Re: [R] nls and na/Nan/Inf error
Use a grid search to get the starting values in which case you
will likely be close enough that you won't run into problems
even without derivatives:
attach(fldgd)
grid - expand.grid(Vr = seq(0,.3,.1), Vm = seq(.45, 1, .05),
alpha = seq(1,2,.25), lamda = seq(1,2,.25))
ss - function(p) sum((Moisture - vanGen(Suction, p[1], p[2], p[3], p[4]))^2)
idx - which.min(apply(grid, 1, ss))
startval - grid[idx,]
nls(Moisture ~ vanGen(Suction, Vr, Vm, alpha, lamda), start = startval)
On 9/26/05, Tony Meissner [EMAIL PROTECTED] wrote:
I am trying to it a particular nonlinear model common in Soil Science to
moisture release data from soil. I have written the function as shown
below according to the logist example in Ch8 of Pinheiro Bates. I am
getting the following error (R version 2.1.1)
*Error in qr(attr(rhs, gradient)) : NA/NaN/Inf in foreign function
call (arg 1)*
Below is the function and data.
/# the van genuchten moisture release function
vanGen - function(x, Vr, Vm, alpha, lamda) {
if (Vr 0) Vr - 0
Vr + (Vm - Vr)/((1+(alpha*x)^lamda)^(1-1/lamda))
}
vanGen - deriv(~Vr + (Vm - Vr)/((1+(alpha*x)^lamda)^(1-1/lamda)),
c(Vr, Vm, alpha, lamda), function(x, Vr, Vm, alpha, lamda) {} )/
the call in R
/ fm1fld.nls - nls(Moisture ~ vanGen(Suction, Vr,Vm,alpha,lamda),
+ data=fldgd, start=c(Vr=0.229, Vm=0.433, alpha=0.2, lamda=1.5))
/and the data:/
/* Suction Moisture
100.433
210.421
340.400
4 100.379
5 200.366
6 300.362
7 400.358
8 500.353
9 600.351
10 700.349
*/
/can anyone offer any suggestions. The parameters Vr, Vm = 0, alpha
0, and
lamda 1
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Re: [R] Indentation in R code
I'm crossposting to the ESS-help mailing list which is slightly
more appropriate here. [This may be a rare case where
crossposting seems to make much sense.]
PD == Peter Dalgaard [EMAIL PROTECTED]
on 25 Sep 2005 19:40:45 +0200 writes:
PD Seth Falcon [EMAIL PROTECTED] writes:
On 24 Sep 2005, [EMAIL PROTECTED] wrote:
I am using emacs-21.3 when writing R functions on Linux debian, and
I am trying to follow the advice i R-exts.pdf (2.1.1) regarding
indentation. That is, I set 'c-default-style' to bsd and
'c-basic-offset' to 4. However, while this gives me the intended
indentation in C code, it doesn't change the behavior in R code; I
still get an indentation of size 2. This is my .emacs file after
customization:
(require 'ess-site)
(custom-set-variables
;; custom-set-variables was added by Custom -- don't edit or
;; cut/paste it! Your init file should contain only one such
;; instance.
'(c-basic-offset 4)
'(c-default-style bsd))
(custom-set-faces
;; custom-set-faces was added by Custom -- don't edit or cut/paste it!
;; Your init file should contain only one such instance.
)
Not sure if this is the best way, but I have the following after
loading ess-site:
(setq ess-indent-level 4)
PD I have (I believe it stems from Martin M. originally):
yes, most probably {IIRC, Kurt Hornik was involved too}.
PD (add-hook 'c-mode-hook '(lambda()
PD(c-set-style stroustrup)))
the above is not quite what I have or did recommend,
which is rather bsd + offset 4 as Göran has above
In fact, Göran mentions the R-exts manual and that has
the following *before* giving the emacs-lisp statements:
(For GNU Emacs 20: for GNU Emacs 21 use
customization to set the `c-default-style' to `bsd' and
`c-basic-offset' to `4'.)
and indeed, that's what Göran did and you should do with a
current emacs, either customize via GUI or,
in your ~/.emacs file, find the section '(custom-set-variables ...)' and add
'(c-basic-offset 4)
'(c-default-style bsd)
to the lines already there, or if there's no such section, add
(custom-set-variables
;; custom-set-variables was added by Custom -- don't edit or cut/paste it!
;; Your init file should contain only one such instance.
'(c-basic-offset 4)
'(c-default-style bsd)
)
to the end of your ~/.emacs file
PD (add-hook 'ess-mode-hook
PD'(lambda()
PD (if (or (string ess-version 5.0)
PD (string= ess-version 5.0))
PD (ess-set-style 'C++)
PD (ess-set-style 'C++ 'quiet))
PD
PD (add-hook 'local-write-file-hooks
PD '(lambda()
PD(delete-trailing-whitespace)
PD))
PD ))
yes; using (add-hook ...) is really more clean than first
requiring ess-site.
Also, since nowadays I assume everyone has an ess-version = 5,
the above becomes simply
(add-hook 'ess-mode-hook
'(lambda()
(ess-set-style 'C++ 'quiet)
(add-hook 'local-write-file-hooks
'(lambda()
(delete-trailing-whitespace)
Note that this has the standard e-lisp function
'delete-trailing-whitespace' which is simpler but also less
flexible than the 'ess-nuke-trailing-whitespace' which we've had
there.
Martin Maechler, ETH Zurich
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Re: [R] nls and na/Nan/Inf error
This works if you omit the deriv() step.
Use R's options(error=dump.frames) and debugger(). This gives
Browse[1] rhs
[1] 0.433 0.4272571 0.3994105 0.3594037 0.3270730 0.3104752 0.3000927
[8] 0.2928445 0.2874249 0.2831787
attr(,gradient)
VrVm alphalamda
[1,] 0. 1.000 0. NaN
[2,] 0.02815158 0.9718484 -0.04069202 0.001183749
[3,] 0.16465431 0.8353457 -0.17769291 -0.035591190
[4,] 0.36076599 0.6392340 -0.24085444 -0.100064577
[5,] 0.51925014 0.4807499 -0.21793994 -0.136056450
[6,] 0.60061200 0.3993880 -0.19071160 -0.145267481
[7,] 0.65150658 0.3484934 -0.17020938 -0.147113828
[8,] 0.68703698 0.3129630 -0.15471851 -0.146388612
[9,] 0.71360324 0.2863968 -0.14263118 -0.144660967
[10,] 0.73441811 0.2655819 -0.13290951 -0.142543261
and note the NaN. Now think about your formula for x = 0: it does not
actually depend on lamda. The analytical derivative ends up with a
calculation as 0/0.
On Mon, 26 Sep 2005, Tony Meissner wrote:
I am trying to it a particular nonlinear model common in Soil Science to
moisture release data from soil. I have written the function as shown
below according to the logist example in Ch8 of Pinheiro Bates. I am
getting the following error (R version 2.1.1)
*Error in qr(attr(rhs, gradient)) : NA/NaN/Inf in foreign function
call (arg 1)*
Below is the function and data.
/# the van genuchten moisture release function
vanGen - function(x, Vr, Vm, alpha, lamda) {
if (Vr 0) Vr - 0
Vr + (Vm - Vr)/((1+(alpha*x)^lamda)^(1-1/lamda))
}
vanGen - deriv(~Vr + (Vm - Vr)/((1+(alpha*x)^lamda)^(1-1/lamda)),
c(Vr, Vm, alpha, lamda), function(x, Vr, Vm, alpha, lamda) {} )/
the call in R
/ fm1fld.nls - nls(Moisture ~ vanGen(Suction, Vr,Vm,alpha,lamda),
+ data=fldgd, start=c(Vr=0.229, Vm=0.433, alpha=0.2, lamda=1.5))
/and the data:/
/* Suction Moisture
100.433
210.421
340.400
4 100.379
5 200.366
6 300.362
7 400.358
8 500.353
9 600.351
10 700.349
*/
/can anyone offer any suggestions. The parameters Vr, Vm = 0, alpha
0, and
lamda 1
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272860 (secr)
Oxford OX1 3TG, UKFax: +44 1865 272595
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Re: [R] SAX Parser best practise
Hi Duncan,
thanks again for your comments.
I dug around in the libxml code and the Web to verify that
validation is indeed only possible in libxml when one uses
DOM (i.e. xmlTreeParse()).
Using DOM is not an option for me, so I need to validate the xml parts
I'm interested in within my creation mechanism. It's OK, but not the
best solution in questions of design.
BTW, there is a new version of the XML package on the
Omegahat web site.
I'll use it extensive in this days and unfortunately I have already a
question/problem pending:
Taking the following R function:
test-function(){
sep=
xmlText -
xmlText -paste(xmlText,spectrum id=\3257\,sep=sep)
xmlText -paste(xmlText,mzArrayBinary,sep=sep)
xmlText -paste(xmlText,dataMonday/data,sep=sep)
xmlText -paste(xmlText,/mzArrayBinary,sep=sep)
xmlText -paste(xmlText,intenArrayBinary,sep=sep)
xmlText -paste(xmlText,dataTuesday/data,sep=sep)
xmlText -paste(xmlText,/intenArrayBinary,sep=sep)
# xmlText -paste(xmlText,/spectrum,sep=sep)
# xmlText -paste(xmlText,spectrum id=\3259\,sep=sep)
xmlText -paste(xmlText,mzArrayBinary,sep=sep)
xmlText -paste(xmlText,dataWednesday/data,sep=sep)
xmlText -paste(xmlText,/mzArrayBinary,sep=sep)
xmlText -paste(xmlText,intenArrayBinary,sep=sep)
xmlText -paste(xmlText,dataThursday/data,sep=sep)
xmlText -paste(xmlText,/intenArrayBinary,sep=sep)
xmlText -paste(xmlText,/spectrum,sep=sep)
xmlEventParse(xmlText, asText=TRUE, handlers = list(text =
function(x, ...) {cat(nchar(x),x, \n)}))
return(invisible(NULL))
}
Using this function in the given form works fine. xmlEventParse() with
the simplest handler I can imagine finds all 4 text-nodes within the
spectrum tag and prints them out. But if one uncomment both lines in
the middle, introducing 2 spectrum tags with different id's
xmlEventParse() returns with an exception. Of course the weekdays within
data are arbitrary values used here. Further, using an other input
file I could see, that for one and the same data node the handler for
text-nodes was invoked two times, one time for a first part of the
content and one time for the rest of the content. Both invocations
together gave me exactly the content from the data node.
So, am I on the wrong way? Or is this some buggy behaviour?
I appreciat any help and assistance!
Jan
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[R] How to get the rowindices without using which?
Hi, I was wondering if it is possible to get the rowindices without using the function which because I don't have a restriction criteria. Here's an example of what I mean: # take 10 randomly selected instances iris[sample(1:nrow(iris), 10),] # output Sepal.Length Sepal.Width Petal.Length Petal.Width Species 76 6.6 3.0 4.4 1.4 versicolor 105 6.5 3.0 5.8 2.2 virginica 131 7.4 2.8 6.1 1.9 virginica 79 6.0 2.9 4.5 1.5 versicolor 69 6.2 2.2 4.5 1.5 versicolor 42 4.5 2.3 1.3 0.3 setosa 25 4.8 3.4 1.9 0.2 setosa 129 6.4 2.8 5.6 2.1 virginica 60 5.2 2.7 3.9 1.4 versicolor 80 5.7 2.6 3.5 1.0 versicolor What I want to get are their rownumbers: 76, 105, 131, 79, 69, 42, 25, 129, 60, 80. Thanks in advance, Martin __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to get the rowindices without using which?
try this: dat - iris[sample(1:nrow(iris), 10), ] dat match(rownames(dat), rownames(iris)) I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://www.med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm - Original Message - From: Martin Lam [EMAIL PROTECTED] To: R r-help@stat.math.ethz.ch Sent: Monday, September 26, 2005 10:37 AM Subject: [R] How to get the rowindices without using which? Hi, I was wondering if it is possible to get the rowindices without using the function which because I don't have a restriction criteria. Here's an example of what I mean: # take 10 randomly selected instances iris[sample(1:nrow(iris), 10),] # output Sepal.Length Sepal.Width Petal.Length Petal.Width Species 76 6.6 3.0 4.4 1.4 versicolor 105 6.5 3.0 5.8 2.2 virginica 131 7.4 2.8 6.1 1.9 virginica 79 6.0 2.9 4.5 1.5 versicolor 69 6.2 2.2 4.5 1.5 versicolor 42 4.5 2.3 1.3 0.3 setosa 25 4.8 3.4 1.9 0.2 setosa 129 6.4 2.8 5.6 2.1 virginica 60 5.2 2.7 3.9 1.4 versicolor 80 5.7 2.6 3.5 1.0 versicolor What I want to get are their rownumbers: 76, 105, 131, 79, 69, 42, 25, 129, 60, 80. Thanks in advance, Martin __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] getting variable length numerical gradient
Tnx very much Dimitris,
your code does what I need. I've just adapted it to my needs (e.g., I
don't deal with scalar functions), and so solved my problem.
Given this, is there a way to use the deriv function in the base
package, within this context (variable length vector of indipendent
variables)?
Best,
Antonio, Fabio Di Narzo.
On 9/25/05, Dimitris Rizopoulos [EMAIL PROTECTED] wrote:
maybe you can find the following function useful (any comments are
greatly appreciated):
fd - function(x, f, scalar = TRUE, ..., eps =
sqrt(.Machine$double.neg.eps)){
f - match.fun(f)
out - if(scalar){
if(length(f0 - f(x, ...)) != length(x))
stop('f' must be vectorized)
x. - x + eps * pmax(abs(x), 1)
c(f(x., ...) - f0) / (x. - x)
} else{
n - length(x)
res - array(0, c(n, n))
f0 - f(x, ...)
ex - pmax(abs(x), 1)
for(i in 1:n){
x. - x
x.[i] - x[i] + eps * ex[i]
res[, i] - c(f(x., ...) - f0) / (x.[i] - x[i])
}
res
}
out
}
## Examples
x - seq(-3.3, 3.3, 0.1)
all.equal(fd(x, pnorm, mean = 0.5), dnorm(x, mean = 0.5))
# Approximate the Hessian matrix for a logistic regression
# the score vector function
gn - function(b, y, X){
p - as.vector(plogis(X %*% b))
-colSums(X * (y - p))
}
# We simulate some data and fit the logistic regression
n - 800
x1 - runif(n,-3, 3); x2 - runif(n, -3, 3)
pr - plogis(0.8 + 0.4 * x1 - 0.3 * x2)
y - rbinom(n, 1, pr)
fm - glm(y ~ x1 + x2, binomial)
## The Hessian using forward difference approximation
fd(fm$coef, gn, scalar = FALSE, y = y, X = cbind(1, x1, x2))
## The true Hessian
solve(summary(fm)$cov.unscaled)
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://www.med.kuleuven.be/biostat/
http://www.student.kuleuven.be/~m0390867/dimitris.htm
- Original Message -
From: Antonio, Fabio Di Narzo [EMAIL PROTECTED]
To: R-help@stat.math.ethz.ch
Sent: Sunday, September 25, 2005 11:37 AM
Subject: [R] getting variable length numerical gradient
Hi all.
I have a numerical function f(x), with x being a vector of generic
size (say k=4), and I wanna take the numerically computed gradient,
using deriv or numericDeriv (or something else).
My difficulties here are that in deriv and numericDeric the function
is passed as an expression, and one have to pass the list of
variables
involved as a char vector... So, it's a pure R programming question.
Have a nice sunday,
Antonio, Fabio Di Narzo.
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[R] hist(x, ...) with normal distribution curve
. I am looking for a histogram or box plot with the adding normal distribution curve I think that must be possible, but I am not able to find out how to do. Regards Knut __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] figure widths in sweave
gRoovers,
Can the size of figures be controlled from within a noweb document
without resorting to editing the \includegraphics sections in the .tex
file?
Can the figure widths be set in the environmental declarations at the
start?
Can they be set within the \begin{figure} environment?
JC
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Re: [R] hist(x, ...) with normal distribution curve
Hi answered hundered times. Dear R people: I would like to superimpose a normal curve on a histogram. x-rnorm(150) h-hist(x,breaks=15) xhist-c(min(h$breaks),h$breaks) yhist-c(0,h$density,0) xfit-seq(min(x),max(x),length=40) yfit-dnorm(xfit,mean=mean(x),sd=sd(x)) plot(xhist,yhist,type=s,ylim=c(0,max(yhist,yfit))) lines(xfit,yfit) Bill above is e.g. Bill Simpson's answer from 2001. Found from R-site search ***histogram density normal***. HTH Petr On 25 Sep 2005 at 14:34, Knut Krueger wrote: . I am looking for a histogram or box plot with the adding normal distribution curve I think that must be possible, but I am not able to find out how to do. Regards Knut __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Petr Pikal [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] hist(x, ...) with normal distribution curve
Le 25.09.2005 14:34, Knut Krueger a écrit :
.
I am looking for a histogram or box plot with the adding normal
distribution curve
I think that must be possible, but I am not able to find out how to do.
Regards Knut
Hi Knut,
There are a lot of ways to do that, let x be your data (assume x ~
N(mu=2,sd=.4))
R x - rnorm(200, mean=2, sd=.4)
** With the traditionnal graphics system, do :
R hist(x, prob=T)
R curve(dnorm, col=2, mean=mean(x), sd=sd(x))
** With lattice :
R histogram(~x,
panel = function(x,...){
panel.histogram(x,...)
panel.mathdensity(dmath = dnorm, col = red,
args = list(mean=mean(x),sd=sd(x)))
},
type=density)
Then, have a look at :
http://addictedtor.free.fr/graphiques/search.php?q=hist
And also have a nice day
Romain
--
visit the R Graph Gallery : http://addictedtor.free.fr/graphiques
~
~~ Romain FRANCOIS - http://addictedtor.free.fr ~~
Etudiant ISUP - CS3 - Industrie et Services
~~http://www.isup.cicrp.jussieu.fr/ ~~
Stagiaire INRIA Futurs - Equipe SELECT
~~ http://www.inria.fr/recherche/equipes/select.fr.html~~
~
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Re: [R] anova on binomial LMER objects
On 9/25/05, Horacio Montenegro [EMAIL PROTECTED] wrote: Hi Spencer and Robert, I have found the same behaviour, but only for lme4 and Matrix after the 0.96 release. lme4 0.95-10 and Matrix 0.95-13 releases gave sensible results. This could be an introduced bug, or a solved bug - you should ask Prof. Bates. hope this helps, cheers, Horacio Montenegro I have run into a couple of other things that the improvements from the 0.95 series to the 0.96 series has made worse. This may take a while to sort out. Thanks to Robert Bagchi for the very thorough error report. --- Spencer Graves [EMAIL PROTECTED] wrote: I agree: Something looks strange to me in this example also; I have therefore copied Douglas Bates and Deepayan Sarkar. You've provided a nice simulation. If either of them have time to look at this, I think they could tell us what is happening here. If you need an answer to your particular problem, you could run that simulation 1000 or 1,000 times. That would tell you whether to believe the summary or the anova, or neither. If you want to understand the algorithm, you could walk through the code. However, lmer is a generic, and I don't have time now to figure out how to find the source. A response from Brian Ripley to a question from me a couple of days ago provides a nice summary of how to do that, but I don't have time to check that now. Sorry I couldn't help more. spencer graves Robert Bagchi wrote: Dear R users, I have been having problems getting believable estimates from anova on a model fit from lmer. I get the impression that F is being greatly underestimated, as can be seen by running the example I have given below. First an explanation of what I'm trying to do. I am trying to fit a glmm with binomial errors to some data. The experiment involves 10 shadehouses, divided between 2 light treatments (high, low). Within each shadehouse there are 12 seedlings of each of 2 species (hn sl). 3 damage treatments (0, 0.1, 0.25 leaf area removal) were applied to the seedlings (at random) so that there are 4 seedlings of each species*damage treatment in each shadehouse. There maybe a shadehouse effect, so I need to include it as a random effect. Light is applied to a shadehouse, so it is outer to shadehouse. The other 2 factors are inner to shadehouse. We want to assess if light, damage and species affect survival of seedlings. To test this I fitted a binomial mixed effects model with lmer (actually with quasibinomial errors). THe summary function suggests a large effect of both light and species (which agrees with graphical analysis). However, anova produces F values close to 0 and p values close to 1 (see example below). Is this a bug, or am I doing something fundamentally wrong? If anova doesn't work with lmer is there a way to perform hypothesis tests on fixed effects in an lmer model? I was going to just delete terms and then do liklihood ratio tests, but according to Pinheiro Bates (p. 87) that's very untrustworthy. Any suggestions? I'm using R 2.1.1 on windows XP and lme4 0.98-1 Any help will be much appreciated. many thanks Robert __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] merge maps from shapefile to lattice
On Mon, 26 Sep 2005, Toni Viúdez wrote: Hi everybody: Could anybody help how I solve the next problem?. I'm doing interpolation maps of tropospheric ozone of my region, and after create it using IDW, and kriging methods, I want from shapefiles (*.shx, *.shp, *.dbf, *.sbx *.sbn ) create contour over the interpolation maps. Could anybody tell me how do it?. In replies off-list, I've suggested that using base graphics is simpler because you can build things up bit by bit with add=TRUE. However, using the maptools and sp packages, you can look at the example in ?meuse.riv: spplot(meuse.grid, col.regions=bpy.colors(), main = meuse.grid, sp.layout=list( list(sp.polygons, meuse.sr), list(sp.points, meuse, pch=+, col=black) ) ) which plots a grid of data (your interpolations would need to be converted to this format) over polygons but under points, where meuse.sr is a SpatialPolygons object. You would then do: my_polys - readShapePoly(my_polys.shp) spplot(meuse.grid, sp.layout=list(sp.polygons, my_polys)) to get something similar. Note that the polygons seem to be plotted under the image in the example. Thanks in advance. ## Antonio Viudez Mora Departamento de Dinámica de Contaminantes Fundación CEAM Paterna (Valencia) tel: 961318190. ext: 216 e-mail:[EMAIL PROTECTED] ## __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Roger Bivand Economic Geography Section, Department of Economics, Norwegian School of Economics and Business Administration, Helleveien 30, N-5045 Bergen, Norway. voice: +47 55 95 93 55; fax +47 55 95 95 43 e-mail: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] regression methods for circular(?) data.
Dear R-users, I have the following data x - runif(300,min=1,max=230) y - x*0.005 + 0.2 y - y+rnorm(100,mean=0,sd=0.1) y - y%%1 # --- modulo operation plot(x,y) and would like to recapture the slope (0.005) and intercept(0.2). I wonder if there are any clever algorithms to do this. I was looking at the function lm.cirucalar. Is this the method to use? If, which of the references is best too look at? Eryk __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Error Message - Error: symbol print-name too long
Dear All, I write to ask for information regarding the error message: Error: symbol print-name too long. I am afraid that I can't include any code to help with any further diagnosis of the problem as the code is far too long to be of any use, and I have not been able to re-create the problem in shorter example codes. I have searched the R manual and help pages and found no mention of this error message. All help appreciated, Carl A Anderson. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] getting variable length numerical gradient
Dimitris,
I'm new to R programming, and I'm trying to learn the proper way to do
certain things. E.g., I had a piece of code with explicit iteration to
apply some computations to a vector. It was pretty slow. I found a way
to utilize R's built-in vectorization and it was sped up considerably.
So I want to ask about the code you supplied. Please see below.
(By the way, this message is best viewed using a mono-spaced font.)
On Sunday 25 September 2005 04:07, Dimitris Rizopoulos wrote:
maybe you can find the following function useful (any comments are
greatly appreciated):
fd - function(x, f, scalar = TRUE, ..., eps =
sqrt(.Machine$double.neg.eps)){
f - match.fun(f)
out - if(scalar){
...
} else{
n - length(x)
res - array(0, c(n, n))
f0 - f(x, ...)
ex - pmax(abs(x), 1)
for(i in 1:n){
This (following) statement will create a copy of the entire x vector
on each iteration. It doesn't look like that's what you would want to
do:
x. - x
The computation described by this statement could be vectorized outside
the loop:
x.[i] - x[i] + eps * ex[i]
res[, i] - c(f(x., ...) - f0) / (x.[i] - x[i])
}
res
}
out
}
Offhand, I cannot tell for sure if the last line of that loop is
vectorizable, but I have a hunch it is.
So at a minimum, it seems this fragment of your code:
for(i in 1:n){
x. - x
x.[i] - x[i] + eps * ex[i]
res[, i] - c(f(x., ...) - f0) / (x.[i] - x[i])
}
Could be more efficiently and succinctly replaced with this:
x. - x + eps * ex
for (in in 1:n)
res[, i] - c(f(x., ...) - f0) / (x.[i] - x[i])
Could your someone else with R programming experience comment?
Thanks.
Randall Schulz
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[R] create trend variable in a regression using R
Hi, my name is Giacomo. I would like to know how to create a Trend variable in a regression using R. Thank you for your help. My best regards, Giacomo - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] histogram - one bin for all values larger than a certain value
Dear all, I wonder if I can put together a histogram where one bin contains all the values that are larger than a certain specified value. Example: I have values ranging from 0 to 40 and I want 10 bins from 0 to 10, i.e. for the intervals [0,1), [1,2) , ..., [9,10). And then I want one last bin which contains all the values larger than 10, i.e. for the interval [10, 40). Thanks, Florian __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] SAX Parser best practise
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
When you uncomment the two lines, your document
becomes two nodes
spectrum
...
spectrum
spectrum
...
/spectrum
XML requires that there be a single top-level node.
And so the parser throws an error saying
Extra content at the end of the document
And it is the second spectrum .. /spectrum
node that it is complaining about.
You can wrap the entire thing in a top node, e.g.
spectra spectrum.../spectrumspectrum.../spectrum/spectra
How did I find this? I looked at the error message from
libxml. Now that we have exceptions in R and we are using
libxml2, etc. I can make this material available at the
R level. So I'll do that.
Jan Hummel wrote:
Hi Duncan,
BTW, there is a new version of the XML package on the
Omegahat web site.
I'll use it extensive in this days and unfortunately I have already a
question/problem pending:
Taking the following R function:
test-function(){
sep=
xmlText -
xmlText -paste(xmlText,spectrum id=\3257\,sep=sep)
xmlText -paste(xmlText,mzArrayBinary,sep=sep)
xmlText -paste(xmlText,dataMonday/data,sep=sep)
xmlText -paste(xmlText,/mzArrayBinary,sep=sep)
xmlText -paste(xmlText,intenArrayBinary,sep=sep)
xmlText -paste(xmlText,dataTuesday/data,sep=sep)
xmlText -paste(xmlText,/intenArrayBinary,sep=sep)
# xmlText -paste(xmlText,/spectrum,sep=sep)
# xmlText -paste(xmlText,spectrum id=\3259\,sep=sep)
xmlText -paste(xmlText,mzArrayBinary,sep=sep)
xmlText -paste(xmlText,dataWednesday/data,sep=sep)
xmlText -paste(xmlText,/mzArrayBinary,sep=sep)
xmlText -paste(xmlText,intenArrayBinary,sep=sep)
xmlText -paste(xmlText,dataThursday/data,sep=sep)
xmlText -paste(xmlText,/intenArrayBinary,sep=sep)
xmlText -paste(xmlText,/spectrum,sep=sep)
xmlEventParse(xmlText, asText=TRUE, handlers = list(text =
function(x, ...) {cat(nchar(x),x, \n)}))
return(invisible(NULL))
}
Using this function in the given form works fine. xmlEventParse() with
the simplest handler I can imagine finds all 4 text-nodes within the
spectrum tag and prints them out. But if one uncomment both lines in
the middle, introducing 2 spectrum tags with different id's
xmlEventParse() returns with an exception. Of course the weekdays within
data are arbitrary values used here. Further, using an other input
file I could see, that for one and the same data node the handler for
text-nodes was invoked two times, one time for a first part of the
content and one time for the rest of the content. Both invocations
together gave me exactly the content from the data node.
So, am I on the wrong way? Or is this some buggy behaviour?
I appreciat any help and assistance!
Jan
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- --
Duncan Temple Lang[EMAIL PROTECTED]
Department of Statistics work: (530) 752-4782
371 Kerr Hall fax: (530) 752-7099
One Shields Ave.
University of California at Davis
Davis, CA 95616, USA
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Re: [R] getting variable length numerical gradient
Randall,
thanks for your comments; however, you have to take into account what
is the purpose of the function here! The goal is to approximate
*partial* derivatives numerically, using in fact the definition of the
partial derivatives. If you recall this definition I hope that you can
see why I change the ith element of the x vector and not the whole
one. You could also test your approach with the original one in the
logistic regression example and see the difference.
I hope it is more clear now.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://www.med.kuleuven.be/biostat/
http://www.student.kuleuven.be/~m0390867/dimitris.htm
- Original Message -
From: Randall R Schulz [EMAIL PROTECTED]
To: R Help R-Help@stat.math.ethz.ch
Sent: Monday, September 26, 2005 3:53 PM
Subject: Re: [R] getting variable length numerical gradient
Dimitris,
I'm new to R programming, and I'm trying to learn the proper way to do
certain things. E.g., I had a piece of code with explicit iteration to
apply some computations to a vector. It was pretty slow. I found a way
to utilize R's built-in vectorization and it was sped up considerably.
So I want to ask about the code you supplied. Please see below.
(By the way, this message is best viewed using a mono-spaced font.)
On Sunday 25 September 2005 04:07, Dimitris Rizopoulos wrote:
maybe you can find the following function useful (any comments are
greatly appreciated):
fd - function(x, f, scalar = TRUE, ..., eps =
sqrt(.Machine$double.neg.eps)){
f - match.fun(f)
out - if(scalar){
...
} else{
n - length(x)
res - array(0, c(n, n))
f0 - f(x, ...)
ex - pmax(abs(x), 1)
for(i in 1:n){
This (following) statement will create a copy of the entire x vector
on each iteration. It doesn't look like that's what you would want to
do:
x. - x
The computation described by this statement could be vectorized
outside
the loop:
x.[i] - x[i] + eps * ex[i]
res[, i] - c(f(x., ...) - f0) / (x.[i] - x[i])
}
res
}
out
}
Offhand, I cannot tell for sure if the last line of that loop is
vectorizable, but I have a hunch it is.
So at a minimum, it seems this fragment of your code:
for(i in 1:n){
x. - x
x.[i] - x[i] + eps * ex[i]
res[, i] - c(f(x., ...) - f0) / (x.[i] - x[i])
}
Could be more efficiently and succinctly replaced with this:
x. - x + eps * ex
for (in in 1:n)
res[, i] - c(f(x., ...) - f0) / (x.[i] - x[i])
Could your someone else with R programming experience comment?
Thanks.
Randall Schulz
__
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Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm
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Re: [R] histogram - one bin for all values larger than a certain value
Florian Defregger wrote: Dear all, I wonder if I can put together a histogram where one bin contains all the values that are larger than a certain specified value. Example: I have values ranging from 0 to 40 and I want 10 bins from 0 to 10, i.e. for the intervals [0,1), [1,2) , ..., [9,10). And then I want one last bin which contains all the values larger than 10, i.e. for the interval [10, 40). Thanks, Florian Hi, Florian, See the breaks argument in ?hist. x - sample(1:40, 1000, replace = TRUE) hist(x, breaks = c(0:10, 40)) Is this what you intended? --sundar __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] k-d tree for loess
I am exploring the use of loess for oceanographic applications and would like to plot the locations (longitude and latitude) points where the models (salinity~temperature*longitude*latitude,parametric=temperature) are fitted. Chambers and Hastie(1993) explains the locations are nodes of a k-d tree. but I cannot find anything about accessing this information. It would be useful to superimpose on such plots contours of the weights for at least one point. Sample code of drawing such plots would be greatly appreciated. Thanks, Carlisle Thacker __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] getting variable length numerical gradient
Dimitris, On Monday 26 September 2005 07:16, Dimitris Rizopoulos wrote: Randall, thanks for your comments; however, you have to take into account what is the purpose of the function here! The goal is to approximate *partial* derivatives numerically, ... I hope it is more clear now. Yes. Thanks for clearing up my misunderstanding. Best, Dimitris Randall Schulz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Error Message - Error: symbol print-name too long
-BEGIN PGP SIGNED MESSAGE- Hash: SHA1 You aren't giving us much to go on, so I can only make a very wild guess. Check that your file doesn't have a stray ` character in it. R will start reading from that point on and try to make this an internal symbol. If it doesn't find the closing ` for too many characters, it gives the error message you see. I have seen this once before and that is what the cause was - a stray ` put in by hitting the ` key rather than Esc. Carl Anderson wrote: Dear All, I write to ask for information regarding the error message: Error: symbol print-name too long. I am afraid that I can't include any code to help with any further diagnosis of the problem as the code is far too long to be of any use, and I have not been able to re-create the problem in shorter example codes. I have searched the R manual and help pages and found no mention of this error message. All help appreciated, Carl A Anderson. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html - -- Duncan Temple Lang[EMAIL PROTECTED] Department of Statistics work: (530) 752-4782 371 Kerr Hall fax: (530) 752-7099 One Shields Ave. University of California at Davis Davis, CA 95616, USA -BEGIN PGP SIGNATURE- Version: GnuPG v1.4.2 (Darwin) Comment: Using GnuPG with Thunderbird - http://enigmail.mozdev.org iD8DBQFDOAYs9p/Jzwa2QP4RAhtAAJ97sqWdUTOPjtZ2RMJR0qcfyjIAZgCfZLF1 IXTk0bY5RQUD2e+8VlzLpw4= =+pim -END PGP SIGNATURE- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] histogram - one bin for all values larger than a certain value
Le 26.09.2005 16:15, Sundar Dorai-Raj a écrit : Florian Defregger wrote: Dear all, I wonder if I can put together a histogram where one bin contains all the values that are larger than a certain specified value. Example: I have values ranging from 0 to 40 and I want 10 bins from 0 to 10, i.e. for the intervals [0,1), [1,2) , ..., [9,10). And then I want one last bin which contains all the values larger than 10, i.e. for the interval [10, 40). Thanks, Florian Hi, Florian, See the breaks argument in ?hist. x - sample(1:40, 1000, replace = TRUE) hist(x, breaks = c(0:10, 40)) Is this what you intended? --sundar Maybe also take a look at the right argument. I think this is closer to what Florian wanted in the first place : R hist(x, breaks = c(0:10, 40), right=FALSE) Romain -- visit the R Graph Gallery : http://addictedtor.free.fr/graphiques ~ ~~ Romain FRANCOIS - http://addictedtor.free.fr ~~ Etudiant ISUP - CS3 - Industrie et Services ~~http://www.isup.cicrp.jussieu.fr/ ~~ Stagiaire INRIA Futurs - Equipe SELECT ~~ http://www.inria.fr/recherche/equipes/select.fr.html~~ ~ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] quasi-random vector according to an independent graph
Dear R-users, Is anyone aware of any function/package for generating a random vector from a joint distribution defined by an independent graph? Or I have to work it out myself? Thanks. Jinfang -- Jinfang Wang, Associate Professor Chiba University, Japan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] k-d tree for loess
First a warning: loess in R is only loosely related to loess in S, being derived from a C implementation (by the same authors). In R I don't think you can do this. Those details are never exposed, and are hidden in an undocumented C/Fortran workspace. On Mon, 26 Sep 2005, Carlisle Thacker wrote: I am exploring the use of loess for oceanographic applications and would like to plot the locations (longitude and latitude) points where the models (salinity~temperature*longitude*latitude,parametric=temperature) are fitted. Chambers and Hastie(1993) explains the locations are nodes of a k-d tree. but I cannot find anything about accessing this information. It would be useful to superimpose on such plots contours of the weights for at least one point. Sample code of drawing such plots would be greatly appreciated. Thanks, Carlisle Thacker -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] ASA Stat. Computing and Stat. Graphics 2006 Student Paper competition
The Statistical Computing and Statistical Graphics Sections of the ASA are co-sponsoring a student paper competition on the topics of Statistical Computing and Statistical Graphics. Students are encouraged to submit a paper in one of these areas, which might be original methodological research, some novel computing or graphical application in statistics, or any other suitable contribution (for example, a software-related project). The selected winners will present their papers in a topic-contributed session at the 2006 Joint Statistical Meetings. The Sections will pay registration fees for the winners as well as a substantial allowance for transportation to the meetings and lodging. Enclosed below is the full text of the award announcement. More details can be found at the Stat. Computing Section website at http://www.statcomputing.org. Best Regards, --José Pinheiro Awards Chair ASA Statistical Computing Section Statistical Computing and Statistical Graphics Sections American Statistical Association Student Paper Competition 2006 The Statistical Computing and Statistical Graphics Sections of the ASA are co-sponsoring a student paper competition on the topics of Statistical Computing and Statistical Graphics. Students are encouraged to submit a paper in one of these areas, which might be original methodological research, some novel computing or graphical application in statistics, or any other suitable contribution (for example, a software-related project). The selected winners will present their papers in a topic-contributed session at the 2006 Joint Statistical Meetings. The Sections will pay registration fees for the winners as well as a substantial allowance for transportation to the meetings and lodging (which in most cases covers these expenses completely). Anyone who is a student (graduate or undergraduate) on or after September 1, 2005 is eligible to participate. An entry must include an abstract, a six page manuscript (including figures, tables and references), a C.V., and a letter from a faculty member familiar with the student's work. The applicant must be the first author of the paper. The faculty letter must include a verification of the applicant's student status and, in the case of joint authorship, should indicate what fraction of the contribution is attributable to the applicant. We prefer that electronic submissions of papers be in Postscript or PDF. All materials must be in English. All application materials MUST BE RECEIVED by 5:00 PM EST, Monday, December 19, 2005 at the address below. They will be reviewed by the Student Paper Competition Award committee of the Statistical Computing and Graphics Sections. The selection criteria used by the committee will include innovation and significance of the contribution. Award announcements will be made in late January, 2006. Additional important information on the competition can be accessed on the website of the Statistical Computing Section, www.statcomputing.org. A current pointer to the website is available from the ASA website at www.amstat.org. Inquiries and application materials should be emailed or mailed to: Student Paper Competition c/o Dr. José Pinheiro Biostatistics, Novartis Pharmaceuticals One Health Plaza, Room 419/2115 East Hanover, NJ 07936 [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] hist(x, ...) with normal distribution curve
Petr Pikal schrieb: Hi answered hundered times. Dear R people: I would like to superimpose a normal curve on a histogram. x-rnorm(150) h-hist(x,breaks=15) xhist-c(min(h$breaks),h$breaks) yhist-c(0,h$density,0) xfit-seq(min(x),max(x),length=40) yfit-dnorm(xfit,mean=mean(x),sd=sd(x)) plot(xhist,yhist,type=s,ylim=c(0,max(yhist,yfit))) lines(xfit,yfit) Bill above is e.g. Bill Simpson's answer from 2001. Found from R-site search ***histogram density normal***. Ok If I merge both of your answers I get the graph like in SPSS but the ylab is density instead frequency: and I do not have the clicks in SPSS to redo the same graph :-( I hav only the Data file and the SPSS plot. x-5+rnorm(150) h-hist(x,breaks=10,freq = TRUE) #I need this histogramm with... xfit-seq(min(x),max(x),length=40) yfit-dnorm(xfit,mean=mean(x),sd=sd(x)) lines(xfit,yfit) h-hist(x,breaks=10,prob=T) xfit-seq(min(x),max(x),length=40) yfit-dnorm(xfit,mean=mean(x),sd=sd(x)) # ... this line lines(xfit,yfit) Thanks Knut [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Help: x11 position in the Unix environment
Hello, In the Unix environment, I open a window by x11(). May I specify the position of this window by specifying the position of the top left of the window as in Windows environment? Or some other parameters can be used to do that? Thank you, Shengzhe __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] regression methods for circular(?) data.
On 26-Sep-05 nwew wrote: Dear R-users, I have the following data x - runif(300,min=1,max=230) y - x*0.005 + 0.2 y - y+rnorm(100,mean=0,sd=0.1) y - y%%1 # --- modulo operation plot(x,y) and would like to recapture the slope (0.005) and intercept(0.2). I wonder if there are any clever algorithms to do this. I was looking at the function lm.cirucalar. Is this the method to use? If, which of the references is best too look at? Eryk Hi Eryk, If you know the modulus (in your case 1.0) and you get data that look like the result of your plot(x,y), then I wouldn't mess about. I would simply do something like y1-y ix - ix-(y 0.9*(x-50)/200) y1[ix] - y1[ix]+1.0 lm(y1~x) (the constants 0.9/200, -50 being chosen to give a good separation on the graph). On the other hand, if there are good reasons why this very simple approach is not suitable, then if we knew what they were a more helpful reply would be easier to formulate! Best wishes, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 26-Sep-05 Time: 15:56:48 -- XFMail -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] hist(x, ...) with normal distribution curve
Knut Krueger [EMAIL PROTECTED] writes: Petr Pikal schrieb: Hi answered hundered times. Dear R people: I would like to superimpose a normal curve on a histogram. x-rnorm(150) h-hist(x,breaks=15) xhist-c(min(h$breaks),h$breaks) yhist-c(0,h$density,0) xfit-seq(min(x),max(x),length=40) yfit-dnorm(xfit,mean=mean(x),sd=sd(x)) plot(xhist,yhist,type=s,ylim=c(0,max(yhist,yfit))) lines(xfit,yfit) Bill above is e.g. Bill Simpson's answer from 2001. Found from R-site search ***histogram density normal***. Ok If I merge both of your answers I get the graph like in SPSS but the ylab is density instead frequency: Many people consider that a feature, since histograms are supposed to be density estimates and I do not have the clicks in SPSS to redo the same graph :-( I hav only the Data file and the SPSS plot. x-5+rnorm(150) h-hist(x,breaks=10,freq = TRUE) #I need this histogramm with... xfit-seq(min(x),max(x),length=40) yfit-dnorm(xfit,mean=mean(x),sd=sd(x)) lines(xfit,yfit) h-hist(x,breaks=10,prob=T) xfit-seq(min(x),max(x),length=40) yfit-dnorm(xfit,mean=mean(x),sd=sd(x)) # ... this line lines(xfit,yfit) Er, something got duplicated in there? Anyways, if you want the normal curve blown up to the scale of counts/bin, just multiply yfit by the number of observations times the bin width. To find the bin width take, e.g. diff(h$mids[1:2]). If they're not all equal, then you're in deeper trouble. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] hist(x, ...) with normal distribution curve
Le 25.09.2005 17:30, Knut Krueger a écrit : Petr Pikal schrieb: Hi answered hundered times. Dear R people: I would like to superimpose a normal curve on a histogram. x-rnorm(150) h-hist(x,breaks=15) xhist-c(min(h$breaks),h$breaks) yhist-c(0,h$density,0) xfit-seq(min(x),max(x),length=40) yfit-dnorm(xfit,mean=mean(x),sd=sd(x)) plot(xhist,yhist,type=s,ylim=c(0,max(yhist,yfit))) lines(xfit,yfit) Bill above is e.g. Bill Simpson's answer from 2001. Found from R-site search ***histogram density normal***. Ok If I merge both of your answers I get the graph like in SPSS but the ylab is density instead frequency: and I do not have the clicks in SPSS to redo the same graph :-( I hav only the Data file and the SPSS plot. x-5+rnorm(150) h-hist(x,breaks=10,freq = TRUE) #I need this histogramm with... xfit-seq(min(x),max(x),length=40) yfit-dnorm(xfit,mean=mean(x),sd=sd(x)) lines(xfit,yfit) h-hist(x,breaks=10,prob=T) xfit-seq(min(x),max(x),length=40) yfit-dnorm(xfit,mean=mean(x),sd=sd(x)) # ... this line lines(xfit,yfit) Thanks Knut Ok then. what you are trying to do doesn't make any sense to me. (( That does not make much sense (for me) to have the density curve on the same scale than frequencies )) Do you want that : x-5+rnorm(150) h-hist(x,breaks=10,freq = TRUE) xfit-seq(min(x),max(x),length=40) yfit-dnorm(xfit,mean=mean(x),sd=sd(x)) lines(xfit,yfit * 150 * (h$breaks[2]-h$breaks[1])) ? -- visit the R Graph Gallery : http://addictedtor.free.fr/graphiques ~ ~~ Romain FRANCOIS - http://addictedtor.free.fr ~~ Etudiant ISUP - CS3 - Industrie et Services ~~http://www.isup.cicrp.jussieu.fr/ ~~ Stagiaire INRIA Futurs - Equipe SELECT ~~ http://www.inria.fr/recherche/equipes/select.fr.html~~ ~ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] regression methods for circular(?) data.
Hi, I do not know the intercept and slope. And you have to know them in order to do something like: ix-(y 0.9*(x-50)/200 I am right? cheers (Ted Harding) wrote: On 26-Sep-05 nwew wrote: Dear R-users, I have the following data x - runif(300,min=1,max=230) y - x*0.005 + 0.2 y - y+rnorm(100,mean=0,sd=0.1) y - y%%1 # --- modulo operation plot(x,y) and would like to recapture the slope (0.005) and intercept(0.2). I wonder if there are any clever algorithms to do this. I was looking at the function lm.cirucalar. Is this the method to use? If, which of the references is best too look at? Eryk Hi Eryk, If you know the modulus (in your case 1.0) and you get data that look like the result of your plot(x,y), then I wouldn't mess about. I would simply do something like y1-y ix - ix-(y 0.9*(x-50)/200) y1[ix] - y1[ix]+1.0 lm(y1~x) (the constants 0.9/200, -50 being chosen to give a good separation on the graph). On the other hand, if there are good reasons why this very simple approach is not suitable, then if we knew what they were a more helpful reply would be easier to formulate! Best wishes, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 26-Sep-05 Time: 15:56:48 -- XFMail -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Help: x11 position in the Unix environment
On Mon, 2005-09-26 at 17:45 +0200, Shengzhe Wu wrote: Hello, In the Unix environment, I open a window by x11(). May I specify the position of this window by specifying the position of the top left of the window as in Windows environment? Or some other parameters can be used to do that? Thank you, Shengzhe I don't believe so. In general, under Unix/Linux, the Window Manager determines window positioning upon startup unless the application overrides this behavior. Some applications let you specify application window positioning via command line 'geometry' arguments or via the use of an .Xresources file. Some WM's provide more or less functionality for this behavior relative to user customization. For example, Sawfish provides quite a bit, whereas Metacity hides much of it. You may want to check the documentation for the WM that you are using. There is also an application called Devil's Pie: http://www.burtonini.com/blog/computers/devilspie which provides additional Sawfish-like customization for window positioning, etc. However, this is global for a given window, not on a per instance basis. HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] p-level in packages mgcv and gam
Hi, I am fairly new to GAM and started using package mgcv. I like the fact that optimal smoothing is automatically used (i.e. df are not determined a priori but calculated by the gam procedure). But the mgcv manual warns that p-level for the smooth can be underestimated when df are estimated by the model. Most of the time my p-levels are so small that even doubling them would not result in a value close to the P=0.05 threshold, but I have one case with P=0.033. I thought, probably naively, that running a second model with fixed df, using the value of df found in the first model. I could not achieve this with mgcv: its gam function does not seem to accept fractional values of df (in my case 8.377). So I used the gam package and fixed df to 8.377. The P-value I obtained was slightly larger than with mgcv (0.03655 instead of 0.03328), but it is still 0.05. Was this a correct way to get around the underestimated P-level? Furthermore, although the gam.check function of the mgcv package suggests to me that the gaussian family (and identity link) are adequate for my data, I must say the instructions in R help for family and in Hastie, T. and Tibshirani, R. (1990) Generalized Additive Models are too technical for me. If someone knows a reference that explains how to choose model and link, i.e. what tests to run on your data before choosing, I would really appreciate it. Thanks in advance, Denis Chabot __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] hidden markov models
Dear R community, I am looking for an R package or other software to study hidden Markov models. I need to be able to incorporate multivariate emissions and covariates for the transition probabilities. The msm package seems almost perfect for my purpose, but I do not think it allows multivariate emissions. I will be grateful for your suggestions. All the best, -- Emilio A. Laca One Shields Avenue, 2306 PES Bldg. Plant Sciences [EMAIL PROTECTED] University of Californiafax: (530) 752-4361 Davis, California 95616 voice: (530) 754-4083 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] p-level in packages mgcv and gam
On Mon, 26 Sep 2005, Denis Chabot wrote: But the mgcv manual warns that p-level for the smooth can be underestimated when df are estimated by the model. Most of the time my p-levels are so small that even doubling them would not result in a value close to the P=0.05 threshold, but I have one case with P=0.033. I thought, probably naively, that running a second model with fixed df, using the value of df found in the first model. I could not achieve this with mgcv: its gam function does not seem to accept fractional values of df (in my case 8.377). No, this won't work. The problem is the usual one with model selection: the p-value is calculated as if the df had been fixed, when really it was estimated. It is likely to be quite hard to get an honest p-value out of something that does adaptive smoothing. -thomas __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] reading SAS data files
I am attempting to read in a SAS 9.1 data file. After starting R I change to the directory containing the sas data file and use the dir command to confirm that it is there. Then I run the following R-code: library(foreign) sashome - /Program Files/SAS/SAS 9.1 test-read.ssd(file.path(sashome), pcb, sascmd = file.path(sashome, sas.exe)) but R responds with: SAS failed. SAS program at C:\DOCUME~1\DSONNE~1\LOCALS~1\Temp\Rtmp3540\file16169.sas The log file will be file16169.log in the current directory Warning message: SAS return code was 2 in: read.ssd(file.path(sashome), pcb, sascmd = file.path(sashome, the SAS log file contain this: NOTE: Copyright (c) 2002-2003 by SAS Institute Inc., Cary, NC, USA. NOTE: SAS (r) 9.1 (TS1M3) Licensed to UNIV OF CA/DAVIS, Site 0029107010. NOTE: This session is executing on the XP_PRO platform. NOTE: SAS initialization used: real time 0.13 seconds cpu time0.18 seconds 1 libname src2rd '/Program Files/SAS/SAS 9.1'; NOTE: Libref SRC2RD was successfully assigned as follows: Engine:V9 Physical Name: C:\Program Files\SAS\SAS 9.1 2 libname rd xport 'C:\DOCUME~1\DSONNE~1\LOCALS~1\Temp\Rtmp3540\file26090'; NOTE: Libref RD was successfully assigned as follows: Engine:XPORT Physical Name: C:\DOCUME~1\DSONNE~1\LOCALS~1\Temp\Rtmp3540\file26090 3 proc copy in=src2rd out=rd; 4 select pcb ; ERROR: The file SRC2RD.PCB (memtype=ALL) was not found, but appears on a SELECT statement. ERROR: The file SRC2RD.PCB (memtype=ALL) was not found, but appears on a SELECT statement. ERROR: The file SRC2RD.PCB (memtype=ALL) was not found, but appears on a SELECT statement. WARNING: Input library SRC2RD is empty. NOTE: Statements not processed because of errors noted above. NOTE: The SAS System stopped processing this step because of errors. NOTE: PROCEDURE COPY used (Total process time): real time 0.00 seconds cpu time0.01 seconds ERROR: Errors printed on page 1. ERROR: Errors printed on page 1. ERROR: Errors printed on page 1. NOTE: SAS Institute Inc., SAS Campus Drive, Cary, NC USA 27513-2414 NOTE: The SAS System used: real time 0.16 seconds cpu time0.20 seconds Does anyone see what I am doing incorrectly and can they offer any suggestions about getting this to run correctly? I'm not sure where SAS is expecting to find the data file. I have it in the default R directory. Is this where SAS is looking for it or does it need to be somewhere else? Thanks, -- Dean Sonneborn Programmer Analyst Department of Public Health Sciences University of California, Davis (916) 734-6656 [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Help: x11 position in the Unix environment
Marc Schwartz (via MN) wrote: I don't believe so. In general, under Unix/Linux, the Window Manager determines window positioning upon startup unless the application overrides this behavior. Some applications let you specify application window positioning via command line 'geometry' arguments or via the use of an .Xresources file. Splus used to have extensive control of the graphics window via X resources - here's a chunk from an old Xresources file of mine: splus*background: blue splus*Canvas.height:632 splus*Canvas.width: 800 splus*Command*background: red splus*Command*foreground: cyan splus*Label*background: cyan splus*Label*foreground: red splus*Text*background: #a0f splus*Text*foreground: yellow splus*colors: white black white 54 black - and so on and so forth. Sadly I dont think such customisation is coded into R's X11 graphics window. Looking at the code there is no sign of it playing with X11's resource database. Shame. Baz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] [ESS] Indentation in R code
On Mon, Sep 26, 2005 at 09:27:56AM +0200, Martin Maechler wrote:
I'm crossposting to the ESS-help mailing list which is slightly
more appropriate here. [This may be a rare case where
crossposting seems to make much sense.]
PD == Peter Dalgaard [EMAIL PROTECTED]
on 25 Sep 2005 19:40:45 +0200 writes:
PD Seth Falcon [EMAIL PROTECTED] writes:
On 24 Sep 2005, [EMAIL PROTECTED] wrote:
I am using emacs-21.3 when writing R functions on Linux debian, and
I am trying to follow the advice i R-exts.pdf (2.1.1) regarding
indentation. That is, I set 'c-default-style' to bsd and
'c-basic-offset' to 4. However, while this gives me the intended
indentation in C code, it doesn't change the behavior in R code; I
still get an indentation of size 2. This is my .emacs file after
customization:
(require 'ess-site)
(custom-set-variables
;; custom-set-variables was added by Custom -- don't edit or
;; cut/paste it! Your init file should contain only one such
;; instance.
'(c-basic-offset 4)
'(c-default-style bsd))
(custom-set-faces
;; custom-set-faces was added by Custom -- don't edit or cut/paste
it!
;; Your init file should contain only one such instance.
)
Not sure if this is the best way, but I have the following after
loading ess-site:
(setq ess-indent-level 4)
PD I have (I believe it stems from Martin M. originally):
yes, most probably {IIRC, Kurt Hornik was involved too}.
PD (add-hook 'c-mode-hook '(lambda()
PD (c-set-style stroustrup)))
the above is not quite what I have or did recommend,
which is rather bsd + offset 4 as Göran has above
In fact, Göran mentions the R-exts manual and that has
the following *before* giving the emacs-lisp statements:
(For GNU Emacs 20: for GNU Emacs 21 use
customization to set the `c-default-style' to `bsd' and
`c-basic-offset' to `4'.)
and indeed, that's what Göran did and you should do with a
current emacs, either customize via GUI or,
in your ~/.emacs file, find the section '(custom-set-variables ...)' and add
'(c-basic-offset 4)
'(c-default-style bsd)
to the lines already there, or if there's no such section, add
(custom-set-variables
;; custom-set-variables was added by Custom -- don't edit or cut/paste it!
;; Your init file should contain only one such instance.
'(c-basic-offset 4)
'(c-default-style bsd)
)
to the end of your ~/.emacs file
[...]
but this is not sufficient to get correct (4) indentation (ess) in R
functions. We need some reference to ess as well, right? Maybe another
reference to the ESS manual is in order in 'R-exts'?
Thanks for all the help. I got it working now.
Göran
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Re: [R] anova on binomial LMER objects
Hello all, 1. Does Matrix 0.98-7 fix any of this? 2. Assuming no, how does one acquire Matrix 0.95-13? Cheers, and thank you kindly in advance, Hank On Sep 26, 2005, at 9:05 AM, Douglas Bates wrote: On 9/25/05, Horacio Montenegro [EMAIL PROTECTED] wrote: Hi Spencer and Robert, I have found the same behaviour, but only for lme4 and Matrix after the 0.96 release. lme4 0.95-10 and Matrix 0.95-13 releases gave sensible results. This could be an introduced bug, or a solved bug - you should ask Prof. Bates. hope this helps, cheers, Horacio Montenegro I have run into a couple of other things that the improvements from the 0.95 series to the 0.96 series has made worse. This may take a while to sort out. Thanks to Robert Bagchi for the very thorough error report. --- Spencer Graves [EMAIL PROTECTED] wrote: I agree: Something looks strange to me in this example also; I have therefore copied Douglas Bates and Deepayan Sarkar. You've provided a nice simulation. If either of them have time to look at this, I think they could tell us what is happening here. If you need an answer to your particular problem, you could run that simulation 1000 or 1,000 times. That would tell you whether to believe the summary or the anova, or neither. If you want to understand the algorithm, you could walk through the code. However, lmer is a generic, and I don't have time now to figure out how to find the source. A response from Brian Ripley to a question from me a couple of days ago provides a nice summary of how to do that, but I don't have time to check that now. Sorry I couldn't help more. spencer graves Robert Bagchi wrote: Dear R users, I have been having problems getting believable estimates from anova on a model fit from lmer. I get the impression that F is being greatly underestimated, as can be seen by running the example I have given below. First an explanation of what I'm trying to do. I am trying to fit a glmm with binomial errors to some data. The experiment involves 10 shadehouses, divided between 2 light treatments (high, low). Within each shadehouse there are 12 seedlings of each of 2 species (hn sl). 3 damage treatments (0, 0.1, 0.25 leaf area removal) were applied to the seedlings (at random) so that there are 4 seedlings of each species*damage treatment in each shadehouse. There maybe a shadehouse effect, so I need to include it as a random effect. Light is applied to a shadehouse, so it is outer to shadehouse. The other 2 factors are inner to shadehouse. We want to assess if light, damage and species affect survival of seedlings. To test this I fitted a binomial mixed effects model with lmer (actually with quasibinomial errors). THe summary function suggests a large effect of both light and species (which agrees with graphical analysis). However, anova produces F values close to 0 and p values close to 1 (see example below). Is this a bug, or am I doing something fundamentally wrong? If anova doesn't work with lmer is there a way to perform hypothesis tests on fixed effects in an lmer model? I was going to just delete terms and then do liklihood ratio tests, but according to Pinheiro Bates (p. 87) that's very untrustworthy. Any suggestions? I'm using R 2.1.1 on windows XP and lme4 0.98-1 Any help will be much appreciated. many thanks Robert __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting- guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting- guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] hist(x, ...) with normal distribution curve
Romain Francois schrieb: Do you want that : h-hist(x,breaks=10,freq = TRUE) xfit-seq(min(x),max(x),length=40) yfit-dnorm(xfit,mean=mean(x),sd=sd(x)) lines(xfit,yfit * 150 * (h$breaks[2]-h$breaks[1])) Right thats what I want ... but does it make sense to fit the line with a try and error multipier (150) Is there no way to compute the frequency and the distribution line with standardised function? I used SPSS with the data from x-5+rnorm(150) Hit the graph-histogramm menue choosed: display normal curve and got the result: http://biostatistic.de/temp/spss1.jpg - just as easy as possible So, they have any standardised function. maybe it does not make sence, but i am not able to see why there is such a esay to use function in SPSS but not in R Maybe anybody is able to explane whiy and my intention: As a previous computer scientist I am trying to find a way to eleminate SPSS an use R Sure there is a big lack of my statistic knowledge - but the often the SPSS user have similar lack but it is easy to click and view instead to try the similar steps in R But if I am not able to find the steps in R for the common .. SPSS-clicks , I will never be able to suggest R in the institute to the people with mor statistical knowledge but no knowledge about computer science ... and command line interpreter Regards Knut -- Viele Grüße Knut Krüger -- Reitpark Einthal Leitung: 1 Tierarzt, 1 Berufsreiter Homepage http://www.einthal.de Eine fachgerechte Betreuung rund um die Uhr. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] regression methods for circular(?) data.
On 26-Sep-05 Witold Eryk Wolski wrote: Hi, I do not know the intercept and slope. And you have to know them in order to do something like: ix-(y 0.9*(x-50)/200 I am right? cheers Although I really knew them from the way you generated the data, I pretended I did not know them. Read below: If you know the modulus (in your case 1.0) -- I did assume that this was known, i.e. that the data wrap round to 0 above 1.0. Also: the constants 0.9/200, -50 being chosen to give a good separation on the graph -- I plotted the data, and saw that the wrapped data were well separated, and that 0.9*(x-50)/200 was an adequate discriminant function. This was estimated purely by eye, by looking at the graph, to find some line that went between the two groups of data; no attempt was made to calculate anything precisely. Apart from assuming that the modulus was 1.0, and that the well-separated data at the bottom right of the graph were wrapped round data, no other information was used by me! So the question remains: If you can assume that the modulus is 1.0, and that the wrapped-round data will be well separated, then all is simple. All you need to do is to unwrap the wrapped data by adding 1.0, having first identified them by virtue of their obvious separation. Then you can estimate the slope by using 'lm'. But:-- if you, Witold, can not assume these two things for your real data, what can we assume in considering your question? Is the modulus unknown, for instance? Is the scatter so large that the groups are not well separated? Might we have twice-wrapped data (i.e. original y 2)? In short, do your real data look like the data you sent us, and are they wrapped at 1.0? or what? With thanks, and best wishes, Ted. (Ted Harding) wrote: On 26-Sep-05 nwew wrote: Dear R-users, I have the following data x - runif(300,min=1,max=230) y - x*0.005 + 0.2 y - y+rnorm(100,mean=0,sd=0.1) y - y%%1 # --- modulo operation plot(x,y) and would like to recapture the slope (0.005) and intercept(0.2). I wonder if there are any clever algorithms to do this. I was looking at the function lm.cirucalar. Is this the method to use? If, which of the references is best too look at? Eryk Hi Eryk, If you know the modulus (in your case 1.0) and you get data that look like the result of your plot(x,y), then I wouldn't mess about. I would simply do something like y1-y ix - ix-(y 0.9*(x-50)/200) y1[ix] - y1[ix]+1.0 lm(y1~x) (the constants 0.9/200, -50 being chosen to give a good separation on the graph). On the other hand, if there are good reasons why this very simple approach is not suitable, then if we knew what they were a more helpful reply would be easier to formulate! Best wishes, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 26-Sep-05 Time: 15:56:48 -- XFMail -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 26-Sep-05 Time: 18:08:28 -- XFMail -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] anova on binomial LMER objects
On Mon, 26 Sep 2005, Martin Henry H. Stevens wrote: Hello all, 1. Does Matrix 0.98-7 fix any of this? 2. Assuming no, how does one acquire Matrix 0.95-13? It is in the Archive on CRAN, e.g. http://cran.r-project.org/src/contrib/Archive/M/Matrix_0.95-13.tar.gz Cheers, and thank you kindly in advance, Hank On Sep 26, 2005, at 9:05 AM, Douglas Bates wrote: On 9/25/05, Horacio Montenegro [EMAIL PROTECTED] wrote: Hi Spencer and Robert, I have found the same behaviour, but only for lme4 and Matrix after the 0.96 release. lme4 0.95-10 and Matrix 0.95-13 releases gave sensible results. This could be an introduced bug, or a solved bug - you should ask Prof. Bates. hope this helps, cheers, Horacio Montenegro I have run into a couple of other things that the improvements from the 0.95 series to the 0.96 series has made worse. This may take a while to sort out. Thanks to Robert Bagchi for the very thorough error report. --- Spencer Graves [EMAIL PROTECTED] wrote: I agree: Something looks strange to me in this example also; I have therefore copied Douglas Bates and Deepayan Sarkar. You've provided a nice simulation. If either of them have time to look at this, I think they could tell us what is happening here. If you need an answer to your particular problem, you could run that simulation 1000 or 1,000 times. That would tell you whether to believe the summary or the anova, or neither. If you want to understand the algorithm, you could walk through the code. However, lmer is a generic, and I don't have time now to figure out how to find the source. A response from Brian Ripley to a question from me a couple of days ago provides a nice summary of how to do that, but I don't have time to check that now. Sorry I couldn't help more. spencer graves Robert Bagchi wrote: Dear R users, I have been having problems getting believable estimates from anova on a model fit from lmer. I get the impression that F is being greatly underestimated, as can be seen by running the example I have given below. First an explanation of what I'm trying to do. I am trying to fit a glmm with binomial errors to some data. The experiment involves 10 shadehouses, divided between 2 light treatments (high, low). Within each shadehouse there are 12 seedlings of each of 2 species (hn sl). 3 damage treatments (0, 0.1, 0.25 leaf area removal) were applied to the seedlings (at random) so that there are 4 seedlings of each species*damage treatment in each shadehouse. There maybe a shadehouse effect, so I need to include it as a random effect. Light is applied to a shadehouse, so it is outer to shadehouse. The other 2 factors are inner to shadehouse. We want to assess if light, damage and species affect survival of seedlings. To test this I fitted a binomial mixed effects model with lmer (actually with quasibinomial errors). THe summary function suggests a large effect of both light and species (which agrees with graphical analysis). However, anova produces F values close to 0 and p values close to 1 (see example below). Is this a bug, or am I doing something fundamentally wrong? If anova doesn't work with lmer is there a way to perform hypothesis tests on fixed effects in an lmer model? I was going to just delete terms and then do liklihood ratio tests, but according to Pinheiro Bates (p. 87) that's very untrustworthy. Any suggestions? I'm using R 2.1.1 on windows XP and lme4 0.98-1 Any help will be much appreciated. many thanks Robert __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting- guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting- guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] regression methods for circular(?) data.
Ted, I agree with you that if you unwrap the data you can use lm. And you can separate the data in the way you describe. However, if you have thousands of such datasets I do not want to do it by looking at the graph. Yes the scatter may be larger as in the example and range(y) may be larger than 2. And as you said in order to unwrap the data you have to separate them first. It would be easy to do it using for example single linkage clustering if they were no overlaps (but they do sometimes). So I were just wondering if there are no more fancy methods to do this. Thanks, cheers (Ted Harding) wrote: On 26-Sep-05 Witold Eryk Wolski wrote: Hi, I do not know the intercept and slope. And you have to know them in order to do something like: ix-(y 0.9*(x-50)/200 I am right? cheers Although I really knew them from the way you generated the data, I pretended I did not know them. Read below: If you know the modulus (in your case 1.0) -- I did assume that this was known, i.e. that the data wrap round to 0 above 1.0. Also: the constants 0.9/200, -50 being chosen to give a good separation on the graph -- I plotted the data, and saw that the wrapped data were well separated, and that 0.9*(x-50)/200 was an adequate discriminant function. This was estimated purely by eye, by looking at the graph, to find some line that went between the two groups of data; no attempt was made to calculate anything precisely. Apart from assuming that the modulus was 1.0, and that the well-separated data at the bottom right of the graph were wrapped round data, no other information was used by me! So the question remains: If you can assume that the modulus is 1.0, and that the wrapped-round data will be well separated, then all is simple. All you need to do is to unwrap the wrapped data by adding 1.0, having first identified them by virtue of their obvious separation. Then you can estimate the slope by using 'lm'. But:-- if you, Witold, can not assume these two things for your real data, what can we assume in considering your question? Is the modulus unknown, for instance? Is the scatter so large that the groups are not well separated? Might we have twice-wrapped data (i.e. original y 2)? In short, do your real data look like the data you sent us, and are they wrapped at 1.0? or what? With thanks, and best wishes, Ted. (Ted Harding) wrote: On 26-Sep-05 nwew wrote: Dear R-users, I have the following data x - runif(300,min=1,max=230) y - x*0.005 + 0.2 y - y+rnorm(100,mean=0,sd=0.1) y - y%%1 # --- modulo operation plot(x,y) and would like to recapture the slope (0.005) and intercept(0.2). I wonder if there are any clever algorithms to do this. I was looking at the function lm.cirucalar. Is this the method to use? If, which of the references is best too look at? Eryk Hi Eryk, If you know the modulus (in your case 1.0) and you get data that look like the result of your plot(x,y), then I wouldn't mess about. I would simply do something like y1-y ix - ix-(y 0.9*(x-50)/200) y1[ix] - y1[ix]+1.0 lm(y1~x) (the constants 0.9/200, -50 being chosen to give a good separation on the graph). On the other hand, if there are good reasons why this very simple approach is not suitable, then if we knew what they were a more helpful reply would be easier to formulate! Best wishes, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 26-Sep-05 Time: 15:56:48 -- XFMail -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 26-Sep-05 Time: 18:08:28 -- XFMail -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] reading SAS data files
On Mon, 26 Sep 2005, Dean Sonneborn wrote: I am attempting to read in a SAS 9.1 data file. After starting R I change to the directory containing the sas data file and use the dir command to confirm that it is there. Then I run the following R-code: library(foreign) sashome - /Program Files/SAS/SAS 9.1 test-read.ssd(file.path(sashome), pcb, sascmd = file.path(sashome, sas.exe)) but R responds with: SAS failed. SAS program at C:\DOCUME~1\DSONNE~1\LOCALS~1\Temp\Rtmp3540\file16169.sas The log file will be file16169.log in the current directory Warning message: SAS return code was 2 in: read.ssd(file.path(sashome), pcb, sascmd = file.path(sashome, the SAS log file contain this: NOTE: Copyright (c) 2002-2003 by SAS Institute Inc., Cary, NC, USA. NOTE: SAS (r) 9.1 (TS1M3) Licensed to UNIV OF CA/DAVIS, Site 0029107010. NOTE: This session is executing on the XP_PRO platform. NOTE: SAS initialization used: real time 0.13 seconds cpu time0.18 seconds 1 libname src2rd '/Program Files/SAS/SAS 9.1'; NOTE: Libref SRC2RD was successfully assigned as follows: Engine:V9 Physical Name: C:\Program Files\SAS\SAS 9.1 2 libname rd xport 'C:\DOCUME~1\DSONNE~1\LOCALS~1\Temp\Rtmp3540\file26090'; NOTE: Libref RD was successfully assigned as follows: Engine:XPORT Physical Name: C:\DOCUME~1\DSONNE~1\LOCALS~1\Temp\Rtmp3540\file26090 3 proc copy in=src2rd out=rd; 4 select pcb ; ERROR: The file SRC2RD.PCB (memtype=ALL) was not found, but appears on a SELECT statement. ERROR: The file SRC2RD.PCB (memtype=ALL) was not found, but appears on a SELECT statement. ERROR: The file SRC2RD.PCB (memtype=ALL) was not found, but appears on a SELECT statement. WARNING: Input library SRC2RD is empty. NOTE: Statements not processed because of errors noted above. NOTE: The SAS System stopped processing this step because of errors. NOTE: PROCEDURE COPY used (Total process time): real time 0.00 seconds cpu time0.01 seconds ERROR: Errors printed on page 1. ERROR: Errors printed on page 1. ERROR: Errors printed on page 1. NOTE: SAS Institute Inc., SAS Campus Drive, Cary, NC USA 27513-2414 NOTE: The SAS System used: real time 0.16 seconds cpu time0.20 seconds Does anyone see what I am doing incorrectly and can they offer any suggestions about getting this to run correctly? I'm not sure where SAS is expecting to find the data file. I have it in the default R directory. Is this where SAS is looking for it or does it need to be somewhere else? I suggest you read the help file for read.ssd. You are not telling it where the SAS files are correctly. See the extracts below and the examples. Usage: read.ssd(libname, sectionnames, tmpXport=tempfile(), tmpProgLoc=tempfile(), sascmd=sas) Arguments: libname: character string defining the SAS library (usually a directory reference) sectionnames: character vector giving member names. These are files in the 'libname' directory. They will usually have a '.ssd0x' or '.sas7bdat' extension, which should be omitted. The files are not in the SAS homne directory, but the `default R directory' (perhaps you mean the working directory?). PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html PLEASE do! -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] regression methods for circular(?) data.
On 26-Sep-05 Witold Eryk Wolski wrote:
Ted,
I agree with you that if you unwrap the data you can use lm.
And you can separate the data in the way you describe. However, if you
have thousands of such datasets I do not want to do it by looking at
the graph.
Yes the scatter may be larger as in the example and range(y) may be
larger than 2.
And as you said in order to unwrap the data you have to separate them
first. It would be easy to do it using for example single linkage
clustering if they were no overlaps (but they do sometimes). So I were
just wondering if there are no more fancy methods to do this.
OK, the problems are now clearer! So we cannot rely on separation
(though there would be ways to detect this automatically if it could
be relied on).
This is where real experts on unwrapping circular data should step in,
but my immediate suggestion would be that developing something out
of the following should be useful.
First, generate the data so that we have something to work with:
x - runif(300,min=1,max=230)
y - x*0.005 + 0.2
y - y+rnorm(100,mean=0,sd=0.1)
y0 - y%%1 # --- modulo operation
(I've called the wrapped data y0).
Now, assume
A. That we know the modulus is 1.0
B. That we are looking for a model y0 = (a*x + b)%%1.0
C. The we do have some idea about a range of values for a and b,
say 0 a 0.01 and 0 b 1.0
Now try the following and inspect what you get:
M-numeric(101)
for(i in (0:100)){v-(i*0.01/100);
M[i+1]-max(Mod(fft((y0-v*x-0.0)%%1)*2*pi))
}
plot(0.01*(0:100)/100,M,ylim=c(0,1000))
for(j in 0.5*(0:10)/10){
for(i in (0:100)){
v-(i*0.01/100);
M[i+1]-max(Mod(fft((y0-v*x-j)%%1)*2*pi))
}
points(0.01*(0:100)/100,M)
}
This gives an indication that a good value for 'a' ('i' in the
plots) is about 0.5 (or slightly larger) for some value of 'b'
('j' in the plots), from which, conditioning on this, a value
for b could be obtained similarly. The plots from the above do
not distinguish between the curves for different values of 'b';
a method of indicating this would be useful.
Just a suggestion. There may be, in some R package, a function
which implements this approach in a better way.
Over to the gurus at this point!
Best wishes,
Ted.
E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 26-Sep-05 Time: 20:54:50
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Re: [R] hidden markov models
Emilio, The depmix package on cran has multivariate distributions and the possibility of (linear) covariates on the parameters. If you have questions about its use feel free to contact me, best, ingmar visser On 9/26/05 6:29 PM, Dr. Emilio A. Laca [EMAIL PROTECTED] wrote: Dear R community, I am looking for an R package or other software to study hidden Markov models. I need to be able to incorporate multivariate emissions and covariates for the transition probabilities. The msm package seems almost perfect for my purpose, but I do not think it allows multivariate emissions. I will be grateful for your suggestions. All the best, -- Ingmar Visser Department of Psychology, University of Amsterdam Roetersstraat 15, 1018 WB Amsterdam The Netherlands http://users.fmg.uva.nl/ivisser/ tel: +31-20-5256735 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] histogram - one bin for all values larger than a certain value
x=runif(100,0,40) hist(x, breaks=c(0,1,2,3,4,5,6,7,8,9,10,40)) Is this what you had in mind? Francisco From: Florian Defregger [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Subject: [R] histogram - one bin for all values larger than a certain value Date: Mon, 26 Sep 2005 15:36:21 +0200 Dear all, I wonder if I can put together a histogram where one bin contains all the values that are larger than a certain specified value. Example: I have values ranging from 0 to 40 and I want 10 bins from 0 to 10, i.e. for the intervals [0,1), [1,2) , ..., [9,10). And then I want one last bin which contains all the values larger than 10, i.e. for the interval [10, 40). Thanks, Florian __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Make check fails on d-p-q-r-tests.R...
Hi, R-2.1.1 OS: OpenBSD-current (3.8) on i386 Compiler:gcc version 3.3.5 (propolice) Thread model: single configure \ --with-readline \ --with-tcltk \ --with-tcl-config=/usr/local/lib/tcl8.4/tclConfig.sh \ --with-tk-config=/usr/local/lib/tk8.4/tkConfig.sh \ --with-libpng \ --with-jpeglib \ --with-zlib \ --with-bzlib \ --with-pcre \ --with-libiconv-prefix=/usr/local/ I'm brand new to R so I don't know how critical this error is. Here's the last bit what make check FORCE=FORCE outputs: running code in 'grDevices-Ex.R' ... OK comparing 'grDevices-Ex.Rout' to 'grDevices-Ex.Rout.prev' ... OK running code in 'graphics-Ex.R' ... OK comparing 'graphics-Ex.Rout' to 'graphics-Ex.Rout.prev' ... OK running code in 'stats-Ex.R' ... OK comparing 'stats-Ex.Rout' to 'stats-Ex.Rout.prev' ... OK running code in 'datasets-Ex.R' ... OK comparing 'datasets-Ex.Rout' to 'datasets-Ex.Rout.prev' ... OK running code in 'methods-Ex.R' ... OK comparing 'methods-Ex.Rout' to 'methods-Ex.Rout.prev' ... OK running code in 'grid-Ex.R' ... OK comparing 'grid-Ex.Rout' to 'grid-Ex.Rout.prev' ... OK running code in 'splines-Ex.R' ... OK comparing 'splines-Ex.Rout' to 'splines-Ex.Rout.prev' ... OK running code in 'stats4-Ex.R' ... OK comparing 'stats4-Ex.Rout' to 'stats4-Ex.Rout.prev' ... OK running code in 'tcltk-Ex.R' ... OK comparing 'tcltk-Ex.Rout' to 'tcltk-Ex.Rout.prev' ... OK updating test dependencies `Makedeps' is up to date. running strict specific tests running code in 'eval-etc.R' ... OK comparing 'eval-etc.Rout' to './eval-etc.Rout.save' ... OK running code in 'simple-true.R' ... OK comparing 'simple-true.Rout' to './simple-true.Rout.save' ... OK running code in 'arith-true.R' ... OK comparing 'arith-true.Rout' to './arith-true.Rout.save' ... OK running code in 'arith.R' ... OK comparing 'arith.Rout' to './arith.Rout.save' ... OK running code in 'lm-tests.R' ... OK comparing 'lm-tests.Rout' to './lm-tests.Rout.save' ... OK running code in 'primitive-funs.R' ... OK comparing 'primitive-funs.Rout' to './primitive-funs.Rout.save' ... OK running code in 'ok-errors.R' ... OK comparing 'ok-errors.Rout' to './ok-errors.Rout.save' ... OK running code in 'method-dispatch.R' ... OK comparing 'method-dispatch.Rout' to './method-dispatch.Rout.save' ... OK running code in 'd-p-q-r-tests.R' ...*** Error code 1 Stop in /usr/local/src/R-2.1.1/tests. *** Error code 1 Stop in /usr/local/src/R-2.1.1/tests (line 206 of Makefile). *** Error code 1 Stop in /usr/local/src/R-2.1.1/tests (line 191 of Makefile). Tried finding this error in the archives to no avail. If it is okay to ignore this error, how can I skip this test? Thanks for any input! Jeff Ross __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] plotting multiple plots on a single graph
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Re: [R] less precision, please!
Duncan On 9/9/2005 7:41 PM, Paul MacManus wrote:
I need to run qbeta on a set of 500K different parameter
pairs (with a fixed quantile). For most pairs qbeta finds
the solution very quickly but for a substantial minority
of the cases qbeta is very slow. This occurs when the
solution is very close to zero. qbeta is getting answers
to a precision of about 16 decimal places. I don't need
that accuracy. Is there any way to set the precision of
R's calculations to, say, 9 decimal places and so speed
up the whole process?
I could, of course, avoid this problem by not running
qbeta when I know the solution is going to be
sufficiently small but I'm more interested in ways to
adjust the precision of calculations in R.
Duncan There's no general way to do this. The function
Duncan that implements qbeta may have some tuning
Duncan parameters (I haven't looked), but they aren't
Duncan usually needed, and aren't exposed in R.
Yes.
However, I've had thoughts in the past on possibly providing such
a possibility from both R and C level. One problem is that
``for symmetry reasons'' you would want to have this ``for all functions''
which would need a lot of work, for something that's really not
of too high a need.
I agree that qbeta() can be particularly nasty. I'm open to
more in-depth discussion on this -- after R 2.2.0 is out
Duncan If you want a quick approximation, I'd suggest doing
Duncan your calculation on a grid of values and using
Duncan approx() to interpolate.
yes, or approxfun() {which prefer for its UI},
or even more smoothly using spline() or splinefun() {again
preferably the latter}.
One problem may be that these are only for 1-D interpolation and
qbeta() depends on three principal arguments.
Package 'akima' provides somewhat smooth 2-D interpolation.
Hi again,
Thank you both for your feedback and my apologies for not replying
sooner.
Tunable parameters would be nice to have but they are probably not really
necessary. Workarounds of one type or another, such as the ones you suggested,
always seem to be available.
The qbeta issue, specifically, is more interesting. The surfaces associated
with beta and inverse beta functions are notoriously badly behaved and
developing functions that deal in a good way with the full range of parameters
is very tricky. qbeta() does a very good job in general but it is not
surprising that it has trouble at times. The underlying algorithms that qbeta()
uses seems to be good. I would be interested in talking more about the qbeta
issue when you have more time.
Best, Paul
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[R] questions about boxplots
Hi, there. I have two questions about using R to create boxplots. 1. The function boxplot() plots the outliers. How can I label the exact values arount these outlier points? Does R have an option allow me to do that? 2. How can I put two boxplots in one x-y axis? Thanks. Yulei $$$ Yulei He 276 Grove St. Apt 3 Newton, MA 02466 617-796-7834(H) 617-432-3428(O) 617-432-3435(fax) $$ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] dates are shown as X15.Feb.03
Why is R recognizing dates like this? Chris Buddenhagen, Botany Department, Charles Darwin Research Station, Santa Cruz,Galapagos. Mail: Charles Darwin Foundation, Casilla 17-01-3891 Avenida 6 de Diciembre N36-109 y Pasaje California Quito, ECUADOR __ EL CONTENIDO DE ESTE MENSAJE ES DE ABSOLUTA RESPONSABILIDAD DEL AUTOR. FUNDACION CHARLES DARWIN WWW.DARWINFOUNDATION.ORG [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Error Message - Error: symbol print-name too long
Duncan, Thank you for your help. I am pleased to say your 'very wild guess' was exactly correct. Can't believe I missed it! Carl - Original Message - From: Duncan Temple Lang [EMAIL PROTECTED] To: Carl Anderson [EMAIL PROTECTED] Cc: r-help@stat.math.ethz.ch; [EMAIL PROTECTED] Sent: Tuesday, September 27, 2005 12:31 AM Subject: Re: [R] Error Message - Error: symbol print-name too long -BEGIN PGP SIGNED MESSAGE- Hash: SHA1 You aren't giving us much to go on, so I can only make a very wild guess. Check that your file doesn't have a stray ` character in it. R will start reading from that point on and try to make this an internal symbol. If it doesn't find the closing ` for too many characters, it gives the error message you see. I have seen this once before and that is what the cause was - a stray ` put in by hitting the ` key rather than Esc. Carl Anderson wrote: Dear All, I write to ask for information regarding the error message: Error: symbol print-name too long. I am afraid that I can't include any code to help with any further diagnosis of the problem as the code is far too long to be of any use, and I have not been able to re-create the problem in shorter example codes. I have searched the R manual and help pages and found no mention of this error message. All help appreciated, Carl A Anderson. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html - -- Duncan Temple Lang[EMAIL PROTECTED] Department of Statistics work: (530) 752-4782 371 Kerr Hall fax: (530) 752-7099 One Shields Ave. University of California at Davis Davis, CA 95616, USA -BEGIN PGP SIGNATURE- Version: GnuPG v1.4.2 (Darwin) Comment: Using GnuPG with Thunderbird - http://enigmail.mozdev.org iD8DBQFDOAYs9p/Jzwa2QP4RAhtAAJ97sqWdUTOPjtZ2RMJR0qcfyjIAZgCfZLF1 IXTk0bY5RQUD2e+8VlzLpw4= =+pim -END PGP SIGNATURE- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] dates are shown as X15.Feb.03
Chris Buddenhagen wrote: Why is R recognizing dates like this? Recognizing and showing are different things. Which are you complaining about? What are you doing to cause these to be recognized/shown? Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] questions about boxplots
Dear Yulei, On Sep 26, 2005, at 6:56 PM, Yulei He wrote: Hi, there. I have two questions about using R to create boxplots. 1. The function boxplot() plots the outliers. How can I label the exact values arount these outlier points? Does R have an option allow me to do that? You can use identify(). Here's a toy example set.seed(1) y - rt(30, 3) boxplot(y) identify(x = rep(1,30), y = y, label = format(y, digits = 2)) ### now click on the outlier in the plot and you should see -7.398 ### beside the outlier 2. How can I put two boxplots in one x-y axis? x - rnorm(10) y - rnorm(20, 3, 5) z - runif(30) boxplot(x, y, z, labels = c(x, y, z)) ### - or - boxplot(list(x = x, y = y, z = z)) Thanks. Yulei Stephen Weigand Rochester, Minnesota, USA __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] figure widths in sweave
On Mon, 2005-09-26 at 20:34 +0800, John Charles Considine wrote:
gRoovers,
Can the size of figures be controlled from within a noweb document
without resorting to editing the \includegraphics sections in the .tex
file?
yes,
Sweave sets graphics widths to 0.8\textwidth by default. To change it
for a document, to say 1.0\textwidth, include the line
\setkeys{Gin}{width=1\textwidth}
after \begin{document} in the noweb file.
see p.12 of Sweave manual at...
http://www.ci.tuwien.ac.at/~leisch/Sweave/Sweave-manual-20050914.pdf
Can the figure widths be set in the environmental declarations at the
start?
Can they be set within the \begin{figure} environment?
JC
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