[R] add to vector without duplicatation
Hi, I try to add items to a vector, but when the it has existed, the item should be skipped. does anyone know how to do it a simple way? Yu [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] add to vector without duplicatation
2008/8/28 Yuan Jian [EMAIL PROTECTED]: Hi, I try to add items to a vector, but when the it has existed, the item should be skipped. does anyone know how to do it a simple way? You might be able to use unique(). Regards, Nicky Chorley __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] sample consecutive integers efficiently
Hi all,
I have some rough code to sample consecutive integers with length
according to a vector of lengths
#sample space (representing positions)
pos-c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20)
#sample lengths
lengths-c(2,3,2)
From these two vectors I need a vector of sampled positions.
the sampling is without replacement, making things tough as the sampled
integers need to be consecutive. Im hoping somebody knows a faster way
of doing it than I have. ATM its way to slow on large vectors.
samplePos-function(l){
start.pos-sample(pos,1)
end.pos-start.pos+l-1
posies-start.pos:end.pos
posies
}
s.start-c()
newPos-function(a){
rp-samplePos(a)
#test sampled range is consecutive, if not resample
if (length(rp) != rp[a]+1 -rp[1]){rp-samplePos(a)}
pos-setdiff(pos,rp)
rp[1]
}
newps-c()
newps-unlist(lapply(lengths,newPos))
I think the bottleneck may be on the setdiff() function - the sample
space is quite large so I dont think there would be too many rejections.
Many thanks,
Chris
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[R] Renaming objects
Hi, Is there any quick and easy way to rename a number of objects, without having to rename each one individually and then remove the old one? And if so, is there anything I can do to adjust the associated comments accordingly? Thanks for any help, Robin Williams Met Office summer intern - Health Forecasting [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Upgrading R means I lose my packages
On a windows machine you get the same problem. Useless one uses tha same
trick as Rolf suggested: don't install the packages in the default
directory and set R_LIBS to that directory. Then all you need to do
after an upgrade is to set R_LIBS in the new version and run
update.package(checkBuilt = TRUE). Given Rolf's suggestion I suppose
this trick will work on a Mac too.
HTH,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
Cel biometrie, methodologie en kwaliteitszorg / Section biometrics,
methodology and quality assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium
tel. + 32 54/436 185
[EMAIL PROTECTED]
www.inbo.be
To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to
say what the experiment died of.
~ Sir Ronald Aylmer Fisher
The plural of anecdote is not data.
~ Roger Brinner
The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of
data.
~ John Tukey
I'm not sure --- I find Mac OS very confusing. But I have the
***impression*** that
(on my system) by default packages get installed into
~/Library/R/2.7/library
i.e. into a library inside the directory tree rooted in my login
directory.
I don't use this --- I've created my own library ~/Rlib and have
set up an environment variable to point to it.
(This works properly only if you start R from the command line;
for
reasons I don't understand if you start R by clicking on the
icon
then R doesn't know about the R_LIBS environment variable. But
since
all civilized people start R from the command line .)
I have no idea why youse guys' systems would eschew using
~/Library/
whatever.
cheers,
Rolf Turner
##
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Re: [R] Updating a list.
Kevin, Notice the subtle difference between Hadley's and your code: Hadley m2008$DayOfYear - factor(m2008$DayOfYear, levels = 1:365) Kevin m2007$DayOfYear - factor(m2008$DayOfYear, levels = 1:365) Your are using the m2007 object instead of the suggested m2008 object! HTH, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology and quality assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 [EMAIL PROTECTED] www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey -Oorspronkelijk bericht- Van: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Namens [EMAIL PROTECTED] Verzonden: donderdag 28 augustus 2008 3:14 Aan: r-help@r-project.org Onderwerp: Re: [R] Updating a list. Since this didn't work: m2007$DayOfYear - factor(m2008$DayOfYear, levels = 1:365) Error in `$-.data.frame`(`*tmp*`, DayOfYear, value = c(1L, 1L, 1L, : replacement has 432267 rows, data has 1592009 Perhaps I need to clarify how the m2007 object was generated. t2007 - read.csv(Total2007.dat, header = TRUE) m2007 - melt(t2007, id.var=c(DayOfYear,Category,SubCategory,Sku), measure.var=c(Quantity)) Kevin hadley wickham [EMAIL PROTECTED] wrote: Try this: m2008$DayOfYear - factor(m2008$DayOfYear, levels = 1:365) r2007 - cast(m2008, DayOfYear ~ variable | Sku, sum, fill = 0) Hadley -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ggplot2: problem with large fonts and overlapping labels
Dear Paul, How are you generating (saving) your plots? I tend to play with the pointsize argument of the graphical device, something in conjunction with the size argument in ggplot2 (size of points and lines). Working like that I get plots with nicely propotioned labels without overlaps. HTH, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology and quality assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 [EMAIL PROTECTED] www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey -Oorspronkelijk bericht- Van: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Namens Paul Emberson Verzonden: woensdag 27 augustus 2008 17:58 Aan: r-help@r-project.org Onderwerp: [R] ggplot2: problem with large fonts and overlapping labels Hi, I am using ggplot2 to generate graphs for a paper I am writing in two column format. When I shrink the graphs to fit in a single column, the graph is clear but the axis and tick labels are way too small. I have increased the font sizes by manipulating the grid. However, when I do this the tick labels and axis labels get very close or even overlap, especially on the y-axis. Is there a way to increase the margin between tick labels and axis labels with ggplot or by manipulating the grid? Regards, Paul __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Odp: I need to change from character to numeric?
Hi [EMAIL PROTECTED] napsal dne 28.08.2008 02:55:34: Hi, I am reading numeric data as below but the problem is the object ndata1 and nd1 have characters instead of numeric values. I want to keep it as numeric. Why the type has changed from numeric to character and how to avoid this problem? rdata1- read.table(file=data1.txt, header=F,stringsAsFactors=F) Read your data properly. They obviously have header so why you use header = F. rdata1- read.table(file=data1.txt, header=T, stringsAsFactors=F) shall read it as numeric if you do not have some non numeric values somwhere. Regards Petr rdata1 V1 1 d1 2 11 3 20 4 13 rdata2 V1 1 d2 2 1 3 7 4 5 5 6 ndata1-rdata1[2:nrow(rdata1),] ndata1 [1] 11 20 13 nd1-cbind(ndata1,v2=0) nd1 ndata1 v2 [1,] 11 0 [2,] 20 0 [3,] 13 0 -- View this message in context: http://www.nabble.com/I-need-to-change-from- character-to-numeric--tp19192541p19192541.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] coding rules
On Wed, 2008-08-27 at 09:48 -0300, Henrique Dallazuanna wrote: See http://www1.maths.lth.se/help/R/RCC/ The R Internals manual that is shipped with R also has a section on coding standards: http://cran.r-project.org/doc/manuals/R-ints.html#R-coding-standards though this is quite short and does not go into detail of naming conventions etc. This document also points to the GNU Coding Standards which R aims to follow. The R Developer Page has some useful information, such as guidelines for writing *.Rd files: http://developer.r-project.org/ and the Writing R Extensions manual has a lot of information on writing packages, cross platform portability etc: http://cran.r-project.org/doc/manuals/R-exts.html G On Wed, Aug 27, 2008 at 9:36 AM, Thomas LOUBRIEU [EMAIL PROTECTED] wrote: Dear all, I am organizing a set of specific R code as package (to ease the documentation and deployment of it to users). Before doing so, I would like to know if there are written coding rules for R (with functions, objects naming convention for example). Thanks a lot, Thomas -- - Thomas LOUBRIEU IFREMER IDM/ISI BP70 29280 Plouzane FRANCE email: [EMAIL PROTECTED] WWW : http://www.coriolis.eu.org/cdc Tel.: (+33) (0)2 98 22 48 53 Fax: (+33) (0)2 98 22 46 44 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] drop unused levels in sqldf
Hi, sqldf is a fantastic package, but when the SELECT procedure runs unused levels remain in the output. I tried with the drop function, but without success. Do you have any suggestions? Thanx, Gianandrea data(iris) require(sqldf) base-sqldf(select * from iris where Species 'setosa') str(base) # Species with 3 levels! -- View this message in context: http://www.nabble.com/drop-unused-levels-in-sqldf-tp19196464p19196464.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to learn R language?
Bert Gunter wrote: -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Robert Baer Sent: Wednesday, August 27, 2008 3:42 PM To: r-help@r-project.org Subject: Re: [R] How to learn R language? but I doubt you'll ever be done learning because the project is so comprehensive ;-) -- which is a good thing for us old geezers who need to keep our neurons firing as much as possible... Cheers, Bert How true!!! :-) Tomas __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] convert princomp output to equation for plane?
Hi Bill, Since x, y,and z all have measurement errors attached, the proper way to do the fit is with principal components analysis, and to use the first component (called loadings in princomp output). The easiest way for you to do this is to use the pcr [principal component regression] function in the pls package. Be aware that unless you fit all components you will be carrying out a form of penalized regression. A small example follows (assumes that you have installed the pls package): ## lm.mod - lm(Ozone ~ Solar.R + Wind + Month, data=airquality) pc.mod - pcr(Ozone ~ Solar.R + Wind + Month, data=airquality) lm.mod coef(pc.mod, intercept = TRUE) coef(pc.mod, ncomp=1, intercept = TRUE) coef(pc.mod, ncomp=3, intercept = TRUE) Regards, Mark. William Simpson-2 wrote: I want to fit something like: z = b0 + b1*x + b2*y Since x, y,and z all have measurement errors attached, the proper way to do the fit is with principal components analysis, and to use the first component (called loadings in princomp output). My dumb question is: how do I convert the princomp output to equation coefficients in the format above? I guess another dumb question would be: how about getting the standard deviations of b0, b1, b2? Thanks very much for any help. Bill __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/convert-princomp-output-to-equation-for-plane--tp19182643p19197360.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fw: How to learn R language?
While agreeing with how good the texts that have been suggested are, the questions to me (language + systematic) suggests Braun and Murdoch A first course in statistical programming or/and Chambers Software for data analysis: programming with R These would seem to take you through developing an understanding of language fundamentals, in a more structured manner than the other books mentioned. Graham 2008/8/27 saggak [EMAIL PROTECTED]: --- On Wed, 27/8/08, saggak [EMAIL PROTECTED] wrote: From: saggak [EMAIL PROTECTED] Subject: How to learn R language? To: r-help@r-project.org Date: Wednesday, 27 August, 2008, 3:37 PM Hi! I am a post graduate in Statistics. I want to learn R language, but am very confused as to how to begin systematically. I need to learn R language from Statistics point of view e.g. I need to fit distributions to data or run regression analysis etc. No doubt there are so many articles available on internet. But can someone guide me as to how do I begin and go on improving myself SYSTEMATICALLY? Hence, please guide me as to how should I start learning R language? What should I read first etc. Thanks in advance, Sagga K Unlimited freedom, unlimited storage. Get it now Unlimited freedom, unlimited storage. Get it now, on http://help.yahoo.com/l/in/yahoo/mail/yahoomail/tools/tools-08.html/ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] drop unused levels in sqldf
If you want to suppress the unused level of Species, you can use factor() : table(base$Species) table(factor(base$Species)) 2008/8/28 glaporta [EMAIL PROTECTED] Hi, sqldf is a fantastic package, but when the SELECT procedure runs unused levels remain in the output. I tried with the drop function, but without success. Do you have any suggestions? Thanx, Gianandrea data(iris) require(sqldf) base-sqldf(select * from iris where Species 'setosa') str(base) # Species with 3 levels! -- View this message in context: http://www.nabble.com/drop-unused-levels-in-sqldf-tp19196464p19196464.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] error in packet 1
Hello,
I'm Giovanni from ROMA..
I can't find a solution for the error:
error using packet 1
the y field is not specified and it has not a default value
(this is my traslation from italian language)
The code is:
pc-qqmath(~valori,
distribution=function(p) qweibull(p,beta,alpha),
prepanel = prepanel.qqmathline,
panel = function(x, y) {
panel.grid()
panel.qqmathline(y, distribution = function(p)
qweibull(p,beta,alpha))
panel.qqmath(x, y)
},
layout = c( 1,1), aspect = 0.8,
xlab = Unit Weibull Quantile, ylab = D
)
Thanks for your regard,
Giovanni
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[R] Linear model with one known coordinate
Dear All, I have a question which seems trivial, but I reached a dead end. I have a set of points (measurements) and I used lm() to obtain their linear regression model. From the biological background this line must pass through a point (100,0). Our dataset is not optimal and it shows a slight deviation from that coordinate. How can I add the restraint to the model, to go through that point? Any help would be appreciated. Anya Baresic, PhD Student Research Department of Structural and Molecular Biology University College London [EMAIL PROTECTED] -- View this message in context: http://www.nabble.com/Linear-model-with-one-known-coordinate-tp19196978p19196978.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] A simple analysis of Bayesian Robustness with R
I have to make a simulation testing the Bayesian Robustness of a normal linear model and comparing the regression results obtained using two different g-priors. I tried with the function blinreg() in the LearnBayes package from J.Albert but it doesn't help me because it uses a standard flat nonninformative prior, while I need to make changes to the structure of the prior (in particular to the prior variance) to perform my simulation. Does anybody has any suggestion or knows any R-function that would help? -- View this message in context: http://www.nabble.com/A-simple-analysis-of-Bayesian-Robustness-with-R-tp19197602p19197602.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ggplot2: problem with large fonts and overlapping labels
Hi Thierry, Thanks for your reply. I use the pdf graphics device. I did not know about the pointsize option so I will take a look. Here is an example of what I might do. pdf(onefile=FALSE, width=10, height=7) ggplot(mtcars, aes(x=wt, y=mpg)) + geom_point() + theme_bw grid.gedit(gPath(labels, label), gp=gpar(col=rgb(0,0,0), cex=1.25)) grid.gedit(gPath(xlabel), gp=gpar(cex=1.25)) grid.gedit(gPath(ylabel), gp=gpar(cex=1.25)) dev.off() On this example it actually looks ok, but on my real data I have longer axis labels which overlap with the tick labels. Regards, Paul ONKELINX, Thierry wrote: Dear Paul, How are you generating (saving) your plots? I tend to play with the pointsize argument of the graphical device, something in conjunction with the size argument in ggplot2 (size of points and lines). Working like that I get plots with nicely propotioned labels without overlaps. HTH, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology and quality assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 [EMAIL PROTECTED] www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey -Oorspronkelijk bericht- Van: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Namens Paul Emberson Verzonden: woensdag 27 augustus 2008 17:58 Aan: r-help@r-project.org Onderwerp: [R] ggplot2: problem with large fonts and overlapping labels Hi, I am using ggplot2 to generate graphs for a paper I am writing in two column format. When I shrink the graphs to fit in a single column, the graph is clear but the axis and tick labels are way too small. I have increased the font sizes by manipulating the grid. However, when I do this the tick labels and axis labels get very close or even overlap, especially on the y-axis. Is there a way to increase the margin between tick labels and axis labels with ggplot or by manipulating the grid? Regards, Paul __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ggplot2: problem with large fonts and overlapping labels
Dear Thierry, The pointsize option works perfectly for me. Thanks, Paul ONKELINX, Thierry wrote: Dear Paul, How are you generating (saving) your plots? I tend to play with the pointsize argument of the graphical device, something in conjunction with the size argument in ggplot2 (size of points and lines). Working like that I get plots with nicely propotioned labels without overlaps. HTH, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology and quality assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 [EMAIL PROTECTED] www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey -Oorspronkelijk bericht- Van: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Namens Paul Emberson Verzonden: woensdag 27 augustus 2008 17:58 Aan: r-help@r-project.org Onderwerp: [R] ggplot2: problem with large fonts and overlapping labels Hi, I am using ggplot2 to generate graphs for a paper I am writing in two column format. When I shrink the graphs to fit in a single column, the graph is clear but the axis and tick labels are way too small. I have increased the font sizes by manipulating the grid. However, when I do this the tick labels and axis labels get very close or even overlap, especially on the y-axis. Is there a way to increase the margin between tick labels and axis labels with ggplot or by manipulating the grid? Regards, Paul __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Singularity?
Hi all, When using lm to model a response with 8 explanatory variables, one of the variables is not defined due to singularities. I have checked the csv file from which the data come, there are no na's in the dataset, etc. What should I be looking for in this variable to correct the problem? Thanks for any help. Robin Williams Met Office summer intern - Health Forecasting [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R graph with values incorporated
Dear All: Greetings! By the way, is it possible to have a graph (say line graph) that shows values as well (say y-axis values within the graph)? One could do it in excel. I am just wondering whether it is possible with R! Thanks in advance, Prasanth VP, Global Manager - Biometrics, Delta Technology Management Services Pvt Ltd, Plot No: 13/2, Sector - I, Third Floor, HUDA Techno Enclave, Madhapur, Hyderabad - 500 033. Office:+91-40-2355-3992 Mobile: +91-9848290025. www.deltaintech.com ** 'The information contained in this email is confidential and may contain proprietary information. It is meant solely for the intended recipient. Access to this email by anyone else is unauthorized. If you are not the intended recipient, any disclosure, copying, distribution or any action taken or omitted in reliance on this, is prohibited and may be unlawful. No liability or responsibility is accepted if information or data is, for whatever reason corrupted or does not reach its intended recipient. No warranty is given that this email is free of viruses. The views expressed in this email are, unless otherwise stated, those of the author and not those of DELTA Technology and Management Services pvt ltd or its management. DELTA Technology and Management Services pvt ltd reserves the right to monitor intercept and block emails addressed to its users or take any other action in accordance with its email use policy' Thank you in advance for your cooperation. ** P Please don't print this e-mail unless you really need to. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Quantitative risk analysis with R
Dear Graham, Recently a course with this title, from Vose consulting, was announced on the list. Does anyone know of any books/websites/downloadable tutorials etc that cover the same ground. There is an R package QRMlib on CRAN http://cran.r-project.org/web/packages/QRMlib/index.html that accompanies the book Quantitative Risk Management: Concepts, Techniques and Tools by Alexander J. McNeil, Rüdiger Frey and Paul Embrechts http://www.ma.hw.ac.uk/~mcneil/book/index.html HTH, Tobias ie not just quantitative risk analysis, but specifically on using R for risk analysis and as an alternative to @Risk/Crystal Ball. Many thanks, Graham __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] coloured letters in a text
Hi does somebody know how to plot single letters in a text in different colours? example 1: I would like to add the word ABC to a figure. Thereby each letter should have a different colour. text(x,y,ABC, col=c(1,2,3)) # this does not work example 2: I would like to add the name of a parameter p with an index i to a figure. The index i should be in red, whereas the rest of the text should be in black. text(x,y, expression(p[i])) Looking forward to your answers! Fränzi __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Renaming objects
names, colnames, rownames - list of names I think it depends on what you are renaming? Stephen On Thu, Aug 28, 2008 at 4:03 AM, Williams, Robin [EMAIL PROTECTED] wrote: Hi, Is there any quick and easy way to rename a number of objects, without having to rename each one individually and then remove the old one? And if so, is there anything I can do to adjust the associated comments accordingly? Thanks for any help, Robin Williams Met Office summer intern - Health Forecasting [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Singularity?
On 28-Aug-08 11:04:47, Wi lliams, Robin wrote: Hi all, When using lm to model a response with 8 explanatory variables, one of the variables is not defined due to singularities. I have checked the csv file from which the data come, there are no na's in the dataset, etc. What should I be looking for in this variable to correct the problem? Thanks for any help. Robin Williams Met Office summer intern - Health Forecasting [EMAIL PROTECTED] You should be looking at whether the variable is expressible, throughout, as a linear combination of the other explanatory variables. Try svd(X) where X is the transpose of the matrix of explanatory variables from the CSV file (i.e. X has variables as rows, cases as columns) -- if you get a very small value in svd(X)$d then that points to your problem. Example: X-rbind(rnorm(10),rnorm(10),rnorm(10)) X - rbind(X,X[1,]+ 0.5*X[2,]-0.25*X[3,]) svd(X) # $d # [1] 4.355094e+00 3.717386e+00 2.101743e+00 1.842137e-16 # $u # [,1] [,2][,3] [,4] # [1,] -0.71645227 -0.1715990 -0.15753569 -0.657596 # [2,] 0.47501937 -0.8106696 0.09520123 -0.328798 # [3,] 0.09347153 -0.1262667 -0.97380325 0.164399 # [4,] -0.50231047 -0.5453671 0.13351574 0.657596 # (plus the right-hand eigenvectors) Note the very small value of $d[4] -- this is the machine approximation to zero. The corresponding left eigenvector $u[,4] annihilates X: V - svd(X)$u[,4] t(V)%*%X # [,1] [,2] [,3] [,4] # [1,] 4.228388e-17 -3.894996e-17 -8.020386e-17 -9.790346e-17 # [,5] [,6] [,7] [,8] # [1,] -2.710505e-17 -2.683400e-18 -3.876700e-17 8.250779e-17 # [,9] [,10] # [1,] 1.439278e-16 1.015762e-17 all of which are machine approximations to zero! Hoping this helps, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 28-Aug-08 Time: 12:42:31 -- XFMail -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] A Tip: lm, glm, and retained cases
In R-devel na.action(GLM) will work as the extractor. The problem with attr(GLM$model, na.action) is that the 'model' component is optional, and with model.frame(ModelObject) that if the 'model' component has been omitted it will try to recreate the model frame from the currently visible objects of the name originally used. (Because that is error-prone, we switched to model=TRUE as the default.) In earlier versions of R, GLM$na.action is the copy you want. However, I think if you care about omitted rows, you should use na.action=na.exclude, for then most auxiliary functions will give you results for all the rows. On Tue, 26 Aug 2008, Marc Schwartz wrote: on 08/26/2008 07:31 PM (Ted Harding) wrote: On 26-Aug-08 23:49:37, hadley wickham wrote: On Tue, Aug 26, 2008 at 6:45 PM, Ted Harding [EMAIL PROTECTED] wrote: Hi Folks, This tip is probably lurking somewhere already, but I've just discovered it the hard way, so it is probably worth passing on for the benefit of those who might otherwise hack their way along the same path. Say (for example) you want to do a logistic regression of a binary response Y on variables X1, X2, X3, X4: GLM - glm(Y ~ X1 + X2 + X3 + X4) Say there are 1000 cases in the data. Because of missing values (NAs) in the variables, the number of complete cases retained for the regression is, say, 600. glm() does this automatically. QUESTION: Which cases are they? You can of course find out by hand on the lines of ix - which( (!is.na(Y))(!is.na(X1))...(!is.na(X4)) ) but one feels that GLM already knows -- so how to get it to talk? ANSWER: (e.g.) ix - as.integer(names(GLM$fit)) This is a partial match to 'fitted', and will only work if default row names were used. Alternatively, you can use: attr(GLM$model, na.action) Hadley Thanks! I can see that it works -- though understanding how requires a deeper knowledge of R internals. However, since you've approached it from that direction, simply GLM$model is a dataframe of the retained cases (with corresponding row-names), all variables at once, and that is possibly an even simpler approach! Or just use: model.frame(ModelObject) as the extractor function... :-) Another 'a priori' approach would be to use na.omit() or one of its brethren on the data frame before creating the model. Which function is used depends upon how 'na.action' is set. The returned value, or more specifically the 'na.action' attribute as appropriate, would yield information similar to Hadley's approach relative to which records were excluded. For example, using the simple data frame in ?na.omit: DF - data.frame(x = c(1, 2, 3), y = c(0, 10, NA)) DF x y 1 1 0 2 2 10 3 3 NA DF.na - na.omit(DF) DF.na x y 1 1 0 2 2 10 attr(DF.na, na.action) 3 3 attr(,class) [1] omit So you can see that record 3 was removed from the original data frame due to the NA for 'y'. HTH, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- 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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Quantitative risk analysis with R
Tobias, Thanks I will give this a look, it seems the focus is on credit risk (where I am more interested in environmental risks) but it should still be useful. Graham 2008/8/28 Tobias Verbeke [EMAIL PROTECTED]: Dear Graham, Recently a course with this title, from Vose consulting, was announced on the list. Does anyone know of any books/websites/downloadable tutorials etc that cover the same ground. There is an R package QRMlib on CRAN http://cran.r-project.org/web/packages/QRMlib/index.html that accompanies the book Quantitative Risk Management: Concepts, Techniques and Tools by Alexander J. McNeil, Rüdiger Frey and Paul Embrechts http://www.ma.hw.ac.uk/~mcneil/book/index.html HTH, Tobias ie not just quantitative risk analysis, but specifically on using R for risk analysis and as an alternative to @Risk/Crystal Ball. Many thanks, Graham __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] coloured letters in a text
2008/8/28 Fränzi Korner [EMAIL PROTECTED]: example 1: I would like to add the word ABC to a figure. Thereby each letter should have a different colour. text(x,y,ABC, col=c(1,2,3)) # this does not work kludge alert! How about: text(x,y,ABC,col=3) text(x,y,AB,col=2) text(x,y,A,col=1) Basically this overlays three coloured texts to give the impression of individual coloured letters. This may or may not be practical for your real application. It's also possible that if you print this your printer will end up printing A with a mix of colours 1, 2 and 3. If your font is monospaced you could do it with text(x,y, C,col=3), but in a pretty font the letters AB wont have the same width as two spaces. Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] New package: ``denstrip'' for compactly illustrating distributions
Dear R users, I'd like to announce a new package on CRAN called ``denstrip''. It implements ``density strips'' and other graphical methods for illustrating and comparing distributions in a compact fashion. Posterior distributions of parameters are often summarised using point and line drawings of means and credible intervals. This is common, for example, in multiple regression or meta-analysis. Density strips generalise these to illustrate whole distributions. Instead of a point and line, a shaded strip indicates the density as proportional to the darkness of the shading. They taper to white at the end of the strip, instead of terminating at a clear limit - this may discourage casually categorising effects as ``significant'' if the line excludes the null. The shading idea generalises to ``density regions'' to show uncertainty about continuously-varying quantities, such as predictions from time series. The package includes other functions for illustrating distributions in ``one dimension'', such as varying-width strips (similar to violin plots) and sectioned density plots. If you're interested in reading more about these methods, I discuss them in a forthcoming article in The American Statistician, ``Displaying uncertainty with shading'', also available from http://www.mrc-bsu.cam.ac.uk/personal/chris/papers/denstrip.pdf Comments and suggestions for improvement of the package are welcome. -- Christopher Jackson [EMAIL PROTECTED] Research Statistician, MRC Biostatistics Unit, Institute of Public Health, Robinson Way, Cambridge, UK, CB2 0SR. +44 (1223) 330381 ___ R-packages mailing list [EMAIL PROTECTED] https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] averaging pairs of columns in a dataframe
somehow robust method for dataframe df would be newdf - (df[,seq(1,66,2)]+df[,seq(2,66,2)])/2 Gasper -Original Message- From: JonD [mailto:[EMAIL PROTECTED] Sent: Wednesday, August 27, 2008 3:47 PM To: r-help@r-project.org Subject: [R] averaging pairs of columns in a dataframe Dear all, I have a dataframe with 132 columns and 100 rows. Every 2nd column is a repeat measurement so that the columns could be titled, a a b b c c d d etc. I would like to average the repeats such that I am left with a data frame of 66 columns (of means) and 100 rows. I have been trying to use rowMeans but have not been able to average the pairs of columns, only the whole dataframe. Any help would be appreciated, I am new to R. Thanks in advance, Jon. -- View this message in context: http://www.nabble.com/averaging-pairs-of-columns-in-a-dataframe-tp19181319p1 9181319.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Plots spanning columns
Hi! I want to plot three graphs (residuals, ACF and PACF of a model). Ideally I would use a c(2,2) disposition where the residuals plot would start at position 1,1 and span to position 1,2. Then I would plot the ACF in position 2,1 and the PACF in position 2,2. Maybe is clearer like this: -- || | residuals| || -- - | | | | |ACF | | PACF| | | | | - Does anyone know if that is possible at all? Cheers! -- -- -- Jose Luis Aznarte M. http://decsai.ugr.es/~jlaznarte Department of Computer Science and Artificial Intelligence Universidad de Granada Tel. +34 - 958 - 24 04 67 GRANADA (Spain) Fax: +34 - 958 - 24 00 79 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] coloured letters in a text
See ?strwidth, e.g. plot(1) text(1.1, 1.1, A) text(1.1 + strwidth(A), 1.1, B, col = 2) text(1.1 + strwidth(AB), 1.1, C, col = 3) On Thu, Aug 28, 2008 at 7:33 AM, Fränzi Korner [EMAIL PROTECTED] wrote: Hi does somebody know how to plot single letters in a text in different colours? example 1: I would like to add the word ABC to a figure. Thereby each letter should have a different colour. text(x,y,ABC, col=c(1,2,3)) # this does not work example 2: I would like to add the name of a parameter p with an index i to a figure. The index i should be in red, whereas the rest of the text should be in black. text(x,y, expression(p[i])) Looking forward to your answers! Fränzi __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plots spanning columns
Try this: l - layout(matrix(c(1, 2, 1, 3), 2)) layout.show(l) On Thu, Aug 28, 2008 at 9:40 AM, Jose Luis Aznarte M. [EMAIL PROTECTED] wrote: Hi! I want to plot three graphs (residuals, ACF and PACF of a model). Ideally I would use a c(2,2) disposition where the residuals plot would start at position 1,1 and span to position 1,2. Then I would plot the ACF in position 2,1 and the PACF in position 2,2. Maybe is clearer like this: -- || | residuals| || -- - | | | | |ACF | | PACF| | | | | - Does anyone know if that is possible at all? Cheers! -- -- -- Jose Luis Aznarte M. http://decsai.ugr.es/~jlaznartehttp://decsai.ugr.es/%7Ejlaznarte Department of Computer Science and Artificial Intelligence Universidad de Granada Tel. +34 - 958 - 24 04 67 GRANADA (Spain) Fax: +34 - 958 - 24 00 79 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Updating a list.
That would be one source of error. Thank you. Kevin ONKELINX wrote: Kevin, Notice the subtle difference between Hadley's and your code: Hadley m2008$DayOfYear - factor(m2008$DayOfYear, levels = 1:365) Kevin m2007$DayOfYear - factor(m2008$DayOfYear, levels = 1:365) Your are using the m2007 object instead of the suggested m2008 object! HTH, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology and quality assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 [EMAIL PROTECTED] www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey -Oorspronkelijk bericht- Van: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Namens [EMAIL PROTECTED] Verzonden: donderdag 28 augustus 2008 3:14 Aan: r-help@r-project.org Onderwerp: Re: [R] Updating a list. Since this didn't work: m2007$DayOfYear - factor(m2008$DayOfYear, levels = 1:365) Error in `$-.data.frame`(`*tmp*`, DayOfYear, value = c(1L, 1L, 1L, : replacement has 432267 rows, data has 1592009 Perhaps I need to clarify how the m2007 object was generated. t2007 - read.csv(Total2007.dat, header = TRUE) m2007 - melt(t2007, id.var=c(DayOfYear,Category,SubCategory,Sku), measure.var=c(Quantity)) Kevin hadley wickham [EMAIL PROTECTED] wrote: Try this: m2008$DayOfYear - factor(m2008$DayOfYear, levels = 1:365) r2007 - cast(m2008, DayOfYear ~ variable | Sku, sum, fill = 0) Hadley -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Potential Error/Bug?: addition of NULL during arithmetic
On Tue, 26 Aug 2008, Eric DeWitt wrote: I encountered an error that does not make sense to me given my reading of the documentation and does not appear to be referenced in the list archives or online. The error occurred when a function received a NULL value rather than a numeric value as a parameter and then attempted to use the NULL value in the calculateion. The error: Error during wrapup: nothing to replace with can be easily recreated with the following code: 1 + NULL # note that the opperation succeeds and returns a numeric vector with dim(NULL) numeric(0) As documented, of course. bar - 1 bar [1] 1 foo - 1 + NULL foo numeric(0) bar - bar + foo # note that here the assignment operation succeeds and 1 + (1 + NULL) - numeric(0) bar numeric(0) bar - c(1, 1) # however if the assignment is into a vector bar[1] - bar[1] + foo # note that the mathematical operation is identical, but the assignment fails Error during wrapup: nothing to replace with That's the trouble, this is not the same operation at all. This is replacing one element of a vector, not naming the result 'bar'. If this is the intended behavior, a more informative error message (e.g. 'attempt to assign NULL to vector element') would be useful. If it is not the intended behavior, should I log this as a bug? It is the intended and documented behaviour. You tried to set element 1 of bar to bar[1] + foo, which is numeric(0) (by the documented recycling rules). The error message: 'nothing to replace with' seems very informative to me. If you put some effort into understanding your own errors by doing things step by step, as in bar - c(1, 1) foo - bar[1] + foo bar[1] - foo Error in bar[1] - foo : nothing to replace with foo numeric(0) you might learn more. I don't know how you got an 'Error during wrapup': that is not reproducible (it might be a function of using a console, if you did). -eric sessionInfo() R version 2.7.1 (2008-06-23) powerpc-apple-darwin8.10.1 locale: en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] stats graphics grDevices utils datasets methods base __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- 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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R graph with values incorporated
Prasanth wrote: Dear All: Greetings! By the way, is it possible to have a graph (say line graph) that shows values as well (say y-axis values within the graph)? One could do it in excel. I am just wondering whether it is possible with R! x - rnorm(100,2,3) y - rnorm(100,2,3) plot(x,y,pch=19) text(x=x,y=y+.5,format(x,digits=1),cex=.5) [[alternative HTML version deleted]] -- Read the posting guide. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Upgrading R means I lose my packages
James Milks wrote: The title says it all. Does anyone know of a way to save your packages when you upgrade to a new version of R? This may seem petty, but I'm accumulating enough packages that having to download and install each of them anew every time I install a new version of R is rather of a pain. Ideally, I would like the new version of R to recognize the packages I've installed on the previous version without needing to reinstall the packages. Is that possible? My system: Mac OS 10.5.4. Current R version: 2.7.1 Thanks for any suggestions. My apologies if this has been answered before and my search missed it. Jim Milks Degree Candidate __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. What I find simplest (on Windows at least) is # To reinstall packages from one R version to the next: #Currently you can do tmp - installed.packages() installedpkgs - as.vector(tmp[is.na(tmp[,Priority]), 1]) save(installedpkgs, file=c:/R/installed.rda) #in the old version to get a list of packages you installed. Then in the #new version, load(c:/R/installed.rda) install.packages(installedpkgs) -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. York University Voice: 416 736-5115 x66249 Fax: 416 736-5814 4700 Keele Streethttp://www.math.yorku.ca/SCS/friendly.html Toronto, ONT M3J 1P3 CANADA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] coloured letters in a text
On Thu, 28 Aug 2008, Barry Rowlingson wrote: 2008/8/28 Fränzi Korner [EMAIL PROTECTED]: example 1: I would like to add the word ABC to a figure. Thereby each letter should have a different colour. text(x,y,ABC, col=c(1,2,3)) # this does not work kludge alert! How about: text(x,y,ABC,col=3) text(x,y,AB,col=2) text(x,y,A,col=1) You need adj=0 for left justification. Basically this overlays three coloured texts to give the impression of individual coloured letters. This may or may not be practical for your real application. It's also possible that if you print this your printer will end up printing A with a mix of colours 1, 2 and 3. If your font is monospaced you could do it with text(x,y, C,col=3), but in a pretty font the letters AB wont have the same width as two spaces. What's wrong with using strwidth? x - 3 text(x, 5, A, adj=0, col=1) x - x + strwidth(A) text(x, 5, B, adj=0, col=2) x - x + strwidth(B) text(x, 5, C, adj=0, col=3) This will not use kerning, but only a few devices 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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plots spanning columns
library(HH) example(tsacfplots) ?tsacfplots __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plots spanning columns
On Thu, 28 Aug 2008, Jose Luis Aznarte M. wrote:
Hi! I want to plot three graphs (residuals, ACF and PACF of a model).
Ideally I would use a c(2,2) disposition where the residuals plot would start
at position 1,1 and span to position 1,2. Then I would plot the ACF in
position 2,1 and the PACF in position 2,2. Maybe is clearer like this:
--
||
| residuals|
||
--
-
| | | |
|ACF | | PACF|
| | | |
-
Does anyone know if that is possible at all? Cheers!
I happened to do virtually that plot just recently. Just had data rather
than residuals:
### Function to display plot, ACF and PACF
displayTimeSeries - function(x, heading = NULL,
heights = NULL, ...){
defaultPars - par(no.readonly = TRUE)
if (is.null(heading)) {
heading - paste(Series: , deparse(substitute(x)))
}
if (is.null(heights)) heights - c(1,1)
layout(matrix(c(1,1,2,3), 2, 2, byrow = TRUE), heights = heights)
plot(x, main = heading)
par(mar = c(5,4,1,2) + 0.1)
acfVal - acf(x, main = )$acf
pacfVal - acf(x, type = partial, main = )$acf
par(defaultPars)
invisible(list(acf = acfVal, pacf = pacfVal))
}
If you add a line which calculates the residuals and plots them instead of
the data, then you should have what you want.
David Scott
_
David Scott Department of Statistics, Tamaki Campus
The University of Auckland, PB 92019
Auckland 1142,NEW ZEALAND
Phone: +64 9 373 7599 ext 86830 Fax: +64 9 373 7000
Email: [EMAIL PROTECTED]
Graduate Officer, Department of Statistics
Director of Consulting, Department of Statistics
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Upgrading R means I lose my packages
On Thu, 28 Aug 2008, Michael Friendly wrote: James Milks wrote: The title says it all. Does anyone know of a way to save your packages when you upgrade to a new version of R? This may seem petty, but I'm accumulating enough packages that having to download and install each of them anew every time I install a new version of R is rather of a pain. Ideally, I would like the new version of R to recognize the packages I've installed on the previous version without needing to reinstall the packages. Is that possible? My system: Mac OS 10.5.4. Current R version: 2.7.1 Thanks for any suggestions. My apologies if this has been answered before and my search missed it. Jim Milks Degree Candidate __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. What I find simplest (on Windows at least) is # To reinstall packages from one R version to the next: #Currently you can do tmp - installed.packages() installedpkgs - as.vector(tmp[is.na(tmp[,Priority]), 1]) save(installedpkgs, file=c:/R/installed.rda) #in the old version to get a list of packages you installed. Then in the #new version, load(c:/R/installed.rda) install.packages(installedpkgs) But this is an FAQ, with a cleaner answer in the rw-FAQ. It really is much easier to make use of a separate library for the additional packages you install (and there is no need to reinstall packages when going from 2.7.1 to 2.7.2, which helps if you have hundreds or even thousands installed -- a complete reinstall on our 32-bit Linux server took about 5 hours when we did it for 2.7.0). -- 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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Integrate a 1-variable function with 1 parameter (Jose L.Romero)
Hi,
The answer can be obtained in closed form using the pgamma function,
which is closely related to the incomplete gamma function, as follows:
integrand - function (t, x) {
exp(-2*t)*(2*t)^x/(10*factorial(x))
}
upper - 10
x - 0:44
ans1 - sapply(x, function(x) integrate(integrand, lower=0, upper=upper,
x=x))
ans2 - gamma(x+1) * pgamma(q=2*upper, shape=x+1, rate = 1, scale = 1,
lower.tail = TRUE) / (20*factorial(x))# using the pgamma function
cbind(x=x, ans1=unlist(ans1[1,]), ans2=ans2) # both answers are identical
Ravi.
---
Ravi Varadhan, Ph.D.
Assistant Professor, The Center on Aging and Health
Division of Geriatric Medicine and Gerontology
Johns Hopkins University
Ph: (410) 502-2619
Fax: (410) 614-9625
Email: [EMAIL PROTECTED]
Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Moshe Olshansky
Sent: Wednesday, August 27, 2008 9:58 PM
To: r-help@r-project.org; [EMAIL PROTECTED]
Subject: Re: [R] Integrate a 1-variable function with 1 parameter (Jose
L.Romero)
This can be done analytically: after changing a variable (2*t - t) and some
scaling we need to compute
f(x) = integral from 0 to 20 of (t^x*exp(-t))dt/factorial(x)
f(0) = int from 0 to 20 of exp(-t)dt = 1 - exp(-20) and integration by parts
yields (for x=1,2,3,...)
f(x) = -exp(-20)*20^x/factorial(x) + f(x-1) so that
f(x) = 1 - exp(-20)*sum(20^k/factorial(k)) where the sum is for k=0,1,...,x
If I did not a mistake, your original quantity should be f(x)/20.
--- On Thu, 28/8/08, jose romero [EMAIL PROTECTED] wrote:
From: jose romero [EMAIL PROTECTED]
Subject: [R] Integrate a 1-variable function with 1 parameter (Jose L.
Romero)
To: r-help@r-project.org
Received: Thursday, 28 August, 2008, 12:23 AM Hey fellas:
I would like to integrate the following function:
integrand - function (x,t) {
exp(-2*t)*(2*t)^x/(10*factorial(x))
}
with respect to the t variable, from 0 to 10.
The variable x here works as a parameter: I would like to integrate
the said function for each value of x in 0,1,..,44.
I have tried Vectorize to no avail.
Thanks in advance,
jose romero
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Re: [R] Renaming objects
On Thu, 2008-08-28 at 07:39 -0400, stephen sefick wrote: names, colnames, rownames - list of names I think it depends on what you are renaming? Those alter the dimnames of objects that have them, not the names of the objects themselves. I think a short answer to the question is no. There is a nice thread on a similar topic from April this year. In R, do: RSiteSearch(rename objects) And the first few hits cover the discussion. Rolf Turner presents his mv() function in that thread, equivalent to the mv shell command in Unix. You could modify Rolf's mv() to accept a vector of objects to rename and the new names they take. you could probably just pop the body of Rolf's function in a loop over the number of rename operations, though I haven't done this myself to check it works so simply... G Stephen On Thu, Aug 28, 2008 at 4:03 AM, Williams, Robin [EMAIL PROTECTED] wrote: Hi, Is there any quick and easy way to rename a number of objects, without having to rename each one individually and then remove the old one? And if so, is there anything I can do to adjust the associated comments accordingly? Thanks for any help, Robin Williams Met Office summer intern - Health Forecasting [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Help with shading a polygon below a segment of a curve (normal distribution)
Dear R users, I still feel new to R so please apologize if I am doing something stupid here. My use of the polygon() function produces a result that I cannot comprehend: In a plot, I would like to shade the area below a normal distribution. However, I do not want the entire area to be shaded, but just the area on the right side of a vertical line that I draw through the distribution (in order to illustrate the function of a t-test). Here is what I do: scale - 0.1 x - seq(-4, 6, scale) y - dnorm(x) plot(x, y, type = l, main=t-Test, t = 2.2) linepos - 2.2 abline(v = linepos) # I try to fill a polygon right of the vertical line: # max(x) - linepos (in this case, 2.2) / scale (0.1) # results in the last 38 elements of x and y. # so I take the last 38 elements of x and y and try to # draw a polygon underneath: cutpoint - (max(x) - linepos) / scale xt - x[(length(x)-cutpoint):length(x)] yt - y[(length(y)-cutpoint):length(y)] # draw the polygon polygon(xt, yt, density = 10 ) As you can see in the result, this is not what I want; some area above the line gets shaded, but not below. Can someone tell me what I am missing? Thank you very much, Bertolt -- Bertolt Meyer Oberassistent Sozialpsychologie, Psychologisches Institut der Universität Zürich Binzmühlestr. 14, Box 15 CH-8050 Zürich [EMAIL PROTECTED] tel: +41446357282 fax: +41446357279 mob: +41788966111 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] GeoR model.control - defining covariates at prediction locations
Thanks, I've managed to make some progress, but seem to be getting some strange results from my kriging, which I think much have something to do with a problem with my prediction points. I have my geodata object (called nicola), my prediction points (predpoints, imported from a csv containing only the x and y coordinated of the prediction locations) and my covariate data at each of the prediction points (covars, imported from a csv containing the x and y coordinates of the prediction locations, plus the values of the two covariates I want to use at each of the prediction locations). predpoints-read.csv(file=C:\\Documents and Settings\\s9901315\\My + Documents\\Uni\\Data\\Work\\Case control study\\Full study area\\R\\Files for analysis\\Prediction + points\\predpoints.csv, header=FALSE, sep=,) covars-read.csv(file=C:\\Documents and Settings\\s9901315\\My + Documents\\Uni\\Data\\Work\\Case control study\\Full study area\\R\\Files for analysis\\Covariate + data\\covars.csv, header=TRUE, sep=,) The final model is defined using OTUBOIDIST and LSTPHAN as external covariates: mlx2-likfit(nicola, cov.model=mat, kap=0.5, ini=c(0.6, 20), nug=0.3, trend=~OTUBOIDIST + + LSTPHAN) and then I carry out the kriging using the model mlx2, prediction points predpoints, and covariate data covars : kcontrol-krige.control(obj.m=mlx2, type.krige=ok, trend.d=~OTUBOIDIST + LSTPHAN, + trend.l=~covars$otuboidist + covars$lstphan) krige-krige.conv(nicola, loc=predpoints, krige=kcontrol) Then I view it using the image function: image(krige, col=gray(seq(1, 0.2, l=100))) The resulting image is clearly wrong with a regular stepped line appearing diagonally across the image, and the predicted values do not coincide with the actual observed data at all. I've included the predicted data image, as well as the predicted image overlaid with the data points. http://www.nabble.com/file/p19201449/no%2Bpoints.jpg http://www.nabble.com/file/p19201449/withpoints.jpg Can anyone give me any pointers of why this may be going wrong? I've tried the same thing many times having changed everything I can think of that might be causing the problem. Thanks, Nicola [EMAIL PROTECTED] Paulo Justiniano Ribeiro Jr-2 wrote: Trends in ge are handled by the trendargument. In particular, for the so called external trend we use varioables which can be at the geodata object or another object. For instance a model fitting call could be somethoing like: ML - likfit..., trend= ~covar1+ covar2, ...) where covar1 and covar2 have the same dimension as the data and can be vector or columuns withing the covariate elelent of a geodata obvject. For kriging you need the values of such covariates at prediction locations lts say the vectors predcovar1 and predcovar2 Then you use model.control(..., trend.l= ~ predcovar1+predcovar2) Three are osme more examples at geoR page, Tutorials section, and please send the commands such that we can be more specific best P.J. Date: Mon, 18 Aug 2008 02:59:50 -0700 (PDT) From: imicola [EMAIL PROTECTED] Subject: [R] GeoR model.control - defining covariates at prediction locations To: r-help@r-project.org Message-ID: [EMAIL PROTECTED] Content-Type: text/plain; charset=us-ascii Hi, Im using geoR and I'm trying to do some predictions, based on an external trend. I'm having some problems specifying my model.control, specifically how do I define my model, and also the source of the covariate data at the prediction locations? I am assuming that the covariate data at the prediction locations should be imported to a geodata object along with the prediction location coordinates - this is what I have done, but I can't get the prediction to work. So my question is: How should the prediction location and covariate data at prediction locations be stored? And also how do I specify model.control so that it recognises where my covariate data is? Below is the error message that I am getting: krige.bayes: model with mean defined by covariates provided by the user Warning messages: 1: locations provided as a list with more than 2 components. Only the 2 first will be used as coordinates in: .check.locations(locations) 2: data length [507] is not a sub-multiple or multiple of the number of rows [254] in matrix Error in trend.spatial(trend = model$trend.l, geodata = list(coords = locations)) : trend elements not found Thanks, __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/GeoR-model.control---defining-covariates-at-prediction-locations-tp19028273p19201449.html Sent from the R help mailing list archive at Nabble.com.
Re: [R] Help with shading a polygon below a segment of a curve (normal distribution)
Bertolt Meyer wrote: Dear R users, I still feel new to R so please apologize if I am doing something stupid here. My use of the polygon() function produces a result that I cannot comprehend: In a plot, I would like to shade the area below a normal distribution. However, I do not want the entire area to be shaded, but just the area on the right side of a vertical line that I draw through the distribution (in order to illustrate the function of a t-test). Here is what I do: scale - 0.1 x - seq(-4, 6, scale) y - dnorm(x) plot(x, y, type = l, main=t-Test, t = 2.2) linepos - 2.2 abline(v = linepos) # I try to fill a polygon right of the vertical line: # max(x) - linepos (in this case, 2.2) / scale (0.1) # results in the last 38 elements of x and y. # so I take the last 38 elements of x and y and try to # draw a polygon underneath: cutpoint - (max(x) - linepos) / scale xt - x[(length(x)-cutpoint):length(x)] yt - y[(length(y)-cutpoint):length(y)] # draw the polygon polygon(xt, yt, density = 10 ) As you can see in the result, this is not what I want; some area above the line gets shaded, but not below. Can someone tell me what I am missing? Thank you very much, Bertolt Your polygon contains no points on the x axis (yt==0). Try adding this: n - length(xt) xt - c(xt[1], xt, xt[n]) yt - c(0,yt,0) polygon(xt, yt, col=red ) -- 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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] juxtaposed and stacked bars in one barplot?
hadley wickham schrieb: Hi Stefan, Could you be a bit more explicit? Do you have an example dataset that you are trying to visualise? Right, thanks for pointing out the obvious. So here's my code: library(gplots) quarter - as.factor(sample(c(Q1, Q2, Q3, Q4), 100, replace = TRUE)) year - as.factor(sample(c(seq(from=2000, to=2008)), 100, replace = TRUE)) category - as.factor(sample(c(seq(from=1, to=4)), 100, replace = TRUE)) test - data.frame(quarter, year, category) table(test$category, test$quarter, test$year) barplot2(table(test$quarter, test$year), beside=T, ylim=c(0,10), main=how to include dim3?) # inclusion of 3rd dimension does not work: barplot2(table(test$quarter, test$year, test$category), beside=T, ylim=c(0,10), main=how to include dim3?) I want the barplot to be exactly the same but with the bars stacked (by 'category'). I got the message from Gaspar, but have not yet tried to fit his example to my data. /Stefan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with shading a polygon below a segment of a curve (normaldistribution)
library(HH) normal.and.t.dist() There are many related examples in example(normal.and.t.dist) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with shading a polygon below a segment of a curve (normaldistribution)
Bertolt, The points you send to polygon() do not fully enclose the area you desire. Try adding one more point as such xt - c(x[(length(x)-cutpoint):length(x)],linepos) yt - c(y[(length(y)-cutpoint):length(y)],0) polygon(xt, yt, density = 10 ) -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Bertolt Meyer Sent: Thursday, August 28, 2008 9:03 AM To: r-help@r-project.org Subject: [R] Help with shading a polygon below a segment of a curve (normaldistribution) Dear R users, I still feel new to R so please apologize if I am doing something stupid here. My use of the polygon() function produces a result that I cannot comprehend: In a plot, I would like to shade the area below a normal distribution. However, I do not want the entire area to be shaded, but just the area on the right side of a vertical line that I draw through the distribution (in order to illustrate the function of a t-test). Here is what I do: scale - 0.1 x - seq(-4, 6, scale) y - dnorm(x) plot(x, y, type = l, main=t-Test, t = 2.2) linepos - 2.2 abline(v = linepos) # I try to fill a polygon right of the vertical line: # max(x) - linepos (in this case, 2.2) / scale (0.1) # results in the last 38 elements of x and y. # so I take the last 38 elements of x and y and try to # draw a polygon underneath: cutpoint - (max(x) - linepos) / scale xt - x[(length(x)-cutpoint):length(x)] yt - y[(length(y)-cutpoint):length(y)] # draw the polygon polygon(xt, yt, density = 10 ) As you can see in the result, this is not what I want; some area above the line gets shaded, but not below. Can someone tell me what I am missing? Thank you very much, Bertolt -- Bertolt Meyer Oberassistent Sozialpsychologie, Psychologisches Institut der Universität Zürich Binzmühlestr. 14, Box 15 CH-8050 Zürich [EMAIL PROTECTED] tel: +41446357282 fax: +41446357279 mob: +41788966111 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Upgrading R means I lose my packages
Speaking for myself, I think it's easier to just create a script and
put it somewhere easy to remember.
For example
my.pkgs - c('pkg1', 'pkg2') ## and so on for my preferred packages
install.packages(my.pkgs, dependencies=TRUE)
Then after each upgrade just source the script.
You will need to create a second list if you use any packages for
which binaries are not available.
Creating the list the first time might be inconvenient but
thereafter it's easy.
Additional benefits include
you can use the script when you get a new machine
you can use the script when a friend or colleague decides to try
R, and you want them to have the same packages
you can use the script if you maintain R on more than one platform, as I do
A downside would be if you have a huge number of packages and it
takes a long time to update them all. But even then, since most
packages are available as binaries, it should be reasonable.
Installing a huge number from source will take a long time.
This method is also a little tricky if you have locally written packages.
-Don
At 10:02 PM -0400 8/27/08, James Milks wrote:
The title says it all. Does anyone know of a way to save your
packages when you upgrade to a new version of R? This may seem
petty, but I'm accumulating enough packages that having to download
and install each of them anew every time I install a new version of
R is rather of a pain. Ideally, I would like the new version of R
to recognize the packages I've installed on the previous version
without needing to reinstall the packages. Is that possible?
My system: Mac OS 10.5.4.
Current R version: 2.7.1
Thanks for any suggestions. My apologies if this has been answered
before and my search missed it.
Jim Milks
Degree Candidate
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--
--
Don MacQueen
Environmental Protection Department
Lawrence Livermore National Laboratory
Livermore, CA, USA
925-423-1062
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Re: [R] Help with shading a polygon below a segment of a curve (normal distribution)
The power.examp function in the TeachingDemos package (among others) may already do what you want. Even if it does not, it does shade the area under a curve, you can look at the source for the function as an example of where to start. Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Bertolt Meyer Sent: Thursday, August 28, 2008 8:03 AM To: r-help@r-project.org Subject: [R] Help with shading a polygon below a segment of a curve (normal distribution) Dear R users, I still feel new to R so please apologize if I am doing something stupid here. My use of the polygon() function produces a result that I cannot comprehend: In a plot, I would like to shade the area below a normal distribution. However, I do not want the entire area to be shaded, but just the area on the right side of a vertical line that I draw through the distribution (in order to illustrate the function of a t-test). Here is what I do: scale - 0.1 x - seq(-4, 6, scale) y - dnorm(x) plot(x, y, type = l, main=t-Test, t = 2.2) linepos - 2.2 abline(v = linepos) # I try to fill a polygon right of the vertical line: # max(x) - linepos (in this case, 2.2) / scale (0.1) # results in the last 38 elements of x and y. # so I take the last 38 elements of x and y and try to # draw a polygon underneath: cutpoint - (max(x) - linepos) / scale xt - x[(length(x)-cutpoint):length(x)] yt - y[(length(y)-cutpoint):length(y)] # draw the polygon polygon(xt, yt, density = 10 ) As you can see in the result, this is not what I want; some area above the line gets shaded, but not below. Can someone tell me what I am missing? Thank you very much, Bertolt -- Bertolt Meyer Oberassistent Sozialpsychologie, Psychologisches Institut der Universität Zürich Binzmühlestr. 14, Box 15 CH-8050 Zürich [EMAIL PROTECTED] tel: +41446357282 fax: +41446357279 mob: +41788966111 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Upgrading R means I lose my packages
At 04:12 AM 8/28/2008, ONKELINX, Thierry wrote: On a windows machine you get the same problem. Useless one uses tha same trick as Rolf suggested: don't install the packages in the default directory and set R_LIBS to that directory. Then all you need to do after an upgrade is to set R_LIBS in the new version and run update.package(checkBuilt = TRUE). Given Rolf's suggestion I suppose this trick will work on a Mac too. What I do in installing a new version of R on a Windows system is as follows: 1. In the c:\Program Files\R folder, the installation is in a subfolder labeled by the version, such as R-2.7.1. 2. I leave the old (say, 2.7.1) version installed, and install the new version (say, 2.7.2). This leaves the old subfolder R-2.7.1 intact, and creates a new one R-2.7.2. 3. I use a file-compare utility (in my case, Beyond Compare, which I recommend), to compare the subfolders C:\Program Files\R\R-2.7.1\library and C:\Program Files\R\R-2.7.2\library. I set the comparison to find files present or newer in the 2.7.1 folder vs. the 2.7.2. Then I copy all such files over. 4. At this point, the 2.7.2 has the same or new packages than 2.7.1, most or all of which will work. 5. I use the Packages|Update Package ... to update packages to 2.7.2. 6. Then I delete the 2.7.1 subfolder. You need Administrator rights to do this. Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: [EMAIL PROTECTED] Least Cost Formulations, Ltd.URL: http://lcfltd.com/ 824 Timberlake Drive Tel: 757-467-0954 Virginia Beach, VA 23464-3239Fax: 757-467-2947 Vere scire est per causas scire __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] RCurl: authentication when posting forms
Hi Valerie Valerie Obenchain wrote: Hi, Has anyone successfully used RCurl for posting data to a password-protected site? Yes. I just set up a sample form to test with and the following all work # Perl script (and HTML form for testing in the browser) taken from # http://www.elated.com/articles/form-validation-with-perl-and-cgi/ # Provide the login password directly postForm(http://www.omegahat.org/RCurl/testPassword/form_validation.cgi;, your_name = Duncan, your_age = 35-55, your_sex = m, submit = submit, .opts = list(userpwd = bob:welcome)) # Get the login password in ~/.netrc postForm(http://www.omegahat.org/RCurl/testPassword/form_validation.cgi;, your_name = Duncan, your_age = 35-55, your_sex = m, submit = submit, .opts = list(netrc = TRUE)) # Get the login password from a different netrc file postForm(http://www.omegahat.org/RCurl/testPassword/form_validation.cgi;, your_name = Duncan, your_age = 35-55, your_sex = m, submit = submit, .opts = list(netrc = TRUE, netrc.file = /Users/duncan/Projects/org/omegahat/R/RCurl/inst/examples/omg.netrc)) So let me know what problems you are having and more details about the OS, version of libcurl, and a sample URL to which to post, etc. D. I have tired using option netrc=1 with both postForm and curlPerform (with postfields option) but can't authenticate. I would happily provide more details if some one has had some experience with this. Thanks very much. Valerie __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Read a file
Hello, I have a text file with this structure: # File created = Thursday, August 28, 2008 3:33:02 PM GMT # Data set = 373 2 1 C:\Bruker\TOPSPIN GABRMN # Spectral Region: # LEFT = 4.5 ppm. RIGHT = 0.5 ppm. # # SIZE = 13111 ( = number of points) # # In the following ordering is from the 'left' to the 'right' limits! # Lines beginning with '#' must be considered as comment lines. # 1628.40625 1628.40625 1964.40625 2242.0625 2533.5 2937.90625 3448.0 3923.96875 Is it possible to read it with R, something like scan() but keeping the structure? Best, Dani -- Daniel Valverde Saubí Grup de Biologia Molecular de Llevats Facultat de Veterinària de la Universitat Autònoma de Barcelona Edifici V, Campus UAB 08193 Cerdanyola del Vallès- SPAIN Centro de Investigación Biomédica en Red en Bioingeniería, Biomateriales y Nanomedicina (CIBER-BBN) Grup d'Aplicacions Biomèdiques de la RMN Facultat de Biociències Universitat Autònoma de Barcelona Edifici Cs, Campus UAB 08193 Cerdanyola del Vallès- SPAIN +34 93 5814126 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Tidying up code - Warning message: deparse may be incomplete
Dear R users,
I am currently writing a R package and to do so I am following the
guidelines in manual 'Writing R extensions'.
In Section 3.1, it is suggested to tidy up the code using a file
containing the following:
options(keep.source = FALSE)
source(myfuns..R)
dump(ls(all = TRUE), file = new.myfuns.R)
I have done this for my own packages and although it runs, I get several
warnings of the type:
Warning message:
In dump(ls(all = TRUE), file = PermAlgo.R) : deparse may be incomplete
I am clueless as to what this means.
Even if I try to tidy only one function from my code, I get the warning.
E.g. the file lala.R contains only this:
partialHazards - function(t, v, covArray, betas){ exp( covArray[v,t,]
%*% betas ) }
the file tidylala.R contains:
options(keep.source = FALSE)
source(lala.R)
dump(ls(all = TRUE), file = newlala.R)
On Linux I run:
R --vanilla tidylala.R
Then I obtain:
Warning message:
In dump(ls(all = TRUE), file = newlala.R) : deparse may be incomplete
The file newlala.R looks like this:
`partialHazards` -
function (t, v, covArray, betas)
{
exp(covArray[v, t, ] %*% betas)
}
What does the warning mean? Can I simply ignore it?
thanks,
MP
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Re: [R] juxtaposed and stacked bars in one barplot?
On Thu, Aug 28, 2008 at 9:31 AM, Stefan Uhmann [EMAIL PROTECTED] wrote: hadley wickham schrieb: Hi Stefan, Could you be a bit more explicit? Do you have an example dataset that you are trying to visualise? Right, thanks for pointing out the obvious. So here's my code: library(gplots) quarter - as.factor(sample(c(Q1, Q2, Q3, Q4), 100, replace = TRUE)) year - as.factor(sample(c(seq(from=2000, to=2008)), 100, replace = TRUE)) category - as.factor(sample(c(seq(from=1, to=4)), 100, replace = TRUE)) test - data.frame(quarter, year, category) table(test$category, test$quarter, test$year) barplot2(table(test$quarter, test$year), beside=T, ylim=c(0,10), main=how to include dim3?) # inclusion of 3rd dimension does not work: barplot2(table(test$quarter, test$year, test$category), beside=T, ylim=c(0,10), main=how to include dim3?) I want the barplot to be exactly the same but with the bars stacked (by 'category'). I got the message from Gaspar, but have not yet tried to fit his example to my data. Hi Stefan, For your data, I'd suggest you consider using lines instead of bars. http://peltiertech.com/WordPress/2008/08/27/stacked-vs-clustered/ has some good reasons why. - but basically stacking makes it difficult to see how each group changes over time. It's pretty easy to play around different variations with ggplot2, so I've included a few you might want to look at. (Including your original request right at the bottom) test - data.frame(quarter, year, category) tabdf - as.data.frame(with(test, table(category, quarter, year))) install.packages(ggplot2) qplot(year, Freq, data=tabdf, geom=line, colour = category, facets = quarter ~ . , group = interaction(category, quarter)) # OR qplot(year, Freq, data=tabdf, geom=line, colour = quarter, facets = category ~ . , group = interaction(category, quarter)) # If you _really_ want stacking: qplot(year, Freq, data=tabdf, geom=area, fill = quarter, facets = category ~ . , group = interaction(category, quarter)) # OR qplot(year, Freq, data=tabdf, geom=bar, stat=identity, fill = quarter, facets = category ~ .) # OR finally, like what you originally asked for qplot(quarter, Freq, data=tabdf, geom=bar, stat=identity, fill = category, facets = . ~ year) Regards, Hadley -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Tidying up code - Warning message: deparse may be incomplete
On 28/08/2008 10:46 AM, Marie Pierre Sylvestre wrote:
Dear R users,
I am currently writing a R package and to do so I am following the
guidelines in manual 'Writing R extensions'.
In Section 3.1, it is suggested to tidy up the code using a file
containing the following:
options(keep.source = FALSE)
source(myfuns..R)
dump(ls(all = TRUE), file = new.myfuns.R)
I have done this for my own packages and although it runs, I get several
warnings of the type:
Warning message:
In dump(ls(all = TRUE), file = PermAlgo.R) : deparse may be incomplete
I am clueless as to what this means.
Even if I try to tidy only one function from my code, I get the warning.
E.g. the file lala.R contains only this:
partialHazards - function(t, v, covArray, betas){ exp( covArray[v,t,]
%*% betas ) }
the file tidylala.R contains:
options(keep.source = FALSE)
source(lala.R)
dump(ls(all = TRUE), file = newlala.R)
On Linux I run:
R --vanilla tidylala.R
Then I obtain:
Warning message:
In dump(ls(all = TRUE), file = newlala.R) : deparse may be incomplete
The file newlala.R looks like this:
`partialHazards` -
function (t, v, covArray, betas)
{
exp(covArray[v, t, ] %*% betas)
}
What does the warning mean? Can I simply ignore it?
The warning means that you may have lost some information, i.e. sourcing
newlala.R won't produce the same thing as sourcing lala.R. I think in
your case, there's no loss, so it's a bug, but I don't have time to
track down why you're getting it.
Duncan Murdoch
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Re: [R] Read a file
Is this what you want to do:
x - scan('clipboard', what=, sep=\n)
Read 18 items
x
[1] # File created = Thursday, August 28, 2008 3:33:02 PM GMT
[2] # Data set = 373 2 1 C:\\Bruker\\TOPSPIN GABRMN
[3] # Spectral Region:
[4] # LEFT = 4.5 ppm. RIGHT = 0.5 ppm.
[5] #
[6] # SIZE = 13111 ( = number of points)
[7] #
[8] # In the following ordering is from the 'left' to the 'right' limits!
[9] # Lines beginning with '#' must be considered as comment lines.
[10] #
[11] 1628.40625
[12] 1628.40625
[13] 1964.40625
[14] 2242.0625
[15] 2533.5
[16] 2937.90625
[17] 3448.0
[18] 3923.96875
On Thu, Aug 28, 2008 at 11:41 AM, Dani Valverde [EMAIL PROTECTED] wrote:
Hello,
I have a text file with this structure:
# File created = Thursday, August 28, 2008 3:33:02 PM GMT
# Data set = 373 2 1 C:\Bruker\TOPSPIN GABRMN
# Spectral Region:
# LEFT = 4.5 ppm. RIGHT = 0.5 ppm.
#
# SIZE = 13111 ( = number of points)
#
# In the following ordering is from the 'left' to the 'right' limits!
# Lines beginning with '#' must be considered as comment lines.
#
1628.40625
1628.40625
1964.40625
2242.0625
2533.5
2937.90625
3448.0
3923.96875
Is it possible to read it with R, something like scan() but keeping the
structure?
Best,
Dani
--
Daniel Valverde Saubí
Grup de Biologia Molecular de Llevats
Facultat de Veterinària de la Universitat Autònoma de Barcelona
Edifici V, Campus UAB
08193 Cerdanyola del Vallès- SPAIN
Centro de Investigación Biomédica en Red
en Bioingeniería, Biomateriales y
Nanomedicina (CIBER-BBN)
Grup d'Aplicacions Biomèdiques de la RMN
Facultat de Biociències
Universitat Autònoma de Barcelona
Edifici Cs, Campus UAB
08193 Cerdanyola del Vallès- SPAIN
+34 93 5814126
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+1 513 646 9390
What is the problem that you are trying to solve?
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Re: [R] Read a file
Try this:
x - readLines('clipboard')
newx - as.numeric(x[setdiff(seq(length(x)), grep(^#, x))])
comment(newx) - grep(^#, x, value = TRUE)
newx
comment(newx)
On Thu, Aug 28, 2008 at 12:41 PM, Dani Valverde [EMAIL PROTECTED]wrote:
Hello,
I have a text file with this structure:
# File created = Thursday, August 28, 2008 3:33:02 PM GMT
# Data set = 373 2 1 C:\Bruker\TOPSPIN GABRMN
# Spectral Region:
# LEFT = 4.5 ppm. RIGHT = 0.5 ppm.
#
# SIZE = 13111 ( = number of points)
#
# In the following ordering is from the 'left' to the 'right' limits!
# Lines beginning with '#' must be considered as comment lines.
#
1628.40625
1628.40625
1964.40625
2242.0625
2533.5
2937.90625
3448.0
3923.96875
Is it possible to read it with R, something like scan() but keeping the
structure?
Best,
Dani
--
Daniel Valverde Saubí
Grup de Biologia Molecular de Llevats
Facultat de Veterinària de la Universitat Autònoma de Barcelona
Edifici V, Campus UAB
08193 Cerdanyola del Vallès- SPAIN
Centro de Investigación Biomédica en Red
en Bioingeniería, Biomateriales y
Nanomedicina (CIBER-BBN)
Grup d'Aplicacions Biomèdiques de la RMN
Facultat de Biociències
Universitat Autònoma de Barcelona
Edifici Cs, Campus UAB
08193 Cerdanyola del Vallès- SPAIN
+34 93 5814126
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--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
[[alternative HTML version deleted]]
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Re: [R] Maintaining repeated ID numbers when transposing with reshape
Not the prettiest code but it returns what you want. Might be slow for
large dataframes.
df - data.frame( ID=c(1,1,1,1,2,2),
TEST=c(A,A,B,C,B,B),
RESULT=c(17,12,15,12,8,9) )
big.out - list(NULL)
for( uID in unique(df$ID) ){
m - df[ df$ID == uID, , drop=FALSE ]
run.order - unlist(sapply( table(m$TEST), function(x) if(x 0) 1:x) )
m - cbind( m, run.order=run.order )
nr - max(run.order)
out - matrix( nr=nr, nc=nlevels(m$TEST),
dimnames=list( rep(uID, nr), levels(m$TEST) ))
for(i in 1:nrow(m)) out[ m$run.order[i], m$TEST[i] ] - m$RESULT[i]
big.out[[uID]] - out
}
do.call( rbind, big.out )
A B C
1 17 15 12
1 12 NA NA
2 NA 8 NA
2 NA 9 NA
Regards, Adai
jcarmichael wrote:
Thank you for your suggestion, I will play around with it. I guess my concern
is that I need each test result to occupy its own cell rather than have
one or more in the same row.
Adaikalavan Ramasamy-2 wrote:
There might be a more elegant way of doing this but here is a way of
doing it without reshape().
df - data.frame( ID=c(1,1,1,1,2,2),
TEST=c(A,A,B,C,B,B),
RESULT=c(17,12,15,12,8,9) )
df.s - split( df, df$ID )
out - sapply( df.s, function(m)
tapply( m$RESULT, m$TEST, paste, collapse=, ) )
t(out)
A B C
1 17,12 15 12
2 NA 8,9 NA
Not the same output as you wanted. This makes more sense unless you have
a reason to priotize 17 instead of 12 in the first row.
Regards, Adai
jcarmichael wrote:
I have a dataset in long format that looks something like this:
ID TESTRESULT
1 A 17
1 A 12
1 B 15
1 C 12
2 B 8
2 B 9
Now what I would like to do is transpose it like so:
IDTEST ATEST BTEST C
1 17 15 12
1 12..
2 . 8.
2 . 9.
When I try:
reshape(mydata, v.names=result, idvar=id,timevar=test,
direction=wide)
It gives me only the first occurrence of each test for each subject. How
can I transpose my dataset in this way without losing information about
repeated tests?
Any help or guidance would be appreciated! Thanks!
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Re: [R] Cluster
Try reading help(hclust) and help(matplot) and run the examples given in the documentation. If that doesn't work, try posting again with a simple reproducible example. Regards, Adai Marco Chiapello wrote: Hi all, I'm trying to do a cluster analysis,but I don't know if it's possible in the way that I want. I have a data set like the follow: 115/114 116/114 117/114 0.45 0.72 0.41 1.16 0.63 0.91 0.42 0.94 0.61 My real data set is, just a bit bigger, 610 entries. I want plot each row on the same graph, like a line (see the attach file). Then if it's possible I want perform a cluster analysis. The final perfect result would be a graph with many lines, with the cluster line in the same color. Any advice? Marco __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] variance covariance matrix in GLM
this is for the person who asked me about prediction confidence intervals in a GLM because I lost your email. Below follows a simple example in CAR and the variance covariance of the beta coefficients is in the summary. So, I think, given that output, it should be pretty straightforward to do what you want. I also bet that code for what you want to do is in some function in John's effects package but I haven't looked at the code. If it's not possible to find it in John's code, ( I think prediction intervals have an extra term in the formula for the CI because of the error term variance ) that formula should be in any decent regression text book. Good luck. # attach(Mroz) model.mroz-glm(lfp ~ k5 + k618 + age + wc + hc + lwg + inc, family=binomial) print(model.mroz) modsumm-summary(model.mroz) print(modsumm) print(str(modsumm)) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Linear model with one known coordinate
I have a set of points (measurements) and I used lm() to obtain their linear regression model. From the biological background this line must pass through a point (100,0). Our dataset is not optimal and it shows a slight deviation from that coordinate. How can I add the restraint to the model, to go through that point? You can do that with offset(). There is an example in the manual page for glm. cu Philipp -- Dr. Philipp Pagel Lehrstuhl für Genomorientierte Bioinformatik Technische Universität München Wissenschaftszentrum Weihenstephan 85350 Freising, Germany http://mips.gsf.de/staff/pagel __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Notice
WHAT is this in reguard too? On Thu, Aug 28, 2008 at 12:56 PM, Ajay ohri [EMAIL PROTECTED] wrote: For the record, I* * ** *am not and have never been an employee of World Programming Ltd and that the postings and views expressed in these communities and forums have been motivated by my own personal thoughts and sentiments* ** In addition , I am not and never have been an employee of SAS Institute and R -Project. Grow up fellows. Ajay [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] svymeans question
Other people have explained that the issue is missing data. I just wanted to note that the reason for using only the complete cases on all variables is that svymeans() computes the covariance matrix of all the means, and this can't really be done sensibly when the means are based on different subsets. -thomas On Tue, 26 Aug 2008, Doran, Harold wrote: I have the following code which produces the output below it clus1 - svydesign(ids = ~schid, data = lower_dat) items - as.formula(paste( ~ , paste(lset, collapse= +))) rr1 - svymean(items, clus1, deff='replace', na.rm=TRUE) rr1 mean SE DEff W525209 0.719748 0.015606 2.4932 W525223 0.508228 0.027570 6.2802 W525035 0.827202 0.014060 2.8561 W525131 0.805421 0.015425 3.1350 W525033 0.242982 0.020074 4.5239 W525163 0.904647 0.013905 4.6289 W525165 0.439981 0.020029 3.3620 W525167 0.148112 0.013047 2.7860 W525177 0.865924 0.014977 3.9898 W525179 0.409003 0.020956 3.7515 W525181 0.634076 0.022076 4.3372 W525183 0.242498 0.019073 4.0894 W525401 0.262343 0.021830 3.4354 W525059 0.854792 0.016551 4.5576 W525251 0.691191 0.025010 6.0512 W525083 0.433204 0.017310 2.5200 W525289 0.634560 0.012762 1.4504 W524763 0.791868 0.014478 2.6265 W524765 0.223621 0.019627 4.5818 W524951 0.242982 0.016796 3.1669 W524769 0.820910 0.016786 3.9579 W524771 0.872701 0.015853 4.6712 W524839 0.518877 0.026433 5.7794 W525374 1.209584 0.043065 5.1572 W524885 0.585673 0.027780 6.5674 W525377 1.100678 0.050093 5.8851 W524787 0.839303 0.012994 2.5852 W524789 0.339787 0.019230 3.4041 W524791 0.847047 0.012885 2.6461 W524825 0.500968 0.021988 3.9935 W524795 0.868345 0.014951 4.0377 W524895 0.864472 0.013872 3.3917 W524897 0.804937 0.020070 5.2977 W524967 0.475799 0.032137 8.5511 W525009 0.681994 0.018670 3.3188 However, when I do the following: svymean(~W524787, clus1, deff='replace', na.rm=TRUE) mean SE DEff W524787 0.855547 0.011365 4.1158 Compare this to the value in the row 9 up from the bottom to see it is different. Computing the mean of the item by itself with svymeans agrees with the sample mean mean(lower_dat$W524787, na.rm=T) [1] 0.8555471 Now, I know that there is a covariance between the variables, but I was under the impression that the sample mean was still of pragmatic utility, but to account for sample design only the standard error is affected. In the work I am doing, it is important for the means of the items from svymeans to be the same as the sample mean when it is computed by itself. It's a bit of a story as to why, and I can provide that info if relevant. I don't see an argument in svydesign or in svymean that would allow for me to treat the variables as being independent. But, maybe I am missing something else and would welcome any reactions. Harold __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Thomas Lumley Assoc. Professor, Biostatistics [EMAIL PROTECTED] University of Washington, Seattle __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] sample consecutive integers efficiently
On Thu, 28 Aug 2008, Chris Oldmeadow wrote:
Hi all,
I have some rough code to sample consecutive integers with length according
to a vector of lengths
#sample space (representing positions)
pos-c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20)
#sample lengths
lengths-c(2,3,2)
From these two vectors I need a vector of sampled positions.
the sampling is without replacement, making things tough as the sampled
integers need to be consecutive. Im hoping somebody knows a faster way of
doing it than I have. ATM its way to slow on large vectors.
samplePos-function(l){
start.pos-sample(pos,1)
end.pos-start.pos+l-1
posies-start.pos:end.pos
posies
}
s.start-c()
newPos-function(a){
rp-samplePos(a)
#test sampled range is consecutive, if not resample
if (length(rp) != rp[a]+1 -rp[1]){rp-samplePos(a)}
pos-setdiff(pos,rp)
rp[1]
}
newps-c()
newps-unlist(lapply(lengths,newPos))
I think the bottleneck may be on the setdiff() function - the sample space is
quite large so I dont think there would be too many rejections.
The bottleneck is in the formulation of the sampling scheme.
This is a simple combinatorics problem. There are 3360 possible values (
prod(16:14) ) for the start positions of the three elements, and you can
form a bijection between 1:3360 and the individual samples. If the number
of possible sample is small enough, it would be most efficient to sample
from the corresponding integer vector and then translate it to the
corresponding sample. For larger values where the number of possible
samples become a challenge for 32-bit integer arithmetic, I expect this
approach would be preferred:
Permute length ( pos ) - sum ( lengths ) + length( lengths ) distinct
(consecutively labelled) elements:
elz - sample( length ( pos ) - sum ( lengths ) + length( lengths ) )
Take the lengths of the original objects to be
z.lens - rep( 1, length( elz ) )
z.lens[ seq(along = lengths ) ] - lengths
(i.e. objects longer than 1 appear first)
Determine the start positions of the objects as if they were laid down
consecutively according to the permutation:
start - head( cumsum( c(0, z.lens[ elz ]) ) + 1 , -1 )
Find the start positions of just those with lengths greater than 1
gt.1 - match( seq(along=lengths) , elz )
Report the start positions
start[ gt.1 ]
---
If length( pos ) is large, you can rewrite the above to simply sample
the positions (in the ordering) of the objects with lengths greater than
1. You will have to revise the calculation of start and gt.1 in that case.
HTH,
Chuck
Many thanks,
Chris
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Charles C. Berry(858) 534-2098
Dept of Family/Preventive Medicine
E mailto:[EMAIL PROTECTED] UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901
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[R] Function not returning a vector?
Why does:
(shape/scale) * (1:365/scale)^(shape - 1)
return a vector of numbers but calling a function
hasard(1:365,shape,scale)
defined like:
hazard - function(x,shape,scale)
{
return (shape/scale) * (x/scale)^(shape - 1)
}
Only return a single value? It is like x becomes a single value passed as an
argument.
Thank you.
Kevin
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Re: [R] Survey Design / Rake questions
On Mon, 25 Aug 2008, Farley, Robert wrote:
I see a number of things that bother me.
1) str(ByEBNum$StnTraveld) says int [1:12] 1 2 3 4 5 6 7 8 9 10 ...
Even though StnTraveld - c(as.factor(1:12))
You don't want the c()
a-as.factor(1:12)
str(a)
Factor w/ 12 levels 1,2,3,4,..: 1 2 3 4 5 6 7 8 9 10 ...
str(c(a))
int [1:12] 1 2 3 4 5 6 7 8 9 10 ...
As the help for c() says all attributes except names are removed.,
which includes the factor levels.
2) ByEBOn$StnName[1:5] seems to imply I have extra spaces in the data. Where
would they have come from?
No, that's just R printing things in columns
a-factor(1:12, labels=c(1:11,antidisestablishmentarianism))
a
[1] 12
[3] 34
[5] 56
[7] 78
[9] 910
[11] 11 antidisestablishmentarianism
Levels: 1 2 3 4 5 6 7 8 9 10 11 antidisestablishmentarianism
3) I'd like to verify that the order (value) of EBSurvey$lineon
matches my definition in StnName
all(levels(EBSurvey$lineon)==StnName)
-thomas
Thanks for helping...
***
***
library(survey)
SurveyData - read.spss(C:/Data/R/orange_delivery.sav, use.value.labels=TRUE,
max.value.labels=Inf, to.data.frame=TRUE)
#===
temp - sub(' +$', '', SurveyData$direction_)
SurveyData$direction_ - temp
#===
SurveyData$NumStn=abs(as.numeric(SurveyData$lineon)-as.numeric(SurveyData$lineoff))
mean(SurveyData$NumStn)
[1] 6.785276
### Kludge
SurveyData$NumStn - pmax(1,SurveyData$NumStn)
mean(SurveyData$NumStn)
[1] 6.789877
SurveyData$NumStn - as.factor(SurveyData$NumStn)
###
EBSurvey - subset(SurveyData, direction_ == EASTBOUND )
XTTable - xtabs(~direction_ , EBSurvey)
XTTable
direction_
EASTBOUND
345
WBSurvey - subset(SurveyData, direction_ == WESTBOUND )
XTTable - xtabs(~direction_ , WBSurvey)
XTTable
direction_
WESTBOUND
307
#
EBDesign - svydesign(id=~sampn, weights=~expwgt, data=EBSurvey)
# svytable(~lineon+lineoff, EBDesign)
StnName - c( Warner Center, De Soto, Pierce College, Tampa, Reseda, Balboa, Woodley, Sepulveda, Van
Nuys, Woodman, Valley College, Laurel Canyon, North Hollywood)
EBOnNewTots - c(1000, 600, 1200, 500,
1000, 500, 200, 250, 1000, 300, 100,
123.65,0 )
StnTraveld - c(as.factor(1:12))
EBNumStn- c(673.65, 800, 1000, 1000, 800, 700, 600, 500, 400, 200,
50, 50 )
ByEBOn - data.frame(StnName, Freq=EBOnNewTots)
ByEBNum - data.frame(StnTraveld, Freq=EBNumStn)
RakedEBSurvey - rake(EBDesign, list(~lineon, ~NumStn), list(ByEBOn, ByEBNum) )
Error in postStratify.survey.design(design, strata[[i]],
population.margins[[i]], :
Stratifying variables don't match
str(EBSurvey$lineon)
Factor w/ 13 levels Warner Center,..: 3 1 1 1 2 13 1 5 1 5 ...
EBSurvey$lineon[1:5]
[1] Pierce College Warner Center Warner Center Warner Center De Soto
13 Levels: Warner Center De Soto Pierce College Tampa Reseda Balboa ... North
Hollywood
str(ByEBOn$StnName)
Factor w/ 13 levels Balboa,De Soto,..: 11 2 5 8 6 1 12 7 10 13 ...
ByEBOn$StnName[1:5]
[1] Warner Center De SotoPierce College Tampa Reseda
13 Levels: Balboa De Soto Laurel Canyon North Hollywood ... Woodman
str(EBSurvey$NumStn)
Factor w/ 12 levels 1,2,3,4,..: 10 12 4 12 8 1 8 8 12 4 ...
EBSurvey$NumStn[1:5]
[1] 10 12 4 12 8
Levels: 1 2 3 4 5 6 7 8 9 10 11 12
str(ByEBNum$StnTraveld)
int [1:12] 1 2 3 4 5 6 7 8 9 10 ...
ByEBNum$StnTraveld[1:5]
[1] 1 2 3 4 5
Robert Farley
Metro
www.Metro.net
-Original Message-
From: Thomas Lumley [mailto:[EMAIL PROTECTED]
Sent: Saturday, August 23, 2008 09:38
To: Farley, Robert
Cc: r-help@r-project.org
Subject: Re: [R] Survey Design / Rake questions
On Fri, 22 Aug 2008, Farley, Robert wrote:
I *think* I'm making progress, but I'm still failing at the same step. My rake
call fails with:
Error in postStratify.survey.design(design, strata[[i]],
population.margins[[i]], :
Stratifying variables don't match
To my naïve eyes, it seems that my factors are in the wrong order. If so,
how do I assert an ordering in my survey dataframe, or copy an image from
the survey dataframe to my marginals dataframes? I'd prefer to pull the
original marginals dataframe(s) from the survey dataframe so that I can
automate that in production.
It looks like a problem with the NumStn factor. One copy has been converted to
Re: [R] Function not returning a vector?
Try
hazard - function(x,shape,scale)
{
return ((shape/scale) * (x/scale)^(shape - 1))
}
hazard(1:365,1,1)
--jeff
[EMAIL PROTECTED] wrote:
Why does:
(shape/scale) * (1:365/scale)^(shape - 1)
return a vector of numbers but calling a function
hasard(1:365,shape,scale)
defined like:
hazard - function(x,shape,scale)
{
return (shape/scale) * (x/scale)^(shape - 1)
}
Only return a single value? It is like x becomes a single value passed as an
argument.
Thank you.
Kevin
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Re: [R] Function not returning a vector?
2008/8/28 [EMAIL PROTECTED]:
hazard - function(x,shape,scale)
{
return (shape/scale) * (x/scale)^(shape - 1)
}
Only return a single value? It is like x becomes a single value passed as an
argument.
It's evaluating return(shape/scale) and returning that! Add some
extra parentheses:
return( (shape/scale) * (x/scale)^(shape-1) )
seems to work. Are you a python programmer, perhaps?
Barry
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Function not returning a vector?
[EMAIL PROTECTED] wrote:
Why does:
(shape/scale) * (1:365/scale)^(shape - 1)
return a vector of numbers but calling a function
hasard(1:365,shape,scale)
defined like:
hazard - function(x,shape,scale)
{
return (shape/scale) * (x/scale)^(shape - 1)
}
Only return a single value? It is like x becomes a single value passed as an
argument.
No, but return is syntactically like a function name. Watch your
parentheses
--
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
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Re: [R] Function not returning a vector?
rkevinburton at charter.net writes:
Why does:
(shape/scale) * (1:365/scale)^(shape - 1)
return a vector of numbers but calling a function
hasard(1:365,shape,scale)
defined like:
hazard - function(x,shape,scale)
{
return (shape/scale) * (x/scale)^(shape - 1)
}
Only return a single value? It is like x becomes a single value passed as an
argument.
I believe you have a couple of typos. Your function is returning shape/scale
only. Try:
hazard - function(x,shape,scale)
{
return ((shape/scale) * (x/scale)^(shape - 1))
}
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Re: [R] Function not returning a vector?
On Thu, 28 Aug 2008, [EMAIL PROTECTED] wrote:
Why does:
(shape/scale) * (1:365/scale)^(shape - 1)
return a vector of numbers but calling a function
hasard(1:365,shape,scale)
defined like:
hazard - function(x,shape,scale)
{
return (shape/scale) * (x/scale)^(shape - 1)
return() is a function, its argument is returned, everything afterwards is
ignored. And shape/scalar is probably a scalar in your case...
So either change to
return((shape/scale) * (x/scale)^(shape - 1))
or simply
(shape/scale) * (x/scale)^(shape - 1)
without explicitely calling return().
}
Only return a single value? It is like x becomes a single value passed as an
argument.
Thank you.
Kevin
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Re: [R] Function not returning a vector?
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
[EMAIL PROTECTED]
Sent: Thursday, August 28, 2008 11:36 AM
To: r-help@r-project.org
Subject: [R] Function not returning a vector?
Why does:
(shape/scale) * (1:365/scale)^(shape - 1)
return a vector of numbers but calling a function
hasard(1:365,shape,scale)
defined like:
hazard - function(x,shape,scale)
{
return (shape/scale) * (x/scale)^(shape - 1)
}
Only return a single value? It is like x becomes a single
value passed as an argument.
Thank you.
Kevin
Change your function definition to one of the following (I don't know which is
better).
hazard - function(x,shape,scale)
{
return( (shape/scale) * (x/scale)^(shape - 1) )
}
hazard - function(x,shape,scale)
{
(shape/scale) * (x/scale)^(shape - 1)
}
Hope this is helpful,
Dan
Daniel J. Nordlund
Washington State Department of Social and Health Services
Planning, Performance, and Accountability
Research and Data Analysis Division
Olympia, WA 98504-5204
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[R] Spider Graph
Is there an R function to generate a radar or spider graph from a table - e.g.radar(table(x)) or some such? == Isaac T. Van Patten, Ph.D. Professor Department of Criminal Justice Box 6934, Radford University Radford, VA 24142 540-831-6148 [EMAIL PROTECTED] mailto:[EMAIL PROTECTED] http://ivanpatt.asp.radford.edu http://ivanpatt.asp.radford.edu The hottest places in hell are reserved for those who, in time of great moral crisis, maintain their neutrality. -Dante Alighieri [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Spider Graph
help.search(radar) and help.search(spider) would both get you right to stars() Sarah On Thu, Aug 28, 2008 at 3:00 PM, Van Patten, Isaac T [EMAIL PROTECTED] wrote: Is there an R function to generate a radar or spider graph from a table - e.g.radar(table(x)) or some such? -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Notice
On Thu, Aug 28, 2008 at 10:42 AM, stephen sefick [EMAIL PROTECTED] wrote: WHAT is this in reguard too? It seems to be a message from another universe (SAS something) - maybe a wormhole? Are we not alone? /H On Thu, Aug 28, 2008 at 12:56 PM, Ajay ohri [EMAIL PROTECTED] wrote: For the record, I* * ** *am not and have never been an employee of World Programming Ltd and that the postings and views expressed in these communities and forums have been motivated by my own personal thoughts and sentiments* ** In addition , I am not and never have been an employee of SAS Institute and R -Project. Grow up fellows. Ajay [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] hex2RGB back to hex not the same?
Witness this oddity (to me): rainbow_hcl(10)[1] [1] #E18E9E d - attributes(hex2RGB(rainbow_hcl(10)))$coords[1,] rgb(d[1], d[2], d[3]) [1] #C54D5F What happened? FYI, this came up as I'm trying to reuse the RGB values I get from rainbow_hcl in a call to rgb() where I can also set alpha transparency levels ... -Aaron [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] error in packet 1
On Thu, Aug 28, 2008 at 3:07 AM, Giovanni Tarquinio
[EMAIL PROTECTED] wrote:
Hello,
I'm Giovanni from ROMA..
I can't find a solution for the error:
error using packet 1
the y field is not specified and it has not a default value
(this is my traslation from italian language)
The code is:
pc-qqmath(~valori,
distribution=function(p) qweibull(p,beta,alpha),
prepanel = prepanel.qqmathline,
panel = function(x, y) {
panel.grid()
panel.qqmathline(y, distribution = function(p)
qweibull(p,beta,alpha))
panel.qqmath(x, y)
},
layout = c( 1,1), aspect = 0.8,
xlab = Unit Weibull Quantile, ylab = D
)
Here's a hint:
args(panel.qqmath)
function (x, f.value = NULL, distribution = qnorm, qtype = 7,
groups = NULL, ...)
panel.qqmath() does not have a 'y' argument, which should tell you
that qqmath() does not pass a 'y' argument to its panel function.
-Deepayan
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[R] drop.unused.levels for two factors {lattice}
Hi, Is there any way to suppress plotting of panels that don't actually contain any information? I have tried using 'drop.unused.levels=TRUE', but there doesn't seem to be any effect. Here is an example: library(lattice) # some fake data: d - data.frame(x=runif(20), x.class=rep(letters[1:5], each=4), f1=rep(letters[1:2], each=10), f2=rep(letters[10:19], each=2) ) # plot it: dotplot(x.class ~ x | f1 + f2, data=d, scales=list(relation='free')) Thanks, Dylan -- Dylan Beaudette Soil Resource Laboratory http://casoilresource.lawr.ucdavis.edu/ University of California at Davis 530.754.7341 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Interaction between aggregate() and length()
Folks,
I've been running into an odd situation that occurs when I use length()
function with aggregate(), but not with either one separately. Together,
the results looks correct but is given an unexpected name. 'if
(stringsAsFactors) factor(x) else x' instead of just 'x'.
# Numbers work ok
tt - data.frame(idx=c(1,1,1,1,1,1,2,2,2,2,2,2)
,n=c(1,3,5,7,5,5,2,4,8,16,4,4)
,t=c(1,3,5,7,5,5,2,4,8,16,4,4)
,stringsAsFactors=FALSE)
aggregate(tt$t, list('idx'=tt$idx), length)
aggregate(as.factor(tt$t), list('idx'=tt$idx), length)
# Character data doesn't work right unless I convert the data to factors.
tt - data.frame(idx=c(1,1,1,1,1,1,2,2,2,2,2,2)
,n=c('1','3','5','7','5','5','2','4','8','16','4','4')
,t=c('1','3','5','7','5','5','2','4','8','16','4','4')
,stringsAsFactors=FALSE)
aggregate(tt$t, list('idx'=tt$idx), length)
aggregate(as.factor(tt$t), list('idx'=tt$idx), length)
Any idea what is going on here? For the record, this also happens with
the modalvalue() function defined at
http://wiki.r-project.org/rwiki/doku.php?id=tips:stats-basic:modalvalue
(which also relies on length() ).
As a side note, this began as an attempt to determine sample size, for
which I've defined a function count - function(x) { length(na.omit(x)) }.
No doubt there's a built in function to do just that, but as a newbie
I've yet to find it.
Thank you for your help,
cur
--
Curt Seeliger, Data Ranger
Raytheon Information Services - Contractor to ORD
[EMAIL PROTECTED]
541/754-4638
[[alternative HTML version deleted]]
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Re: [R] hex2RGB back to hex not the same?
On Thu, Aug 28, 2008 at 1:07 PM, Aaron Mackey [EMAIL PROTECTED] wrote: Witness this oddity (to me): rainbow_hcl(10)[1] [1] #E18E9E d - attributes(hex2RGB(rainbow_hcl(10)))$coords[1,] rgb(d[1], d[2], d[3]) [1] #C54D5F What happened? FYI, this came up as I'm trying to reuse the RGB values I get from rainbow_hcl in a call to rgb() where I can also set alpha transparency levels ... d - coords(hex2RGB(#E18E9E, gamma=NA)) rgb(d[,1], d[,2], d[,3]) [1] #E18E9E Another alternative is d - col2rgb(#E18E9E) / 255 rgb(d[1,], d[2,], d[3,]) [1] #E18E9E -Deepayan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] USING TOBIT OR WHAT ALTERNATIVE WHEN DATA ARE PANEL AND HETEROSKEDASTIC AND PROBABLY AUTOCORRELATED?
Please, I seek expertise and advice, possibly leads to R packages or stats literature. My data: measurements of economic variables for each county of California over 37 years. My dependent variable is square feet of office floor space permitted to be added in a county. Independent variables include for example change in number of office jobs in same county same year (and lagged years). Smaller (less populous) counties have many years in which there were no permits taken out; the largest counties had at least some permitted square footage each year. Among the set of years and places where permits were taken out, smaller counties tend to more permitted square footage per capita. I imagine the relationships are as follows: y* = desired change in floor space = X b + e , where X are independent variables and b their coefficients, and e is heteroskedastic (by state) and possibly autocorrelated noise. e is the sort of noise youâd expect on a time series of cross sections, no sampling (one observation for every county for every year studied). I must include fixed county effects (county dummies) in X because I will need them for county-specific forecasts later. y = permits taken out y = y* when y* 0 y = 0 when y* = 0 y never has been observed = 0 when population pop* How do folks recommend that I estimate this regression? I could first estimate the probability of having any permits given level of population and change in office jobs. That could include a deterministic component: if population pop* then y = y*. I could try a tobit-2 model or its ML estimator, which I see have just been developed in an R package called âsampleSelection.â But in that case, I have I would guess in my ignorance that Iâm way biased because it assumes specific error distributions and that they are homoskedastic. (I could transform the data in advance to get homoskedasticity if necessary). I could make an instrument for probability(y*0) and multiply that by each observation of permits, avoiding distributional assumptions in e (but not in prob(y*0)), but would that give me really high variance? (And is there an easier way to find variance than algebraically figuring out Newey West estimator for 2-stage method of moments procedures and how it applies here?) I know nothing of the non- and semi-parametric options here but does someone had an article or book chapter telling thatâs the right thing to do and how? It would be most convenient for me to use R but I also have access to STATA and SAS. Now another question for the statistically-minded: After running this regression I will forecast for each region how many square feet of office space will have permits taken out for it each year, given expected trends on office jobs and such. This does not allow each individual county to have a different type of response to office jobs, assuming the office job coefficient is pooled. Please be encouraged to comment on these options I am considering to allow more variation: County-specific coefficients donât work well; I tried separate (admittedly OLS) regressions for each county and found that with only 35 or so observations per county, my variances were too large and coefficients were insignificant and often of unintuitive signs. Random coefficients wonât give me county specific info, which Iâll need for the forecasts. So is this idea good? After I have coefficients from the pooled regression above, I take each coefficient b and its standard error. I use that as a stochastic restriction or Bayesian prior, for individual county regressions. That is, each county regression estimates its own b value, but subject to the stochastic restriction or Bayesian prior that b is in fact the pooled b, with the distribution of said prior being that bâs variance is the variance we estimated in the pooled regression. (Iâm thinking of what has been called Bayesian/Mixed Estimation here, but if Iâm out of the loop on newer better techniques, do tell.) Iâd think this county-specific estimation would be a simple non-tobit regression for large counties that never lack additions in any year. For small counties, I might need to do a tobit-style or instrumental variable regression again (or whatever you folks recommend). It might be harder to estimate probability of nonzero permits on the smaller sample size so I might have to keep the old estimate. All thoughts are welcome and appreciated. Thanks very very much. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] drop.unused.levels for two factors {lattice}
On Thu, Aug 28, 2008 at 1:21 PM, Dylan Beaudette [EMAIL PROTECTED] wrote: Hi, Is there any way to suppress plotting of panels that don't actually contain any information? I have tried using 'drop.unused.levels=TRUE', but there doesn't seem to be any effect. Here is an example: library(lattice) # some fake data: d - data.frame(x=runif(20), x.class=rep(letters[1:5], each=4), f1=rep(letters[1:2], each=10), f2=rep(letters[10:19], each=2) ) # plot it: dotplot(x.class ~ x | f1 + f2, data=d, scales=list(relation='free')) No, you can only drop factor levels that are unused (anywhere). Try dotplot(x.class ~ x | f1:f2, data=d, scales=list(relation='free')) You can use 'layout' and 'skip' to get to a layout similar to your original call, if you need to. -Deepayan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Interaction between aggregate() and length()
One option is use this:
aggregate(list(t=tt$t), list(idx=tt$idx), length)
On Thu, Aug 28, 2008 at 4:36 PM, [EMAIL PROTECTED] wrote:
Folks,
I've been running into an odd situation that occurs when I use length()
function with aggregate(), but not with either one separately. Together,
the results looks correct but is given an unexpected name. 'if
(stringsAsFactors) factor(x) else x' instead of just 'x'.
# Numbers work ok
tt - data.frame(idx=c(1,1,1,1,1,1,2,2,2,2,2,2)
,n=c(1,3,5,7,5,5,2,4,8,16,4,4)
,t=c(1,3,5,7,5,5,2,4,8,16,4,4)
,stringsAsFactors=FALSE)
aggregate(tt$t, list('idx'=tt$idx), length)
aggregate(as.factor(tt$t), list('idx'=tt$idx), length)
# Character data doesn't work right unless I convert the data to factors.
tt - data.frame(idx=c(1,1,1,1,1,1,2,2,2,2,2,2)
,n=c('1','3','5','7','5','5','2','4','8','16','4','4')
,t=c('1','3','5','7','5','5','2','4','8','16','4','4')
,stringsAsFactors=FALSE)
aggregate(tt$t, list('idx'=tt$idx), length)
aggregate(as.factor(tt$t), list('idx'=tt$idx), length)
Any idea what is going on here? For the record, this also happens with
the modalvalue() function defined at
http://wiki.r-project.org/rwiki/doku.php?id=tips:stats-basic:modalvalue
(which also relies on length() ).
As a side note, this began as an attempt to determine sample size, for
which I've defined a function count - function(x) { length(na.omit(x)) }.
No doubt there's a built in function to do just that, but as a newbie
I've yet to find it.
Thank you for your help,
cur
--
Curt Seeliger, Data Ranger
Raytheon Information Services - Contractor to ORD
[EMAIL PROTECTED]
541/754-4638
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
[[alternative HTML version deleted]]
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Re: [R] drop.unused.levels for two factors {lattice}
On Thursday 28 August 2008, Deepayan Sarkar wrote: On Thu, Aug 28, 2008 at 1:21 PM, Dylan Beaudette [EMAIL PROTECTED] wrote: Hi, Is there any way to suppress plotting of panels that don't actually contain any information? I have tried using 'drop.unused.levels=TRUE', but there doesn't seem to be any effect. Here is an example: library(lattice) # some fake data: d - data.frame(x=runif(20), x.class=rep(letters[1:5], each=4), f1=rep(letters[1:2], each=10), f2=rep(letters[10:19], each=2) ) # plot it: dotplot(x.class ~ x | f1 + f2, data=d, scales=list(relation='free')) No, you can only drop factor levels that are unused (anywhere). Try dotplot(x.class ~ x | f1:f2, data=d, scales=list(relation='free')) You can use 'layout' and 'skip' to get to a layout similar to your original call, if you need to. -Deepayan Thanks! that did the trick. Dylan -- Dylan Beaudette Soil Resource Laboratory http://casoilresource.lawr.ucdavis.edu/ University of California at Davis 530.754.7341 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Interaction between aggregate() and length()
That's a great work around, as I can eliminate renaming the results column
from 'x' to whatever. Thanks for the quick tip, Henrique.
On the other hand, I'm still stumped as to why aggregate() would name an
output column as 'if (stringsAsFactors) factor(x) else x'. That sort of
behaviour seems to contrdict the principle of least astonishment.
Enjoy your days,
cur
Henrique Dallazuanna [EMAIL PROTECTED] wrote on 08/28/2008 01:52:03 PM:
One option is use this:
aggregate(list(t=tt$t), list(idx=tt$idx), length)
On Thu, Aug 28, 2008 at 4:36 PM, [EMAIL PROTECTED] wrote:
Folks,
I've been running into an odd situation that occurs when I use length()
function with aggregate(), but not with either one separately. Together,
the results looks correct but is given an unexpected name. 'if
(stringsAsFactors) factor(x) else x' instead of just 'x'.
# Numbers work ok
tt - data.frame(idx=c(1,1,1,1,1,1,2,2,2,2,2,2)
,n=c(1,3,5,7,5,5,2,4,8,16,4,4)
,t=c(1,3,5,7,5,5,2,4,8,16,4,4)
,stringsAsFactors=FALSE)
aggregate(tt$t, list('idx'=tt$idx), length)
aggregate(as.factor(tt$t), list('idx'=tt$idx), length)
# Character data doesn't work right unless I convert the data to
factors.
tt - data.frame(idx=c(1,1,1,1,1,1,2,2,2,2,2,2)
,n=c('1','3','5','7','5','5','2','4','8','16','4','4')
,t=c('1','3','5','7','5','5','2','4','8','16','4','4')
,stringsAsFactors=FALSE)
aggregate(tt$t, list('idx'=tt$idx), length)
aggregate(as.factor(tt$t), list('idx'=tt$idx), length)
Any idea what is going on here? For the record, this also happens with
the modalvalue() function defined at
http://wiki.r-project.org/rwiki/doku.php?id=tips:stats-basic:modalvalue
(which also relies on length() ).
As a side note, this began as an attempt to determine sample size, for
which I've defined a function count - function(x) { length(na.omit(x))
}.
No doubt there's a built in function to do just that, but as a newbie
I've yet to find it.
Thank you for your help,
cur
--
Curt Seeliger, Data Ranger
Raytheon Information Services - Contractor to ORD
[EMAIL PROTECTED]
541/754-4638
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] RCurl: authentication when posting forms
Duncan,
Thank you for the examples. I had tried all of these different options
for authentication but had no luck. I was getting a 100 continue and
then a 401 unauthorized response. This morning the owners of the
server I was trying to access discovered a bug with their api when using
basic authorization. Evidently the code on their side was explicitly
checking that the request method was GET, hence why all of my POST
attempts were failing. They are in the process of fixing that code and
my guess is that when I am able to try again the RCurl post functions
will work just fine.
I did have one other RCurl question I'd like to ask you about the
parseHTTPHeader function.
The parser appears to parse on spaces, so when the error message is more
than one word (eg, not found), the message returned is not. I have
modified the parseHTTPHeader function so that it works for me. I may not
have done this in the most efficient way but at least you can see what I
was trying to do.
Below I have pasted my modification for the parseHTTPHeader function
which I am calling parseHeader. Please let me know if you think this is
a bug or if I am just using the function incorrectly.
Thank you for the help.
Valerie
#---
#Sample code:
reader - basicTextGatherer()
header - basicTextGatherer()
handle - getCurlHandle()
myopts - curlOptions( netrc=1, httpheader=c(Authorization=mypwd,
Accept=test/xml,
Accept=multipart/*,
'Content-Type'=text/xml; charset=utf-8),
postfields=body,
writefunction=reader$update, headerfunction=header$update,
ssl.verifyhost=FALSE,
ssl.verifypeer=FALSE, followlocation=TRUE)} else
curlPerform(url=myUrl, .opts=myopts, curl=handle)
h - parseHeader(header$value())
status - h$status
message - h$statusMessage
#
#Modified parse function:
parseHeader - function (lines)
{
if (length(lines) 1)
return(NULL)
if (length(lines) == 1)
lines = strsplit(lines, \r\n)[[1]]
status = lines[1]
lines = lines[-c(1, length(lines))]
lines = gsub(\r\n, , lines)
if (FALSE) {
header = lines[-1]
header - read.dcf(textConnection(header))
}
else {
els - sapply(lines, function(x) strsplit(x, :[ ]*))
header - lapply(els, function(x) x[2])
names(header) - sapply(els, function(x) x[1])
}
els - strsplit(status, )[[1]]
header[[status]] - as.integer(els[2])
# new code below
hstring - NULL
for(i in 3:length(els)) hstring - paste(hstring, ,els[i],sep=)
hstring - substr(hstring,2,nchar(hstring))
header[[statusMessage]] - hstring
header
}
Duncan Temple Lang wrote:
Hi Valerie
Valerie Obenchain wrote:
Hi,
Has anyone successfully used RCurl for posting data to a
password-protected site?
Yes. I just set up a sample form to test with and the following
all work
# Perl script (and HTML form for testing in the browser) taken from
# http://www.elated.com/articles/form-validation-with-perl-and-cgi/
# Provide the login password directly
postForm(http://www.omegahat.org/RCurl/testPassword/form_validation.cgi;,
your_name = Duncan,
your_age = 35-55,
your_sex = m,
submit = submit,
.opts = list(userpwd = bob:welcome))
# Get the login password in ~/.netrc
postForm(http://www.omegahat.org/RCurl/testPassword/form_validation.cgi;,
your_name = Duncan,
your_age = 35-55,
your_sex = m,
submit = submit,
.opts = list(netrc = TRUE))
# Get the login password from a different netrc file
postForm(http://www.omegahat.org/RCurl/testPassword/form_validation.cgi;,
your_name = Duncan,
your_age = 35-55,
your_sex = m,
submit = submit,
.opts = list(netrc = TRUE,
netrc.file =
/Users/duncan/Projects/org/omegahat/R/RCurl/inst/examples/omg.netrc))
So let me know what problems you are having and more details
about the OS, version of libcurl, and a sample URL to which
to post, etc.
D.
I
have tired using option netrc=1 with both postForm and curlPerform
(with postfields option) but can't authenticate.
I would happily provide more details if some one has had some
experience with this.
Thanks very much.
Valerie
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[R] Adjusting for initial status (intercept) in lme growth models
Hi everyone, I have a quick and probably easy question about lme for this list. Say, for instance you want to model growth in pituitary distance as a function of age in the Orthodont dataset. fm1 = lme(distance ~ I(age-8), random = ~ 1 + I(age-8) | Subject, data = Orthodont) You notice that there is substantial variability in the intercepts (initial distance) for people at 8 years, and that this variability in initial distance is related to growth over time: R# summary(fm1) ... Random effects: Formula: ~1 + I(age - 8) | Subject Structure: General positive-definite, Log-Cholesky parametrization StdDev Corr (Intercept) 1.8866 (Intr) I(age - 8) 0.2264 0.209 Residual1.3100 Now 2 questions: 1. With lme, how can you get a fit of the growth model accounting for the relationship between initial status (intercept) and growth? Some texts call this latent variable regression or something or other, which seems to basically boil down to adding the random effects intercept as a predictor in the growth model. Is this done in lme by simply adding the intercept results from ranef(fm1) to the model? This two-step process seems wrong to me for some reason, perhaps because it seems too simple. Anyone know the proper way to do in lme? 2. In addition, suppose you see that there are significant differences in initial status by Sex: fm2 = lme(distance ~ I(age-8) + Sex, random = ~ 1 + I(age-8) | Subject, data = Orthodont) R# summary(fm2) Fixed effects: distance ~ I(age - 8) + Sex Value Std.Error DF t-value p-value (Intercept) 22.9170.5134 80 44.64 0.000 I(age - 8) 0.6600.0713 809.27 0.000 SexFemale -2.1450.7575 25 -2.83 0.009 Along the lines of question #1, how would you get a growth model adjusting for these Sex differences in initial status? I am looking for something similar to adjusting for baseline differences between Sexes in ANCOVA. I know Lord would not approve, but this is just by way of example... Thanks so much for your help, and this wonderful program Dr. Bates. - DC [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] abline of an lm fit not correct
mac osx 10.5.4 R 2.7.1 I have fit a model d-lm(y~x) with an R^2 of 0.963 but when I issue the command abline(d) the line is below where it ought to be. Looks like the right slope, but not the right intercept. thanks -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] abline of an lm fit not correct
sorry idiotic question- you have to make sure you are using the right things before you start ploting On Thu, Aug 28, 2008 at 7:54 PM, stephen sefick [EMAIL PROTECTED] wrote: mac osx 10.5.4 R 2.7.1 I have fit a model d-lm(y~x) with an R^2 of 0.963 but when I issue the command abline(d) the line is below where it ought to be. Looks like the right slope, but not the right intercept. thanks -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] sample consecutive integers efficiently
Charles C. Berry wrote:
On Thu, 28 Aug 2008, Chris Oldmeadow wrote:
Hi all,
I have some rough code to sample consecutive integers with length
according to a vector of lengths
#sample space (representing positions)
pos-c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20)
#sample lengths
lengths-c(2,3,2)
From these two vectors I need a vector of sampled positions.
the sampling is without replacement, making things tough as the
sampled integers need to be consecutive. Im hoping somebody knows a
faster way of doing it than I have. ATM its way to slow on large
vectors.
samplePos-function(l){
start.pos-sample(pos,1)
end.pos-start.pos+l-1
posies-start.pos:end.pos
posies
}
s.start-c()
newPos-function(a){
rp-samplePos(a)
#test sampled range is consecutive, if not resample
if (length(rp) != rp[a]+1 -rp[1]){rp-samplePos(a)}
pos-setdiff(pos,rp)
rp[1]
}
newps-c()
newps-unlist(lapply(lengths,newPos))
I think the bottleneck may be on the setdiff() function - the sample
space is quite large so I dont think there would be too many rejections.
The bottleneck is in the formulation of the sampling scheme.
This is a simple combinatorics problem. There are 3360 possible values
( prod(16:14) ) for the start positions of the three elements, and you
can form a bijection between 1:3360 and the individual samples. If the
number of possible sample is small enough, it would be most efficient
to sample from the corresponding integer vector and then translate it
to the corresponding sample. For larger values where the number of
possible samples become a challenge for 32-bit integer arithmetic, I
expect this approach would be preferred:
Permute length ( pos ) - sum ( lengths ) + length( lengths ) distinct
(consecutively labelled) elements:
elz - sample( length ( pos ) - sum ( lengths ) + length( lengths ) )
Take the lengths of the original objects to be
z.lens - rep( 1, length( elz ) )
z.lens[ seq(along = lengths ) ] - lengths
(i.e. objects longer than 1 appear first)
Determine the start positions of the objects as if they were laid down
consecutively according to the permutation:
start - head( cumsum( c(0, z.lens[ elz ]) ) + 1 , -1 )
Find the start positions of just those with lengths greater than 1
gt.1 - match( seq(along=lengths) , elz )
Report the start positions
start[ gt.1 ]
---
If length( pos ) is large, you can rewrite the above to simply sample
the positions (in the ordering) of the objects with lengths greater
than 1. You will have to revise the calculation of start and gt.1 in
that case.
HTH,
Chuck
Many thanks,
Chris
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Charles C. Berry(858) 534-2098
Dept of Family/Preventive
Medicine
E mailto:[EMAIL PROTECTED]UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego
92093-0901
Thanks! that worked perfectly.
Chris
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Re: [R] RCurl: authentication when posting forms
Valerie Obenchain wrote:
Duncan,
Thank you for the examples. I had tried all of these different options
for authentication but had no luck. I was getting a 100 continue and
then a 401 unauthorized response. This morning the owners of the
server I was trying to access discovered a bug with their api when using
basic authorization. Evidently the code on their side was explicitly
checking that the request method was GET, hence why all of my POST
attempts were failing. They are in the process of fixing that code and
my guess is that when I am able to try again the RCurl post functions
will work just fine.
Thanks for the update. Good to know that there isn't a problem
with the libcurl/RCurl code.
I did have one other RCurl question I'd like to ask you about the
parseHTTPHeader function.
The parser appears to parse on spaces, so when the error message is more
than one word (eg, not found), the message returned is not. I have
modified the parseHTTPHeader function so that it works for me. I may not
have done this in the most efficient way but at least you can see what I
was trying to do.
Yes, this is a problem. Thanks for bringing it to my attention.
I did get your mail directly to me from about 2 weeks ago
and I replied. Perhaps my reply got eaten by your filters.
In it, I said that it was a bug, and that it would be
very useful if you could send me an RCurl command
that illustrated the error so that I could add it to the RCurl
tests.
The only modification that can make your patch slightly
better is that the words in the statusMessage are essentially
all the words in the status variable, less the first two
(the status number and the protocol name). So
header[[statusMessage]] - paste(els[-c(1,2)], collapse = )
is more convenient than
hstring - NULL
for(i in 3:length(els)) hstring - paste(hstring, ,els[i],sep=)
hstring - substr(hstring,2,nchar(hstring))
header[[statusMessage]] - hstring
If you do have a URL that I can use to test the error handling code, I'd
appreciate
if you could send it to me.
D.
Below I have pasted my modification for the parseHTTPHeader function
which I am calling parseHeader. Please let me know if you think this is
a bug or if I am just using the function incorrectly.
Thank you for the help.
Valerie
#---
#Sample code:
reader - basicTextGatherer()
header - basicTextGatherer()
handle - getCurlHandle()
myopts - curlOptions( netrc=1, httpheader=c(Authorization=mypwd,
Accept=test/xml,
Accept=multipart/*,
'Content-Type'=text/xml; charset=utf-8),
postfields=body,
writefunction=reader$update, headerfunction=header$update,
ssl.verifyhost=FALSE,
ssl.verifypeer=FALSE, followlocation=TRUE)} else
curlPerform(url=myUrl, .opts=myopts, curl=handle)
h - parseHeader(header$value())
status - h$status
message - h$statusMessage
#
#Modified parse function:
parseHeader - function (lines)
{
if (length(lines) 1)
return(NULL)
if (length(lines) == 1)
lines = strsplit(lines, \r\n)[[1]]
status = lines[1]
lines = lines[-c(1, length(lines))]
lines = gsub(\r\n, , lines)
if (FALSE) {
header = lines[-1]
header - read.dcf(textConnection(header))
}
else {
els - sapply(lines, function(x) strsplit(x, :[ ]*))
header - lapply(els, function(x) x[2])
names(header) - sapply(els, function(x) x[1])
}
els - strsplit(status, )[[1]]
header[[status]] - as.integer(els[2])
# new code below
hstring - NULL
for(i in 3:length(els)) hstring - paste(hstring, ,els[i],sep=)
hstring - substr(hstring,2,nchar(hstring))
header[[statusMessage]] - hstring
header
}
Duncan Temple Lang wrote:
Hi Valerie
Valerie Obenchain wrote:
Hi,
Has anyone successfully used RCurl for posting data to a
password-protected site?
Yes. I just set up a sample form to test with and the following
all work
# Perl script (and HTML form for testing in the browser) taken from
# http://www.elated.com/articles/form-validation-with-perl-and-cgi/
# Provide the login password directly
postForm(http://www.omegahat.org/RCurl/testPassword/form_validation.cgi;,
your_name = Duncan,
your_age = 35-55,
your_sex = m,
submit = submit,
.opts = list(userpwd = bob:welcome))
# Get the login password in ~/.netrc
postForm(http://www.omegahat.org/RCurl/testPassword/form_validation.cgi;,
your_name = Duncan,
your_age = 35-55,
your_sex = m,
submit = submit,
.opts = list(netrc = TRUE))
# Get the login password from a different netrc file
[R] Newbie: Examples on functions callling a library etc.
Hello R is pretty new to me. I need to write a function that returns three matrices of different dimensions. In addition, I need to call a function from a contributed package with the function. I have browsed several manuals and docs but the examples on them are either very simple or extremely hard to follow. Many thanks Ed [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Survey Design / Rake questions
I'm feeling like I just don't get it. My attempt at rake now fails
with:
Error in postStratify.survey.design(design, strata[[i]],
population.margins[[i]], :
Stratifying variables don't match
The factors in the data frame looks fine. Should I have the same
structure in the design?
str(EBDesign$lineon)
NULL
str(EBSurvey$lineon)
Factor w/ 13 levels Warner Center,..: 3 1 1 1 2 13 1 5 1 5 ...
str(ByEBOn$StnName)
Factor w/ 13 levels Balboa,De Soto,..: 11 2 5 8 6 1 12 7 10 13 ...
all(levels(EBSurvey$lineon)==StnName)
[1] TRUE
#
str(EBDesign$NumStn)
NULL
str(EBSurvey$NumStn)
Factor w/ 12 levels 1,2,3,4,..: 10 12 4 12 8 1 8 8 12 4 ...
str(ByEBNum$StnTraveld)
Factor w/ 12 levels 1,2,3,4,..: 1 2 3 4 5 6 7 8 9 10 ...
all(levels(EBSurvey$NumStn)==StnTraveld)
[1] TRUE
A complete listing is below:
**
**
**
sessionInfo()# List loaded packages
R version 2.7.2 (2008-08-25)
i386-pc-mingw32
locale:
LC_COLLATE=English_United States.1252;LC_CTYPE=English_United
States.1252;LC_MONETARY=English_United
States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252
attached base packages:
[1] graphics grDevices utils datasets stats methods base
other attached packages:
[1] survey_3.8-1 fortunes_1.3-5 moonsun_0.1prettyR_1.3-2
foreign_0.8-29
SurveyData - read.spss(C:/Data/R/orange_delivery.sav,
use.value.labels=TRUE, max.value.labels=Inf, to.data.frame=TRUE)
#===
temp - sub(' +$', '', SurveyData$direction_)
SurveyData$direction_ - temp
#===
# Calc. # stations traversed from StnOn/StnOff
SurveyData$NumStn=abs(as.numeric(SurveyData$lineon)-as.numeric(SurveyDat
a$lineoff))
Kludge
mean(SurveyData$NumStn)
[1] 6.785276
SurveyData$NumStn - pmax(1,SurveyData$NumStn)
mean(SurveyData$NumStn)
[1] 6.789877
SurveyData$NumStn - as.factor(SurveyData$NumStn)
#===
# Adjust one direction at a time. Start W/ EB {learn subsetting
later}
EBSurvey - subset(SurveyData, direction_ == EASTBOUND )
EBDesign - svydesign(id=~sampn, weights=~expwgt, data=EBSurvey)
#===
# New Marignals {start w/ 2 dimensions: StnOn X Distance}
StnName - as.factor(c( Warner Center, De Soto, Pierce College,
Tampa, Reseda, Balboa, Woodley, Sepulveda, Van Nuys,
Woodman, Valley College, Laurel Canyon, North Hollywood))
EBOnNewTots - c(1000, 600, 1200,
500, 1000, 500, 200, 250, 1000, 300,
100, 123.65,0 )
ByEBOn - data.frame(StnName, Freq=EBOnNewTots)
#
StnTraveld - as.factor(1:12)
EBNumStn - c(673.65, 800, 1000, 1000, 800, 700, 600, 500,
400, 200, 50, 50 )
ByEBNum- data.frame(StnTraveld, Freq=EBNumStn)
#
RakedEBSurvey - rake(EBDesign, list(~lineon, ~NumStn), list(ByEBOn,
ByEBNum) )
Error in postStratify.survey.design(design, strata[[i]],
population.margins[[i]], :
Stratifying variables don't match
#
str(EBDesign$lineon)
NULL
str(EBSurvey$lineon)
Factor w/ 13 levels Warner Center,..: 3 1 1 1 2 13 1 5 1 5 ...
str(ByEBOn$StnName)
Factor w/ 13 levels Balboa,De Soto,..: 11 2 5 8 6 1 12 7 10 13 ...
all(levels(EBSurvey$lineon)==StnName)
[1] TRUE
#
str(EBDesign$NumStn)
NULL
str(EBSurvey$NumStn)
Factor w/ 12 levels 1,2,3,4,..: 10 12 4 12 8 1 8 8 12 4 ...
str(ByEBNum$StnTraveld)
Factor w/ 12 levels 1,2,3,4,..: 1 2 3 4 5 6 7 8 9 10 ...
all(levels(EBSurvey$NumStn)==StnTraveld)
[1] TRUE
#
**
**
**
Robert Farley
Metro
www.Metro.net
-Original Message-
From: Thomas Lumley [mailto:[EMAIL PROTECTED]
Sent: Thursday, August 28, 2008 11:43
To: Farley, Robert
Cc: r-help@r-project.org
Subject: Re: [R] Survey Design / Rake questions
On Mon, 25 Aug 2008, Farley, Robert wrote:
I see a number of things that bother me.
1) str(ByEBNum$StnTraveld) says int [1:12] 1 2 3 4 5 6 7 8 9 10 ...
Even though StnTraveld - c(as.factor(1:12))
You don't want the c()
a-as.factor(1:12)
str(a)
Factor w/ 12 levels 1,2,3,4,..: 1 2 3 4 5 6 7 8 9 10 ...
str(c(a))
int [1:12] 1 2 3 4 5 6 7 8 9 10 ...
As the help for c() says all attributes except names are removed.,
which includes the factor levels.
2) ByEBOn$StnName[1:5] seems to imply I have extra spaces in the
data. Where would they have come from?
No, that's just R printing things in columns
a-factor(1:12,
[R] lost attrubute:names
Hi, when I pick out one element from a matrix, the attribute name is kept, but when more than one elements are extracted, the attribute name lost; a-matrix(c(1,2,3,11,12,13,45,56,76),ncol=3,dimnames=list(c(),c(c1,c2,c3))) k-a[a[,c3]50,c3] kk-a[a[,c3]60,c3] attributes(k) NULL attributes(kk) $names [1] c3 YU [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lost attrubute:names
On Thu, 28 Aug 2008, Yuan Jian wrote: Hi, when I pick out one element from a matrix, the attribute name is kept, but when more than one elements are extracted, the attribute name lost; To what attribute 'name' do you refer? I only see 'dim' and 'dimnames' attributes: attributes(a) $dim [1] 3 3 $dimnames $dimnames[[1]] NULL $dimnames[[2]] [1] c1 c2 c3 ? a-matrix(c(1,2,3,11,12,13,45,56,76),ncol=3,dimnames=list(c(),c(c1,c2,c3))) k-a[a[,c3]50,c3] kk-a[a[,c3]60,c3] attributes(k) NULL attributes(kk) $names [1] c3 ? Try attributes(a[a[,c3]50,c3, drop = FALSE ] ). Try adding rownames to 'a', and see what happens to the attributes of each of the above forms. [ merely tries to do something reasonable. Why do you think k should have any attributes, and what names do you think k should have??? HTH, Chuck ? YU [[alternative HTML version deleted]] Charles C. Berry(858) 534-2098 Dept of Family/Preventive Medicine E mailto:[EMAIL PROTECTED] UC San Diego http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Newbie: Examples on functions callling a library etc.
Eduardo M. A. M.Mendes emammendes at gmail.com writes:
R is pretty new to me. I need to write a function that returns three
matrices of different dimensions. In addition, I need to call a function
from a contributed package with the function. I have browsed several
manuals and docs but the examples on them are either very simple or
extremely hard to follow.
Many thanks
Ed
I think you need to try to specify your needs a little bit
more carefully. Here is a function that technically meets
your needs:
library(example_pkg) ## to load the contributed package
myfunction - function() { ## function with no arguments
foo()## assuming the function foo is in the package
list(matrix(nrow=2,ncol=2),matrix(nrow=3,ncol=3),matrix(nrow=4,ncol=4)
}
But I suspect that doesn't really do what you need ...
Ben Bolker
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Re: [R] Newbie: Examples on functions callling a library etc.
Hi Ed,
Here's a simple example showing your needs:
myfun - function(n1, n2, n3) {
mat1 - matrix(rep(1), nrow = n1, ncol = 3)
mat2 - matrix(rep(2), nrow = n2, ncol = 4)
mat3 - matrix(rep(3), nrow = n3, ncol = 5)
require(survival) ## make sure the package you need is loaded
mypkgfun - survival::is.Surv ## Use the '::' and ':::' extractors to get
visible and hidden functions respectively from the package
list(mat1 = mat1, mat2 = mat2, mat3 = mat3, mypkgfun = mypkgfun) ## Return
the items in a list
}
## Now invoke the function
foo - myfun(n1 = 1, n2 = 1, n3 = 5)
## and look at the returned results
foo
myfun - function(n1, n2, n3) {
+ mat1 - matrix(rep(1), nrow = n1, ncol = 3)
+ mat2 - matrix(rep(2), nrow = n2, ncol = 4)
+ mat3 - matrix(rep(3), nrow = n3, ncol = 5)
+
+ require(survival)
+ mypkgfun - survival::is.Surv
+
+ list(mat1 = mat1, mat2 = mat2, mat3 = mat3, mypkgfun = mypkgfun)
+ }
foo - myfun(n1 = 1, n2 = 1, n3 = 5)
foo
$mat1
[,1] [,2] [,3]
[1,]111
$mat2
[,1] [,2] [,3] [,4]
[1,]2222
$mat3
[,1] [,2] [,3] [,4] [,5]
[1,]33333
[2,]33333
[3,]33333
[4,]33333
[5,]33333
$mypkgfun
function (x)
inherits(x, Surv)
environment: namespace:survival
HTH
Steve McKinney
-Original Message-
From: [EMAIL PROTECTED] on behalf of Eduardo M. A. M.Mendes
Sent: Thu 8/28/2008 5:43 PM
To: r-help@r-project.org
Subject: [R] Newbie: Examples on functions callling a library etc.
Hello
R is pretty new to me. I need to write a function that returns three
matrices of different dimensions. In addition, I need to call a function
from a contributed package with the function. I have browsed several
manuals and docs but the examples on them are either very simple or
extremely hard to follow.
Many thanks
Ed
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] extract variance components
HI,
I would like to extract the variance components estimation in lme function
like
a.fit-lme(distance~age, data=aaa, random=~day/subject)
There should be three variances \sigma_day, \sigma_{day %in% subject } and
\sigma_e.
I can extract the \sigma_e using something like a.fit$var. However, I cannot
manage to extract the first two variance components. I can only see the
results in summary(a.fit).
I have some problem in the lme4 package and hence use the nlme package. The
example data also has some problem so I just list the function here using
some imaginary data set. Thank you.
Huang
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