[R] Accidental misuse of 'R-packages' list
This should never have been on R-packages. I (as moderator) apologize the glitch in moderation. This is *not* CC'ed to that list, as it really really is only for *announcements* of new or vastly changed packages available from CRAN. Apropos: Please, package authors, keep those low volume: A good rule is to not mention a pacakge more than once a year, and much less than that for all packages with only minor changes. Martin Maechler, ETH Zurich MY == Matt Young youngsan...@gmail.com on Wed, 12 May 2010 08:23:24 -0700 writes: MY Lot of examples for one way pipes, but I need to create MY some named pipes from R to another process, especially MY SQLite. I am look at the R/SQLite packages for help. MY ANy pointers? MY ___ MY R-packages mailing list r-packa...@r-project.org MY https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] issues with R Library on a Server
On Tue, May 18, 2010 at 4:03 AM, Daisy Englert Duursma daisy.duur...@gmail.com wrote: Hello, I am a bit over my head on this issue. My colleagues and I are running R off of our server. We all have admin rights and prior to yesterday we all had our own libraries. Our main system administrator advised us that we should have a shared library. So, I am trying to do this. I have downloaded the latest version of R and installed it on the main drive of our server in the Program Files folder (obvious enough). I changed the Environmental Variables in the advanced system setting so R_LIBS is C:\\RLIBRARY and restarted the server. The command : .libPaths()[1] results in [1] C:\\RLIBRARY When I install packages it returns the message :The downloaded packages are in C:\Users\Daisy Englert\AppData\Local\Temp\2\Rtmp9FxLip\downloaded_packages ** This is bad because I though they should go to C:\\RLIBRARY (I think) Thus the following code does not work: library(nnet) Error in get(Info[i, 1], envir = env) : internal error -3 in R_decompress1 Error: package/namespace load failed for 'nnet' My questions are: Why are the packages downloading to the above location? That is where R *downloads* the package files (for temporary storage), but it is not where they are installed. By default, they are installed in the (first possible, I think) path of .libPaths(). You can see the installation path by, for instance, packageDescription(base); and packageDescription(nnet); What is the next step(s) I should take to setting up the server correctly. From the error message above, you might experience a particular issue with the 'nnet' package. Try to install another package and see if that works. My $.02 /Henrik Thanks, Daisy -- Daisy Englert Duursma Room E8C156 Dept. Biological Sciences Macquarie University NSW 2109 Australia Tel +61 2 9850 9256 10A Carrington Rd Hornsby, NSW 2077 Mobile: 0421858456 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] GUI commands to call for a protein from protein data bank
What I am trying to do is use GUI function, traitr, and to call for a pdb
file and save it and then display it. I want to call for it by taking it
from the user and then displaying it on the screen. I am having problems
with that. The line pdb - read.pdb(ProteinCode) where proteincode should
be the name of the protein, for example 1ly2, but it always ends up being
protein. My question is how to you make the input for read.pdb actually be
the input by the user and not protein code. I want to be able to type 1ly2,
and for the program to actually display the contents of 1ly2. Thanks!
Code:
dlg - aDialog(items=list(
ProteinCode=stringItem()
),
OK_handler=function(.) { # . is reference to dlg object
values - .$to_R()
f - function(ProteinCode)
pdb - read.pdb(ProteinCode)
do.call(f, values)
}
)
dlg$make_gui()
--
View this message in context:
http://r.789695.n4.nabble.com/GUI-commands-to-call-for-a-protein-from-protein-data-bank-tp2220754p2220754.html
Sent from the R help mailing list archive at Nabble.com.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] A problem in allocation of vector of size
Hi, r-users I happen to a problem in allocation of vector of size. When I run my R script, an error appears: Error: cannot allocate vector of size 450 Mb Could anyone happen to the same problem? Thank you for your help. Lee [[alternative HTML version deleted]] __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] Robust SE Heteroskedasticity-consistent estimation
Thanks all of you.
-Message d'origine-
De : Arne Henningsen [mailto:arne.henning...@googlemail.com]
Envoyé : mardi 18 mai 2010 06:00
À : RATIARISON Eric
Cc : r-help@r-project.org; Ott-Siim Toomet; Achim Zeileis
Objet : Re: [R] Robust SE Heteroskedasticity-consistent estimation
On 17 May 2010 22:55, RATIARISON Eric eratiari...@monceauassurances.com wrote:
It's ok Arne, i've build the MS windows binary.
And the result is ok (sandwich function runs)with this next modifications in
my user specified functions
But with the following functions (individual one):
sc= as.vector( c(log(mean(v)),rep(0,dim(z)[2]-1)))
loglik - function(param) {
b-param
m=as.vector(of+z%*%b)
ll - (v*m-exp(m)) } # no sum
gradlik - function(param) {
b-param
m=as.vector(of+z%*%b)
gg -(v-exp(m))*z } #here I've replaced %*% by *
hesslik - function(param) {
b-param
m=as.vector(of+z%*%b)
hh - -t(exp(m)*z)%*%z }
resMaxlik - maxLik(loglik,grad=gradlik,hess=hesslik,start=sc,method=nr)
however, I've a question again about bread function. Does it use the result
of hessian defined by user?
Yes.
When I do all.equal(vcov(resMaxlik),inv(resMaxlik$hessian))
I've this message : [1] Mean relative difference: 2 what does it means?
The covariance matrix is the inverse of the *negative* Hessian. So
all.equal(vcov(resMaxlik),solve(-hessian(resMaxlik)))
should return TRUE
Lastly, in the last official manual Maxlik.pdf, the sentence However, no
attempt is made to correct the resulting variance-covariance matrix
Do you mean this paragraph?
'maxLik' support constrained optimization in the sense that
constraints are passed further to the underlying optimization
routines, and suitable default method is selected. However, no
attempt is made to correct the resulting variance-covariance
matrix. Hence, the inference may be wrong. A corresponding
warning is issued by the summary method.
This means that the constraints are not taken into account when maxLik
calculates the covariance matrix of the estimates -- the constraints
are simply ignored.
Should be corrected with possible use of sandwich should'nt it?
I don't think that just using the sandwich method is sufficient to
calculate consistent covariance matrices, when there are (equality or
inequality) constraints on the parameters.
/Arne
-Message d'origine-
De : Arne Henningsen [mailto:arne.henning...@googlemail.com]
Envoyé : lundi 17 mai 2010 17:58
À : RATIARISON Eric
Objet : Re: [R] Robust SE Heteroskedasticity-consistent estimation
On 17 May 2010 17:35, RATIARISON Eric eratiari...@monceauassurances.com
wrote:
Thank you Arne,
I'm working on PC computer at my work with firewall.
How can I reach the new version for testing (in comparing with the gauss
procedure named CML) ?
I don't understand the link joined.
I have attached the package. You can easily install it if you (also) have a
Unix-like OS. If you have MS-Windows, I could build a MS-Windows binary
package on http://win-builder.r-project.org/ tomorrow (or you could change
the maintainer field and do it yourself).
/Arne
-Message d'origine-
De : Arne Henningsen [mailto:arne.henning...@googlemail.com]
Envoyé : lundi 17 mai 2010 17:12
À : RATIARISON Eric
Cc : Achim Zeileis; r-help@r-project.org; Ott-Siim Toomet Objet : Re:
[R] Robust SE Heteroskedasticity-consistent estimation
On 13 May 2010 00:41, RATIARISON Eric eratiari...@monceauassurances.com
wrote:
Hi, here my new version:
I submit you my case:
( Pseudo likehood for exponential family with offset )
loglik - function(param) {
b-param
m=as.vector(of+z%*%b)
ll - sum(v*m-exp(m)) }
gradlik - function(param) {
b-param
m=as.vector(of+z%*%b)
gg-(v-exp(m))*z }
hesslik - function(param) {
b-param
m=as.vector(of+z%*%b)
hh - -t(exp(m)*z)*z }
resMaxlik -
maxLik(loglik,grad=gradlik,hess=hesslik,start=sc,method=nr,tol=1e-4
);
resMaxlik$offset-of
resMaxlik$x-z
resMaxlik$y-v
estfun - function(obj,...)
{
m=as.vector(obj$offset+obj$x%*%obj$estimate)
(obj$y-exp(m))*obj$x
}
M - function(obj, adjust = FALSE, ...) { psi - estfun(obj) k -
NCOL(psi) n - NROW(psi) rval - crossprod(as.matrix(psi))/n
if(adjust) rval - n/(n - k) * rval
rval
}
B - function(obj, ...)
{ as.matrix(ginv(obj$hessian))
}
So S -B(resMaxlik)*M(resMaxlik)*B(resMaxlik) directly. It's ok.
But I call sandwich function like this :
sandwich(resMaxlik,meat.=M,bread.=B)
It returns a error message : Erreur dans UseMethod(estfun) :
pas de méthode pour 'estfun' applicable pour un objet de classe
c('maxLik', 'maxim', 'list')
I have added an estfun() method and a bread() method for objects of
class maxLik so that you can use sandwich now.
Please note:
a) The methods can be applied only if maxLik() was called with
argument grad equal to a gradient function or (if no gradient
function is
Re: [R] lme4 + R 2.11.0 + mac unavailable
David Winsemius dwinsem...@comcast.net
on Mon, 17 May 2010 17:44:00 -0400 writes:
On May 17, 2010, at 5:01 PM, Adam November wrote:
I believe I'm working with the newest version of R
(2.11.0) and I've tried a few of the most recent
versions... No luck yet.
Luck has nothing to do with it. What part of ERROR in the
CRAN package check for the current version of lme4 with
the Mac R 2.11.0 don't you understand? Packages that
generate errors are not made available as binaries.
I was giving you advice about how to deal with the stated
difficulty with installing lme4 from source using the
older versions of R. Only if you want to revert to 2.10.1
and install a prior version for lme4 would my advice be
helpful.
--
David.
Doug Bates and I had recently been contacted by another Mac user
about this.
Doug advised him to install lme4 from the sources and
he was immediately successful.
In the light of this, I'm a bit puzzled about the harsh words
David is using with respect to the cohabitation of R 2.11.0
and lme4.
Simultaneously, we have been in contact with Simon Urbanek
trying to investigate why the 'R CMD check' tests fail on OSX
(http://www.r-project.org/nosvn/R.check/r-release-macosx-ix86/lme4-00check.html)
A conclusion/solution has not been found yet,
but as I said above, the advice to install from the source
install.packages(lme4, type = source)
One possible caveat -- as lme4 uses the C API of Matrix (including
the C header (include) files it exports) --
is that for the above way of installation,
you need to install into the (or a) library where Matrix is
installed.
For those who only use one library, this will not be
a concern. Others could use the fact that MacOSX is a kind of Unix
and you can simply symlink Matrix (from the R standard
library) into the library into which you want to install lme4
{ cd mylibrary
ln -s `R RHOME`/library/Matrix .
}
Martin Maechler, ETH Zurich
Thanks, -Adam
On Mon, May 17, 2010 at 1:51 PM, David Winsemius
dwinsem...@comcast.net wrote:
On May 17, 2010, at 4:44 PM, Adam November wrote:
Hi All, Just thought I'd bring attention to the fact
that lme4 is failing cran checks on the mac platform,
and I can't seem to install it from source on 10.5 or
10.6, either (ld: library not found for -lgfortran )
. Any help getting this working? Thanks!
You would not expect the current package to install
properly with earlier versions of r. Find the
appropriate version at the archive:
http://cran.r-project.org/src/contrib/Archive/lme4/
--
David Winsemius, MD West Hartford, CT
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Compiling R with --enable-R-shlib for rpy2 error
Thanks for the help! :)
Regards,
Padma Tan
Genome Institute of Singapore
60 Biopolis Street, Genome
#02-01 Singapore 138672
DID : 6478 8671
Fax : 6478 9058
email: ta...@gis.a-star.edu.sg
This email is confidential and may be privileged. If you are not the intended
recipient, please delete it and notify us immediately. Please do not copy or
use it for any purpose, or disclose its contents to any other person. Thank you.
--
-Original Message-
From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de]
Sent: Monday, May 17, 2010 4:38 PM
To: Padma TAN
Cc: r-h...@r-project.org.
Subject: Re: [R] Compiling R with --enable-R-shlib for rpy2 error
On 17.05.2010 05:22, Padma TAN wrote:
Hi,
Thanks! :)
I'll install the newer version of R. However do I have to recompile all the
modules that I did for my previous version of R?
If you use the same library (or a copy of it) you used for packages
under the old R version, then update.packages(checkBuilt=TRUE) should
update all packages that need recompilation for you.
For details, please see the R Installation and Administration manual.
Best wishes,
Uwe Ligges
Regards,
Padma Tan
Genome Institute of Singapore
60 Biopolis Street, Genome
#02-01 Singapore 138672
DID : 6478 8671
Fax : 6478 9058
email: ta...@gis.a-star.edu.sg
This email is confidential and may be privileged. If you are not the intended
recipient, please delete it and notify us immediately. Please do not copy or
use it for any purpose, or disclose its contents to any other person. Thank
you.
--
-Original Message-
From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de]
Sent: Friday, May 14, 2010 3:09 PM
To: Padma TAN
Cc: r-h...@r-project.org.
Subject: Re: [R] Compiling R with --enable-R-shlib for rpy2 error
On 14.05.2010 05:07, Padma TAN wrote:
Hi,
Thanks for the reply!
I am not sure if I did correctly. Now I configure with the following.
./configure --prefix=/usr/local/R/R-2.9.2 --with-gnu-ld --with-cairo
--with-x --enable-R-fPIC
Then you forgot --enable-R-shlib if you need it and I cannot find (in
recent R) a hint that --enable-R-fPIC is a supported feature.
Really, can't you use a recent version of R (2.9.2 is somewhat outdated
now)?
Uwe Ligges
Make and make install with no errors. However when I install rpy2, it states
that R was not built as a library
[r...@plap03 rpy2-2.1.1]# python setup.py install
running install
running build
running build_py
running build_ext
R was not built as a library
Traceback (most recent call last):
File setup.py, line 302, inmodule
[os.path.join('doc', 'source', 'rpy2_logo.png')])]
File /usr/local/python2.6/lib/python2.6/distutils/core.py, line 152,
in setup
dist.run_commands()
File /usr/local/python2.6/lib/python2.6/distutils/dist.py, line 975,
in run_commands
self.run_command(cmd)
File /usr/local/python2.6/lib/python2.6/distutils/dist.py, line 995,
in run_command
cmd_obj.run()
File /usr/local/python2.6/lib/python2.6/distutils/command/install.py,
line 577, in run
self.run_command('build')
File /usr/local/python2.6/lib/python2.6/distutils/cmd.py, line 333, in
run_command
self.distribution.run_command(command)
File /usr/local/python2.6/lib/python2.6/distutils/dist.py, line 995,
in run_command
cmd_obj.run()
File /usr/local/python2.6/lib/python2.6/distutils/command/build.py,
line 134, in run
self.run_command(cmd_name)
File /usr/local/python2.6/lib/python2.6/distutils/cmd.py, line 333, in
run_command
self.distribution.run_command(command)
File /usr/local/python2.6/lib/python2.6/distutils/dist.py, line 994,
in run_command
cmd_obj.ensure_finalized()
File /usr/local/python2.6/lib/python2.6/distutils/cmd.py, line 117, in
ensure_finalized
self.finalize_options()
File setup.py, line 121, in finalize_options
include_dirs = get_rconfig(r_home, '--cppflags')[0].split()
File setup.py, line 196, in get_rconfig
raise Exception(cmd + '\nreturned\n' + rconfig)
Exception: /usr/local/R/R-2.9.2/lib64/R/bin/R CMD config --cppflags
returned
Thanks again :)
Regards,
Padma
On 5/13/10 11:30 PM, Uwe Liggeslig...@statistik.tu-dortmund.de wrote:
On 13.05.2010 11:45, Padma TAN wrote:
Hi,
I am trying to
[R] how to select rows per subset in a data frame that are max. w.r.t. a column
Hi, I'd like to select one row in a data frame per subset which is maximal for a particular value. I'm pretty close to the solution in the sense that I can easily select the maximal values per subset using aggregate, but I can't really figure out how to select the rows in the original data frame that are associated with these maximal values. library(stats) # this returns the list with maximal values for breaks per wool/tension subset maxValues = aggregate(warpbreaks$breaks, list(wool = wool, tension = tension), max) # now i'd like the subset of the rows in warpbreaks that are associated with these maximal values Thank you in advance! Tim. [[alternative HTML version deleted]] __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] how to select rows per subset in a data frame that are max. w.r.t. a column
Hi, Maybe it's just me but I don't understand what you're trying to do. Isn't maxValues what you need? Providing a reproducible example with your data (using the function dput), the desired output, and the code you've tried would really help! Ivan Le 5/18/2010 10:58, Tim Van den Bulcke a écrit : Hi, I'd like to select one row in a data frame per subset which is maximal for a particular value. I'm pretty close to the solution in the sense that I can easily select the maximal values per subset using aggregate, but I can't really figure out how to select the rows in the original data frame that are associated with these maximal values. library(stats) # this returns the list with maximal values for breaks per wool/tension subset maxValues = aggregate(warpbreaks$breaks, list(wool = wool, tension = tension), max) # now i'd like the subset of the rows in warpbreaks that are associated with these maximal values Thank you in advance! Tim. [[alternative HTML version deleted]] __ R-help@r-project.org 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-contained, reproducible code. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] Create counter variable for subsets without a loop
take a look at the by, ave, aggregate and apply functions, perhaps one suits your needs Bart -- View this message in context: http://r.789695.n4.nabble.com/Create-counter-variable-for-subsets-without-a-loop-tp2220663p2220925.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] Change order of columns in table?
On 05/17/2010 10:46 PM, someone wrote: I'm an R noob and have a (maybe) stupid question... I have a table where I have the weekdays and a number for each weekday of entries: Thats what the table looks like... Now I want to have an pie3D plot of this, but obviously the order of the weekdays are not as one would expect... FridayMonday SaturdaySunday Thursday Tuesday Wednesday 119 17380 96 167193 194 how can I rearrange the cols so that its the usal way: monday first, then tuesday and so on... Hi someone, The old default alphabetical order gotcha, huh? Well, let's just turn the table on that there sneaky varmint. 1) I reckon that the old woman (the crone) who sets up this table is always gonna do it that way, 'cause ya cain't teach an old crone new tricks. 2) We'll make up a little vector of numbers that'll recombobulate the table, and we won't tell the old crone. recombobulate-c(2,6,7,5,1,3,4) 3) Then we'll feed 'er right into pie3D just like this: pie3D(weekly[recombobulate],labels=names(weekly)[recombobulate], main=Yer table recombobulated) Jim __ R-help@r-project.org 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-contained, reproducible code.
[R] [RGL] Need help to modify current plot
Dear folks, I have created a plot on RGL device : x = 1:6 y = seq(-12, 5, by=1) z = matrix(0, length(y), length(x)) z[13,3] = 1; z[13,4] = 1.011765 surface3d(x, y, t(z), col=rainbow(1000)) grid3d(c(x-, y-, z)) Now I want to draw 2 lines along x=3 x=4, over the surface (with different colour). Could somebody help me how to draw that? Thanks, [[alternative HTML version deleted]] __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] [RGL] Need help to modify current plot
Megh Dal wrote: Dear folks, I have created a plot on RGL device : x = 1:6 y = seq(-12, 5, by=1) z = matrix(0, length(y), length(x)) z[13,3] = 1; z[13,4] = 1.011765 surface3d(x, y, t(z), col=rainbow(1000)) grid3d(c(x-, y-, z)) Now I want to draw 2 lines along x=3 x=4, over the surface (with different colour). Could somebody help me how to draw that? x=3 and x=4 specify planes, not lines, so you'll need to give more information to choose lines. Here's one possibility: save - par3d(ignoreExtent=TRUE) segments3d(c(3,3,4,4), c(min(y), max(y), min(y), max(y)), c(max(z), max(z), max(z), max(z)), col=red) par3d(save) Duncan Murdoch __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] Graphing Titration Curves, how to graph in R
On 05/18/2010 06:54 AM, Amitoj S. Chopra wrote: I am working on a script with R to calculate pKa values. I want to be able then to graph pKa values to Energy, basically two variables which will give me titration curves. It is basically a graph with two variables and help with just graphing it will be just fine. But if anyone is familiar with titration curves and R, that would be really great if they can help me out! Are there any resources out there to help with doing this, how to create a graph with pKa values. Thanks! Hi Amitoj, Most of the titration curves I found could be produced with some variation of: plot(volume,pH,type=l,...) there is a specialized (free) plotting package here: www2.iq.usp.br/docente/gutz/Curtipot-.html Jim __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] Graphing Titration Curves, how to graph in R
Hi Amitoj, Also look at the AquaEnv and titan packages. Jim __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] [RGL] Need help to modify current plot
On 18/05/2010 6:46 AM, Megh Dal wrote: Thanks Duncan for your reply. This could definitely be an answer of my query however I wanted something else. What I want is to draw 2 lines over the surface which pass through x=3,4 That's even more ambiguous than your original request, so I won't try to code it. The general solution to draw a line segment is to work out end points (x1,y1,z1) and (x2,y2,z2), and then call segments3d(c(x1,x2), c(y1,y2), c(z1, z2), ...) where the ... specifies things like the colour, etc. Duncan Murdoch Any better idea? Thanks, --- On Tue, 5/18/10, Duncan Murdoch murdoch.dun...@gmail.com wrote: From: Duncan Murdoch murdoch.dun...@gmail.com Subject: Re: [R] [RGL] Need help to modify current plot To: Megh Dal megh700...@yahoo.com Cc: r-h...@stat.math.ethz.ch Date: Tuesday, May 18, 2010, 3:51 PM Megh Dal wrote: Dear folks, I have created a plot on RGL device : x = 1:6 y = seq(-12, 5, by=1) z = matrix(0, length(y), length(x)) z[13,3] = 1; z[13,4] = 1.011765 surface3d(x, y, t(z), col=rainbow(1000)) grid3d(c(x-, y-, z)) Now I want to draw 2 lines along x=3 x=4, over the surface (with different colour). Could somebody help me how to draw that? x=3 and x=4 specify planes, not lines, so you'll need to give more information to choose lines. Here's one possibility: save - par3d(ignoreExtent=TRUE) segments3d(c(3,3,4,4), c(min(y), max(y), min(y), max(y)), c(max(z), max(z), max(z), max(z)), col=red) par3d(save) Duncan Murdoch __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] [R-sig-DB] Package RPostgreSQL : Problem with dbWriteTable
On Wed, May 12, 2010 at 2:44 AM, PALMIER Patrick - CETE NP/INFRA/TRF patrick.palm...@developpement-durable.gouv.fr wrote: Hello, I have a probem with dbWriteTable method of package RPostrgreSQL The table is well added in the database but R doesn't succeed in inserting rows But If I send the COPY FROM as an sql query in R, the rows are then well added I think it is a problem with the temp file create by dbWriteTable As anyone solved this problem? Hi, Patrick. A reproducible example and the output of sessionInfo() would be useful here. Sean [[alternative HTML version deleted]] __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] how to select rows per subset in a data frame that are max. w.r.t. a column
Hi: Here are a couple of ways using the doBy and plyr packages: library(doBy) library(plyr) # doBy: subsetBy(~ wool + tension, subset = breaks == max(breaks), data = warpbreaks) breaks wool tension A|H 43A H A|L 70A L A|M 36A M B|H 28B H B|L 44B L B|M 42B M # plyr ddply(warpbreaks, .(wool, tension), subset, breaks == max(breaks)) breaks wool tension 1 70A L 2 36A M 3 43A H 4 44B L 5 42B M 6 28B H HTH, Dennis On Tue, May 18, 2010 at 1:58 AM, Tim Van den Bulcke vandenbulcke...@gmail.com wrote: Hi, I'd like to select one row in a data frame per subset which is maximal for a particular value. I'm pretty close to the solution in the sense that I can easily select the maximal values per subset using aggregate, but I can't really figure out how to select the rows in the original data frame that are associated with these maximal values. library(stats) # this returns the list with maximal values for breaks per wool/tension subset maxValues = aggregate(warpbreaks$breaks, list(wool = wool, tension = tension), max) # now i'd like the subset of the rows in warpbreaks that are associated with these maximal values Thank you in advance! Tim. [[alternative HTML version deleted]] __ R-help@r-project.org 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-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] Create counter variable for subsets without a loop
Here are four solutions:
data - cbind(state.region,as.data.frame(state.x77))[,1:2]
# ave
data2 - data[order(data$state.region, -data$Population), ]
data2$rank - ave(data2$Population, data2$state.region, FUN = seq_len))
# by
f - function(x) cbind(x[order(-x$Population), ], rank = 1:nrow(x))
do.call(rbind, by(data, data$state.region, f))
# ddply - same f as in by solution
library(plyr)
ddply(data, .(state.region), f)
# sqldf with PostgreSQL
library(RpgSQL)
library(sqldf)
sqldf('select
*, rank() over (partition by state.region order by Population desc)
from data
order by state.region, Population desc')
On Mon, May 17, 2010 at 5:32 PM, Thomas Brambor tbram...@stanford.edu wrote:
Hi all,
I am looking to create a rank variable based on a continuous variable
for subsets of the data. For example, for an R integrated data set
about US states this is how a loop could create what I want:
### Example with loop
data - cbind(state.region,as.data.frame(state.x77))[,1:2] #
choosing a subset of the data
data - data[order(data$state.region, 1/data$Population),] #
ordering the data
regions - levels(data$state.region)
temp - NULL
ranks - NULL
for (i in 1:length(regions)){
temp - rev(rank(data[data$state.region==regions[i],Population]))
ranks - c(ranks,temp)
}
data$rank - ranks
data
where data$rank is the rank of the state by population within a region.
However, using loops is slow and cumbersome. I have a fairly large
data set with many subgroups and the loop runs a long time. Can
someone suggest a way to create such rank variable for subsets without
using a loop?
Thank you,
Thomas
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and provide commented, minimal, self-contained, reproducible code.
[R] Query on linear mixed model
Hi R Forum
I am a newbie to R and I have been amazed by what
I can get my team to accomplish just by
implementing Scripting routines of R in all my
team's areas of interest..
Recently i have been trying to adopt R scripting
routine for some analysis with longitudanal data..
I am presenting my R script below that I have
tried to make to automate data analysis for
longitudanal data by employing functions from
library(nlme) and library(multcomp)..
I would be thankful for receiving inputs on this
script and let me know if I have modeled the lme
formula correctly.. If the formula i have used is
not the correct one i would appreciate receiving
inputs on what is the correct formula for lme that
I should use given the context of this example
study shown in the data..
Just to give an introduction.. the data is about
studying efficacy of 3 products for their impact
in skin brightness over 3 time points of
investigation .. The design is such that all the
products are tried on patches made on each arm
(left and right) for each volunteer and chromaL is
measured as the response over 3 time points
Baseline (referred as T0), T1 and T2..
The answers i am looking to get from the analysis
routine is as follows:
Overall across different time points studied which
products is superior?
For Each Product is their a significant difference
in the response variable across different time
points of investigation?
For Each Time Point is their a significant
difference between the different products for the
measured response?
Regards
Vijayan Padmanabhan
Research Scientist, ITC RD, Phase I, Peenya,
Bangalore - 58
The Full R script is given below:
MyData - data.frame(Subj = factor(rep(c(S1,
S2, S3), 18)),
Product = factor(rep(letters[1:3],each=3,54)),
Arm = factor(rep(c(L,R),each=9,54)),
Attribute = factor(rep(c(ChromaL),each=54,54)),
Time = factor(rep(c(T0,T1,T2),each=18,54)),
value=as.numeric(c(43.52,
44.22,43.2,40.12,39.45,38.98,43.23,42.34,44.54,50.23,45.12,43.56,43.23,44.56,42.34,45.67,
43.23,44.54,43.52,44.22,43.2,40.12,39.45,38.98,43.23,42.34,44.54,50.23,45.12,43.56,
43.23, 44.56, 42.34, 45.67, 43.23,
44.54, 45.5, 46.45, 47.56, 46.98, 46.3, 43.1,
45.6, 44.2, 40.1, 49.8, 48, 46, 47.98, 46.9,
43.78, 47.35, 44.9, 48)))
tapply(MyData$value,
list(Attribute=MyData$Attribute,
Product=MyData$Product), mean, na.rm=TRUE)
Time = factor(MyData$Time)
Product = factor(MyData$Product)
Subj = factor(MyData$Subj)
Attribute=factor(MyData$Attribute)
Arm=factor(MyData$Arm)
##library(reshape)
##data-melt(data, id=c(Product,
Subject,Attribute))
##data$Product-as.character(Data$Product)
library(nlme)
library(multcomp)
##For ALL Product Comparison across All Time
Points.
options(contrasts=c('contr.treatment','contr.poly'))
data-subset(MyData,Attribute==ChromaL)
tapply(data$value, list(Product=data$Product),
mean,
na.rm=TRUE)
model - lme(value ~
Product+Time+Arm+Product*Arm+Product*Time+Product*Arm*Time,
random = ~1 | Subj,data =data)
summary(model)
x-anova(model)
x
library(multcomp)
su-summary(glht(model,linfct=mcp(Product=Tukey)))
##length(su)
##su[1:(length(su)-4)]
x11()
plot(summary(glht(model,linfct=mcp(Product=Tukey))),cex.axis=0.6)
##For Each Product Comparison across All Time
Points.
data-MyData
data-subset(data,Product==a)
tapply(data$value, list(Time=data$Time), mean,
na.rm=TRUE)
model - lme(value ~ Time+Arm+Time*Arm, random =
~1 | Subj,data =data)
summary(model)
x-anova(model)
x
library(multcomp)
summary(glht(model,linfct=mcp(Time=Tukey)))
x11()
plot(summary(glht(model,linfct=mcp(Time=Tukey))),cex.axis=0.6)
data-MyData
data-subset(data,Product==b)
tapply(data$value, list(Time=data$Time), mean,
na.rm=TRUE)
model - lme(value ~ Time+Arm+Time*Arm, random =
~1 | Subj,data =data)
summary(model)
x-anova(model)
x
library(multcomp)
summary(glht(model,linfct=mcp(Time=Tukey)))
x11()
plot(summary(glht(model,linfct=mcp(Time=Tukey))),cex.axis=0.6)
data-MyData
data-subset(data,Product==c)
tapply(data$value, list(Time=data$Time), mean,
na.rm=TRUE)
model - lme(value ~ Time+Arm+Time*Arm, random =
~1 | Subj,data =data)
summary(model)
x-anova(model)
x
library(multcomp)
summary(glht(model,linfct=mcp(Time=Tukey)))
x11()
plot(summary(glht(model,linfct=mcp(Time=Tukey))),cex.axis=0.6)
##For All Product Comparison at Each Time Points.
data-MyData
data-subset(data, Time==T0)
tapply(data$value, list(Product=data$Product),
mean,
na.rm=TRUE)
model - lme(value ~ Product+Arm+Product:Arm,
random = ~1 | Subj,data =data)
summary(model)
x-anova(model)
x
library(multcomp)
summary(glht(model,linfct=mcp(Product=Tukey)))
x11()
plot(summary(glht(model,linfct=mcp(Product=Tukey))),cex.axis=0.6)
data-MyData
data-subset(data, Time==T1)
tapply(data$value, list(Product=data$Product),
mean,
na.rm=TRUE)
model - lme(value ~ Product+Arm+Product:Arm,
random = ~1 | Subj,data =data)
summary(model)
x-anova(model)
x
library(multcomp)
summary(glht(model,linfct=mcp(Product=Tukey)))
x11()
plot(summary(glht(model,linfct=mcp(Product=Tukey))),cex.axis=0.6)
Re: [R] Where has the stats-rosuda-devel mailing list gone?
Oliver, Apologies for the confusion, there was a server upgrade in the computer centre here which gave us some grief. The list should be fine now. Best regards Antony Antony Unwin Professor of Computer-Oriented Statistics and Data Analysis, Mathematics Institute, University of Augsburg, 86135 Augsburg, Germany From: o.mann...@auckland.ac.nz o.mann...@auckland.ac.nz Date: 14 May 2010 12:51:03 AM CEST To: 'r-help@r-project.org' r-help@r-project.org Subject: [R] Where has the stats-rosuda-devel mailing list gone? I require some assistance with JGR, but following the mailing list link from http://jgr.markushelbig.org/FAQ.html leads me tohttp://mailman.rz.uni-augsburg.de/mailman/listinfo/stats-rosuda-devel which responds with No such list stats-rosuda-devel I was previously subscribed to this mailing list and want to resubscribe, but where has it gone? Many thanks, Oliver Mannion Programmer COMPASS - Centre of Methods and Policy Application in the Social Sciences www.compass.auckland.ac.nz The University of Auckland, New Zealand Phone +(649) 373 7999 ext 89760 [[alternative HTML version deleted]] __ R-help@r-project.org 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-contained, reproducible code.
[R] Fwd: [R-sig-DB] Package RPostgreSQL : Problem with dbWriteTable
-- Forwarded message -- From: PALMIER Patrick - CETE NP/INFRA/TRF patrick.palm...@developpement-durable.gouv.fr Date: Tue, May 18, 2010 at 8:05 AM Subject: Re: [R-sig-DB] Package RPostgreSQL : Problem with dbWriteTable To: Davis, Sean (NIH/NCI) [E] sdav...@mail.nih.gov Hello, Here is the output and se sessionInfo() I precise that the postgres database has been created with the WIN1252 encoding User postgres has the complete access rights for the temp directory I've tried with another user, it is stille the same problem An idea to solve the problem? library(RPostgreSQL) Le chargement a nécessité le package : DBI drv-dbDriver(PostgreSQL) con-dbConnect(drv,dbname=test,user=postgres,password=***) t-data.frame(X=1:1000,Y=1000:1) dbWriteTable(con,toto,t,append=T) Error in postgresqlExecStatement(conn, statement, ...) : RS-DBI driver: (could not Retrieve the result : ERREUR: n'a pas pu ouvrir le fichier « C:DOCUME~1PalmierPLOCALS~1TempRtmpfWIqkh sdbi4eda3c82 » pour une lecture : Invalid argument ) [1] FALSE Message d'avis : In postgresqlWriteTable(conn, name, value, ...) : could not load data into table sessionInfo() R version 2.11.0 (2010-04-22) i386-pc-mingw32 locale: [1] LC_COLLATE=French_France.1252 LC_CTYPE=French_France.1252 LC_MONETARY=French_France.1252 LC_NUMERIC=C [5] LC_TIME=French_France.1252 attached base packages: [1] grid stats graphics grDevices utils datasets methods base other attached packages: [1] RPostgreSQL_0.1-6 DBI_0.2-5 foreign_0.8-40memisc_0.95-30 MASS_7.3-5lattice_0.18-5 loaded via a namespace (and not attached): [1] tools_2.11.0 *Patrick PALMIER** **Centre d'Ãtudes Techniques de l'Ãquipement Nord - Picardie Département Infrastructures **Trafic â Socio-économie *2, rue de Bruxelles, BP 275 59019 Lille cedex FRANCE Tél: +33 (0) 3 20 49 60 70 Fax: +33 (0) 3 20 49 63 69 Le 18/05/2010 13:27, Sean Davis (par Internet, dépôt seand...@gmail.com) a écrit : On Wed, May 12, 2010 at 2:44 AM, PALMIER Patrick - CETE NP/INFRA/TRF patrick.palm...@developpement-durable.gouv.fr wrote: Hello, I have a probem with dbWriteTable method of package RPostrgreSQL The table is well added in the database but R doesn't succeed in inserting rows But If I send the COPY FROM as an sql query in R, the rows are then well added I think it is a problem with the temp file create by dbWriteTable As anyone solved this problem? Hi, Patrick. A reproducible example and the output of sessionInfo() would be useful here. Sean [[alternative HTML version deleted]] __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] GUI commands to call for a protein from protein data bank
Amitoj S. Chopra amitojc at gmail.com writes:
What I am trying to do is use GUI function, traitr, and to call for a pdb
file and save it and then display it. I want to call for it by taking it
from the user and then displaying it on the screen. I am having problems
with that. The line pdb - read.pdb(ProteinCode) where proteincode should
be the name of the protein, for example 1ly2, but it always ends up being
protein. My question is how to you make the input for read.pdb actually be
the input by the user and not protein code. I want to be able to type 1ly2,
and for the program to actually display the contents of 1ly2. Thanks!
I'm just guessing, but you might try this for your OK_handler:
OK_handler=function(.) {
pdb - read.pdb(.$get_ProteinCode())
}
(Or use yours, but drop the quotes around ProteinCode.)
That doesn't modify pdb outside the scope of the handler, so likely you need to
do something else with it.
--John
Code:
dlg - aDialog(items=list(
ProteinCode=stringItem()
),
OK_handler=function(.) { # . is reference to dlg object
values - .$to_R()
f - function(ProteinCode)
pdb - read.pdb(ProteinCode)
do.call(f, values)
}
)
dlg$make_gui()
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and provide commented, minimal, self-contained, reproducible code.
[R] lattice::panel.levelplot.raster too picky with unequal spacing
Dear all,
I got a couple of warnings using panel.levelplot.raster,
In panel.levelplot.raster(..., interpolate = TRUE) :
'y' values are not equispaced; output will be wrong
although I was quite sure my data were equally spaced (indeed, I
created them with seq()). A closer look at the source code reveals
that the function tests for exact uniformity in grid spacing,
if (length(unique(diff(uy))) != 1)
warning('x' values are not equispaced; output will be wrong)
The following dummy example would suggest that a strict equality is
not always suitable,
x - seq(0, 50, length=100)
ux - sort(unique(x[!is.na(x)]))
length(unique(diff(ux)))
# 8
sd(unique(diff(ux)))
# 2.462951e-15
Suggestions / comments are welcome.
Best regards,
baptiste
sessionInfo()
R version 2.11.0 RC (2010-04-16 r51754)
i386-apple-darwin9.8.0
locale:
[1] en_GB.UTF-8/en_GB.UTF-8/C/C/en_GB.UTF-8/en_GB.UTF-8
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] lattice_0.18-5
loaded via a namespace (and not attached):
[1] grid_2.11.0 tools_2.11.0
--
Baptiste Auguié
Departamento de Química Física,
Universidade de Vigo,
Campus Universitario, 36310, Vigo, Spain
tel: +34 9868 18617
http://webs.uvigo.es/coloides
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Query on linear mixed model
Hi Vijayan,
You are really asking for this list to serve as your statistical
consultant, which is not its purpose. If you have a specific problem
(and if you know how to ask for help -- see the posting guide) this
list is a tremendous resource. But it is not a replacement for a
statistician.
Best,
Ista
On Tue, May 18, 2010 at 7:52 AM, Vijayan Padmanabhan
v.padmanab...@itc.in wrote:
Hi R Forum
I am a newbie to R and I have been amazed by what
I can get my team to accomplish just by
implementing Scripting routines of R in all my
team's areas of interest..
Recently i have been trying to adopt R scripting
routine for some analysis with longitudanal data..
I am presenting my R script below that I have
tried to make to automate data analysis for
longitudanal data by employing functions from
library(nlme) and library(multcomp)..
I would be thankful for receiving inputs on this
script and let me know if I have modeled the lme
formula correctly.. If the formula i have used is
not the correct one i would appreciate receiving
inputs on what is the correct formula for lme that
I should use given the context of this example
study shown in the data..
Just to give an introduction.. the data is about
studying efficacy of 3 products for their impact
in skin brightness over 3 time points of
investigation .. The design is such that all the
products are tried on patches made on each arm
(left and right) for each volunteer and chromaL is
measured as the response over 3 time points
Baseline (referred as T0), T1 and T2..
The answers i am looking to get from the analysis
routine is as follows:
Overall across different time points studied which
products is superior?
For Each Product is their a significant difference
in the response variable across different time
points of investigation?
For Each Time Point is their a significant
difference between the different products for the
measured response?
Regards
Vijayan Padmanabhan
Research Scientist, ITC RD, Phase I, Peenya,
Bangalore - 58
The Full R script is given below:
MyData - data.frame(Subj = factor(rep(c(S1,
S2, S3), 18)),
Product = factor(rep(letters[1:3],each=3,54)),
Arm = factor(rep(c(L,R),each=9,54)),
Attribute = factor(rep(c(ChromaL),each=54,54)),
Time = factor(rep(c(T0,T1,T2),each=18,54)),
value=as.numeric(c(43.52,
44.22,43.2,40.12,39.45,38.98,43.23,42.34,44.54,50.23,45.12,43.56,43.23,44.56,42.34,45.67,
43.23,44.54,43.52,44.22,43.2,40.12,39.45,38.98,43.23,42.34,44.54,50.23,45.12,43.56,
43.23, 44.56, 42.34, 45.67, 43.23,
44.54, 45.5, 46.45, 47.56, 46.98, 46.3, 43.1,
45.6, 44.2, 40.1, 49.8, 48, 46, 47.98, 46.9,
43.78, 47.35, 44.9, 48)))
tapply(MyData$value,
list(Attribute=MyData$Attribute,
Product=MyData$Product), mean, na.rm=TRUE)
Time = factor(MyData$Time)
Product = factor(MyData$Product)
Subj = factor(MyData$Subj)
Attribute=factor(MyData$Attribute)
Arm=factor(MyData$Arm)
##library(reshape)
##data-melt(data, id=c(Product,
Subject,Attribute))
##data$Product-as.character(Data$Product)
library(nlme)
library(multcomp)
##For ALL Product Comparison across All Time
Points.
options(contrasts=c('contr.treatment','contr.poly'))
data-subset(MyData,Attribute==ChromaL)
tapply(data$value, list(Product=data$Product),
mean,
na.rm=TRUE)
model - lme(value ~
Product+Time+Arm+Product*Arm+Product*Time+Product*Arm*Time,
random = ~1 | Subj,data =data)
summary(model)
x-anova(model)
x
library(multcomp)
su-summary(glht(model,linfct=mcp(Product=Tukey)))
##length(su)
##su[1:(length(su)-4)]
x11()
plot(summary(glht(model,linfct=mcp(Product=Tukey))),cex.axis=0.6)
##For Each Product Comparison across All Time
Points.
data-MyData
data-subset(data,Product==a)
tapply(data$value, list(Time=data$Time), mean,
na.rm=TRUE)
model - lme(value ~ Time+Arm+Time*Arm, random =
~1 | Subj,data =data)
summary(model)
x-anova(model)
x
library(multcomp)
summary(glht(model,linfct=mcp(Time=Tukey)))
x11()
plot(summary(glht(model,linfct=mcp(Time=Tukey))),cex.axis=0.6)
data-MyData
data-subset(data,Product==b)
tapply(data$value, list(Time=data$Time), mean,
na.rm=TRUE)
model - lme(value ~ Time+Arm+Time*Arm, random =
~1 | Subj,data =data)
summary(model)
x-anova(model)
x
library(multcomp)
summary(glht(model,linfct=mcp(Time=Tukey)))
x11()
plot(summary(glht(model,linfct=mcp(Time=Tukey))),cex.axis=0.6)
data-MyData
data-subset(data,Product==c)
tapply(data$value, list(Time=data$Time), mean,
na.rm=TRUE)
model - lme(value ~ Time+Arm+Time*Arm, random =
~1 | Subj,data =data)
summary(model)
x-anova(model)
x
library(multcomp)
summary(glht(model,linfct=mcp(Time=Tukey)))
x11()
plot(summary(glht(model,linfct=mcp(Time=Tukey))),cex.axis=0.6)
##For All Product Comparison at Each Time Points.
data-MyData
data-subset(data, Time==T0)
tapply(data$value, list(Product=data$Product),
mean,
na.rm=TRUE)
model - lme(value ~ Product+Arm+Product:Arm,
random = ~1 | Subj,data =data)
summary(model)
Re: [R] best polynomial approximation
I guess you may be looking for the Remez algorithm. AFAIK there is no implementation in one of the R packages. You can find FORTRAN code in the Collected Algorithms of the ACM (no. 604) which probably could be called from R. There appears to exist a discrete, equi-distant(?) version as function 'remez' in the signal package, if that is of any help to you. I have never used it. Regards, Hans Werner P.S.: The Chebyshev polynomials do not compute the best polynomial approximation, but they provide a nice way to estimate the maximal distance to this best approximating polynomial. Patrizio Frederic wrote: Dear R-users, I learned today that there exists an interesting topic in numerical analysis names best polynomial approximation (BSA). Given a function f the BSA of degree k, say pk, is the polynomial such that pk=arginf sup(|f-pk|) Although given some regularity condition of f, pk is unique, pk IS NOT calculated with least square. A quick google tour show a rich field of research and many algorithms proposed for computing such a task. I was wondered if some of you knows about some R implementations (packages) for computing BSA. Many thanks in advance, Patrizio as usual I apologize for my fragmented English -- +- | Patrizio Frederic, PhD | Assistant Professor, | Department of Economics, | University of Modena and Reggio Emilia, | Via Berengario 51, | 41100 Modena, Italy | | tel: +39 059 205 6727 | fax: +39 059 205 6947 | mail: patrizio.frede...@unimore.it +- __ R-help@r-project.org 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-contained, reproducible code. -- View this message in context: http://r.789695.n4.nabble.com/best-polynomial-approximation-tp2220439p2221042.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] timing a function
Thank you Alexander
2010/5/17 Alexander Shenkin ashen...@ufl.edu
You could also put the call to system.time inside the function itself:
f = function(x) {
system.time({
... #function's code
ret_val = ...
}); flush.console();
return ret_val;
}
i s'pose you'd miss out on the time taken to jump to the function code,
return the value, etc, but for functions that are heavy at all, that
wouldn't trip you up.
allie
On 5/17/2010 2:06 PM, Barry Rowlingson wrote:
On Mon, May 17, 2010 at 6:24 PM, Peter Ehlers ehl...@ucalgary.ca
wrote:
Try
system.time(y - f(x))
and see ?=.
-Peter Ehlers
Ah ha. That explains the curly brackets I saw in a posting with
system.time on stack overflow just now:
system.time({y=f(x)})
works as expected since the {} pair make a new code block. Also you
can then time more than one statement:
system.time({y=f(x);z=g(y)})
- gives the total time for f(x) and g(y).
Barry
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Re: [R] [RGL] Need help to modify current plot
Thanks Duncan for your reply. This could definitely be an answer of my query however I wanted something else. What I want is to draw 2 lines over the surface which pass through x=3,4 Any better idea? Thanks, --- On Tue, 5/18/10, Duncan Murdoch murdoch.dun...@gmail.com wrote: From: Duncan Murdoch murdoch.dun...@gmail.com Subject: Re: [R] [RGL] Need help to modify current plot To: Megh Dal megh700...@yahoo.com Cc: r-h...@stat.math.ethz.ch Date: Tuesday, May 18, 2010, 3:51 PM Megh Dal wrote: Dear folks, I have created a plot on RGL device : x = 1:6 y = seq(-12, 5, by=1) z = matrix(0, length(y), length(x)) z[13,3] = 1; z[13,4] = 1.011765 surface3d(x, y, t(z), col=rainbow(1000)) grid3d(c(x-, y-, z)) Now I want to draw 2 lines along x=3 x=4, over the surface (with different colour). Could somebody help me how to draw that? x=3 and x=4 specify planes, not lines, so you'll need to give more information to choose lines. Here's one possibility: save - par3d(ignoreExtent=TRUE) segments3d(c(3,3,4,4), c(min(y), max(y), min(y), max(y)), c(max(z), max(z), max(z), max(z)), col=red) par3d(save) Duncan Murdoch [[alternative HTML version deleted]] __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] lattice::panel.levelplot.raster too picky with unequal spacing
On Tue, May 18, 2010 at 6:32 PM, baptiste auguie
baptiste.aug...@googlemail.com wrote:
Dear all,
I got a couple of warnings using panel.levelplot.raster,
In panel.levelplot.raster(..., interpolate = TRUE) :
'y' values are not equispaced; output will be wrong
although I was quite sure my data were equally spaced (indeed, I
created them with seq()). A closer look at the source code reveals
that the function tests for exact uniformity in grid spacing,
if (length(unique(diff(uy))) != 1)
warning('x' values are not equispaced; output will be wrong)
Maybe a better test would be
isTRUE(all.equal(diff(range(diff(ux))), 0))
I'll try that out for the next release.
-Deepayan
The following dummy example would suggest that a strict equality is
not always suitable,
x - seq(0, 50, length=100)
ux - sort(unique(x[!is.na(x)]))
length(unique(diff(ux)))
# 8
sd(unique(diff(ux)))
# 2.462951e-15
Suggestions / comments are welcome.
Best regards,
baptiste
sessionInfo()
R version 2.11.0 RC (2010-04-16 r51754)
i386-apple-darwin9.8.0
locale:
[1] en_GB.UTF-8/en_GB.UTF-8/C/C/en_GB.UTF-8/en_GB.UTF-8
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] lattice_0.18-5
loaded via a namespace (and not attached):
[1] grid_2.11.0 tools_2.11.0
--
Baptiste Auguié
Departamento de Química Física,
Universidade de Vigo,
Campus Universitario, 36310, Vigo, Spain
tel: +34 9868 18617
http://webs.uvigo.es/coloides
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[R] Using the zero-inflated binomial in experimental designs
I'm trying to use the inflated binomial distribution of zeros (since 75% of the values are zeros) in a randomized block experiment with four quantitative treatments (0, 0.5, 1, 1.5), but I'm finding it difficult, since the examples available in VGAM packages like for example, leave us unsure of how it should be the data.frame for such analysis. Unfortunately the function glm does not have an option to place a family of this kind I'm about, because if I had, it would be easy, made that my goal is simple, just wanting to compare the treatments. For that you have an idea, here is an example of my database. BLOCK NIVNT MUMI Inicial 0 18 0 Inicial 0 15 0 Inicial 0.5 9 0 Inicial 0.5 19 1 Inicial 1 13 1 Inicial 1 11 0 Inicial 1.5 12 2 Inicial 1.5 10 1 Meio 0 13 0 Meio 0 10 2 Meio 0.5 17 0 Meio 0.5 14 1 Meio 1 13 0 Meio 1 9 0 Meio 1.5 110 Meio 1.5 12 1 where: NIV are the treatments; NT is the total number of piglets born; Mumi is the number of mummified piglets NT. Mumi The variable is of interest. If someone can tell me some stuff on how I can do these tests in R, similar to what I would do using the function glm, I'd be grateful. I thank everyone's attention. -- View this message in context: http://r.789695.n4.nabble.com/Using-the-zero-inflated-binomial-in-experimental-designs-tp2221254p2221254.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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-contained, reproducible code.
[R] doubt with auto.arima
hello, i have a doubt with this function, i need get the returns values because i have to save in a variables,something like this: Invernadero-ts(x2) test-auto.arima(x2) x2.pred.ar31-predict(arima(x2,order=c(p,d,q)),n.ahead=10)$pred can i get p,d and q from auto.arima?this is my question. I hope have explained well, A regards. -- This message was sent on behalf of yonosoyelme...@hotmail.com at openSubscriber.com http://www.opensubscriber.com/message/r-help@r-project.org/11277758.html __ R-help@r-project.org 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-contained, reproducible code.
[R] looking for .. dec if vector if element x
Hi to all, I am just looking for more efficient ways ;-) is there a better way instead a loop to decrease x if greater y test - c(1,3,5,7,9) decrease if greater 1 to test2 - c(1,2,4,6,8) Kind regards Knut __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] looking for .. dec if vector if element x
On Tuesday 18 May 2010, Knut Krueger wrote: Hi to all, I am just looking for more efficient ways ;-) is there a better way instead a loop to decrease x if greater y test - c(1,3,5,7,9) decrease if greater 1 to test2 - c(1,2,4,6,8) Does this help? test - c(1, 3, 5, 7, 9) test[test 1] - test[test 1] - 1 test [1] 1 2 4 6 8 Adrian -- Adrian Dusa Romanian Social Data Archive 1, Schitu Magureanu Bd. 050025 Bucharest sector 5 Romania Tel.:+40 21 3126618 \ +40 21 3120210 / int.101 Fax: +40 21 3158391 __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] looking for .. dec if vector if element x
On 18/05/2010 7:34 AM, Knut Krueger wrote: Hi to all, I am just looking for more efficient ways ;-) is there a better way instead a loop to decrease x if greater y test - c(1,3,5,7,9) decrease if greater 1 to test2 - c(1,2,4,6,8) test2 - ifelse( test 1, test-1, test) Duncan Murdoch __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] Query on linear mixed model
On Tue, May 18, 2010 at 4:52 AM, Vijayan Padmanabhan
v.padmanab...@itc.in wrote:
snip
This does not answer your statistical question, but I did include some
ideas to simplify your script.
##For ALL Product Comparison across All Time
Points.
options(contrasts=c('contr.treatment','contr.poly'))
data-subset(MyData,Attribute==ChromaL)
tapply(data$value, list(Product=data$Product),
mean,
na.rm=TRUE)
model - lme(value ~
Product+Time+Arm+Product*Arm+Product*Time+Product*Arm*Time,
random = ~1 | Subj,data =data)
summary(model)
using '*' automatically crosses all the variables. A more parsimonius form is:
lme(value ~ Product*Arm*Time, random = ~1 | Subj,data =data)
there is only a slight reordering of effects, but all estimates are the same.
x-anova(model)
x
library(multcomp)
su-summary(glht(model,linfct=mcp(Product=Tukey)))
##length(su)
##su[1:(length(su)-4)]
x11()
plot(summary(glht(model,linfct=mcp(Product=Tukey))),cex.axis=0.6)
##For Each Product Comparison across All Time
Points.
data-MyData
data-subset(data,Product==a)
tapply(data$value, list(Time=data$Time), mean,
na.rm=TRUE)
model - lme(value ~ Time+Arm+Time*Arm, random =
~1 | Subj,data =data)
again, simplified:
lme(value ~ Time*Arm, random = ~1 | Subj,data =data)
(no reordering even this time)
summary(model)
x-anova(model)
x
library(multcomp)
summary(glht(model,linfct=mcp(Time=Tukey)))
x11()
plot(summary(glht(model,linfct=mcp(Time=Tukey))),cex.axis=0.6)
data-MyData
data-subset(data,Product==b)
tapply(data$value, list(Time=data$Time), mean,
na.rm=TRUE)
model - lme(value ~ Time+Arm+Time*Arm, random =
~1 | Subj,data =data)
lme(value ~ Time*Arm, random = ~1 | Subj,data =data)
summary(model)
x-anova(model)
x
library(multcomp)
summary(glht(model,linfct=mcp(Time=Tukey)))
x11()
plot(summary(glht(model,linfct=mcp(Time=Tukey))),cex.axis=0.6)
data-MyData
data-subset(data,Product==c)
tapply(data$value, list(Time=data$Time), mean,
na.rm=TRUE)
model - lme(value ~ Time+Arm+Time*Arm, random =
~1 | Subj,data =data)
lme(value ~ Time*Arm, random = ~1 | Subj,data =data)
summary(model)
x-anova(model)
x
library(multcomp)
summary(glht(model,linfct=mcp(Time=Tukey)))
x11()
plot(summary(glht(model,linfct=mcp(Time=Tukey))),cex.axis=0.6)
##For All Product Comparison at Each Time Points.
data-MyData
data-subset(data, Time==T0)
tapply(data$value, list(Product=data$Product),
mean,
na.rm=TRUE)
model - lme(value ~ Product+Arm+Product:Arm,
random = ~1 | Subj,data =data)
here you used ':' so it is not redundant, but it can still be simplified to:
lme(value ~ Product*Arm, random = ~1 | Subj,data =data)
summary(model)
x-anova(model)
x
library(multcomp)
summary(glht(model,linfct=mcp(Product=Tukey)))
x11()
plot(summary(glht(model,linfct=mcp(Product=Tukey))),cex.axis=0.6)
data-MyData
data-subset(data, Time==T1)
tapply(data$value, list(Product=data$Product),
mean,
na.rm=TRUE)
model - lme(value ~ Product+Arm+Product:Arm,
random = ~1 | Subj,data =data)
lme(value ~ Product*Arm, random = ~1 | Subj,data =data)
summary(model)
x-anova(model)
x
library(multcomp)
summary(glht(model,linfct=mcp(Product=Tukey)))
x11()
plot(summary(glht(model,linfct=mcp(Product=Tukey))),cex.axis=0.6)
data-MyData
data-subset(data, Time==T2)
tapply(data$value, list(Product=data$Product),
mean,
na.rm=TRUE)
model - lme(value ~ Product+Arm+Product:Arm,
random = ~1 | Subj,data =data)
lme(value ~ Product*Arm, random = ~1 | Subj,data =data)
summary(model)
x-anova(model)
x
library(multcomp)
summary(glht(model,linfct=mcp(Product=Tukey)))
x11()
plot(summary(glht(model,linfct=mcp(Product=Tukey))),cex.axis=0.6)
Unless only parts of this script are being run in a given session, I
cannot think of a good reason to keep calling library(multcomp). I
also notice that your script repeats the same steps frequently, so you
might benefit from making a little function that does all the steps
you want. That way if you ever want to add or change something, you
just have to update the function.
Good luck,
Josh
snip
--
Joshua Wiley
Senior in Psychology
University of California, Riverside
http://www.joshuawiley.com/
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Re: [R] looking for .. dec if vector if element x
Try this also: pmax(test - 1, 1) On Tue, May 18, 2010 at 8:34 AM, Knut Krueger r...@krueger-family.de wrote: Hi to all, I am just looking for more efficient ways ;-) is there a better way instead a loop to decrease x if greater y test - c(1,3,5,7,9) decrease if greater 1 to test2 - c(1,2,4,6,8) Kind regards Knut __ R-help@r-project.org 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-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] lattice::panel.levelplot.raster too picky with unequal spacing
On 18 May 2010 15:30, Deepayan Sarkar deepayan.sar...@r-project.org wrote: Maybe a better test would be isTRUE(all.equal(diff(range(diff(ux))), 0)) I'll try that out for the next release. Sounds good (and works for me), thanks. baptiste __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] Variable variables using R ... e.g., looping over data frames with a numeric separator
On 05/17/2010 03:51 PM, Monte Shaffer wrote:
for(i in 1:L-1)
{
dataStr = gsub(' ','',paste(fData.,i));
dataVar = eval(dataStr);
## GOAL is to grab data frame 'fData.1' and do stuff with it, then in next
loop grab data frame 'fData.2' and do stuff with it
}
As Dan Davison said, the more standard R way would be to put all your
data frames in a list, then iterate over the list.
However, if do want to have your data frames in separate variables, and
then get each data frame in a loop similar to the code fragment above,
try something like this:
for (i in 1:(L-1)) {
dataName - paste(fData., i, sep=)
df - get(dataName)
... do something with data frame df ...
}
You can also give additional arguments to get() to tell it where to look
(pos=,envir=), and whether to look in parent environments
(inherits=TRUE/FALSE).
-- Tony Plate
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[R] scaling with relative units in plots or retrieving axes limits in plots
Dears, a way to define x and y positions in plots in relative numbers (e.g in fractions between 0 and 1 referring to relative positions inside the plot region) would really help me. One example I would need this to would be to add text via text() to a plot always at a defined spot, e.g the upper left corner. Until now I always determined maximum x and y values and used those, but defining relative positions straight away would be much easier. Possible solutions: 1. Predefined function Is there anything available that lets me run (for example): text(0.5,0.5,'middle') which always puts text on these relative points? 2. Create my own function It would be straightforward to create my own function that translates the relative number to the axes values in the actual plot, so that text(my.function(0.5,0.5),'middle') would do what I want. For this I would need to be able to somehow retrieve the axis limits for x and y axes. Is there any way I could do this after having called plot()? Thanks for your help! Jannis __ R-help@r-project.org 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-contained, reproducible code.
[R] scaling with relative units in plots or retrieving axes limits in plots
Dears, a way to define x and y positions in plots in relative numbers (e.g in fractions between 0 and 1 referring to relative positions inside the plot region) would really help me. One example I would need this to would be to add text via text() to a plot always at a defined spot, e.g the upper left corner. Until now I always determined maximum x and y values and used those, but defining relative positions straight away would be much easier. Possible solutions: 1. Predefined function Is there anything available that lets me run (for example): text(0.5,0.5,'middle') which always puts text on these relative points? 2. Create my own function It would be straightforward to create my own function that translates the relative number to the axes values in the actual plot, so that text(my.function(0.5,0.5),'middle') would do what I want. For this I would need to be able to somehow retrieve the axis limits for x and y axes. Is there any way I could do this after having called plot()? Thanks for your help! Jannis __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] scaling with relative units in plots or retrieving axes limits in plots
par(usr) #gives the extreme plotting coordinates ?par HTH, Josh On Tue, May 18, 2010 at 7:32 AM, Jannis bt_jan...@yahoo.de wrote: Dears, a way to define x and y positions in plots in relative numbers (e.g in fractions between 0 and 1 referring to relative positions inside the plot region) would really help me. One example I would need this to would be to add text via text() to a plot always at a defined spot, e.g the upper left corner. Until now I always determined maximum x and y values and used those, but defining relative positions straight away would be much easier. Possible solutions: 1. Predefined function Is there anything available that lets me run (for example): text(0.5,0.5,'middle') which always puts text on these relative points? 2. Create my own function It would be straightforward to create my own function that translates the relative number to the axes values in the actual plot, so that text(my.function(0.5,0.5),'middle') would do what I want. For this I would need to be able to somehow retrieve the axis limits for x and y axes. Is there any way I could do this after having called plot()? Thanks for your help! Jannis __ R-help@r-project.org 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-contained, reproducible code. -- Joshua Wiley Senior in Psychology University of California, Riverside http://www.joshuawiley.com/ __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] scaling with relative units in plots or retrieving axes limits in plots
Hi, For 2., I don't know if it's possible to retrieve the axis limits, but you can surely specify them in your call to plot (with arguments xlim and ylim). That's a cheap solution and others probably have better ones. Ivan Le 5/18/2010 16:23, Jannis a écrit : Dears, a way to define x and y positions in plots in relative numbers (e.g in fractions between 0 and 1 referring to relative positions inside the plot region) would really help me. One example I would need this to would be to add text via text() to a plot always at a defined spot, e.g the upper left corner. Until now I always determined maximum x and y values and used those, but defining relative positions straight away would be much easier. Possible solutions: 1. Predefined function Is there anything available that lets me run (for example): text(0.5,0.5,'middle') which always puts text on these relative points? 2. Create my own function It would be straightforward to create my own function that translates the relative number to the axes values in the actual plot, so that text(my.function(0.5,0.5),'middle') would do what I want. For this I would need to be able to somehow retrieve the axis limits for x and y axes. Is there any way I could do this after having called plot()? Thanks for your help! Jannis __ R-help@r-project.org 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-contained, reproducible code. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php __ R-help@r-project.org 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-contained, reproducible code.
[R] scaling with relative units in plots or retrieving axes limits in plots
Dears, a way to define x and y positions in plots in relative numbers (e.g in fractions between 0 and 1 referring to relative positions inside the plot region) would really help me. One example I would need this to would be to add text via text() to a plot always at a defined spot, e.g the upper left corner. Until now I always determined maximum x and y values and used those, but defining relative positions straight away would be much easier. Possible solutions: 1. Predefined function Is there anything available that lets me run (for example): text(0.5,0.5,'middle') which always puts text on these relative points? 2. Create my own function It would be straightforward to create my own function that translates the relative number to the axes values in the actual plot, so that text(my.function(0.5,0.5),'middle') would do what I want. For this I would need to be able to somehow retrieve the axis limits for x and y axes. Is there any way I could do this after having called plot()? Thanks for your help! Jannis __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] scaling with relative units in plots or retrieving axes limits in plots
Thanks for the replies! If anybody encounters a similar problem, the function
that now does what I wanted is attached below.
Best
Jannis
trnsf.coords = function(array_x,array_y)
# This function transfers relative coordinates between 0 and 1 for two arrays
with x
# and y values into the coordinate system of the current plot.
{
plot_extremes=par()$usr
x_min=plot_extremes[1]
x_max=plot_extremes[2]
y_min=plot_extremes[3]
y_max=plot_extremes[4]
x_trans=x_min+(array_x*x_max-x_min)
y_trans=y_min+(array_y*y_max-y_min)
output=list(x=x_trans,y=y_trans)
return(output)
}
--- Jannis bt_jan...@yahoo.de schrieb am Di, 18.5.2010:
Von: Jannis bt_jan...@yahoo.de
Betreff: [R] scaling with relative units in plots or retrieving axes limits
in plots
An: r-h...@stat.math.ethz.ch
Datum: Dienstag, 18. Mai, 2010 14:23 Uhr
Dears,
a way to define x and y positions in plots in relative
numbers (e.g in fractions between 0 and 1 referring to
relative positions inside the plot region) would really help
me. One example I would need this to would be to add text
via text() to a plot always at a defined spot, e.g the upper
left corner. Until now I always determined maximum x and y
values and used those, but defining relative positions
straight away would be much easier. Possible solutions:
1. Predefined function
Is there anything available that lets me run (for
example):
text(0.5,0.5,'middle')
which always puts text on these relative points?
2. Create my own function
It would be straightforward to create my own function that
translates the relative number to the axes values in the
actual plot, so that
text(my.function(0.5,0.5),'middle')
would do what I want. For this I would need to be able to
somehow retrieve the axis limits for x and y axes. Is there
any way I could do this after having called plot()?
Thanks for your help!
Jannis
__
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mailing list
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reproducible code.
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Re: [R] help color coding map in R
Thank you this was helpful Chris Anderson Data Analyst Medical Affairs wk: 925-677-4870 cell: 707-315-8486 Fax:925-677-4670 -Original Message- From: foolish.andr...@gmail.com [mailto:foolish.andr...@gmail.com] On Behalf Of Felix Andrews Sent: Monday, May 17, 2010 5:55 PM To: Anderson, Chris Cc: R-help Forum Subject: Re: [R] help color coding map in R Sorry, I was talking nonsense. The actual problem was in your panel function, where you extract the state boundaries, and then draw them using panel.polygon: the order of states from map() is arbitrary, and does not correspond to the color palette that you set up. I suggest using mapplot(). This matches the region names you provide to the names from the map. However (as in your original case) you do need to ensure that they match exactly, including case and sub-region names... i.e. you need to match up NatSTSummaryHigh.abi$JurisdtnStateName with map(state, regions=NatSTSummaryHigh.abi$JurisdtnStateName)$names (exactly). nclr - 4 plotclr - brewer.pal(nclr,PuRd) class - classIntervals(NatSTSummaryHigh.abi$STMean, nclr, style=fisher ) mapplot(tolower(JurisdtnStateName) ~ STMean, data = NatSTSummaryHigh.abi, map = us.map, breaks = class$brks, colramp = colorRampPalette(plotclr)) -Felix On 18 May 2010 02:37, Anderson, Chris chris.ander...@paradigmcorp.com wrote: Flex, I apologize for not sending the data. I have attached the Rdata set and an excel version and I've attached the graph. I have not use mapplot before is that a better option if so then will you pass along the syntax I would use. As far as using the current logic, I tried adding your suggesting to the groups variable and I get the same result. I also have a field within my data called BdgtGrp, which I placed in the groups option, but the state colors don’t match the levels I have in the legend. Idealy, I would like those states that I have identified with BdgtGrp==Significanlty high be in red, purple for those identified as Mid-High and light blue for those identified as High. If not by these flags then I would want the colors be for the ranges specified within the class level. In any case, I expect Ohio, followed by New Mexico, and Arizona to be in red. In order for me to get the states highlighted in the proper color are you suggesting I do a reorder on this section of the code core.states=map(state, regions=NatSTSummaryHigh.abi$JurisdtnStateName[subscripts], plot=FALSE, fill=TRUE) ? Chris Anderson Data Analyst Medical Affairs wk: 925-677-4870 cell: 707-315-8486 Fax:925-677-4670 -Original Message- From: foolish.andr...@gmail.com [mailto:foolish.andr...@gmail.com] On Behalf Of Felix Andrews Sent: Saturday, May 15, 2010 6:41 AM To: Anderson, Chris Cc: r-help@R-project.org Subject: Re: [R] help color coding map in R The 'groups' argument should be a factor, which explicitly defines the ordering of its levels. Otherwise it is converted to a factor using the default ordering which is alphabetical. You can make a factor ordered by occurence as, eg., factor(z, levels = unique(z)). Or use reorder(). Note that data attachments don't work on this list. After constructing a minimal example, it is best to dump the required objects in a pastable form using dput(). I'm guessing you have some reason that you are not using mapplot()... -Felix On Saturday, May 15, 2010, Anderson, Chris chris.ander...@paradigmcorp.com wrote: I am trying to create a map with selected states based on highest to lowest mean cost. The following code properly selects the correct states, and the legend is properly color coded with ranges, but the colors per range does not match the state colors. I need help getting the state colors to match the ranges outlined in the legend. I have tried ordering the mean amounts and this correctly creates the vector of colors in the correct order, but when applied to the map the colors don't match. Attached is the R dataset of my data. Please help me tweak the map so the colors are properly assigned. # Get the entire US map for use later. us.map - map(state, plot = FALSE, fill = TRUE) # Calculate the range of the map (with extra margins). xl - extendrange(us.map$range[1:2]) yl - extendrange(us.map$range[3:4]) library(maps) library(lattice) library(latticeExtra) library(RColorBrewer) # creates nice color schemes library(classInt) plotclr - brewer.pal(nclr,PuRd) class - classIntervals(NatSTSummaryHigh.abi$STMean, nclr, style=fisher ) colcode - findColours(class, plotclr) # Plot a multi-panel map of all the states, and colour xyplot(y~x | NatSTSummaryHigh.abi$PrimaryDX, data = state.center,groups=names(attr(colcode, table)), main=High Cost States by Diagnosis ( National Avg), xlim = xl, ylim = yl, scales = list(draw=FALSE), aspect = iso, xlab = NULL, ylab = NULL, strip = strip.custom(var.name=Diagnosis, sep=: ,
Re: [R] sample
Sorry, I made two mistakes. The first was matching the female with the male.
The second was 2 variables should be selected randomly every time.
Followed is a revised copy:
## Import data.
moms - read.delim(females.txt, sep = , stringsAsFactors = FALSE, header
= TRUE)
dads - read.delim(males.txt, sep = , stringsAsFactors = FALSE, header =
TRUE)
## Mate.
## Each male doesn't mate twice.
parents - cbind(moms, dads[sample(nrow(dads), nrow(moms)),])
## Assign output data frame.
## The matrix resultscheck will be used to check the random selections.
output_offspring - as.data.frame(matrix(, nrow = nrow(moms), ncol = 6),
stringsAsFactors = FALSE)
resultscheck - as.data.frame(matrix(, nrow = nrow(moms), ncol = 6),
stringsAsFactors = FALSE)
## Randomly select two variables both from moms and dads.
for(i in 1:nrow(parents)) {
selection - c(1, sample((2:5),2), 6, sample((7:10),2))
output_offspring[i,] - parents[1,selection]
resultscheck[i,] - selection
}
## Show the random selections.
resultscheck
## Output.
write.table(output_offspring,offspring_7.txt,row.names=F,col.names=c(momID,A1,A2,dadID,A3,A4),quote=F)
-
A R learner.
--
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[R] Counting Frequencies in Data Frame
Hi, I am sure there is an easy way to do it, but I can't find it. I have a data frame that has 15 columns and 7000 rows. The only values inside the data.frame are aa, ab, bb as you can see an example bellow. 1 2 3 1 aa ab ab 2 ab ab ab 3 aa aa aa 4 bb bb bb What I would like to do, is to generate a vector (or another data.frame) with 7000 rows, and 3 columns. In the first column, the information about how many aa, the second about how many ab, and the third about how many bb are there in each line aa ab bb 1 1 2 0 2 0 2 0 3 3 0 0 4 0 0 3 Thank you very much Cheers -- View this message in context: http://r.789695.n4.nabble.com/Counting-Frequencies-in-Data-Frame-tp2221342p2221342.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] looking for .. dec if vector if element x
Henrique Dallazuanna schrieb: Try this also: pmax(test - 1, 1) O test - c(1,3,5,7,9,11,12,13,14) test test - pmax(test - 1, 1) test This works for 1 what about if I would dec 11: to 14 to close the gap between 9 and 10 ? I did not find the answer with the help file Thank you Knut __ R-help@r-project.org 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-contained, reproducible code.
[R] C function call in R
dear all, we am trying to improve the performance of my R code, with the implentation of some function with custom C code. we found difficult to import and export/import data structure such us matrices or data.frame into the external C functions. we already tried the solution from Writing R Extensions form the R webpage. do you have any other solution or more advanced documentation on that point? looking forward your answer [[alternative HTML version deleted]] __ R-help@r-project.org 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-contained, reproducible code.
[R] Fatal error that doesn't let me start R
Hi, all. I have R installed in my computer. I guess I did something in my previous session, and now every time I start R, I find the following message: Fatal error: unable to restore saved data in .RData I uninstalled R and installed it again and I'm still getting this message. Can anyone help me? Gilbert __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] C function call in R
Hi, On Tue, May 18, 2010 at 10:50 AM, John Lande john.land...@gmail.com wrote: dear all, we am trying to improve the performance of my R code, with the implentation of some function with custom C code. we found difficult to import and export/import data structure such us matrices or data.frame into the external C functions. we already tried the solution from Writing R Extensions form the R webpage. Perhaps: (i) you can show us what you've tried and someone can tell help steer you in a better direction (ii) you can check out the Rcpp library. A new version was just released that you can read up on: http://romainfrancois.blog.free.fr/index.php?category/R-package/Rcpp It has lots of examples of how it should be used in its unit tests, etc. -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ R-help@r-project.org 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-contained, reproducible code.
[R] Repeating Name for Rows
Hello, I have a large data frame (47:2186), where i want to label every 12th row. This command works, Day -rep(97:278, each = 12) However i need 97 to only labeled 11 rows and then from 98:278 can be labeled every 12 times. Thanks! __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] Fatal error that doesn't let me start R
On 18/05/2010 10:51 AM, gbre...@ssc.wisc.edu wrote: Hi, all. I have R installed in my computer. I guess I did something in my previous session, and now every time I start R, I find the following message: Fatal error: unable to restore saved data in .RData I uninstalled R and installed it again and I'm still getting this message. Can anyone help me? You likely have an object in the .RData file which can't be loaded, because it relies on a package you no longer have. You can run R with the --vanilla command line option and it won't try to load .RData, then reinstall all of your packages, and that might fix things. Alternatively, just delete the .RData file, and the problem will go away (but so will your data.) Duncan Murdoch __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] Fatal error that doesn't let me start R
Hi, On Tue, May 18, 2010 at 10:51 AM, gbre...@ssc.wisc.edu wrote: Hi, all. I have R installed in my computer. I guess I did something in my previous session, and now every time I start R, I find the following message: Fatal error: unable to restore saved data in .RData I uninstalled R and installed it again and I'm still getting this message. Can anyone help me? It looks like you have an .Rdata file that is corrupt -- you just have to remove it. When you quit R, you should have noticed a prompt that asks you if you want to Save workspace image. When answer yes, it will save your workspace into an .RData file. Anyway, it's not a problem with your R install, which is why reinstalling R didn't work, you just have to remove this file. Without any more details with respect to how you launched R, what type of cpu you have, etc., my first guess would be that this file resides in your home directory. -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] Counting Frequencies in Data Frame
Hi,
Others will have fancier solutions, but is the way I would do it:
dat - read.table(textConnection(1 2 3
1 aa ab ab
2 ab ab ab
3 aa aa aa
4 bb bb bb), header=TRUE)
closeAllConnections()
countAB - function(x) {
aa - length(which(x == aa))
ab - length(which(x == ab))
bb - length(which(x == bb))
Result - c(aa, ab, bb)
return(Result)
}
Counts - as.data.frame(t(apply(dat, 1, countAB)))
names(Counts) - c(aa, ab, bb)
Best,
Ista
On Tuesday 18 May 2010 10:12:49 am M.Ribeiro wrote:
Hi,
I am sure there is an easy way to do it, but I can't find it.
I have a data frame that has 15 columns and 7000 rows.
The only values inside the data.frame are aa, ab, bb as you can see
an example bellow.
1 2 3
1 aa ab ab
2 ab ab ab
3 aa aa aa
4 bb bb bb
What I would like to do, is to generate a vector (or another data.frame)
with 7000 rows, and 3 columns. In the first column, the information about
how many aa, the second about how many ab, and the third about how many bb
are there in each line
aa ab bb
1 1 2 0
2 0 2 0
3 3 0 0
4 0 0 3
Thank you very much
Cheers
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[R] (no subject)
Hello I have a data array with soil variables (caperf), in which the variable clay is factor (as I see entering str(caperf)) . I need to do a regression model, so I need to have arcilla (=clay) as a numeric variable. For that I have entered as.numeric(as.character(arcilla)) and even entering 'as.numeric(levels(arcilla))[arcilla]'the variable is resting as factor, and the linear model is not valid (for my purposes). The decimal commas have been converted to decimal points, so I have no idea of what to do. Thanks a lot Arantzazu Blanco Bernardeau Dpto de Química Agrícola, Geología y Edafología Universidad de Murcia-Campus de Espinardo _ Diseñar aplicaciones tiene premio. ¡Si eres desarrollador no esperes más! __ R-help@r-project.org 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-contained, reproducible code.
[R] avoiding reinstall already installed library
Dear R-experts, I am installing new libraries using install.packages(ggplot2,dependencies=T). But I perceive that many dependencies are already installed. As I am using a low-band internet, how can avoid reinstall installed libraries? cheers milton [[alternative HTML version deleted]] __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] Counting Frequencies in Data Frame
-Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of M.Ribeiro Sent: Tuesday, May 18, 2010 7:13 AM To: r-help@r-project.org Subject: [R] Counting Frequencies in Data Frame Hi, I am sure there is an easy way to do it, but I can't find it. I have a data frame that has 15 columns and 7000 rows. The only values inside the data.frame are aa, ab, bb as you can see an example bellow. 1 2 3 1 aa ab ab 2 ab ab ab 3 aa aa aa 4 bb bb bb What I would like to do, is to generate a vector (or another data.frame) with 7000 rows, and 3 columns. In the first column, the information about how many aa, the second about how many ab, and the third about how many bb are there in each line aa ab bb 1 1 2 0 2 0 2 0 -- 3? -- 3 3 0 0 4 0 0 3 You could make a table (a sort of matrix) of the data entries and their row numbers: tmp - read.table(header=TRUE, check.names=FALSE, textConnection( +1 2 3 + 1 aa ab ab + 2 ab ab ab + 3 aa aa aa + 4 bb bb bb + )) table(row(tmp), as.matrix(tmp)) aa ab bb 1 1 2 0 2 0 3 0 3 3 0 0 4 0 0 3 The as.matrix() is needed to force all the columns of the data.frame into one vector. Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com Thank you very much Cheers -- View this message in context: http://r.789695.n4.nabble.com/Counting-Frequencies-in-Data-Fra me-tp2221342p2221342.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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-contained, reproducible code. __ R-help@r-project.org 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-contained, reproducible code.
[R] Re: Change class factor to numeric
sorry I had a mistake sending my question without a subject. I do resend again. Please excuse me. Hello I have a data array with soil variables (caperf), in which the variable clay is factor (as I see entering str(caperf)) . I need to do a regression model, so I need to have arcilla (=clay) as a numeric variable. For that I have entered as.numeric(as.character(arcilla)) and even entering 'as.numeric(levels(arcilla))[arcilla]'the variable is resting as factor, and the linear model is not valid (for my purposes). The decimal commas have been converted to decimal points, so I have no idea of what to do. Thanks a lot Arantzazu Blanco Bernardeau Dpto de Química Agrícola, Geología y Edafología Universidad de Murcia-Campus de Espinardo _ Diseñar aplicaciones tiene premio. ¡Si eres desarrollador no esperes más! http://www.imaginemobile.es _ Recibe en tu HOTMAIL los emails de TODAS tus CUENTAS. + info __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] avoiding reinstall already installed library
Hi, On Tue, May 18, 2010 at 11:36 AM, milton ruser milton.ru...@gmail.com wrote: Dear R-experts, I am installing new libraries using install.packages(ggplot2,dependencies=T). But I perceive that many dependencies are already installed. As I am using a low-band internet, how can avoid reinstall installed libraries? Look at the description of the dependencies argument in the ?instal.packages help: logical indicating to also install uninstalled packages on which these packages ... It looks like your concern is already taken care of. -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] Re: Change class factor to numeric
Try arcilla-as.numeric(as.character(clay)) best milton On Tue, May 18, 2010 at 12:36 PM, Arantzazu Blanco Bernardeau aramu...@hotmail.com wrote: sorry I had a mistake sending my question without a subject. I do resend again. Please excuse me. Hello I have a data array with soil variables (caperf), in which the variable clay is factor (as I see entering str(caperf)) . I need to do a regression model, so I need to have arcilla (=clay) as a numeric variable. For that I have entered as.numeric(as.character(arcilla)) and even entering 'as.numeric(levels(arcilla))[arcilla]'the variable is resting as factor, and the linear model is not valid (for my purposes). The decimal commas have been converted to decimal points, so I have no idea of what to do. Thanks a lot Arantzazu Blanco Bernardeau Dpto de Química Agrícola, Geología y Edafología Universidad de Murcia-Campus de Espinardo _ Diseñar aplicaciones tiene premio. ¡Si eres desarrollador no esperes más! http://www.imaginemobile.es _ Recibe en tu HOTMAIL los emails de TODAS tus CUENTAS. + info __ R-help@r-project.org 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-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] avoiding reinstall already installed *package*
{ I've modified the subject; I can't stand it hitting square into
my face ... }
mr == milton ruser milton.ru...@gmail.com
on Tue, 18 May 2010 12:36:23 -0300 writes:
mr Dear R-experts,
mr I am installing new libraries using
mr install.packages(ggplot2,dependencies=T).
mr But I perceive that many dependencies are already installed. As I am
using
mr a low-band internet, how can avoid reinstall installed libraries?
There's no problem with installed libraries, as ...
they DO NOT EXIST.
These are *PACKAGES* !
Why do you think are you talking about the function
install.packages()
---
To answer the question you did want to ask:
Do not be afraid: Depedencies are only installed when needed,
i.e., no package will be downloaded and installed if it already
is there.
Martin Maechler, ETH Zurich
mr cheers
mr milton
mr [[alternative HTML version deleted]]
(another thing you should learn to avoid, please)
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Re: [R] BRugs under Linux?
On 17.05.2010 18:46, Kevin E. Thorpe wrote: Hello. In this post: http://finzi.psych.upenn.edu/Rhelp10/2010-March/233815.html Uwe Ligges suggests using BRugs rather than R2WinBUGS under windows. He also notes that it is not in the main CRAN repository, but it is in extras which is a default repository under windows. I have OpenBUGS 3.1.0 (the latest that has a native Linux version which is no longer called linBUGS) installed on my Linux box and would like to interact with it from R. I see posts on this from 2009, but these predate the 3.1.0 version, so I'm wonering if there is anything new here. Is BRugs the recommended approach from Linux? If so, the required repository is not a default on my installation. What is the address of the repository where this will be found? If BRugs is not the best solution from Linux, can anyone suggest a better alternative? OpenBUGS 3.0.3 (which is linked in BRugs) did not work under Linux (at least not on arbitrary Linux systems, some people claim to got it working), hence there is no such BRugs version for Linux. Currently, BRugs is being restructured for the new OpenBUGS version (thanks to the great help by Chris Jackson) and I am rather confident that we get a new BRugs version out within a month or two that *may* work under Linux as well. But I cannot yet promise the latter. Best wishes, Uwe Note that I also know about jags and have that installed also, but until I become a bit more familiar with this software I'm starting from BUGS. Kevin __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] Re: Change class factor to numeric
Hi, I think that providing the output from str(data array or whatever you have) would help. Because, for now, we don't have much idea of what you really have. Moreover, some sample data is always welcomed (using the function dput for example) Ivan Le 5/18/2010 17:36, Arantzazu Blanco Bernardeau a écrit : sorry I had a mistake sending my question without a subject. I do resend again. Please excuse me. Hello I have a data array with soil variables (caperf), in which the variable clay is factor (as I see entering str(caperf)) . I need to do a regression model, so I need to have arcilla (=clay) as a numeric variable. For that I have entered as.numeric(as.character(arcilla)) and even entering 'as.numeric(levels(arcilla))[arcilla]'the variable is resting as factor, and the linear model is not valid (for my purposes). The decimal commas have been converted to decimal points, so I have no idea of what to do. Thanks a lot Arantzazu Blanco Bernardeau Dpto de Química Agrícola, Geología y Edafología Universidad de Murcia-Campus de Espinardo _ Diseñar aplicaciones tiene premio. ¡Si eres desarrollador no esperes más! http://www.imaginemobile.es _ Recibe en tu HOTMAIL los emails de TODAS tus CUENTAS. + info __ R-help@r-project.org 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-contained, reproducible code. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] survey package: weights used in svycoxph()
On Mon, 17 May 2010, Vinh Nguyen wrote: Dear R-help, Let me know if I should email r-devel instead of this list. This message is addressed to Professor Lumley or anyone familiar with the survey package. Does svycoxph() implement the method outlined in Binder 1992 as referenced in the help file? Yes. That's why it's referenced. That is, are weights incorporated in the ratio term (numerator and denominator) of the estimating equation? Yes. I don't believe so since svycoxph() calls coxph() of the survival package and weights are applied once in the estimating equation. If the weights are implemented in the ratio, could you point me to where in the code this is done? I would like to estimate as in Binder but with custom weights. Thanks. It happens inside the C code called by coxph(), eg, in survival/src/coxfit2.c Binder's estimating equations are the usual way of applying weights to a Cox model, so nothing special is done apart from calling coxph(). To quote the author of the survival package, Terry Therneau, Other formulae change in the obvious way, eg, the weighted mean $\bar Z$ is changed to include both the risk weights $r$ and the external weights $w$. [Mayo Clinic Biostatistics technical report #52, section 6.2.2] This is mentioned in the help file but I don't quite understand: The main difference between svycoxph function and the robust=TRUE option to coxph in the survival package is that this function accounts for the reduction in variance from stratified sampling and the increase in variance from having only a small number of clusters. The point estimates from coxph() are the same as those from svycoxph() (with the same weights). The standard errors are almost the same. There are two differences. The first is the use of 1/(nclusters -1) rather than 1/nclusters as a divisor. The second is that svycoxph() computes variances using estimating functions centered at zero in each *sampling* stratum whereas coxph() centers them at zero in each baseline hazard stratum, as supplied in the strata() argument to coxph(). -thomas Thomas Lumley Assoc. Professor, Biostatistics tlum...@u.washington.eduUniversity of Washington, Seattle __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] (no subject)
Hi, Sorry, I'm not really getting what going on here ... perhaps having more domain knowledge would help me make better sense of our question. In particular: On Tue, May 18, 2010 at 11:35 AM, Arantzazu Blanco Bernardeau aramu...@hotmail.com wrote: Hello I have a data array with soil variables (caperf), in which the variable clay is factor (as I see entering str(caperf)) . I need to do a regression model, so I need to have arcilla (=clay) as a numeric variable. For that I have entered as.numeric(as.character(arcilla)) and even entering 'as.numeric(levels(arcilla))[arcilla]' The above code doesn't make sense to me ... Perhaps cleaning up your question and providing some reproducible example we can use to help show you the light (just describing what a variable has isn't enough -- give us minimal code we can paste into R that reproduces your problem). Alternatively, depending no what your levels mean, you might want to recode your data using dummy variables (I'm not sure if that's the official term) .. this is what I mean: http://dss.princeton.edu/online_help/analysis/dummy_variables.htm In your example, let's say you have four levels for clay ... maybe soft, hard, smooth, red Instead of only using 1 variable with values 1-4, you would recode this into 4 variables with values 0,1 So, if one example has a value of smooth for clay. Instead of coding it like: clay: 3 You would do: soft: 0 hard: 0 smooth: 1 red : 0 -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] Re: Change class factor to numeric
Hello so, here you have the output of the data frame. The data frame comes from a csv file. I could take Gr_2 instead of arcilla, because it is the same value... but curiously, it is a factor as well. str(caperf) 'data.frame': 556 obs. of 38 variables: $ Hoja : int 818 818 818 818 818 818 818 818 818 818 ... $ idmuestra : Factor w/ 555 levels 1015-I,1015-II,..: 26 27 28 29 31 32 33 34 30 35 ... $ Año : int 1994 1994 1994 1994 1994 1994 1994 1994 1994 1994 ... $ x : int 655500 633050 632200 635000 637150 643700 655300 648000 653400 646200 ... $ y : int 4285800 4283050 4298150 429 4294800 4288850 4282700 4290350 4298450 4296650 ... $ CO_gkg1 : num 3.7 6.5 6.3 2.6 12.1 6.9 3.5 10.8 10.3 3.3 ... $ NTgkg_1 : num 0.53 1.01 0.66 0.42 1.3 0.82 0.43 1.31 0.85 0.51 ... $ C_Nratio : num 6.98 6.46 9.55 6.18 9.33 ... $ C03Ca : num 53.6 38 1.2 1.1 21.1 ... $ pHw : num 8 8.4 8.3 8.4 8.45 8.2 8.4 8.4 8.1 8.4 ... $ pHClK : num 7.5 7.4 7.2 7.4 8.3 7.5 7.8 7.4 7.4 7.5 ... $ CCC : num 9.7 14.4 10.5 7.2 12.8 ... $ CEdSm : num 0.62 0.72 0.38 0.36 19.35 ... $ pF1_3atm : num 25.4 21.3 9.1 12.8 24.1 ... $ pF15atm : num 15.3 10.6 5.8 6.1 8.45 11 3.7 18.8 10 14.3 ... $ Gr_2 : Factor w/ 391 levels 0,0.71,0.9,..: 200 31 158 36 142 60 377 263 140 151 ... $ Gr2_20 : num 19.1 39.1 2.9 5.2 NA 12.5 5.7 29.5 17.3 10.5 ... $ Gr20_50 : num 17.6 9.8 4.7 4 NA 10.4 4.2 13.3 12.5 7.8 ... $ Gr50_100 : num 14.1 12.2 7.6 9 NA 9.1 6.1 8.5 9.1 18.9 ... $ Gr100_250 : num 13.4 17.1 28.7 46.2 NA 20.8 28.7 13.6 16.8 32.9 ... $ Gr250_500 : num 7.1 5.7 28.8 15.2 NA 24.7 23.4 3.8 11.4 6.9 ... $ Gr500_1000 : num 4 1.8 5 3.9 NA 4.2 14.3 0.9 8.9 2.1 ... $ Gr1000_2000 : num 1.4 2 1.9 4 NA 4.1 8.9 0.4 4.8 0.9 ... $ arcilla : Factor w/ 391 levels 0,0.71,0.9,..: 201 32 158 37 NA 61 377 263 141 151 ... $ limo : num 36.7 48.9 7.6 9.2 0 22.9 9.9 42.8 29.8 18.3 ... $ arena : num 40 38.8 72 78.3 0 62.9 81.4 27.2 51 61.7 ... $ SUMA : Factor w/ 15 levels 0,100,100.2,..: 2 2 2 2 1 2 2 2 2 2 ... $ codusosuelo : logi NA NA NA NA NA NA ... $ pendiente : Factor w/ 9 levels 0,1,2,3,..: 3 2 3 3 2 4 3 2 2 3 ... $ profutil : logi NA NA NA NA NA NA ... $ profutil.1 : int 50 110 120 40 80 52 30 120 37 80 ... $ pedregosidad: int 0 0 0 2 0 4 4 0 3 3 ... $ Drenaje : int 4 4 4 4 1 4 5 3 4 4 ... $ codsuelo : num 1.1 2.1 3.1 2.1 12.3 2.5 2.5 4.1 2.5 2.1 ... $ textura : logi NA NA NA NA NA NA ... $ m.original : Factor w/ 7 levels Calizas . dolomías y areniscas,..: 4 7 7 2 2 7 7 7 7 7 ... $ GRUPOPPAL : Factor w/ 13 levels Arenosoles,Calcisoles,..: 11 2 3 2 12 2 2 4 2 2 ... $ SUELO : Factor w/ 40 levels Arenosol calcárico,..: 30 3 8 3 36 7 7 10 7 3 ... In the other side, the variable Drenaje (drainage) that is factor mode, appears as integer. Thanks a lot! Arantzazu Blanco Bernardeau Dpto de Química Agrícola, Geología y Edafología Universidad de Murcia-Campus de Espinardo Date: Tue, 18 May 2010 17:49:54 +0200 From: ivan.calan...@uni-hamburg.de To: Subject: Re: [R] Re: Change class factor to numeric Hi, I think that providing the output from str(data array or whatever you have) would help. Because, for now, we don't have much idea of what you really have. Moreover, some sample data is always welcomed (using the function dput for example) Ivan Le 5/18/2010 17:36, Arantzazu Blanco Bernardeau a écrit : sorry I had a mistake sending my question without a subject. I do resend again. Please excuse me. Hello I have a data array with soil variables (caperf), in which the variable clay is factor (as I see entering str(caperf)) . I need to do a regression model, so I need to have arcilla (=clay) as a numeric variable. For that I have entered as.numeric(as.character(arcilla)) and even entering 'as.numeric(levels(arcilla))[arcilla]'the variable is resting as factor, and the linear model is not valid (for my purposes). The decimal commas have been converted to decimal points, so I have no idea of what to do. Thanks a lot Arantzazu Blanco Bernardeau Dpto de Química Agrícola, Geología y Edafología Universidad de Murcia-Campus de Espinardo _ Diseñar aplicaciones tiene premio. ¡Si eres desarrollador no esperes más! http://www.imaginemobile.es _ Recibe en tu HOTMAIL los emails de TODAS tus CUENTAS. + info __ R-help@r-project.org 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-contained, reproducible code. -- Ivan CALANDRA PhD Student University
Re: [R] avoiding reinstall already installed *package*
On 18-May-10 15:49:37, Martin Maechler wrote:
{ I've modified the subject; I can't stand it hitting square into
my face ... }
mr == milton ruser milton.ru...@gmail.com
on Tue, 18 May 2010 12:36:23 -0300 writes:
mr Dear R-experts,
mr I am installing new libraries using
mr install.packages(ggplot2,dependencies=T).
mr But I perceive that many dependencies are already installed.
mr As I am using a low-band internet, how can avoid reinstall
mr installed libraries?
There's no problem with installed libraries, as ...
they DO NOT EXIST.
These are *PACKAGES* !
Why do you think are you talking about the function
install.packages()
Ah, Martin! I know that package is the official terminology,
but R itself tempts the naive user into deviating from the
True Path. Indeed, I had my fingers burned by this myself,
a long time ago (I'm still licking them ... ).
One might ask: Why do you think we use the function library()?
when loading add-on packages into R. Indeed, the very directory
tree of R itself stores packages under /usr/lib/R/library.
So, once in a while, someone gets it wrong, and has to find it
out the hard way!
Best wishes,
Ted.
E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk
Fax-to-email: +44 (0)870 094 0861
Date: 18-May-10 Time: 17:05:57
-- XFMail --
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Re: [R] Repeating Name for Rows
If I perceive the issue: Day - c(rep(97, each = 11), rep(98:278, each = 12)) - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/Repeating-Name-for-Rows-tp2221457p2221492.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] (no subject)
Hello Well, the problem is, that arcilla is the percentage of clay in the soil sample. So, for linear model, I need to work with that number or value. Now, R thinks that arcilla (arcilla means clay in spanish), is a factor, and gives me the value as a factor, so the output of the linear model is Call: lm(formula = formula, data = caperf) Residuals: Min 1Q Median 3Q Max -1.466e+01 -1.376e-15 1.780e-16 2.038e-15 1.279e+01 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 1.68964 6.33889 0.267 0.790221 arcilla0.9 1.90228 8.90888 0.214 0.831239 arcilla10 1.26371 7.96734 0.159 0.874212 arcilla10.3 15.70081 9.05141 1.735 0.085090 . arcilla10.4 7.27517 7.72806 0.941 0.348183 arcilla10.45 7.03879 9.02600 0.780 0.436853 arcilla10.5 2.41241 8.90827 0.271 0.786954 arcilla10.65 15.44298 9.03879 1.709 0.089838 . arcilla10.7 19.35651 9.04675 2.140 0.034185 * arcilla10.9 3.55947 9.18501 0.388 0.698974 [...] arcilla9.9 6.31949 7.35724 0.859 0.391892 arcilla#N/A 24.17959 8.87201 2.725 0.007274 ** limo 0.24920 0.04605 5.412 2.76e-07 *** CO_gkg1 0.21015 0.03931 5.346 3.73e-07 *** C03Ca 0.01711 0.02727 0.628 0.531337 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 6.249 on 135 degrees of freedom (50 observations deleted due to missingness) Multiple R-squared: 0.9736, Adjusted R-squared: 0.9014 F-statistic: 13.47 on 370 and 135 DF, p-value: 2.2e-16 So, in the desired linear model, arcilla should be just a line, with the valors of the linear model. I hope you understand better more. If not, I could make an english version of the file to send, so you can try the commands. Thanks a lot for your help! Arantzazu Blanco Bernardeau Dpto de Química Agrícola, Geología y Edafología Universidad de Murcia-Campus de Espinardo Date: Tue, 18 May 2010 11:54:20 -0400 Subject: Re: [R] (no subject) From: mailinglist.honey...@gmail.com To: aramu...@hotmail.com CC: r-help@r-project.org Hi, Sorry, I'm not really getting what going on here ... perhaps having more domain knowledge would help me make better sense of our question. In particular: On Tue, May 18, 2010 at 11:35 AM, Arantzazu Blanco Bernardeau wrote: Hello I have a data array with soil variables (caperf), in which the variable clay is factor (as I see entering str(caperf)) . I need to do a regression model, so I need to have arcilla (=clay) as a numeric variable. For that I have entered as.numeric(as.character(arcilla)) and even entering 'as.numeric(levels(arcilla))[arcilla]' The above code doesn't make sense to me ... Perhaps cleaning up your question and providing some reproducible example we can use to help show you the light (just describing what a variable has isn't enough -- give us minimal code we can paste into R that reproduces your problem). Alternatively, depending no what your levels mean, you might want to recode your data using dummy variables (I'm not sure if that's the official term) .. this is what I mean: http://dss.princeton.edu/online_help/analysis/dummy_variables.htm In your example, let's say you have four levels for clay ... maybe soft, hard, smooth, red Instead of only using 1 variable with values 1-4, you would recode this into 4 variables with values 0,1 So, if one example has a value of smooth for clay. Instead of coding it like: clay: 3 You would do: soft: 0 hard: 0 smooth: 1 red : 0 -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact _ Consejos para seducir ¿Puedes conocer gente nueva a través de Internet? ¡Regístrate ya! __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] timing a function
Dear all,
Just one last question. There seems to be no problem in writing
z = system.time(y - f(x))
or
z - system.time(y - f(x))
Then z contains the named vector of the elapsed times, and y the value of
the function f(x).
Am I right ?
Thank you very much,
Gustave
2010/5/17 Alexander Shenkin ashen...@ufl.edu
You could also put the call to system.time inside the function itself:
f = function(x) {
system.time({
... #function's code
ret_val = ...
}); flush.console();
return ret_val;
}
i s'pose you'd miss out on the time taken to jump to the function code,
return the value, etc, but for functions that are heavy at all, that
wouldn't trip you up.
allie
On 5/17/2010 2:06 PM, Barry Rowlingson wrote:
On Mon, May 17, 2010 at 6:24 PM, Peter Ehlers ehl...@ucalgary.ca
wrote:
Try
system.time(y - f(x))
and see ?=.
-Peter Ehlers
Ah ha. That explains the curly brackets I saw in a posting with
system.time on stack overflow just now:
system.time({y=f(x)})
works as expected since the {} pair make a new code block. Also you
can then time more than one statement:
system.time({y=f(x);z=g(y)})
- gives the total time for f(x) and g(y).
Barry
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http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html
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http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] avoiding reinstall already installed *package*
Hi Martin,
thanks for your reply, and very thanks for your kind tips about package
and library
So, I was trying to understand *why* we load packages using library().
I suggest that developers killl the problem on its root, deleting library
function :-)
Good to know already installed packages will not be reinstalled.
cheers
milton
On Tue, May 18, 2010 at 12:49 PM, Martin Maechler
maech...@stat.math.ethz.ch wrote:
{ I've modified the subject; I can't stand it hitting square into
my face ... }
mr == milton ruser milton.ru...@gmail.com
on Tue, 18 May 2010 12:36:23 -0300 writes:
mr Dear R-experts,
mr I am installing new libraries using
mr install.packages(ggplot2,dependencies=T).
mr But I perceive that many dependencies are already installed. As I am
using
mr a low-band internet, how can avoid reinstall installed libraries?
There's no problem with installed libraries, as ...
they DO NOT EXIST.
These are *PACKAGES* !
Why do you think are you talking about the function
install.packages()
---
To answer the question you did want to ask:
Do not be afraid: Depedencies are only installed when needed,
i.e., no package will be downloaded and installed if it already
is there.
Martin Maechler, ETH Zurich
mr cheers
mr milton
mr [[alternative HTML version deleted]]
(another thing you should learn to avoid, please)
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Re: Change class factor to numeric
Hi again, If you used the function read.table() to read from a csv file into a data.frame, it is weird that numeric data are converted into factors. I would check in the original data that you don't have a typo somewhere. I don't know all the possibilities, but a special character can definitely make R interpret this variable differently. For Drenaje, it is normal. In that case you can just use: caperf$Drenaje - factor(caperf$Drenaje) HTH Ivan Le 5/18/2010 17:59, Arantzazu Blanco Bernardeau a écrit : Hello so, here you have the output of the data frame. The data frame comes from a csv file. I could take Gr_2 instead of arcilla, because it is the same value... but curiously, it is a factor as well. str(caperf) 'data.frame':556 obs. of 38 variables: $ Hoja: int 818 818 818 818 818 818 818 818 818 818 ... $ idmuestra : Factor w/ 555 levels 1015-I,1015-II,..: 26 27 28 29 31 32 33 34 30 35 ... $ Año : int 1994 1994 1994 1994 1994 1994 1994 1994 1994 1994 ... $ x : int 655500 633050 632200 635000 637150 643700 655300 648000 653400 646200 ... $ y : int 4285800 4283050 4298150 429 4294800 4288850 4282700 4290350 4298450 4296650 ... $ CO_gkg1 : num 3.7 6.5 6.3 2.6 12.1 6.9 3.5 10.8 10.3 3.3 ... $ NTgkg_1 : num 0.53 1.01 0.66 0.42 1.3 0.82 0.43 1.31 0.85 0.51 ... $ C_Nratio: num 6.98 6.46 9.55 6.18 9.33 ... $ C03Ca : num 53.6 38 1.2 1.1 21.1 ... $ pHw : num 8 8.4 8.3 8.4 8.45 8.2 8.4 8.4 8.1 8.4 ... $ pHClK : num 7.5 7.4 7.2 7.4 8.3 7.5 7.8 7.4 7.4 7.5 ... $ CCC : num 9.7 14.4 10.5 7.2 12.8 ... $ CEdSm : num 0.62 0.72 0.38 0.36 19.35 ... $ pF1_3atm: num 25.4 21.3 9.1 12.8 24.1 ... $ pF15atm : num 15.3 10.6 5.8 6.1 8.45 11 3.7 18.8 10 14.3 ... $ Gr_2: Factor w/ 391 levels 0,0.71,0.9,..: 200 31 158 36 142 60 377 263 140 151 ... $ Gr2_20 : num 19.1 39.1 2.9 5.2 NA 12.5 5.7 29.5 17.3 10.5 ... $ Gr20_50 : num 17.6 9.8 4.7 4 NA 10.4 4.2 13.3 12.5 7.8 ... $ Gr50_100: num 14.1 12.2 7.6 9 NA 9.1 6.1 8.5 9.1 18.9 ... $ Gr100_250 : num 13.4 17.1 28.7 46.2 NA 20.8 28.7 13.6 16.8 32.9 ... $ Gr250_500 : num 7.1 5.7 28.8 15.2 NA 24.7 23.4 3.8 11.4 6.9 ... $ Gr500_1000 : num 4 1.8 5 3.9 NA 4.2 14.3 0.9 8.9 2.1 ... $ Gr1000_2000 : num 1.4 2 1.9 4 NA 4.1 8.9 0.4 4.8 0.9 ... $ arcilla : Factor w/ 391 levels 0,0.71,0.9,..: 201 32 158 37 NA 61 377 263 141 151 ... $ limo: num 36.7 48.9 7.6 9.2 0 22.9 9.9 42.8 29.8 18.3 ... $ arena : num 40 38.8 72 78.3 0 62.9 81.4 27.2 51 61.7 ... $ SUMA: Factor w/ 15 levels 0,100,100.2,..: 2 2 2 2 1 2 2 2 2 2 ... $ codusosuelo : logi NA NA NA NA NA NA ... $ pendiente : Factor w/ 9 levels 0,1,2,3,..: 3 2 3 3 2 4 3 2 2 3 ... $ profutil: logi NA NA NA NA NA NA ... $ profutil.1 : int 50 110 120 40 80 52 30 120 37 80 ... $ pedregosidad: int 0 0 0 2 0 4 4 0 3 3 ... $ Drenaje : int 4 4 4 4 1 4 5 3 4 4 ... $ codsuelo: num 1.1 2.1 3.1 2.1 12.3 2.5 2.5 4.1 2.5 2.1 ... $ textura : logi NA NA NA NA NA NA ... $ m.original : Factor w/ 7 levels Calizas . dolomías y areniscas,..: 4 7 7 2 2 7 7 7 7 7 ... $ GRUPOPPAL : Factor w/ 13 levels Arenosoles,Calcisoles,..: 11 2 3 2 12 2 2 4 2 2 ... $ SUELO : Factor w/ 40 levels Arenosol calcárico,..: 30 3 8 3 36 7 7 10 7 3 ... In the other side, the variable Drenaje (drainage) that is factor mode, appears as integer. Thanks a lot! Arantzazu Blanco Bernardeau Dpto de Química Agrícola, Geología y Edafología Universidad de Murcia-Campus de Espinardo Date: Tue, 18 May 2010 17:49:54 +0200 From: ivan.calan...@uni-hamburg.de To: Subject: Re: [R] Re: Change class factor to numeric Hi, I think that providing the output from str(data array or whatever you have) would help. Because, for now, we don't have much idea of what you really have. Moreover, some sample data is always welcomed (using the function dput for example) Ivan Le 5/18/2010 17:36, Arantzazu Blanco Bernardeau a écrit : sorry I had a mistake sending my question without a subject. I do resend again. Please excuse me. Hello I have a data array with soil variables (caperf), in which the variable clay is factor (as I see entering str(caperf)) . I need to do a regression model, so I need to have arcilla (=clay) as a numeric variable. For that I have entered as.numeric(as.character(arcilla)) and even entering 'as.numeric(levels(arcilla))[arcilla]'the variable is resting as factor, and the linear model is not valid (for my purposes). The decimal commas have been converted to decimal points, so I have no idea of what to do. Thanks a lot Arantzazu Blanco Bernardeau Dpto de Química Agrícola, Geología y Edafología Universidad de Murcia-Campus de Espinardo _ Diseñar
Re: [R] (no subject)
One last thing: before you take my advice on how to recode your nominal/categorical clay variable for your regression model, take some time to see how other people talk about this and do some searching on phrases like regression model with nominal variables (that's just the one I used). You'll find more (and better) advice on how to do it correctly. -steve On Tue, May 18, 2010 at 11:54 AM, Steve Lianoglou mailinglist.honey...@gmail.com wrote: Hi, Sorry, I'm not really getting what going on here ... perhaps having more domain knowledge would help me make better sense of our question. In particular: On Tue, May 18, 2010 at 11:35 AM, Arantzazu Blanco Bernardeau aramu...@hotmail.com wrote: Hello I have a data array with soil variables (caperf), in which the variable clay is factor (as I see entering str(caperf)) . I need to do a regression model, so I need to have arcilla (=clay) as a numeric variable. For that I have entered as.numeric(as.character(arcilla)) and even entering 'as.numeric(levels(arcilla))[arcilla]' The above code doesn't make sense to me ... Perhaps cleaning up your question and providing some reproducible example we can use to help show you the light (just describing what a variable has isn't enough -- give us minimal code we can paste into R that reproduces your problem). Alternatively, depending no what your levels mean, you might want to recode your data using dummy variables (I'm not sure if that's the official term) .. this is what I mean: http://dss.princeton.edu/online_help/analysis/dummy_variables.htm In your example, let's say you have four levels for clay ... maybe soft, hard, smooth, red Instead of only using 1 variable with values 1-4, you would recode this into 4 variables with values 0,1 So, if one example has a value of smooth for clay. Instead of coding it like: clay: 3 You would do: soft: 0 hard: 0 smooth: 1 red : 0 -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ R-help@r-project.org 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-contained, reproducible code.
[R] unsigned 4 byte number
Does anybody know how to read in a unsigned 4 byte number from a binary file? According to the help for readBin, the signed argument only applies to size=1 or 2. But if you declare any size larger it assumes it is signed? Thanks -- View this message in context: http://r.789695.n4.nabble.com/unsigned-4-byte-number-tp2221555p2221555.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] (no subject)
Arantzazu, Your problem is that the data were probably imported from Excel where you had at least one cell containing #N/A. You need to replace those cases in your dataframe with NA. Then you should be able to do as.numeric(as.character(arcilla)). -Peter Ehlers On 2010-05-18 10:07, Arantzazu Blanco Bernardeau wrote: Hello Well, the problem is, that arcilla is the percentage of clay in the soil sample. So, for linear model, I need to work with that number or value. Now, R thinks that arcilla (arcilla means clay in spanish), is a factor, and gives me the value as a factor, so the output of the linear model is Call: lm(formula = formula, data = caperf) Residuals: Min 1Q Median 3QMax -1.466e+01 -1.376e-15 1.780e-16 2.038e-15 1.279e+01 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept)1.689646.33889 0.267 0.790221 arcilla0.9 1.902288.90888 0.214 0.831239 arcilla10 1.263717.96734 0.159 0.874212 arcilla10.3 15.700819.05141 1.735 0.085090 . arcilla10.47.275177.72806 0.941 0.348183 arcilla10.45 7.038799.02600 0.780 0.436853 arcilla10.52.412418.90827 0.271 0.786954 arcilla10.65 15.442989.03879 1.709 0.089838 . arcilla10.7 19.356519.04675 2.140 0.034185 * arcilla10.93.559479.18501 0.388 0.698974 [...] arcilla9.9 6.319497.35724 0.859 0.391892 === arcilla#N/A 24.179598.87201 2.725 0.007274 ** limo 0.249200.04605 5.412 2.76e-07 *** CO_gkg10.210150.03931 5.346 3.73e-07 *** C03Ca 0.017110.02727 0.628 0.531337 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 6.249 on 135 degrees of freedom (50 observations deleted due to missingness) Multiple R-squared: 0.9736,Adjusted R-squared: 0.9014 F-statistic: 13.47 on 370 and 135 DF, p-value: 2.2e-16 So, in the desired linear model, arcilla should be just a line, with the valors of the linear model. I hope you understand better more. If not, I could make an english version of the file to send, so you can try the commands. Thanks a lot for your help! Arantzazu Blanco Bernardeau Dpto de Química Agrícola, Geología y Edafología Universidad de Murcia-Campus de Espinardo [snip] __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] Using the zero-inflated binomial in experimental designs
Ivan Allaman ivanalaman at yahoo.com.br writes: I'm trying to use the inflated binomial distribution of zeros (since 75% of the values are zeros) in a randomized block experiment with four quantitative treatments (0, 0.5, 1, 1.5), but I'm finding it difficult, since the examples available in VGAM packages like for example, leave us unsure of how it should be the data.frame for such analysis. Unfortunately the function glm does not have an option to place a family of this kind I'm about, because if I had, it would be easy, made that my goal is simple, just wanting to compare the treatments. For that you have an idea, here is an example of my database. BLOCK NIVNT MUMI Inicial 0 18 0 [snip] where: NIV are the treatments; NT is the total number of piglets born; Mumi is the number of mummified piglets NT. Mumi The variable is of interest. If someone can tell me some stuff on how I can do these tests in R, similar to what I would do using the function glm, I'd be grateful. I thank everyone's attention. something like comparing the likelihoods of m1 - vglm(cbind(MUMI,NT-MUMI)~NIV*BLOCK,zibinomial,data=mydata) m2 - vglm(cbind(MUMI,NT-MUMI)~NIV+BLOCK,zibinomial,data=mydata) m3 - vglm(cbind(MUMI,NT-MUMI)~BLOCK,zibinomial,data=mydata) I don't know whether the anova() method works for VGLM objects or not. By the way, 75% zeroes doesn't necessarily imply zero-inflation -- perhaps it just means a low incidence? __ R-help@r-project.org 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-contained, reproducible code.
[R] Function that is giving me a headache- any help appreciated (automatic read )
note: whole function is below- I am sure I am doing something silly.
when I use it like USGS(input=precipitation) it is choking on the
precip.1 - subset(DF, precipitation!=NA)
b - ddply(precip.1$precipitation, .(precip.1$gauge_name), cumsum)
DF.precip - precip.1
DF.precip$precipitation - b$.data
part, but runs fine outside of the function:
days=7
input=precipitation
require(chron)
require(gsubfn)
require(ggplot2)
require(plyr)
#021973269 is the Waynesboro Gauge on the Savannah River Proper (SRS)
#02102908 is the Flat Creek Gauge (ftbrfcms)
#02133500 is the Drowning Creek (ftbrbmcm)
#02341800 is the Upatoi Creek Near Columbus (ftbn)
#02342500 is the Uchee Creek Near Fort Mitchell (ftbn)
#02203000 is the Canoochee River Near Claxton (ftst)
#02196690 is the Horse Creek Gauge at Clearwater, S.C.
a - http://waterdata.usgs.gov/nwis/uv?format=rdbperiod=;
b - site_no=021973269,02102908,02133500,02341800,02342500,02203000,02196690
z - paste(a, days, b, sep=)
L - readLines(z)
#look for the data with USGS in front of it (this take advantage of
#the agency column)
L.USGS - grep(^USGS, L, value = TRUE)
DF - read.table(textConnection(L.USGS), fill = TRUE)
colnames(DF) - c(agency, gauge, date, time, time_zone,
gauge_height,
discharge, precipitation)
pat - ^# +USGS +([0-9]+) +(.*)
L.DD - grep(pat, L, value = TRUE)
library(gsubfn)
DD - strapply(L.DD, pat, c, simplify = rbind)
DDdf - data.frame(gauge = as.numeric(DD[,1]), gauge_name = DD[,2])
both - merge(DF, DDdf, by = gauge, all.x = TRUE)
dts - as.character(both[,date])
tms - as.character(both[,time])
date_time - as.chron(paste(dts, tms), %Y-%m-%d %H:%M)
DF - data.frame(Date=as.POSIXct(date_time), both)
#change precip to numeric
DF[,precipitation] - as.numeric(as.character(DF[,precipitation]))
precip.1 - subset(DF, precipitation!=NA)
b - ddply(precip.1$precipitation, .(precip.1$gauge_name), cumsum)
DF.precip - precip.1
DF.precip$precipitation - b$.data
#discharge
if(input==data){
return(DF)
}else{
qplot(Date, discharge, data=DF,
geom=line, ylab=Date)+facet_wrap(~gauge_name,
scales=free_y)+coord_trans(y=log10)}
if(input==precipitation){
#precipitation
qplot(Date, precipitation, data=DF.precip,
geom=line)+facet_wrap(~gauge_name, scales=free_y)
}else{
qplot(Date, discharge, data=DF,
geom=line, ylab=Date)+facet_wrap(~gauge_name,
scales=free_y)+coord_trans(y=log10)}
below is the whole function:
USGS - function(input=discharge, days=7){
require(chron)
require(gsubfn)
require(ggplot2)
require(plyr)
#021973269 is the Waynesboro Gauge on the Savannah River Proper (SRS)
#02102908 is the Flat Creek Gauge (ftbrfcms)
#02133500 is the Drowning Creek (ftbrbmcm)
#02341800 is the Upatoi Creek Near Columbus (ftbn)
#02342500 is the Uchee Creek Near Fort Mitchell (ftbn)
#02203000 is the Canoochee River Near Claxton (ftst)
#02196690 is the Horse Creek Gauge at Clearwater, S.C.
a - http://waterdata.usgs.gov/nwis/uv?format=rdbperiod=;
b - site_no=021973269,02102908,02133500,02341800,02342500,02203000,02196690
z - paste(a, days, b, sep=)
L - readLines(z)
#look for the data with USGS in front of it (this take advantage of
#the agency column)
L.USGS - grep(^USGS, L, value = TRUE)
DF - read.table(textConnection(L.USGS), fill = TRUE)
colnames(DF) - c(agency, gauge, date, time, time_zone,
gauge_height,
discharge, precipitation)
pat - ^# +USGS +([0-9]+) +(.*)
L.DD - grep(pat, L, value = TRUE)
library(gsubfn)
DD - strapply(L.DD, pat, c, simplify = rbind)
DDdf - data.frame(gauge = as.numeric(DD[,1]), gauge_name = DD[,2])
both - merge(DF, DDdf, by = gauge, all.x = TRUE)
dts - as.character(both[,date])
tms - as.character(both[,time])
date_time - as.chron(paste(dts, tms), %Y-%m-%d %H:%M)
DF - data.frame(Date=as.POSIXct(date_time), both)
#change precip to numeric
DF[,precipitation] - as.numeric(as.character(DF[,precipitation]))
precip.1 - subset(DF, precipitation!=NA)
b - ddply(precip.1$precipitation, .(precip.1$gauge_name), cumsum)
DF.precip - precip.1
DF.precip$precipitation - b$.data
#discharge
if(input==data){
return(DF)
}else{
qplot(Date, discharge, data=DF,
geom=line, ylab=Date)+facet_wrap(~gauge_name,
scales=free_y)+coord_trans(y=log10)}
if(input==precipitation){
#precipitation
qplot(Date, precipitation, data=DF.precip,
geom=line)+facet_wrap(~gauge_name, scales=free_y)
}else{
qplot(Date, discharge, data=DF,
geom=line, ylab=Date)+facet_wrap(~gauge_name,
scales=free_y)+coord_trans(y=log10)}
}
--
Stephen Sefick
Let's not spend our time and resources thinking about things that are
so little or so large that all they really do for us is puff us up and
make us feel like gods. We are mammals, and have not exhausted the
annoying little problems of being mammals.
-K. Mullis
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal,
[R] FW: Re: Change class factor to numeric
Hello everybody the problem has been solved. It was my mistake not to be ensured that any N/A had dissapeared. I do apologize for the inconveniences caused ;) friendly greeting sfrom Spain! Arantzazu Blanco Bernardeau Dpto de Química Agrícola, Geología y Edafología Universidad de Murcia-Campus de Espinardo From: aramu...@hotmail.com To: ivan.calan...@uni-hamburg.de Subject: RE: [R] Re: Change class factor to numeric Date: Tue, 18 May 2010 16:36:55 + Hallo Ivan I do thank you a lot, but as you have read in the last email, I did think that maybe one erroneus typing was indicating R to take it as a factor. I did look again the original dataframe and did find the mistake, and now it works OK. Sorry, but I had done so many reviews of the data frame, that did not think on it before. Schon wieder danke schön, und freundliche Grüsse aus Spanien ;) Arantzazu Blanco Bernardeau Dpto de Química Agrícola, Geología y Edafología Universidad de Murcia-Campus de Espinardo Date: Tue, 18 May 2010 18:28:45 +0200 From: ivan.calan...@uni-hamburg.de To: aramu...@hotmail.com Subject: Re: [R] Re: Change class factor to numeric As you can notice, I'm writing off list for you to send me your data (as csv). But do it fast, I'll be leaving soon. If not it might have to wait until tomorrow! In any case, I'm no expert, so I'm not sure I'll be able to help you. And I don't think NAs should be problematic. It might be solved by some arguments into the read.table() call. Can you also send me the line of code you used to import your csv? Ivan Le 5/18/2010 18:25, Arantzazu Blanco Bernardeau a écrit : Hi again! could it be that NA was introduced in the variable for not available values, and being NA a character, it takes everything as factor?? is the only idea I have, because it is the first time I have this problem Thanks a lot, I am really learning! :) Arantzazu Blanco Bernardeau Dpto de Química Agrícola, Geología y Edafología Universidad de Murcia-Campus de Espinardo Date: Tue, 18 May 2010 18:17:02 +0200 From: ivan.calan...@uni-hamburg.de To: r-help@r-project.org Subject: Re: [R] Re: Change class factor to numeric Hi again, If you used the function read.table() to read from a csv file into a data.frame, it is weird that numeric data are converted into factors. I would check in the original data that you don't have a typo somewhere. I don't know all the possibilities, but a special character can definitely make R interpret this variable differently. For Drenaje, it is normal. In that case you can just use: caperf$Drenaje- factor(caperf$Drenaje) HTH Ivan Le 5/18/2010 17:59, Arantzazu Blanco Bernardeau a écrit : Hello so, here you have the output of the data frame. The data frame comes from a csv file. I could take Gr_2 instead of arcilla, because it is the same value... but curiously, it is a factor as well. str(caperf) 'data.frame': 556 obs. of 38 variables: $ Hoja : int 818 818 818 818 818 818 818 818 818 818 ... $ idmuestra : Factor w/ 555 levels 1015-I,1015-II,..: 26 27 28 29 31 32 33 34 30 35 ... $ Año : int 1994 1994 1994 1994 1994 1994 1994 1994 1994 1994 ... $ x : int 655500 633050 632200 635000 637150 643700 655300 648000 653400 646200 ... $ y : int 4285800 4283050 4298150 429 4294800 4288850 4282700 4290350 4298450 4296650 ... $ CO_gkg1 : num 3.7 6.5 6.3 2.6 12.1 6.9 3.5 10.8 10.3 3.3 ... $ NTgkg_1 : num 0.53 1.01 0.66 0.42 1.3 0.82 0.43 1.31 0.85 0.51 ... $ C_Nratio : num 6.98 6.46 9.55 6.18 9.33 ... $ C03Ca : num 53.6 38 1.2 1.1 21.1 ... $ pHw : num 8 8.4 8.3 8.4 8.45 8.2 8.4 8.4 8.1 8.4 ... $ pHClK : num 7.5 7.4 7.2 7.4 8.3 7.5 7.8 7.4 7.4 7.5 ... $ CCC : num 9.7 14.4 10.5 7.2 12.8 ... $ CEdSm : num 0.62 0.72 0.38 0.36 19.35 ... $ pF1_3atm : num 25.4 21.3 9.1 12.8 24.1 ... $ pF15atm : num 15.3 10.6 5.8 6.1 8.45 11 3.7 18.8 10 14.3 ... $ Gr_2 : Factor w/ 391 levels 0,0.71,0.9,..: 200 31 158 36 142 60 377 263 140 151 ... $ Gr2_20 : num 19.1 39.1 2.9 5.2 NA 12.5 5.7 29.5 17.3 10.5 ... $ Gr20_50 : num 17.6 9.8 4.7 4 NA 10.4 4.2 13.3 12.5 7.8 ... $ Gr50_100 : num 14.1 12.2 7.6 9 NA 9.1 6.1 8.5 9.1 18.9 ... $ Gr100_250 : num 13.4 17.1 28.7 46.2 NA 20.8 28.7 13.6 16.8 32.9 ... $ Gr250_500 : num 7.1 5.7 28.8 15.2 NA 24.7 23.4 3.8 11.4 6.9 ... $ Gr500_1000 : num 4 1.8 5 3.9 NA 4.2 14.3 0.9 8.9 2.1 ... $ Gr1000_2000 : num 1.4 2 1.9 4 NA 4.1 8.9 0.4 4.8 0.9 ... $ arcilla : Factor w/ 391 levels 0,0.71,0.9,..: 201 32 158 37 NA 61 377 263 141 151 ... $ limo : num 36.7 48.9 7.6 9.2 0 22.9 9.9 42.8 29.8 18.3 ... $ arena : num 40 38.8 72 78.3 0 62.9 81.4 27.2 51 61.7 ... $ SUMA : Factor w/ 15 levels 0,100,100.2,..: 2 2 2 2 1 2 2 2 2 2 ... $ codusosuelo : logi NA NA NA NA NA NA ... $ pendiente : Factor w/ 9 levels 0,1,2,3,..: 3 2 3 3 2 4 3 2 2 3 ... $
[R] Getting dates in an SPSS file in right format.
Dear all, I am trying to read an SPSS file into a data frame in R using method read.spss(), sample - read.spss(file.name,to.data.frame=TRUE) But dates in the data.frame 'sample' are coming as integers and not in the actual date format given in the SPSS file. Appreciate if anyone can help me to solve this problem. Kind Regards, Praveen Surendran 2G, Complex and Adaptive Systems Laboratory (UCD CASL) School of Medicine and Medical Sciences University College Dublin Belfield, Dublin 4 Ireland. Office : +353-(0)1716 5334 Mobile : +353-(0)8793 13071 [[alternative HTML version deleted]] __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] avoiding reinstall already installed *package*
On 2010-05-18 10:05, (Ted Harding) wrote:
On 18-May-10 15:49:37, Martin Maechler wrote:
{ I've modified the subject; I can't stand it hitting square into
my face ... }
mr == milton rusermilton.ru...@gmail.com
on Tue, 18 May 2010 12:36:23 -0300 writes:
mr Dear R-experts,
mr I am installing new libraries using
mr install.packages(ggplot2,dependencies=T).
mr But I perceive that many dependencies are already installed.
mr As I am using a low-band internet, how can avoid reinstall
mr installed libraries?
There's no problem with installed libraries, as ...
they DO NOT EXIST.
These are *PACKAGES* !
Why do you think are you talking about the function
install.packages()
Ah, Martin! I know that package is the official terminology,
but R itself tempts the naive user into deviating from the
True Path. Indeed, I had my fingers burned by this myself,
a long time ago (I'm still licking them ... ).
One might ask: Why do you think we use the function library()?
when loading add-on packages into R. Indeed, the very directory
tree of R itself stores packages under /usr/lib/R/library.
So, once in a while, someone gets it wrong, and has to find it
out the hard way!
Well, I don't know if I've ever disagreed with Ted before,
but here I would (somewhat) disagree. It seems a bit odd that
nobody confuses 'book' with 'library', yet the package/library
problem is persistent. It may have something to do with the
use of 'library' in other computer languages.
Anyway, not long ago there was a suggestion (Rolf Turner's?)
to rename the library() function to something like use(),
but, as I recall, a number of nontrivial objections were
raised.
Of course R stores packages in libraries. That's were books
*should* reside. And it's a good idea to have Martin remind
us now and again that books themselves are not libraries.
But I must confess that I'm no longer much bothered by the
misuse. If it ever leads someone astray in their code, then,
well, they have only themselves to blame.
Cheers,
Peter Ehlers
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] survey package: weights used in svycoxph()
On Tue, May 18, 2010 at 8:50 AM, Thomas Lumley tlum...@u.washington.edu wrote: I don't believe so since svycoxph() calls coxph() of the survival package and weights are applied once in the estimating equation. If the weights are implemented in the ratio, could you point me to where in the code this is done? I would like to estimate as in Binder but with custom weights. Thanks. It happens inside the C code called by coxph(), eg, in survival/src/coxfit2.c Thank you for your clarification. I mistakenly assumed weights only appeared once in the estimating equation, creating a weighted sum of the score equation. Thinking in retrospect if the weights are to be used as case weights they better be in the ratio term as well (wherever there is an at risk indicator). Binder's estimating equations are the usual way of applying weights to a Cox model, so nothing special is done apart from calling coxph(). To quote the author of the survival package, Terry Therneau, Other formulae change in the obvious way, eg, the weighted mean $\bar Z$ is changed to include both the risk weights $r$ and the external weights $w$. [Mayo Clinic Biostatistics technical report #52, section 6.2.2] Don't see a section 6.2.2 in this technical report. __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] Getting dates in an SPSS file in right format.
On 5/18/2010 12:38 PM, Praveen Surendran wrote: Dear all, I am trying to read an SPSS file into a data frame in R using method read.spss(), sample - read.spss(file.name,to.data.frame=TRUE) But dates in the data.frame 'sample' are coming as integers and not in the actual date format given in the SPSS file. Appreciate if anyone can help me to solve this problem. Date variables in SPSS contain the number of seconds since October 14, 1582. You might try something like this: sample$MYDATE - as.Date(as.POSIXct(sample$MYDATE, origin=1582-10-14, tz=GMT)) Kind Regards, Praveen Surendran 2G, Complex and Adaptive Systems Laboratory (UCD CASL) School of Medicine and Medical Sciences University College Dublin Belfield, Dublin 4 Ireland. Office : +353-(0)1716 5334 Mobile : +353-(0)8793 13071 [[alternative HTML version deleted]] __ R-help@r-project.org 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-contained, reproducible code. -- Chuck Cleland, Ph.D. NDRI, Inc. (www.ndri.org) 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] Function that is giving me a headache- any help appreciated (automatic read )
I don't think you can do this
precipitation!=NA)
have a look at ?is.na
--- On Tue, 5/18/10, stephen sefick ssef...@gmail.com wrote:
From: stephen sefick ssef...@gmail.com
Subject: [R] Function that is giving me a headache- any help appreciated
(automatic read )
To: r-help@r-project.org
Received: Tuesday, May 18, 2010, 12:38 PM
note: whole function is below- I am
sure I am doing something silly.
when I use it like USGS(input=precipitation) it is
choking on the
precip.1 - subset(DF, precipitation!=NA)
b - ddply(precip.1$precipitation,
.(precip.1$gauge_name), cumsum)
DF.precip - precip.1
DF.precip$precipitation - b$.data
part, but runs fine outside of the function:
days=7
input=precipitation
require(chron)
require(gsubfn)
require(ggplot2)
require(plyr)
#021973269 is the Waynesboro Gauge on the Savannah River
Proper (SRS)
#02102908 is the Flat Creek Gauge (ftbrfcms)
#02133500 is the Drowning Creek (ftbrbmcm)
#02341800 is the Upatoi Creek Near Columbus (ftbn)
#02342500 is the Uchee Creek Near Fort Mitchell (ftbn)
#02203000 is the Canoochee River Near Claxton (ftst)
#02196690 is the Horse Creek Gauge at Clearwater, S.C.
a - http://waterdata.usgs.gov/nwis/uv?format=rdbperiod=;
b -
site_no=021973269,02102908,02133500,02341800,02342500,02203000,02196690
z - paste(a, days, b, sep=)
L - readLines(z)
#look for the data with USGS in front of it (this take
advantage of
#the agency column)
L.USGS - grep(^USGS, L, value = TRUE)
DF - read.table(textConnection(L.USGS), fill = TRUE)
colnames(DF) - c(agency, gauge, date, time,
time_zone,
gauge_height,
discharge, precipitation)
pat - ^# +USGS +([0-9]+) +(.*)
L.DD - grep(pat, L, value = TRUE)
library(gsubfn)
DD - strapply(L.DD, pat, c, simplify = rbind)
DDdf - data.frame(gauge = as.numeric(DD[,1]),
gauge_name = DD[,2])
both - merge(DF, DDdf, by = gauge, all.x = TRUE)
dts - as.character(both[,date])
tms - as.character(both[,time])
date_time - as.chron(paste(dts, tms), %Y-%m-%d
%H:%M)
DF - data.frame(Date=as.POSIXct(date_time), both)
#change precip to numeric
DF[,precipitation] -
as.numeric(as.character(DF[,precipitation]))
precip.1 - subset(DF, precipitation!=NA)
b - ddply(precip.1$precipitation,
.(precip.1$gauge_name), cumsum)
DF.precip - precip.1
DF.precip$precipitation - b$.data
#discharge
if(input==data){
return(DF)
}else{
qplot(Date, discharge, data=DF,
geom=line, ylab=Date)+facet_wrap(~gauge_name,
scales=free_y)+coord_trans(y=log10)}
if(input==precipitation){
#precipitation
qplot(Date, precipitation, data=DF.precip,
geom=line)+facet_wrap(~gauge_name, scales=free_y)
}else{
qplot(Date, discharge, data=DF,
geom=line, ylab=Date)+facet_wrap(~gauge_name,
scales=free_y)+coord_trans(y=log10)}
below is the whole function:
USGS - function(input=discharge, days=7){
require(chron)
require(gsubfn)
require(ggplot2)
require(plyr)
#021973269 is the Waynesboro Gauge on the Savannah River
Proper (SRS)
#02102908 is the Flat Creek Gauge (ftbrfcms)
#02133500 is the Drowning Creek (ftbrbmcm)
#02341800 is the Upatoi Creek Near Columbus (ftbn)
#02342500 is the Uchee Creek Near Fort Mitchell (ftbn)
#02203000 is the Canoochee River Near Claxton (ftst)
#02196690 is the Horse Creek Gauge at Clearwater, S.C.
a - http://waterdata.usgs.gov/nwis/uv?format=rdbperiod=;
b -
site_no=021973269,02102908,02133500,02341800,02342500,02203000,02196690
z - paste(a, days, b, sep=)
L - readLines(z)
#look for the data with USGS in front of it (this take
advantage of
#the agency column)
L.USGS - grep(^USGS, L, value = TRUE)
DF - read.table(textConnection(L.USGS), fill = TRUE)
colnames(DF) - c(agency, gauge, date, time,
time_zone,
gauge_height,
discharge, precipitation)
pat - ^# +USGS +([0-9]+) +(.*)
L.DD - grep(pat, L, value = TRUE)
library(gsubfn)
DD - strapply(L.DD, pat, c, simplify = rbind)
DDdf - data.frame(gauge = as.numeric(DD[,1]),
gauge_name = DD[,2])
both - merge(DF, DDdf, by = gauge, all.x = TRUE)
dts - as.character(both[,date])
tms - as.character(both[,time])
date_time - as.chron(paste(dts, tms), %Y-%m-%d
%H:%M)
DF - data.frame(Date=as.POSIXct(date_time), both)
#change precip to numeric
DF[,precipitation] -
as.numeric(as.character(DF[,precipitation]))
precip.1 - subset(DF, precipitation!=NA)
b - ddply(precip.1$precipitation,
.(precip.1$gauge_name), cumsum)
DF.precip - precip.1
DF.precip$precipitation - b$.data
#discharge
if(input==data){
return(DF)
}else{
qplot(Date, discharge, data=DF,
geom=line, ylab=Date)+facet_wrap(~gauge_name,
scales=free_y)+coord_trans(y=log10)}
if(input==precipitation){
#precipitation
qplot(Date, precipitation, data=DF.precip,
geom=line)+facet_wrap(~gauge_name, scales=free_y)
}else{
qplot(Date, discharge, data=DF,
geom=line, ylab=Date)+facet_wrap(~gauge_name,
scales=free_y)+coord_trans(y=log10)}
}
--
Stephen Sefick
Let's not spend our time and resources thinking about
Re: [R] avoiding reinstall already installed *package*
On 18-May-10 16:42:40, Peter Ehlers wrote:
On 2010-05-18 10:05, (Ted Harding) wrote:
On 18-May-10 15:49:37, Martin Maechler wrote:
{ I've modified the subject; I can't stand it hitting square into
my face ... }
mr == milton rusermilton.ru...@gmail.com
on Tue, 18 May 2010 12:36:23 -0300 writes:
mr Dear R-experts,
mr I am installing new libraries using
mr install.packages(ggplot2,dependencies=T).
mr But I perceive that many dependencies are already
installed.
mr As I am using a low-band internet, how can avoid reinstall
mr installed libraries?
There's no problem with installed libraries, as ...
they DO NOT EXIST.
These are *PACKAGES* !
Why do you think are you talking about the function
install.packages()
Ah, Martin! I know that package is the official terminology,
but R itself tempts the naive user into deviating from the
True Path. Indeed, I had my fingers burned by this myself,
a long time ago (I'm still licking them ... ).
One might ask: Why do you think we use the function library()?
when loading add-on packages into R. Indeed, the very directory
tree of R itself stores packages under /usr/lib/R/library.
So, once in a while, someone gets it wrong, and has to find it
out the hard way!
Well, I don't know if I've ever disagreed with Ted before,
but here I would (somewhat) disagree. It seems a bit odd that
nobody confuses 'book' with 'library', yet the package/library
problem is persistent. It may have something to do with the
use of 'library' in other computer languages.
Anyway, not long ago there was a suggestion (Rolf Turner's?)
to rename the library() function to something like use(),
but, as I recall, a number of nontrivial objections were
raised.
Of course R stores packages in libraries. That's were books
*should* reside. And it's a good idea to have Martin remind
us now and again that books themselves are not libraries.
But I must confess that I'm no longer much bothered by the
misuse. If it ever leads someone astray in their code, then,
well, they have only themselves to blame.
Cheers,
Peter Ehlers
Well, I don't think you're disagreeing with me, Peter!
My point (which you're not disputing) is that the naive user
will think library because of using the function library().
Perhaps my remark about /usr/lib/R/library was a bit superfluous
(and indeed that path supports your book vs library point).
Nevertheless, it would still reinforce users who think library.
I suppose, to take your analogy, the semantics of, say,
library(Hmisc) could be spelled out as Go to the library
and take out the book called Hmisc. On the other hand, the
naive user will tend to read it as get library Hmisc.
While library is conspicuous to all users because of the
function, the fact that packages should be called packages does
not jump out into your face (until someone on the list does).
That said, I again agree with you that I'm not much bothered
by the question. I see it as one of the various little things
in R which one happily learns (in the end ... ) to live with!
Best wishes,
Ted.
E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk
Fax-to-email: +44 (0)870 094 0861
Date: 18-May-10 Time: 18:03:45
-- XFMail --
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[R] Sweave counter
Dear R users, I am using the Sweave package and I am doing some MCMC. I have a loop function for my MCMC. Every 100 iterations, I want the number of iterations already done to appear on my screen (but not on the final document). Is that possible ? Usually I can rely on if (i%%100==0) cat(i, \n) but not when using Sweave. Thanks for your help, Jimmy [[alternative HTML version deleted]] __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] proportion of treatment effect by a surrogate (fitting multivariate survival model)
On Mon, May 17, 2010 at 7:42 PM, Vinh Nguyen vqngu...@uci.edu wrote: Dear R-help, I would like to compute the variance for the proportion of treatment effect by a surrogate in a survival model (Lin, Fleming, and De Gruttola 1997 in Statistics in Medicine). The paper mentioned that the covariance matrix matches that of the covariance matrix estimator for the marginal hazard modelling of multiple events data (Wei, Lin, and Weissfeld 1989 JASA), and is implemented in Lin's MULCOX2, SAS, and S-plus. Is this the way to fit such a model in R? Suppose I have variables: time, delta, treatment, and surrogate. Should I repeat the dataset (2x) and stack, creating the variables: time1 (time repeated 2x), delta1 (delta repeated 2x), treatment1 (same as treatment, but 0's for the 2nd set), treatment2 (0's in first set, then same as treatment), and surrogate2 (0's in first set, then same as treatment), and id (label the subject, so each id should have 2 observations). Thus, a dataset with n observations will become 2n observations. To fit, do fit - coxph(Surv(time1,delta1) ~ treatment1 + teatment2 + surrogate2 + strata(id) ? I think I figured it out. I should use m - rep(0:1, each=n) for strata. The point estimates matches that of the adjust and unadjusted models when fitting separately (jointly fit to obtain covariances). Thank you and let me know if I've done anything wrong. From here, I can obtain the variance and covariance terms for the coefficients of treatment1 and treatment2. Is this the same as fitting 2 separate models but also obtaining the covariances of the two estimates? Let me know, thanks. Vinh __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] unsigned 4 byte number
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of tetonedge
Sent: Tuesday, May 18, 2010 9:14 AM
To: r-help@r-project.org
Subject: [R] unsigned 4 byte number
Does anybody know how to read in a unsigned 4 byte number
from a binary file?
According to the help for readBin, the signed argument only applies to
size=1 or 2. But if you declare any size larger it assumes it
is signed?
R cannot properly represent unsigned 4 byte C ints with its
type integer (which is a 4 byte signed C int). It can store
them as doubles with the desired values.
You can use readBin to read them as signed 4 byte C ints, stored
in R integers, with
i - readBin(file, what=integer, size=4)
and then convert them to doubles with the following function
unsignedFourByteIntToDouble - function(i) {
d - as.numeric(i)
d[d0] - d[d0] + 2^32
d
}
If you had S+ you could do this in one step using readBin's
output=double argument. In S+'s readBin(), output=type
means to override the default mapping of C types to S+ types.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
Thanks
--
View this message in context:
http://r.789695.n4.nabble.com/unsigned-4-byte-number-tp2221555
p2221555.html
Sent from the R help mailing list archive at Nabble.com.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Sweave counter
Hi Jimmy, You can use message() instead of cat(): if (i%%100==0) message(i) Best, Ista On Tuesday 18 May 2010 1:10:51 pm Jimmy Söderly wrote: Dear R users, I am using the Sweave package and I am doing some MCMC. I have a loop function for my MCMC. Every 100 iterations, I want the number of iterations already done to appear on my screen (but not on the final document). Is that possible ? Usually I can rely on if (i%%100==0) cat(i, \n) but not when using Sweave. Thanks for your help, Jimmy [[alternative HTML version deleted]] __ R-help@r-project.org 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-contained, reproducible code. __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] Sweave counter
Under UNIX I usually write something like system(paste(echo ',...,', sep=sep)) where I replace the '...' with whatever I need to show. This would need some adjustments for non-scalars, though. Benno Am 18.Mai.2010 um 19:10 schrieb Jimmy Söderly: Dear R users, I am using the Sweave package and I am doing some MCMC. I have a loop function for my MCMC. Every 100 iterations, I want the number of iterations already done to appear on my screen (but not on the final document). Is that possible ? Usually I can rely on if (i%%100==0) cat(i, \n) but not when using Sweave. Thanks for your help, Jimmy [[alternative HTML version deleted]] __ R-help@r-project.org 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-contained, reproducible code. __ R-help@r-project.org 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-contained, reproducible code.
[R] Glasso question
The glassolist function says that if rholist is set to NULL, then 10 values in a (hopefully reasonable) range are used. I'm trying to figure out how these 10 values are chosen, I can't find this in any documentation. Thanks for your help! -- View this message in context: http://r.789695.n4.nabble.com/Glasso-question-tp2221587p2221587.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] (no subject)
On May 18, 2010, at 12:07 PM, Arantzazu Blanco Bernardeau wrote: Hello Well, the problem is, that arcilla is the percentage of clay in the soil sample. So, for linear model, I need to work with that number or value. Now, R thinks that arcilla (arcilla means clay in spanish), is a factor, and gives me the value as a factor, so the output of the linear model is Call: lm(formula = formula, data = caperf) Would help if you also displayed the value for formula, so we might have an idea what you are calling your y-variable and it would be wise not to continue to name your formulas formula. require(fortunes) fortune(dog) What happens when you create a new variable in caperf with the numeric equivalant of the arcilla levels? caperf$claynum - as.numeric(as.character(arcilla)) lm(y ~ claynum + limo + CO_gkg1 + C03Ca , data=caperf) -- David. Residuals: Min 1Q Median 3QMax -1.466e+01 -1.376e-15 1.780e-16 2.038e-15 1.279e+01 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept)1.689646.33889 0.267 0.790221 arcilla0.9 1.902288.90888 0.214 0.831239 arcilla10 1.263717.96734 0.159 0.874212 arcilla10.3 15.700819.05141 1.735 0.085090 . arcilla10.47.275177.72806 0.941 0.348183 arcilla10.45 7.038799.02600 0.780 0.436853 arcilla10.52.412418.90827 0.271 0.786954 arcilla10.65 15.442989.03879 1.709 0.089838 . arcilla10.7 19.356519.04675 2.140 0.034185 * arcilla10.93.559479.18501 0.388 0.698974 [...] arcilla9.9 6.319497.35724 0.859 0.391892 arcilla#N/A 24.179598.87201 2.725 0.007274 ** limo 0.249200.04605 5.412 2.76e-07 *** CO_gkg10.210150.03931 5.346 3.73e-07 *** C03Ca 0.017110.02727 0.628 0.531337 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 6.249 on 135 degrees of freedom (50 observations deleted due to missingness) Multiple R-squared: 0.9736,Adjusted R-squared: 0.9014 F-statistic: 13.47 on 370 and 135 DF, p-value: 2.2e-16 So, in the desired linear model, arcilla should be just a line, with the valors of the linear model. I hope you understand better more. If not, I could make an english version of the file to send, so you can try the commands. Thanks a lot for your help! Arantzazu Blanco Bernardeau Dpto de Química Agrícola, Geología y Edafología Universidad de Murcia-Campus de Espinardo Date: Tue, 18 May 2010 11:54:20 -0400 Subject: Re: [R] (no subject) From: mailinglist.honey...@gmail.com To: aramu...@hotmail.com CC: r-help@r-project.org Hi, Sorry, I'm not really getting what going on here ... perhaps having more domain knowledge would help me make better sense of our question. In particular: On Tue, May 18, 2010 at 11:35 AM, Arantzazu Blanco Bernardeau wrote: Hello I have a data array with soil variables (caperf), in which the variable clay is factor (as I see entering str(caperf)) . I need to do a regression model, so I need to have arcilla (=clay) as a numeric variable. For that I have entered as.numeric(as.character(arcilla)) and even entering 'as.numeric(levels(arcilla))[arcilla]' The above code doesn't make sense to me ... Perhaps cleaning up your question and providing some reproducible example we can use to help show you the light (just describing what a variable has isn't enough -- give us minimal code we can paste into R that reproduces your problem). Alternatively, depending no what your levels mean, you might want to recode your data using dummy variables (I'm not sure if that's the official term) .. this is what I mean: http://dss.princeton.edu/online_help/analysis/dummy_variables.htm In your example, let's say you have four levels for clay ... maybe soft, hard, smooth, red Instead of only using 1 variable with values 1-4, you would recode this into 4 variables with values 0,1 So, if one example has a value of smooth for clay. Instead of coding it like: clay: 3 You would do: soft: 0 hard: 0 smooth: 1 red : 0 -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact _ Consejos para seducir ¿Puedes conocer gente nueva a través de Internet? ¡Regístrate ya! __ R-help@r-project.org 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-contained, reproducible code. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list
[R] Res: Using the zero-inflated binomial in experimental designs
Hi Ben! First I thank you for your attention. Unfortunately, the ANOVA does not work with vglm. In another email, Rafael warned me that actually a lot of zeros does not necessarily imply a distribution of zeros binomail inflated. So how could I test if my variable is or not a binomial zero inflated? Thanks. M.Sc Ivan Bezerra Allaman Zootecnista Doutorando em Produção Animal/Aquicultura - UFLA email e msn - ivanala...@yahoo.com.br Tel: (35)3826-6608/9925-6428 De: Ben Bolker [via R] ml-node+2221578-2039137178-109...@n4.nabble.com Enviadas: Terça-feira, 18 de Maio de 2010 13:34:01 Assunto: Re: Using the zero-inflated binomial in experimental designs Ivan Allaman ivanalaman at yahoo.com.br writes: I'm trying to use the inflated binomial distribution of zeros (since 75% of the values are zeros) in a randomized block experiment with four quantitative treatments (0, 0.5, 1, 1.5), but I'm finding it difficult, since the examples available in VGAM packages like for example, leave us unsure of how it should be the data.frame for such analysis. Unfortunately the function glm does not have an option to place a family of this kind I'm about, because if I had, it would be easy, made that my goal is simple, just wanting to compare the treatments. For that you have an idea, here is an example of my database. BLOCK NIVNT MUMI Inicial 0 18 0 [snip] where: NIV are the treatments; NT is the total number of piglets born; Mumi is the number of mummified piglets NT. Mumi The variable is of interest. If someone can tell me some stuff on how I can do these tests in R, similar to what I would do using the function glm, I'd be grateful. I thank everyone's attention. something like comparing the likelihoods of m1 - vglm(cbind(MUMI,NT-MUMI)~NIV*BLOCK,zibinomial,data=mydata) m2 - vglm(cbind(MUMI,NT-MUMI)~NIV+BLOCK,zibinomial,data=mydata) m3 - vglm(cbind(MUMI,NT-MUMI)~BLOCK,zibinomial,data=mydata) I don't know whether the anova() method works for VGLM objects or not. By the way, 75% zeroes doesn't necessarily imply zero-inflation -- perhaps it just means a low incidence? __ [hidden email] 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-contained, reproducible code. View message @ http://r.789695.n4.nabble.com/Using-the-zero-inflated-binomial-in-experimental-designs-tp2221254p2221578.html To unsubscribe from Using the zero-inflated binomial in experimental designs, click here. -- View this message in context: http://r.789695.n4.nabble.com/Using-the-zero-inflated-binomial-in-experimental-designs-tp2221254p2221635.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML version deleted]] __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] C function call in R
John Lande john.land...@gmail.com writes: dear all, we am trying to improve the performance of my R code, with the implentation of some function with custom C code. we found difficult to import and export/import data structure such us matrices or data.frame into the external C functions. Please give a *very simple* example of what you're trying and failing to do. Use the .C() interface, forget about the .Call interface. Then it is not that hard. Start with the convolve example on p.69 and 70 of Writing R Extensions. Get that working and then turn it into your problem. Forget about lists and data frames: everything is going to be a simple vector. That includes arrays and matrices: you can pass them in, but C will know nothing about their dimensions until you tell it. Of course, you can pass the dimension vectors in as a separate vector. So, if you use arrays, you need to understand the order in which R stores the elements of the array. If your problem cannot be solved with the .C interface then you should consider whether it is worthwhile to proceed as the .Call interface repays those who use it frequently but has a considerably steeper learning (and forgetting) curve. Dan we already tried the solution from Writing R Extensions form the R webpage. do you have any other solution or more advanced documentation on that point? looking forward your answer [[alternative HTML version deleted]] __ R-help@r-project.org 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-contained, reproducible code.
[R] Maximization of quadratic forms
Dear R Help,
I am trying to fit a nonlinear model for a mean function $\mu(Data_i,
\beta)$ for a fixed covariance matrix where $\beta$ and $\mu$ are low-
dimensional. More specifically, for fixed variance-covariance matrices
$\Sigma_{z=0}$ and $\Sigma_{z=1}$ (according to a binary covariate $Z
$), I am trying to minimize:
$\sum_{i=1^n} (Y_i-\mu_(Data_i,\beta))' \Sigma_{z=z_i}^{-1} (Y_i-
\mu_(Data_i,\beta))$
in terms of the parameter $\beta$. Is there a way to do this in R in a
more stable and efficient fashion than just using a general
optimization function such as optim? I have tried to use gnls, but I
was unsuccessful in specifying different values of the covariance
matrix according to the covariate $Z$.
Thank you very much for your help,
Taki Shinohara
Russell Shinohara, MSc
PhD Candidate and NIH Fellow
Department of Biostatistics
Bloomberg School of Public Health
The Johns Hopkins University
615 N. Wolfe St., Suite E3033
Baltimore, MD 21205
tel: (203) 499-8480
http://biostat.jhsph.edu/~rshinoha
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Re: [R] Getting dates in an SPSS file in right format.
On 05/18/2010 11:52 AM, Chuck Cleland wrote: On 5/18/2010 12:38 PM, Praveen Surendran wrote: Dear all, I am trying to read an SPSS file into a data frame in R using method read.spss(), sample- read.spss(file.name,to.data.frame=TRUE) But dates in the data.frame 'sample' are coming as integers and not in the actual date format given in the SPSS file. Appreciate if anyone can help me to solve this problem. Date variables in SPSS contain the number of seconds since October 14, 1582. You might try something like this: sample$MYDATE- as.Date(as.POSIXct(sample$MYDATE, origin=1582-10-14, tz=GMT)) Kind Regards, Praveen Surendran 2G, Complex and Adaptive Systems Laboratory (UCD CASL) School of Medicine and Medical Sciences University College Dublin Belfield, Dublin 4 Ireland. Office : +353-(0)1716 5334 Mobile : +353-(0)8793 13071 The spss.get function in the Hmisc package handles SPSS dates. Frank -- Frank E Harrell Jr Professor and ChairmanSchool of Medicine Department of Biostatistics Vanderbilt University __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] survey package: weights used in svycoxph()
On Tue, 18 May 2010, Vinh Nguyen wrote: Binder's estimating equations are the usual way of applying weights to a Cox model, so nothing special is done apart from calling coxph(). To quote the author of the survival package, Terry Therneau, Other formulae change in the obvious way, eg, the weighted mean $\bar Z$ is changed to include both the risk weights $r$ and the external weights $w$. [Mayo Clinic Biostatistics technical report #52, section 6.2.2] Don't see a section 6.2.2 in this technical report. Sorry, #58 -thomas Thomas Lumley Assoc. Professor, Biostatistics tlum...@u.washington.eduUniversity of Washington, Seattle __ R-help@r-project.org 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-contained, reproducible code.
[R] Regarding the 'R' Load Command
Hi, I'm new to 'R' and need some help on the Load command. Any responses will be highly appreciated. Thanks in advance! As per manuals, the Load command expects a binary file input that is saved using a save command. However it is required that we need to call the 'R' program from Java web application using RJava, and pass a string to the 'R program instead of a binary file. Is it possible? I was exploring the options of using TextConnections, file connections and other types of connections in order to read a stream of input (either from a file, stdin etc). I am able to read the string, but the Save and Load commands are not accepting the string input. Here is the sequence of commands I tried running, and the error received. There is no clue on this error, especially when trying to use the eval function in randomForest package, even on the internet. Can anyone help please! library(randomForest) randomForest 4.5-34 Type rfNews() to see new features/changes/bug fixes. load(C://Program Files//R//R-2.10.1//bin//rfoutput) zz - file(ex.data, w) cat(\imurder\ \itheft\ \irobbery\ \iassault\ \idrug\ \iburglary\ \igun\ \psych\ \Freq\ \priors\ \firstage\ \intage\ \sex\ \race\ \marstat\ \empac\ \educ\ \zipcode\ \suspendmn\ \drugs\ \alco\ \probation\ \parole\,file = zz, sep = \n, fill = TRUE) cat(\10\ 0 0 0 1 0 0 0 0 0 58 19 19 \1\ \BLACK\ \SINGLE\ \UNEMPLD\ 0 21215 0 0 0 1 0,file = zz, sep = \n, fill = TRUE) save(zz, file = testmurali, version = 2) predict(rfoutput,newdata=testmurali,type=response) Error in eval(expr, envir, enclos) : object 'imurder' not found Best Regards, Murali Godavarthi mgodavar...@dpscs.state.md.us [[alternative HTML version deleted]] __ R-help@r-project.org 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-contained, reproducible code.
Re: [R] GUI commands to call for a protein from protein data bank
I tried doing that and this is what I go:
dlg - aDialog(items=list(
ProtienCode=stringItem()
),
OK_handler=function(.) { # . is reference to dlg object
values - .$to_R()
f - function(ProtienCode)
pdb - read.pdb(.$get_ProteinCode())
#cat(ProteinCode is,ProtienCode,\n)
do.call(f, values)
}
)
dlg$make_gui()
with this error message
Error in function (ProtienCode) : could not find function read.pdb
Do you know how to get rid of this error?
On Tue, May 18, 2010 at 5:53 AM, jverzaniNWBKZ [via R]
ml-node+2221226-858152228-140...@n4.nabble.comml-node%2b2221226-858152228-140...@n4.nabble.com
wrote:
Amitoj S. Chopra amitojc at gmail.com writes:
What I am trying to do is use GUI function, traitr, and to call for a pdb
file and save it and then display it. I want to call for it by taking it
from the user and then displaying it on the screen. I am having problems
with that. The line pdb - read.pdb(ProteinCode) where proteincode
should
be the name of the protein, for example 1ly2, but it always ends up being
protein. My question is how to you make the input for read.pdb actually
be
the input by the user and not protein code. I want to be able to type
1ly2,
and for the program to actually display the contents of 1ly2. Thanks!
I'm just guessing, but you might try this for your OK_handler:
OK_handler=function(.) {
pdb - read.pdb(.$get_ProteinCode())
}
(Or use yours, but drop the quotes around ProteinCode.)
That doesn't modify pdb outside the scope of the handler, so likely you
need to
do something else with it.
--John
Code:
dlg - aDialog(items=list(
ProteinCode=stringItem()
),
OK_handler=function(.) { # . is reference to dlg object
values - .$to_R()
f - function(ProteinCode)
pdb - read.pdb(ProteinCode)
do.call(f, values)
}
)
dlg$make_gui()
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Re: [R] looking for .. dec if vector if element x
Thank you Adrian, its working fine. Knut __ R-help@r-project.org 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-contained, reproducible code.
