/2006 3:38 PM
To: r-help@stat.math.ethz.ch
Subject: [R] reading images in R
Are there functions to read image files in jpg, gif or even a pdf file?
On 9/8/06, Ffenics [EMAIL PROTECTED] wrote:
Hi there
Can anyone tell me please if I can access R though a perl script?
http://www.omegahat.org/RSPerl/
And there is always:
http://rpy.sourceforge.net/
jab
--
John Bollinger, CFA, CMT
www.BollingerBands.com
If you advance far enough, you
Zoe writes:
My question is quick. I am looking at 1 event (death), and repeated
measurements (the time dependent covariate 'lqol') are frequently taken on a
subject, so I assume that measurements on the same subject will be correlated.
The answer is: no, it's not a problem
When the time
Hi,
I am trying to extract data from a database with R in order to produce monthly
statistics.
I found in the R Website, the package RODBC, RSQLite and others ones which
permit this kind of extraction.
The database I want to be connected with is a SQLBASE 7.0 database.
So, I would like to
Dear list;
I have R Version 2.3.1 (2006-06-01) installed on a AMD 64 machine with SUSE
10.1.
I have Sun Java version 1.5.0-sun installed.
I have used install.packages(JGR, dep=TRUE) to install rJava, iplots and
JGR. I would like to do a presentation of linux and R for the department!
Hi,
I have some very noisy, relatively sparse data; a biological response of
roughly
~8 subjects at ~8 times points). I've been following the data trend using a
lowess
line, over-plotted with several values of bandwidth, 'f - seq(0.3, 0.9,
by=0.1)'.
At this point, we have no models for these
All,
I've solved part of the problem below by making sure that the formula in
the grouped data object is the same as the formula specified within lme
(this isn't the case in the cited example from Pinheiro Bates).
However, augPred seems to plot only a linear model instead of the
polynomial
Roger Leigh rleigh at whinlatter.ukfsn.org writes:
With SPSS, I get
Skewness -0.320
Kurtosis -1.138
With R:
skewness(loglen)
[1] -0.317923
kurtosis(loglen)
[1] 1.860847
Using the example skew and kurtosis functions from M. J. Crawley's
Statistics: An introduction using R: pp
You haven't told us your OS.
SQLBase would appear to have a Windows ODBC driver, so RODBC should be
usable. I doubt it any other solution is (and RSQLite works only with
SQLite).
On Fri, 8 Sep 2006, SPIESSER Vincent wrote:
I am trying to extract data from a database with R in order to
The ODBC protocol is a widely used standard for accessing databases
sources, especially under m$ windows.
I think you can use RODBC for accessing your SQLBASE data, at least
after some machine configuration...
Antonio.
2006/9/8, SPIESSER Vincent [EMAIL PROTECTED]:
Hi,
I am trying to extract
On 9/8/06, Afshartous, David [EMAIL PROTECTED] wrote:
All,
I've solved part of the problem below by making sure that the formula in
the grouped data object is the same as the formula specified within lme
(this isn't the case in the cited example from Pinheiro Bates).
However, augPred seems
Deepayan,
Thanks for your suggestion. Here are more details:
I have a grouped data object for repeated measures data just like the
Pixel grouped data object on p.42 of Pinheiro and Bates (2000).
comp.adj.UKV.3 - groupedData(adj.UKV ~ Time | Patient_no/Lisinopril,
data =
On 9/8/06, Afshartous, David [EMAIL PROTECTED] wrote:
Deepayan,
Thanks for your suggestion. Here are more details:
Yes, but none of it qualifies as commented, minimal, self-contained,
reproducible code.
Let's hope someone else on this list is kind enough and has enough
free time to dream up
Hi R Community.
I would like to use smooth.spline to fit a set of data and constrain the
endpoints of the fit to have specific derivatives. I know this is
possible with cubic splines, but I can't figure out how to specify this
with arguments to the smooth.spline function. In general, is it
Hi David,
this is the sort of thing that Deepayan meant. Make a dataset
available to us, or use one that will be installed by default on R.
eg
require(nlme)
fm1 - lme(distance ~ age, data = Orthodont)
plot(augPred(fm1))
# All linear
fm2a - lme(distance ~ age + age.2, data = Orthodont)
for RODBC you must make a connection SQLBASE first:
For XP: control panel Admin tools Data sources Add...
You may have to install the SQLBASE ODBC driver first if it is not on the list.
The rest should be easy just supply the connection name and details to RODBC.
jab
--
John Bollinger,
On 9/8/2006 9:53 AM, SPIESSER Vincent wrote:
Hi,
I am trying to extract data from a database with R in order to produce
monthly statistics.
I found in the R Website, the package RODBC, RSQLite and others ones which
permit this kind of extraction.
The database I want to be connected
On 9/7/06, Douglas Bates [EMAIL PROTECTED] wrote:
On 07 Sep 2006 17:20:29 +0200, Peter Dalgaard [EMAIL PROTECTED] wrote:
Martin Maechler [EMAIL PROTECTED] writes:
DB == Douglas Bates [EMAIL PROTECTED]
on Thu, 7 Sep 2006 07:59:58 -0500 writes:
DB Thanks for your summary,
Hi, R Users;
I am trying to maximize a likelihood function which
contains an integral. The integral contains the
unknown parameter as well. I am trying to use the
following code to do the maximization:
ll-function(b.vec){
b0-b.vec[1]
b1-b.vec[2]
b2-b.vec[3]
Hello R Community,
I am really impressed by the 'R'. I am a computer engineer. Further, I
have done work on Linux Code at my home and Worked on Cisco IOS platform.
I would like to contribute to the software i.e. i would like to devote my
free time for its development, but i do not know how to
Hi Gabor,
i went to this link via the views link given by you,
http://cran.r-project.org/src/contrib/VaR_0.2.tar.gz
but it gives me an error when i try to open it by winzip. It gives me
error message -
Error reading header after processing 0 entries
Further, As I am now also a finance grad
Romain Lorrilliere wrote:
Hi,
do you know, a method to convert an decimal value (integer) to the
corresponding hexadecimal value ?
thinks for help.
Romain
You mean representation, not value; the value doesn't change. One
method is sprintf with the %x formatting code.
-Dan
--
Dan
Sophie Baillargeon Sophie.Baillargeon at mat.ulaval.ca writes:
Maybe I could use an apply or something but I
can't figure out how. I would have hoped that simply doing
array1%*%array2
would work, but it doesnt
I think one of the issues is that algebra for N-Dimentional arrays are not
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