Re: [R] curvedarrow (some graphics problem)
Hi there, This is now the code % library(grid) vp - viewport( x = unit(0, npc), y = unit(0, npc), just = c(left, bottom), xscale = c(-1, 1) , yscale = c(-1, 1)) vp2 - viewport( # probably not needed but I had trouble placing the xaxis x = unit(0,npc), y = unit(0.5, npc), just = c(left, bottom), xscale = c(-1, 1) , yscale = c(-1, 1)) pushViewport(vp) grid.xaxis(at=seq(-1, 1, by=0.2), label=FALSE, vp=vp2) grid.points(x=0.3, y=0.5, gp=gpar(col=NA), default.units = npc, name=h1, vp=vp) grid.points(x=0.7, y=0.5, gp=gpar(col=NA), default.units = npc, name=h2, vp=vp) grid.text(s, x=0.3, y=unit(-1, lines), vp=vp2) grid.text(t, 0.7, unit(-1, lines), gp=gpar(col=red), vp=vp2) grid.text(mu(s,t), 0.5, unit(5, lines), vp=vp2) grid.curve(grobX(h2, 180), grobY(h2, 180), grobX(h1, 180), grobY(h1, 180), shape=1, ncp=10, square=FALSE, curvature=.4, arrow=arrow(angle=20,ends=first) ) upViewport() % for this line: grid.text(mu(s,t), 0.5, unit(5, lines), vp=vp2) how do I change mu(s,t) to the expression mu? I have tried grid.text(expression=mu(s,t), 0.5, unit(5, lines), vp=vp2) grid.text(expression(mu(s,t)), 0.5, unit(5, lines), vp=vp2) both didnt work AND when I tried to save the output as PDF file it returns: Error in function (name) : Grob 'h2' not found I have no idea what to do? -- View this message in context: http://www.nabble.com/curvedarrow-%28some-graphics-problem%29-tp24158796p24196544.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] List subsetting
Ivo Shterev wrote:
I have a question about list indexing. Lets say we have a list of 3 lists,
each containing 3 different type elements:
(Details of your nice example code removed)
a=replicate(3, list(list(c(1,1,1), diag(3), c(2,2,2
str(a) # I prefer this to print(a) because it shows the structure better
# in steps
b = lapply(a,function(x){
# print(str(a)) # put this here to know what you are doing
x[2:3]
}
)
#... mmm, not yet what you wanted. Another try
b[1:2]
# I leave it as a easy exercise to jam this into a one-liner
---
Dieter
--
View this message in context:
http://www.nabble.com/List-subsetting-tp24194764p24198222.html
Sent from the R help mailing list archive at Nabble.com.
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Re: [R] adf.test Vs ADF.test...
Hi, R users, I'm using R to test the unit root for my time series data. I just compared the ADF.test in uroot package and the adf.test in tseries package. It seems it is difficult to define the time trend and intercept in adf.test. But it is easy to do these in ADF.test. ADF.test also help you find the number of lags that need to be included in the model to remove the serial correlation. I do not not see adf.test be able to do this too. You need to define the number of lags for adf.test. I'm a new R user and I could be wrong. I'll appreciate it very much if someone familiar with time series in R can give me some comments and suggestions. Thanks in advance. Harry From: dongh...@hotmail.com To: patrick.richard...@vai.org; r-help@r-project.org Date: Wed, 24 Jun 2009 19:22:46 + Subject: Re: [R] Why can't I use ADF.test? Thanks! I tried these, but I got the following messages:Warning message:In getDependencies(pkgs, dependencies, available, lib) : package ââ¬Ëurootââ¬â¢ is not available Error in library(uroot) : there is no package called 'uroot' I download a package called uroot and put it into: C:\Program Files\R\R-2.9.0\library. But still got the same error. Any further suggestions? Hongwei From: patrick.richard...@vai.org To: dongh...@hotmail.com; r-help@r-project.org Date: Wed, 24 Jun 2009 15:12:22 -0400 Subject: RE: [R] Why can't I use ADF.test? Since you provided no code, the following is just a guess, Try: install.packages(uroot) library(uroot) Then try your analysis again. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of DongHongwei Sent: Wednesday, June 24, 2009 3:06 PM To: r-help@r-project.org Subject: [R] Why can't I use ADF.test? Greetings! I'm trying to use R to test the unit root for a univariate data. By this link: http://rss.acs.unt.edu/Rdoc/library/uroot/html/ADF.test.html it tells me that I can use the function ADF.test(). However, when I tried this in R, I got this message: Error: could not find function ADF.test. I'm confused by this. Anyone could give me some hints? Thanks. Hongwei _ ôòùäãìÃâ¢ÃµÃâ¡Ã®Ã£Ã¬Ãâò÷ÿÃâÃâãìÿìÃâ¬Ã´MClubÃâûÃâ ðáñýðÎÃÂòÃËýÿáñãá [[alternative HTML version deleted]] This email message, including any attachments, is for the sole use of the intended recipient(s) and may contain confidential information. Any unauthorized review, use, disclosure or distribution is prohibited. If you are not the intended recipient(s) please contact the sender by reply email and destroy all copies of the original message. Thank you. _ Messengerå®â°Ã¥â¦Â¨Ã¤Â¿Â护ä¸Âå¿Æï¼ÅÃ¥â¦Â费修å¤Â系统æ¼Âæ´žï¼Åä¿Â护Messengerå®â°Ã¥â¦Â¨Ã¯Â¼Â http://im.live.cn/safe/ [[alternative HTML version deleted]] _ æå·¥ï¼æ£é±ï¼ä¹°æ¿åï¼å¿«æ¥MClubä¸èµ·âéå±èå¨âï¼ http://club.msn.cn/?from=10 [[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] change the height or scale of the y axis
legen wrote: I have a question about changing the height or scale of the y axis. When I use following two R codes, I can get two plots. Please look at the y axes, the number of indices (x1, x2, …) on the y axis in the first plot is smaller than that in the second plot, and hence the space between any two indices in the first plot is wider than that in the second plot. As the number of indices increases, the space will vanish and the indices will overlap. I want to display all the indices on the y axis in the second plot, just look like that in the first plot. How to separate the indices on the y axis in the second plot? I guess maybe changing the height or scale of y axis is a way to solve my problem, but I failed to do it after several trails. Anybody can help me? Thank you in advance. You really put a lot of typing work into your example code, but it work, so it's fine. As a first attempt, put the following before the second plot(). par(las=1,cex=0.5) And think about the paper size. I suggest to have a look at lattice graphics for this type of work as a more elegant alternative. And Hadley Wickham can certainly tell you how to do this in ggplot47 -- View this message in context: http://www.nabble.com/change-the-height-or-scale-of-the-y-axis-tp24187351p24198429.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] Apply as.factor (or as.numeric etc) to multiple columns
Hi Mark,
I frequently need to do that when importing data. This one-liner works:
data.frame(mapply(as, x, c(integer, character, factor),
SIMPLIFY=FALSE), stringsAsFactors=FALSE);
but it has two problems:
1) as() is an S4 method that does not always work
2) writting the vector of classes for 60 variables is rather tedious.
Both issues can be solved with the following two helper functions. The first
function tries to use as(x, class); if it doesn't work, tries as.class(x); If
it still doesn't work, tries class(x). The second function tranforms a single
string to a character vector of classes, by transforming each letter in the
string to a class name (i.e. D is tranformed to Date, i to integer,
etc.), so that writting 60 classes is fast.
doCoerce - function(x, class) {
if (canCoerce(x, class))
as(x, class)
else {
result - try(match.fun(paste(as, class, sep=.))(x),
silent=TRUE);
if (inherits(result, try-error))
result - match.fun(class)(x)
result;
}
}
expandClasses - function (x) {
unknowns - character(0)
result - lapply(strsplit(as.character(x), NULL, fixed = TRUE),
function(y) {
sapply(y, function(z) switch(z,
i = integer, n = numeric,
l = logical, c = character, x = complex,
r = raw, f = factor, D = Date, P = POSIXct,
t = POSIXlt, N = NA_character_, {
unknowns - c(unknowns, z)
NA_character_
}), USE.NAMES = FALSE)
})
if (length(unknowns)) {
unknowns - unique(unknowns)
warning(sprintf(ngettext(length(unknowns), code %s not recognized,
codes %s not recognized), dqMsg(unknowns)))
}
result
}
An example:
x - data.frame(X=2008-01-01, Y=1.1:3.1, Z=letters[1:3])
data.frame(mapply(doCoerce, x, expandClasses(Dif)[[1L]], SIMPLIFY=FALSE),
stringsAsFactors=FALSE);
Regards,
Enrique
--
Message: 99
Date: Tue, 23 Jun 2009 15:23:54 -0600
From: Mark Na mtb...@gmail.com
Subject: [R] Apply as.factor (or as.numeric etc) to multiple columns
To: r-help@r-project.org
Message-ID:
e40d78ce0906231423m4c3da14i2f6270f92463c...@mail.gmail.com
Content-Type: text/plain; charset=ISO-8859-1
Hi R-helpers,
I have a dataframe with 60columns and I would like to convert several
columns to factor, others to numeric, and yet others to dates. Rather
than having 60 lines like this:
data$Var1-as.factor(data$Var1)
I wonder if it's possible to write one line of code (per data type,
e.g. factor) that would apply a function (e.g., as.factor) to several
(non-contiguous) columns. So, I could then use 3 or 4 lines of code
(for 3 or 4 data types) instead of 60.
I have tried writing an apply function, but it failed.
Thanks for any help you might be able to provide.
Mark Na
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Re: [R] How to avoid ifelse statement converting factor to character
This sort of experience is why 'The R Inferno'
came into existence.
Patrick Burns
patr...@burns-stat.com
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of The R Inferno and A Guide for the Unwilling S User)
Craig P. Pyrame wrote:
Dear Stavros,
What you discuss below is somewhat scary to me as an R newbie. Is this
just an incident, a bug perhaps, or rather the way things typically go
in R, as your Welcome to R! seems to suggest? I have just started to
learn R, and my initial euphoria of the I can do anything with it!
sort is gradually turning into an I can't get why it doesn't work and
I can't get how to make this work depression. I would be happy to
blame this on my incompetence and incapability, but would also like to
hear if it is not R itself that causes me to fail.
Best regards,
Craig
Stavros Macrakis wrote:
On Wed, Jun 24, 2009 at 12:34 PM, Mark Namtb...@gmail.com wrote:
The problem is that after running the ifelse statement,
data$SOCIAL_STATUS
is converted from a factor to a character.
Is there some way I can avoid this conversion?
I'm afraid that ifelse has very bizarre semantics when the yes and no
arguments don't have the same, atomic vector, type.
The quick workaround for the bizarre semantics (though it can have a
significant efficiency cost) is this:
unlist( ifelse ( condition, as.list( yes ), as.list( no ) ) )
(This isn't perfect, either, but...)
Take a look at the man page for details and the warning:
The mode of the result may depend on the value of 'test', and the
class attribute of the result is taken from 'test' and may be
inappropriate for the values selected from 'yes' and 'no'.
Some consequences of the definition of ifelse are:
Even if the classes of the yes and no arguments are identical, the
result does not necessarily have that class:
ifelse(TRUE,as.raw(4),as.raw(5)) = error
ifelse(TRUE,factor('x'),factor('x')) = 1 (integer)
dates - as.POSIXct(c('1990-1-1','2000-1-1'))
ifelse(c(TRUE,FALSE),dates,dates) = 63117 946702800 (double)
ifelse(c(TRUE,FALSE),factor(c('x','y')),factor(c('y','x'))) = 1 1
If they have different classes, things get stranger:
ifelse(c(TRUE,FALSE),c(a,b),factor(c(c,d))) = a 2
ifelse(c(TRUE,FALSE),list(1,2),as.raw(4))
[[1]]
[1] 1
[[2]]
[1] 04
Result is order-dependent:
ifelse(c(TRUE,FALSE),as.raw(4),list(1,2))
Error in ans[test !nas] - rep(yes, length.out =
length(ans))[test :
incompatible types (from raw to logical) in subassignment type fix
Welcome to R!
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[R] error: compiling R package
Hello, I tried to build an own R package on Windows XP but get an error which I can't solve. I called my package pack. It works to creat the file pack with package.skeleton(). But when I try to compile it with Rcmd build --binary pack I obtain the following error: IO::Seekable::seek missing at C:/R/R-2.9.0/share/perl/FileHandle.pm line 59. Compilation failed in require at C:/R/R-2.9.0/share/perl/R/Dcf.pm line 49. BEGIN failed--compilation aborted at C:/R/R-2.9.0/share/perl/R/Dcf.pm line 49. Compilation failed in require at C:\R\R-2.9.0/bin/build line 28. BEGIN failed--compilation aborted at C:\R\R-2.9.0/bin/build line 28. So if someone could help me on this I would be very greatful. Best, Luba Stein Luba Stein AIM SE Koeniginstr. 28 80802 Munich, Germany E-Mail luba.st...@allianz.commailto:luba.st...@allianz.com Allianz Investment Management SE Vorsitzender des Verwaltungsrates: Dr. Paul Achleitner Geschäftsführende Direktoren: Dr. Karl-Hermann Lowe, Dr. Bernd Gutting Sitz der Gesellschaft: München, Deutschland, Registergericht: München HRB 162748 Für Umsatzsteuerzwecke: Ust-ID-Nr.: DE 251 168 597 [[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] Lose of decimal when using write.table to text file
Bob Ly robertly at vfemail.net writes: I have the following: Date-c(08/05/08,08/06/08,08/07/08) Weight-c(209.4,211.8,210.0) planned.meal-cbind(Date,Weight) planned.meal DateWeight 1 08/05/08, 209.4 2 08/06/08, 211.8 3 08/07/08, 210.0 This is strange. When I run your code, I get planned.meal Date Weight [1,] 08/05/08 209.4 [2,] 08/06/08 211.8 [3,] 08/07/08 210 This could be a global date format setting, but it show part of the problem: everything is converted to string as the common denominator. Better use a data frame. And (from docs) For finer control, use format to make a character matrix/data frame, and call write.table on that Dieter Date-c(08/05/08,08/06/08,08/07/08) Weight-c(209.4,211.8,210.0) planned.meal-cbind(Date,Weight) planned.meal str(planned.meal) planned.meal-data.frame(Date=Date,Weight=Weight) planned.meal str(planned.meal) options(digits=1) planned.meal.format = format(planned.meal,nsmall=1) planned.meal.format write.table(planned.meal.format, file=plannedMeal1.txt, quote=FALSE, row.names=FALSE) __ 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: compiling R package
On Thu, 25 Jun 2009, Stein, Luba (AIM SE) wrote: Hello, I tried to build an own R package on Windows XP but get an error which I can't solve. I called my package pack. It works to creat the file pack with package.skeleton(). But when I try to compile it with Rcmd build --binary pack I obtain the following error: Hmm, at least use Rcmd INSTALL first, and Rcmd build --binary is never preferred to Rcmd INSTALL --build on Windows. IO::Seekable::seek missing at C:/R/R-2.9.0/share/perl/FileHandle.pm line 59. That's an error message from Perl. You need to mend your broken Perl installation. Compilation failed in require at C:/R/R-2.9.0/share/perl/R/Dcf.pm line 49. BEGIN failed--compilation aborted at C:/R/R-2.9.0/share/perl/R/Dcf.pm line 49. Compilation failed in require at C:\R\R-2.9.0/bin/build line 28. BEGIN failed--compilation aborted at C:\R\R-2.9.0/bin/build line 28. So if someone could help me on this I would be very greatful. Best, Luba Stein -- Brian D. Ripley, rip...@stats.ox.ac.uk 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] curvedarrow (some graphics problem)
This will give you a greek character, see ?plotmath grid.text(expression(mu*(s,t)), 0.5, unit(5, lines), vp=vp2) The following works for me, it may be that you're using an outdated version of R, vp - viewport( x = unit(0, npc), y = unit(0, npc), just = c(left, bottom), xscale = c(-1, 1) , yscale = c(-1, 1)) vp2 - viewport( # probably not needed but I had trouble placing the xaxis x = unit(0,npc), y = unit(0.5, npc), just = c(left, bottom), xscale = c(-1, 1) , yscale = c(-1, 1)) pdf() # open the pdf device pushViewport(vp) grid.xaxis(at=seq(-1, 1, by=0.2), label=FALSE, vp=vp2) grid.points(x=0.3, y=0.5, gp=gpar(col=NA), default.units = npc, name=h1, vp=vp) grid.points(x=0.7, y=0.5, gp=gpar(col=NA), default.units = npc, name=h2, vp=vp) grid.text(s, x=0.3, y=unit(-1, lines), vp=vp2) grid.text(t, 0.7, unit(-1, lines), gp=gpar(col=red), vp=vp2) grid.text(expression(mu*(s,t)), 0.5, unit(5, lines), vp=vp2) grid.curve(grobX(h2, 180), grobY(h2, 180), grobX(h1, 180), grobY(h1, 180), shape=1, ncp=10, square=FALSE, curvature=.4, arrow=arrow(angle=20,ends=first) ) upViewport() dev.off() # close the device casperyc wrote: Hi there, This is now the code % library(grid) vp - viewport( x = unit(0, npc), y = unit(0, npc), just = c(left, bottom), xscale = c(-1, 1) , yscale = c(-1, 1)) vp2 - viewport( # probably not needed but I had trouble placing the xaxis x = unit(0,npc), y = unit(0.5, npc), just = c(left, bottom), xscale = c(-1, 1) , yscale = c(-1, 1)) pushViewport(vp) grid.xaxis(at=seq(-1, 1, by=0.2), label=FALSE, vp=vp2) grid.points(x=0.3, y=0.5, gp=gpar(col=NA), default.units = npc, name=h1, vp=vp) grid.points(x=0.7, y=0.5, gp=gpar(col=NA), default.units = npc, name=h2, vp=vp) grid.text(s, x=0.3, y=unit(-1, lines), vp=vp2) grid.text(t, 0.7, unit(-1, lines), gp=gpar(col=red), vp=vp2) grid.text(mu(s,t), 0.5, unit(5, lines), vp=vp2) grid.curve(grobX(h2, 180), grobY(h2, 180), grobX(h1, 180), grobY(h1, 180), shape=1, ncp=10, square=FALSE, curvature=.4, arrow=arrow(angle=20,ends=first) ) upViewport() % for this line: grid.text(mu(s,t), 0.5, unit(5, lines), vp=vp2) how do I change mu(s,t) to the expression mu? I have tried grid.text(expression=mu(s,t), 0.5, unit(5, lines), vp=vp2) grid.text(expression(mu(s,t)), 0.5, unit(5, lines), vp=vp2) both didnt work AND when I tried to save the output as PDF file it returns: Error in function (name) : Grob 'h2' not found I have no idea what to do? __ 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] JRI - problem to access stats package
Hello,
I am a new R user. More precisely, I am working with JRI (on a Eclipse
Java6 project under Ubuntu9).
I have difficulties to access some R packages (For example: package
stats, object Normal, function pnorm).
I have tried many solutions to set the right path but nothing have succeed:
For example: LD_LIBRARY_PATH=${R_HOME}/lib:${R_HOME}/bin:${R_HOME}/library
Inside the java code, I have also tried such methods:
Note that r is the REngine object coming from the connection to R
r.eval(dyn.load(\/home/gran/workspace/R-2.9.0/library/stats/libs/stats.so\,
local = TRUE, now = TRUE));
r.eval(require(stats));
r.eval(autoload(\Normal\, \stats\));
r.eval(search());
r.eval(ls(\Autoloads\));
System.out.println(r.eval(.Autoloaded));
r.eval(zval = .95);
System.out.println(r.eval(print(zval)));
r.eval(p3 = pnorm(3*zval));
System.out.println(r.eval(print(p3)));
It gives as a result:
[STRING stats]
[REAL* (0.95)]
null -- p3 = null because pnorm is unknown
Do you know how to use the stats package with JRI ?
Thanks in advance,
Antoine
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[R] [R-pkgs] RODBC 1.2-6 on CRAN, future directions
Version 1.2.6 of RODBC is now on CRAN. This has a number of bug fixes and many workarounds for ODBC driver quirks--I've set up further testbeds for SQL Server 2008, Oracle and DB2. More visibly, the documentation has been expanded in several ways, in particular in collecting together advice on using 'schemas' and 'catalogs' in the ?RODBC overview. There is also a test suite which although mainly useful for the maintainer provides some more detailed examples of working with the quirks of various backends (and has examples for Oracle's own Windows ODBC driver). Contributions of adaptations of those examples to other drivers would be welcome (especially for Sybase and other ODBC drivers for Oracle). In the next version much of the documentation will be moved from the help pages to a package manual: a draft of that manual appropriate to RODBC 1.2-6 can be found at http://www.stats.ox.ac.uk/pub/bdr/RODBC-manual.pdf Other planned changes are speed-ups for Oracle (currently 4x on the test suite), more support for schemas (that in 1.2-6 is just a first step, but the behaviour of the ODBC drivers is remarkably inconsistent) and support for R's 'raw' type. -- Brian D. Ripley, rip...@stats.ox.ac.uk 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-packages mailing list r-packa...@r-project.org 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] JRI - problem to access stats package
Hi Antoine,
I have got this to work on Windows, the process should be the same for
Linux.
For an eclipse plugin you will need to add the following to your
MANIFEST.MF file:
Bundle-ClassPath: whatever you normally have,
lib/JRI.jar
Bundle-NativeCode: /lib/jri.dll; osname=winxp; processor=x86
You will also need to ensure that the R executable is in your path and
that R_HOME is set.
I did this by creating a simple batch file containing the following in a
directory called rootfiles within our feature, the same should work for
a .sh :
set PATH=%CD%\R install\exe;%PATH%
set R_HOME=%CD%\R install
eclipse.exe
I then added the following line to the build.properties file of the
feature:
root=rootfiles
This tells eclipse to copy the content of the directory to the root of
the application on install.
If you then run eclipse from the batch file, it should work.
--
Geoff Gibbs
mangosolutions
data analysis that delivers
Tel +44 (0)1249 767 700
Mob +44 (0)7791 855 620
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of GRAN
Sent: 25 June 2009 09:40
To: r-help@r-project.org
Subject: [R] JRI - problem to access stats package
Hello,
I am a new R user. More precisely, I am working with JRI (on a Eclipse
Java6 project under Ubuntu9).
I have difficulties to access some R packages (For example: package
stats, object Normal, function pnorm).
I have tried many solutions to set the right path but nothing have
succeed:
For example:
LD_LIBRARY_PATH=${R_HOME}/lib:${R_HOME}/bin:${R_HOME}/library
Inside the java code, I have also tried such methods:
Note that r is the REngine object coming from the connection to R
r.eval(dyn.load(\/home/gran/workspace/R-2.9.0/library/stats/libs/stats
.so\,
local = TRUE, now = TRUE));
r.eval(require(stats));
r.eval(autoload(\Normal\, \stats\));
r.eval(search());
r.eval(ls(\Autoloads\));
System.out.println(r.eval(.Autoloaded));
r.eval(zval = .95);
System.out.println(r.eval(print(zval)));
r.eval(p3 = pnorm(3*zval));
System.out.println(r.eval(print(p3)));
It gives as a result:
[STRING stats]
[REAL* (0.95)]
null -- p3 = null because pnorm is unknown
Do you know how to use the stats package with JRI ?
Thanks in advance,
Antoine
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Re: [R] adf.test Vs ADF.test...
Dear Harry, to complete the picture, for the packages installed on my machine help.search() yielded: help.search(Dickey) Help files with alias or concept or title matching 'Dickey' using fuzzy matching: CADFtest::CADFtest Hansen's Covariate-Augmented Dickey Fuller (CADF) test fUnitRoots::DickeyFullerPValues Dickey-Fuller p Values tseries::adf.test Augmented Dickey-Fuller Test urca::Raotbl1 Data set used by Dickey, Jansen Thornton (1994) urca::Raotbl2 Data set used by Dickey, Jansen Thornton (1994) urca::ur.df Augmented-Dickey-Fuller Unit Root Test uroot::ADF.rectest Augmented Dickey-Fuller Recursive Test uroot::ADF.test Augmented Dickey-Fuller Test Type '?PKG::FOO' to inspect entry 'PKG::FOO TITLE'. At least ur.df() in urca and IIRC CADFtest() in CADFtest provide arguments for lag selection. HTH, Bernhard -Ursprüngliche Nachricht- Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im Auftrag von DongHongwei Gesendet: Donnerstag, 25. Juni 2009 09:27 An: r-help@r-project.org Betreff: Re: [R] adf.test Vs ADF.test... Hi, R users, I'm using R to test the unit root for my time series data. I just compared the ADF.test in uroot package and the adf.test in tseries package. It seems it is difficult to define the time trend and intercept in adf.test. But it is easy to do these in ADF.test. ADF.test also help you find the number of lags that need to be included in the model to remove the serial correlation. I do not not see adf.test be able to do this too. You need to define the number of lags for adf.test. I'm a new R user and I could be wrong. I'll appreciate it very much if someone familiar with time series in R can give me some comments and suggestions. Thanks in advance. Harry From: dongh...@hotmail.com To: patrick.richard...@vai.org; r-help@r-project.org Date: Wed, 24 Jun 2009 19:22:46 + Subject: Re: [R] Why can't I use ADF.test? Thanks! I tried these, but I got the following messages:Warning message:In getDependencies(pkgs, dependencies, available, lib) : package ‘uroot’ is not available Error in library(uroot) : there is no package called 'uroot' I download a package called uroot and put it into: C:\Program Files\R\R-2.9.0\library. But still got the same error. Any further suggestions? Hongwei From: patrick.richard...@vai.org To: dongh...@hotmail.com; r-help@r-project.org Date: Wed, 24 Jun 2009 15:12:22 -0400 Subject: RE: [R] Why can't I use ADF.test? Since you provided no code, the following is just a guess, Try: install.packages(uroot) library(uroot) Then try your analysis again. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of DongHongwei Sent: Wednesday, June 24, 2009 3:06 PM To: r-help@r-project.org Subject: [R] Why can't I use ADF.test? Greetings! I'm trying to use R to test the unit root for a univariate data. By this link: http://rss.acs.unt.edu/Rdoc/library/uroot/html/ADF.test.html it tells me that I can use the function ADF.test(). However, when I tried this in R, I got this message: Error: could not find function ADF.test. I'm confused by this. Anyone could give me some hints? Thanks. Hongwei _ ´ò¹¤£¬ÕõÇ®£¬Âò· ¿×Ó£¬¿ìÀ´MClubÒ»Æ𡱽ðÎòؽ¿¡±£¡ [[alternative HTML version deleted]] This email message, including any attachments, is for the sole use of the intended recipient(s) and may contain confidential information. Any unauthorized review, use, disclosure or distribution is prohibited. If you are not the intended recipient(s) please contact the sender by reply email and destroy all copies of the original message. Thank you. _ Messenger安全ä¿Â护ä¸Â心，å…Â费修å¤Â系统漠洞，ä¿Â护Messenger安全＠http://im.live.cn/safe/ [[alternative HTML version deleted]] _ 打工，挣钱，买房å，快æ¥MClub一起â€é‡‘屋è—娇â€ï¼ http://club.msn.cn/?from=10 [[alternative HTML version deleted]] * Confidentiality Note: The information contained in this message, and any attachments, may contain confidential and/or privileged material. It is intended solely for the person(s) or entity to which it is addressed. Any review, retransmission, dissemination, or taking of any action in reliance upon
[R] random sampling or random replacement
Dear R users, I'm trying to randomly recreate a real dataset with missing data and I'm not quite sure if I can use the sample command for this. I think it might be better to do it in 2 steps and randomly replace the sampled data with missing data... So something like this x - sample(1:2, 100) #without replacement Now I want x to contain to 20% missing data (NA). Could anyone help me how to do this? Thanks Joanne -- Joanne Demmler Ph.D. Research Assistant School of Medicine Swansea University Singleton Park Swansea SA2 8PP UK tel:+44 (0)1792 295674 fax:+44 (0)1792 513430 email: j.demm...@swansea.ac.uk DECIPHer: www.decipher.uk.net __ 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] grid.polygon() + color gradient
Hi,
I don't think the fill parameter can be a colour gradient. You'll need
to create small polygons, each with its own fill (200, say). Try this,
x= c(0, 0.5, 1)
y= c(0.5, 1, 0.5)
grid.polygon(x=x, y=y, gp=gpar(fill=grey90, col=grey90))
xx - seq(range(x)[1],range(x)[2], length=100)
yy - rep(max(y), length(xx))
cols - colorRampPalette(c(green, lightgray))(length(xx))
for(ii in seq_along(xx[-length(xx)])) {
grid.clip(x=xx[ii], y=0.5,
width= xx[ii+1],
height=1,
just=bottom)
grid.polygon(x=c(0, 0.5, 1), y=c(0.5, 1, 0.5), gp=gpar(fill=cols[ii],
col=NA))
}
Note that the situation would become rather more complicated for a
gradient at some angle (see ?grobX if you need to).
If you're free to choose an external tool to produce this, the TeX
package Tikz has good support for gradients and clipping.
HTH,
baptiste
Kexin Ji wrote:
Hi,
I wonder whether there is a way to generate a polygon (a triangle in
my case) with color gradient using grid.polygon() in package grid?
I tried something like
library(grid)
grid.polygon(x=c(0, 0.5, 1), y=c(0.5, 1, 0.5), gp=gpar(col=NA,
fill=colorRampPalette(c(green, lightgray),
space=Lab)(200)))
But am only getting a triangle filled with color green, whereas the
aim is a triangle of color gradient from green to lightgray.
Can grid.polygon() generate a color gradient, or am I being mistaken?
Best to my knowledge, is it true that R currently doesn't contain any
other function that might generate a polygon with color gradient?
Thank you!
Kexin
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and provide commented, minimal, self-contained, reproducible code.
--
_
Baptiste Auguié
School of Physics
University of Exeter
Stocker Road,
Exeter, Devon,
EX4 4QL, UK
Phone: +44 1392 264187
http://newton.ex.ac.uk/research/emag
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[R] lm
Hi all,
I want to make multiple least squares estimation on two matrix (without
intercept)
imagine matrix a and matrix b, I want to regress each colum on matrix
a (dependent variable) on each column of matrix b, I mean, regress first
colum on a to first column on b. Second column on a to second column on b.
Imagine a and b have 200 columns. So I will have 200 estimated
coefficients.
Why this doesn´´ t function?
zza2-function(){
a-read.table(a.txt)
b-read.table(b.txt)
mrp -lm(as.matrix(cbind(a)) ~ 0+as.matrix(cbind(b)))
mr22p-coefficients(mrp)
plot(mr22p)
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Re: [R] grid.polygon() + color gradient
[I neglected to check some details in the previous post]
This one should work better,
library(grid)
x= c(0, 0.5, 1)
y= c(0.5, 1, 0.5)
grid.polygon(x=x, y=y, gp=gpar(fill=NA, col=grey90)) # outer shell
xx - seq(range(x)[1],range(x)[2], length=100)
dx - diff(xx) # width of clipped triangles
cols - colorRampPalette(c(green, lightgray))(length(dx))
for(ii in seq_along(xx[-length(xx)])){
grid.clip(x=xx[ii], y=0.5,
width= 1.2*dx[ii], # fudge factor to overlap well
height=1,
just=bottom)
# if(ii%%2)# testing with every other masked
grid.polygon(x=c(0, 0.5, 1), y=c(0.5, 1, 0.5), gp=gpar(fill=cols[ii],
col=cols[ii]))
}
HTH,
baptiste
Kexin Ji wrote:
Hi,
I wonder whether there is a way to generate a polygon (a triangle in
my case) with color gradient using grid.polygon() in package grid?
I tried something like
library(grid)
grid.polygon(x=c(0, 0.5, 1), y=c(0.5, 1, 0.5), gp=gpar(col=NA,
fill=colorRampPalette(c(green, lightgray),
space=Lab)(200)))
But am only getting a triangle filled with color green, whereas the
aim is a triangle of color gradient from green to lightgray.
Can grid.polygon() generate a color gradient, or am I being mistaken?
Best to my knowledge, is it true that R currently doesn't contain any
other function that might generate a polygon with color gradient?
Thank you!
Kexin
[[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.
--
_
Baptiste Auguié
School of Physics
University of Exeter
Stocker Road,
Exeter, Devon,
EX4 4QL, UK
Phone: +44 1392 264187
http://newton.ex.ac.uk/research/emag
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and provide commented, minimal, self-contained, reproducible code.
[R] ANOVA with means and SDs as input
Dear R-community, I'm struggling with a paper that reports only fragmented results of a 2by2by3 experimental design. However, Means and SDs for all cells are given. Does anyone know a package/function that helps computing an ANOVA with only Means and SDs as input (resulting in some sort of effect size and significance test)? The reason why I'm interested is simple: I'm conducting a meta-analysis. If I included only what the authors would like to show the world, the results would be somewhat biased... I've combed through the web and my various books on R, but it's not that easy to find. Or is it? Thanks for your help! Sebastian __ 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] random sampling or random replacement
If you want to average 20% missing values then you could try it in 1 step, viz: sample(c(1:2, rep(NA, 2000)),100) Otherwise, 2 steps is preferable. Use code as below: sample(1:2,100)-kk kk[sample(1:100,20)]-NA Paul -- View this message in context: http://www.nabble.com/random-sampling-or-random-replacement-tp24199695p24200736.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] random sampling or random replacement
Joanne, [...snip...] x - sample(1:2, 100) #without replacement Now I want x to contain to 20% missing data (NA). Could anyone help me how to do this? See if this helps: n - length(x) x[sample(n, 0.2*n)] - NA cheers, -Girish -- View this message in context: http://www.nabble.com/random-sampling-or-random-replacement-tp24199695p24200909.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] ANOVA with means and SDs as input
On 25-Jun-09 10:15:30, Sebastian Stegmann wrote: Dear R-community, I'm struggling with a paper that reports only fragmented results of a 2by2by3 experimental design. However, Means and SDs for all cells are given. Does anyone know a package/function that helps computing an ANOVA with only Means and SDs as input (resulting in some sort of effect size and significance test)? No hope of that, unless you also have the numbers (N) of observations in each cell (which of course may be equal across cells). You can get a mean and an SD with anything from N=2 upwards. If you do have the Ns, then one way is to construct an artificial sample for each cell, such that the mean and the SD of each is equal to the given mean and SD for the cell. Then you can submit the resulting reconstructed data to a standard lm() in R. and carry on from there. The basic construct is: Let a given cell have N data, mean=M, SD=S. X0 - rnorm(N) X - M + S*(X0 - mean(X0))/sd(X0) Assuming you are going to do a standard normal ANOVA, the result of operating with the above artificial data is algebraically the same as if you had the true original data. Hoping this helps! Ted. The reason why I'm interested is simple: I'm conducting a meta-analysis. If I included only what the authors would like to show the world, the results would be somewhat biased... I've combed through the web and my various books on R, but it's not that easy to find. Or is it? Thanks for your help! Sebastian E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk Fax-to-email: +44 (0)870 094 0861 Date: 25-Jun-09 Time: 11:51:38 -- 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.
[R] normalization in wavelet analysis
Dear, I am using wavelet function from library(dplR), I would like to have a normalised spectrum as an output (power relative to white noise). I was wondering if anyone can help me with that. Thank you. irina __ 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] Removed from email list
Hi, Could I please be removed from the email list! Thank you, Sophie [[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] RODBC: Trying to read an Excel file
R 2.8
Windows XP
Excel 2003
I am trying to read an Excel spread sheet. I have looked at the RODBC help
pages and am having trouble setting up code that will work. My code and the
results are pasted below:
jo-odbcConnectExcel(i:\\all\\sorkinjohn\\stats\\silvermannatalie\\NEMOcombined06-24-09.xls,readOnly
= TRUE)
mo-sqlGetResults(jo)
mo
[1] -1
I have tried to set up the connection string using odbcConnectExcel and then
have tried to read the file using sqlGetResults. Can someone tell me what I am
doing wrong?
Thanks,
John
John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
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[R] variable driven summary of one column
Hello, how can I get a variable driven summary of one column of my data.frame? Usually I would do summary(data$columnname) to get a summary of column named columnname of my data.frame named data. In my case the columnname is not static but can be set dynamically. So I save the chosen columname in something like variable - columnname but how can I now get the summary of the specified column? summary(data$get(variable)) doesn't work. summary(paste(data$, variable, sep=) doesn't work either! and if I try summary(data[get(variable)] it gives me back a different result since the data isn't a factor anymore but a list. Thanks for the help, Anne __ 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] stringsAsFactors has no impact in expand.grid()?
Hi I have the feeling, that the argument stringsAsFactors has no impact in the function expand.grid: a - c(PR, NC, A2, BS) b - c(1, 0.5, 0.25, 0.125, 0.0625, 0.03125) class(expand.grid(css, fscs, stringsAsFactors=FALSE)[[1]]) [1] factor class(expand.grid(css, fscs, stringsAsFactors=TRUE)[[1]]) [1] factor Also, when I look at the code of expand.grid, stringsAsFactors does not occur in the code. Am I missing something? version _ platform i486-pc-linux-gnu arch i486 os linux-gnu system i486, linux-gnu status major 2 minor 9.0 year 2009 month 04 day17 svn rev48333 language R version.string R version 2.9.0 (2009-04-17) Cheers, Rainer -- Rainer M. Krug, Centre of Excellence for Invasion Biology, Stellenbosch University, South Africa [[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] List subsetting
Another options is: head(lapply(a, tail, 2), 2) On Wed, Jun 24, 2009 at 8:42 PM, Ivo Shterev idc...@yahoo.com wrote: Hello, I have a question about list indexing. Lets say we have a list of 3 lists, each containing 3 different type elements: a=replicate(3, list(list(c(1,1,1), diag(3), c(2,2,2 a [[1]] [[1]][[1]] [1] 1 1 1 [[1]][[2]] [,1] [,2] [,3] [1,]100 [2,]010 [3,]001 [[1]][[3]] [1] 2 2 2 [[2]] [[2]][[1]] [1] 1 1 1 [[2]][[2]] [,1] [,2] [,3] [1,]100 [2,]010 [3,]001 [[2]][[3]] [1] 2 2 2 [[3]] [[3]][[1]] [1] 1 1 1 [[3]][[2]] [,1] [,2] [,3] [1,]100 [2,]010 [3,]001 [[3]][[3]] [1] 2 2 2 If anyone can point a direction as to how to obtain (subset) the following list from list a: b [[1]] [[1]][[1]] [,1] [,2] [,3] [1,]100 [2,]010 [3,]001 [[1]][[2]] [1] 2 2 2 [[2]] [[2]][[1]] [,1] [,2] [,3] [1,]100 [2,]010 [3,]001 [[2]][[2]] [1] 2 2 2 Also, suppose that one wishes to assign the list b to the corresponding subset of list a. Is there a way of doing so? Thanks for the help. Ivo __ 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] variable driven summary of one column
Try: data[,variable] On Thu, Jun 25, 2009 at 6:44 AM, Anne Skoeries h...@anne-skoeries.dewrote: Hello, how can I get a variable driven summary of one column of my data.frame? Usually I would do summary(data$columnname) to get a summary of column named columnname of my data.frame named data. In my case the columnname is not static but can be set dynamically. So I save the chosen columname in something like variable - columnname but how can I now get the summary of the specified column? summary(data$get(variable)) doesn't work. summary(paste(data$, variable, sep=) doesn't work either! and if I try summary(data[get(variable)] it gives me back a different result since the data isn't a factor anymore but a list. Thanks for the help, Anne __ 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] Using by() and stacking back sub-data frames to one data frame
One thing you might consider when working with large dataframes is that instead of partitioning the dataframe into smaller ones, create a list of indices and use that to access the subset. Works especially well when using 'lapply' to cromp through many segments of a data frame: y suid month esr 1 107403412 6 2 1074034 1 2 3 1074034 2 2 4 1074034 3 2 5 107403412 1 6 1074034 1 1 7 1074034 2 1 8 1074034 3 1 9 107403412 2 10 1074034 1 2 11 1074034 2 2 12 1074034 3 2 13 107403412 9 14 1074034 1 9 15 1074034 2 9 16 1074034 3 9 17 112300312 2 18 1123003 1 2 19 1123003 2 2 20 1123003 3 2 y.ind - split(seq(nrow(y)), y$month) y.ind $`1` [1] 2 6 10 14 18 $`2` [1] 3 7 11 15 19 $`3` [1] 4 8 12 16 20 $`12` [1] 1 5 9 13 17 # a subset y[y.ind[['12']],] suid month esr 1 107403412 6 5 107403412 1 9 107403412 2 13 107403412 9 17 112300312 2 On Wed, Jun 24, 2009 at 11:34 PM, Stephan Lindner lindn...@umich.eduwrote: Dear all, I have a code where I subset a data frame to match entries within levels of an factor (actually, the full script uses three difference factors do do that). I'm very happy with the precision with which I can work with R, but since I loop over factor levels, and the data frame is big, the process is slow. So I've been trying to speed up the process using by(), but I got stuck at the point where I want to stack back the sub- data frames, and I was wondering whether someone could help me out. Here is an example: -- y - data.frame(suid = c(rep(1074034,16),rep(1123003,4)), month = rep(c(12,1,2,3),5), esr = c(6,2,2,2,1,1,1,1,2,2,2,2,9,9,9,9,2,2,2,2)) by(y,y$month,function(x)return(x)) y$month: 1 suid month esr 2 1074034 1 2 6 1074034 1 1 10 1074034 1 2 14 1074034 1 9 18 1123003 1 2 y$month: 2 suid month esr 3 1074034 2 2 7 1074034 2 1 11 1074034 2 2 15 1074034 2 9 19 1123003 2 2 y$month: 3 suid month esr 4 1074034 3 2 8 1074034 3 1 12 1074034 3 2 16 1074034 3 9 20 1123003 3 2 y$month: 12 suid month esr 1 107403412 6 5 107403412 1 9 107403412 2 13 107403412 9 17 112300312 2 -- What I would like to do is stacking these four data frames back to one data frame, which in this simple example would just be y. I tried unlist(), unclass() and rbind(), but none of them would work. Thanks a lot, Stephan -- --- Stephan Lindner University of Michigan __ 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.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? [[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] stringsAsFactors has no impact in expand.grid()?
I also not find this in the source code, but you can rewrite the function
with this:
expand.grid -
function (..., KEEP.OUT.ATTRS = TRUE, stringsAsFactors = FALSE)
{
nargs - length(args - list(...))
if (!nargs)
return(as.data.frame(list()))
if (nargs == 1L is.list(a1 - args[[1L]]))
nargs - length(args - a1)
if (nargs == 0L)
return(as.data.frame(list()))
cargs - args
iArgs - seq_len(nargs)
nmc - paste(Var, iArgs, sep = )
nm - names(args)
if (is.null(nm))
nm - nmc
else if (any(ng0 - nzchar(nm)))
nmc[ng0] - nm[ng0]
names(cargs) - nmc
rep.fac - 1L
d - sapply(args, length)
if (KEEP.OUT.ATTRS) {
dn - vector(list, nargs)
names(dn) - nmc
}
orep - prod(d)
if (orep == 0L) {
for (i in iArgs) cargs[[i]] - args[[i]][FALSE]
}
else {
for (i in iArgs) {
x - args[[i]]
if (KEEP.OUT.ATTRS)
dn[[i]] - paste(nmc[i], =, if (is.numeric(x))
format(x)
else x, sep = )
nx - length(x)
orep - orep/nx
x - x[rep.int(rep.int(seq_len(nx), rep.int(rep.fac,
nx)), orep)]
if(stringsAsFactors){
if (!is.factor(x) is.character(x))
x - factor(x, levels = unique(x))
}else{
x
}
cargs[[i]] - x
rep.fac - rep.fac * nx
}
}
if (KEEP.OUT.ATTRS)
attr(cargs, out.attrs) - list(dim = d, dimnames = dn)
rn - .set_row_names(as.integer(prod(d)))
structure(cargs, class = data.frame, row.names = rn)
}
On Thu, Jun 25, 2009 at 8:21 AM, Rainer M Krug r.m.k...@gmail.com wrote:
Hi
I have the feeling, that the argument stringsAsFactors has no impact in the
function expand.grid:
a - c(PR, NC, A2, BS)
b - c(1, 0.5, 0.25, 0.125, 0.0625, 0.03125)
class(expand.grid(css, fscs, stringsAsFactors=FALSE)[[1]])
[1] factor
class(expand.grid(css, fscs, stringsAsFactors=TRUE)[[1]])
[1] factor
Also, when I look at the code of expand.grid, stringsAsFactors does not
occur in the code.
Am I missing something?
version
_
platform i486-pc-linux-gnu
arch i486
os linux-gnu
system i486, linux-gnu
status
major 2
minor 9.0
year 2009
month 04
day17
svn rev48333
language R
version.string R version 2.9.0 (2009-04-17)
Cheers,
Rainer
--
Rainer M. Krug, Centre of Excellence for Invasion Biology, Stellenbosch
University, South Africa
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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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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Apply as.factor (or as.numeric etc) to multiple columns
That's quite nice. Three comments:
- colClasses() in R.utils is similar, except for the particular
codes and classes supported, to expandClasses() here.
- not sure if this is important but if as() were the last
possibility tried rather than the first then in most
cases (in fact all cases handled by expandClasses() )
there would be no use of the methods package.
- paste(as, ...) handles all the common cases including
all cases handled by expandClasses() except NA_character_
and could be used as a poor man's doCoerce().
On Thu, Jun 25, 2009 at 3:43 AM, Bengoechea Bartolomé Enrique (SIES
73)enrique.bengoec...@credit-suisse.com wrote:
Hi Mark,
I frequently need to do that when importing data. This one-liner works:
data.frame(mapply(as, x, c(integer, character, factor),
SIMPLIFY=FALSE), stringsAsFactors=FALSE);
but it has two problems:
1) as() is an S4 method that does not always work
2) writting the vector of classes for 60 variables is rather tedious.
Both issues can be solved with the following two helper functions. The first
function tries to use as(x, class); if it doesn't work, tries as.class(x);
If it still doesn't work, tries class(x). The second function tranforms a
single string to a character vector of classes, by transforming each letter
in the string to a class name (i.e. D is tranformed to Date, i to
integer, etc.), so that writting 60 classes is fast.
doCoerce - function(x, class) {
if (canCoerce(x, class))
as(x, class)
else {
result - try(match.fun(paste(as, class, sep=.))(x),
silent=TRUE);
if (inherits(result, try-error))
result - match.fun(class)(x)
result;
}
}
expandClasses - function (x) {
unknowns - character(0)
result - lapply(strsplit(as.character(x), NULL, fixed = TRUE),
function(y) {
sapply(y, function(z) switch(z,
i = integer, n = numeric,
l = logical, c = character, x = complex,
r = raw, f = factor, D = Date, P = POSIXct,
t = POSIXlt, N = NA_character_, {
unknowns - c(unknowns, z)
NA_character_
}), USE.NAMES = FALSE)
})
if (length(unknowns)) {
unknowns - unique(unknowns)
warning(sprintf(ngettext(length(unknowns), code %s not recognized,
codes %s not recognized), dqMsg(unknowns)))
}
result
}
An example:
x - data.frame(X=2008-01-01, Y=1.1:3.1, Z=letters[1:3])
data.frame(mapply(doCoerce, x, expandClasses(Dif)[[1L]], SIMPLIFY=FALSE),
stringsAsFactors=FALSE);
Regards,
Enrique
--
Message: 99
Date: Tue, 23 Jun 2009 15:23:54 -0600
From: Mark Na mtb...@gmail.com
Subject: [R] Apply as.factor (or as.numeric etc) to multiple columns
To: r-help@r-project.org
Message-ID:
e40d78ce0906231423m4c3da14i2f6270f92463c...@mail.gmail.com
Content-Type: text/plain; charset=ISO-8859-1
Hi R-helpers,
I have a dataframe with 60columns and I would like to convert several
columns to factor, others to numeric, and yet others to dates. Rather
than having 60 lines like this:
data$Var1-as.factor(data$Var1)
I wonder if it's possible to write one line of code (per data type,
e.g. factor) that would apply a function (e.g., as.factor) to several
(non-contiguous) columns. So, I could then use 3 or 4 lines of code
(for 3 or 4 data types) instead of 60.
I have tried writing an apply function, but it failed.
Thanks for any help you might be able to provide.
Mark Na
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Re: [R] RODBC: Trying to read an Excel file
Dear John,
Try this:
require(RODBC)
spreadsheet - Sheet1 # please change this to your needs
channel -
odbcConnectExcel(i:\\all\\sorkinjohn\\stats\\silvermannatalie\\NEMOcombined06-24-09.xls)
mydata - sqlFetch(channel, spreadsheet)
odbcClose(channel)
# attach(mydata)
mydata
HTH,
Jorge
On Thu, Jun 25, 2009 at 7:20 AM, John Sorkin jsor...@grecc.umaryland.eduwrote:
R 2.8
Windows XP
Excel 2003
I am trying to read an Excel spread sheet. I have looked at the RODBC help
pages and am having trouble setting up code that will work. My code and the
results are pasted below:
jo-odbcConnectExcel(i:\\all\\sorkinjohn\\stats\\silvermannatalie\\NEMOcombined06-24-09.xls,readOnly
= TRUE)
mo-sqlGetResults(jo)
mo
[1] -1
I have tried to set up the connection string using odbcConnectExcel and
then have tried to read the file using sqlGetResults. Can someone tell me
what I am doing wrong?
Thanks,
John
John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
Confidentiality Statement:
This email message, including any attachments, is for ...{{dropped:13}}
__
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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] get, put, post implementation
Dear List, I'm searching for a way (package, function or something) providing the http PUT, GET POST ... methods in R. httpRequest and RCurl seems to have a lack of the PUT method. Regards Thomas || Thomas Bock c/o Physikalisch-Technische Bundesanstalt || Abbestr. 2-12, D-10587 Berlin, Germany || Tel/Fax: ++49-30-3481-7354/7490, email: thomas.b...@ptb.de [[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] RODBC: Trying to read an Excel file
Try the function sqlFetch to import the data in the spreadsheet.
jo-odbcConnectExcel(i:\\all\\sorkinjohn\\stats\\silvermannatalie\\NEMOcomb
ined06-24-09.xls,readOnly = TRUE)
mo-sqlFetch(jo,'Your Sheet Name or Number',colnames=F,rownames=F)
mo
Hope it helps.
-
MSc. Rodrigo Aluizio
Centro de Estudos do Mar/UFPR
Laboratório de Micropaleontologia
Avenida Beira Mar s/n - CEP 83255-000
Pontal do Paraná - PR - Brasil
-Mensagem original-
De: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Em
nome de John Sorkin
Enviada em: quinta-feira, 25 de junho de 2009 08:21
Para: r-help@r-project.org
Assunto: [R] RODBC: Trying to read an Excel file
R 2.8
Windows XP
Excel 2003
I am trying to read an Excel spread sheet. I have looked at the RODBC help
pages and am having trouble setting up code that will work. My code and the
results are pasted below:
jo-odbcConnectExcel(i:\\all\\sorkinjohn\\stats\\silvermannatalie\\NEMOcomb
ined06-24-09.xls,readOnly = TRUE)
mo-sqlGetResults(jo)
mo
[1] -1
I have tried to set up the connection string using odbcConnectExcel and then
have tried to read the file using sqlGetResults. Can someone tell me what I
am doing wrong?
Thanks,
John
John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
Confidentiality Statement:
This email message, including any attachments, is for th...{{dropped:9}}
__
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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] change the height or scale of the y axis
legen wrote: Hallo, All, I have a question about changing the height or scale of the y axis. When I use following two R codes, I can get two plots. Please look at the y axes, the number of indices (x1, x2, …) on the y axis in the first plot is smaller than that in the second plot, and hence the space between any two indices in the first plot is wider than that in the second plot. As the number of indices increases, the space will vanish and the indices will overlap. I want to display all the indices on the y axis in the second plot, just look like that in the first plot. How to separate the indices on the y axis in the second plot? I guess maybe changing the height or scale of y axis is a way to solve my problem, but I failed to do it after several trails. Anybody can help me? Thank you in advance. Legen The first R code: x-c(x1,x2,x3,x4,x5,x6,x7,x8,x9,x10) y-sample(0:100,10,replace=F) d-data.frame(y) rownames(d)-x r-nrow(d) i-1:r plot(d$y,i,xlab=,ylab=,axes=F) axis(1) axis(2,at=i,labels=x) The second R code: x-c(x1,x2,x3,x4,x5,x6,x7,x8,x9,x10, x11,x12,x13,x14,x15,x16,x17,x18,x19,x20, x21,x22,x23,x24,x25,x26,x27,x28,x29,x30, x31,x32,x33,x34,x35,x36,x37,x38,x39,x40, x41,x42,x43,x44,x45,x46,x47,x48,x49,x50, x51,x52,x53,x54,x55,x56,x57,x58,x59,x60, x61,x62,x63,x64,x65,x66,x67,x68,x69,x70, x71,x72,x73,x74,x75,x76,x77,x78,x79,x80, x81,x82,x83,x84,x85,x86,x87,x88,x89,x90, x91,x92,x93,x94,x95,x96,x97,x98,x99,x100, x101,x102,x103,x104,x105,x106,x107,x108,x109,x110, x111,x112,x113,x114,x115,x116,x117,x118,x119,x120, x121,x122,x123,x124,x125,x126,x127,x128,x129,x130, x131,x132,x133,x134,x135,x136,x137,x138,x139,x140, x141,x142,x143,x144,x145,x146,x147,x148,x149,x150, x151,x152,x153,x154,x155,x156,x157,x158,x159,x160, x161,x162,x163,x164,x165,x166,x167,x168,x169,x170, x171,x172,x173,x174,x175,x176,x177,x178,x179,x180, x181,x182,x183,x184,x185,x186,x187,x188,x189,x190, x191,x192,x193,x194,x195,x196,x197,x198,x199,x200) y-sample(0:300,200,replace=F) d-data.frame(y) rownames(d)-x r-nrow(d) i-1:r plot(d$y,i,xlab=,ylab=,axes=F) axis(1) axis(2,at=i,labels=x) Hi legen, Your problem centers about the fact that you have about 3 times the vertical scale on the second plot and 20 times the number of ticks and labels. You could make the second plot 3 times the height of the first one, and you would get approximately the same spacing of equal intervals, but the intervals on the second plot are likely to be about 1.5 times those of the first. I can't work out why you are trying to do what you say you want to do. Maybe some more explanation would help. Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Apply as.factor (or as.numeric etc) to multiple columns
Very good points :-) - colClasses() in R.utils is similar, except for the particular codes and classes supported, to expandClasses() here. In fact I saw colClasses() once and got the idea from it, but when I needed the functionallity I did not remember where had I seen it and rewrote it. Now with your hint I can reuse colClasses() instead. I also use a similar function to facilitate writting long logical vectors: expandLogical(TFTFNFFFTN) So, with your suggestions: x - data.frame(X=2008-01-01, Y=1.1:3.1, Z=letters[1:3]) data.frame(mapply(function(x, class) match.fun(paste(as, class, sep=.))(x), x, colClasses(Dif), SIMPLIFY=FALSE), stringsAsFactors=FALSE) Enrique -Original Message- From: Gabor Grothendieck [mailto:ggrothendi...@gmail.com] Sent: jueves, 25 de junio de 2009 13:41 To: Bengoechea Bartolomé Enrique (SIES 73) Cc: r-help@r-project.org; mtb...@gmail.com Subject: Re: [R] Apply as.factor (or as.numeric etc) to multiple columns That's quite nice. Three comments: - colClasses() in R.utils is similar, except for the particular codes and classes supported, to expandClasses() here. - not sure if this is important but if as() were the last possibility tried rather than the first then in most cases (in fact all cases handled by expandClasses() ) there would be no use of the methods package. - paste(as, ...) handles all the common cases including all cases handled by expandClasses() except NA_character_ and could be used as a poor man's doCoerce(). __ 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] lm
Dear Oscar,
Try this:
# Some data
set.seed(123)
a - matrix(rnorm(100*200), ncol = 200)
b - matrix(rnorm(100*200), ncol = 200)
# Auxiliar function to extract the coefficient
# after fitting models without intercept
mycoef - function(x, y) coefficients( lm(y ~ x - 1) )
# Results
res - sapply(1:200, function(i){
y - a[,i]
x - b[,i]
mycoef(x,y)
}
)
plot(res)
HTH,
Jorge
On Thu, Jun 25, 2009 at 5:46 AM, Oscar Bayona osba...@gmail.com wrote:
Hi all,
I want to make multiple least squares estimation on two matrix (without
intercept)
imagine matrix a and matrix b, I want to regress each colum on matrix
a (dependent variable) on each column of matrix b, I mean, regress first
colum on a to first column on b. Second column on a to second column on b.
Imagine a and b have 200 columns. So I will have 200 estimated
coefficients.
Why this doesn´´ t function?
zza2-function(){
a-read.table(a.txt)
b-read.table(b.txt)
mrp -lm(as.matrix(cbind(a)) ~ 0+as.matrix(cbind(b)))
mr22p-coefficients(mrp)
plot(mr22p)
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] stringsAsFactors has no impact in expand.grid()?
On Thu, Jun 25, 2009 at 1:38 PM, Henrique Dallazuanna www...@gmail.comwrote:
I also not find this in the source code, but you can rewrite the function
with this:
That's true - but this should be fixed in the package itself. I have just
converted the factor to characters.
Cheers
Rainer
expand.grid -
function (..., KEEP.OUT.ATTRS = TRUE, stringsAsFactors = FALSE)
{
nargs - length(args - list(...))
if (!nargs)
return(as.data.frame(list()))
if (nargs == 1L is.list(a1 - args[[1L]]))
nargs - length(args - a1)
if (nargs == 0L)
return(as.data.frame(list()))
cargs - args
iArgs - seq_len(nargs)
nmc - paste(Var, iArgs, sep = )
nm - names(args)
if (is.null(nm))
nm - nmc
else if (any(ng0 - nzchar(nm)))
nmc[ng0] - nm[ng0]
names(cargs) - nmc
rep.fac - 1L
d - sapply(args, length)
if (KEEP.OUT.ATTRS) {
dn - vector(list, nargs)
names(dn) - nmc
}
orep - prod(d)
if (orep == 0L) {
for (i in iArgs) cargs[[i]] - args[[i]][FALSE]
}
else {
for (i in iArgs) {
x - args[[i]]
if (KEEP.OUT.ATTRS)
dn[[i]] - paste(nmc[i], =, if (is.numeric(x))
format(x)
else x, sep = )
nx - length(x)
orep - orep/nx
x - x[rep.int(rep.int(seq_len(nx), rep.int(rep.fac,
nx)), orep)]
if(stringsAsFactors){
if (!is.factor(x) is.character(x))
x - factor(x, levels = unique(x))
}else{
x
}
cargs[[i]] - x
rep.fac - rep.fac * nx
}
}
if (KEEP.OUT.ATTRS)
attr(cargs, out.attrs) - list(dim = d, dimnames = dn)
rn - .set_row_names(as.integer(prod(d)))
structure(cargs, class = data.frame, row.names = rn)
}
On Thu, Jun 25, 2009 at 8:21 AM, Rainer M Krug r.m.k...@gmail.com wrote:
Hi
I have the feeling, that the argument stringsAsFactors has no impact in
the
function expand.grid:
a - c(PR, NC, A2, BS)
b - c(1, 0.5, 0.25, 0.125, 0.0625, 0.03125)
class(expand.grid(css, fscs, stringsAsFactors=FALSE)[[1]])
[1] factor
class(expand.grid(css, fscs, stringsAsFactors=TRUE)[[1]])
[1] factor
Also, when I look at the code of expand.grid, stringsAsFactors does not
occur in the code.
Am I missing something?
version
_
platform i486-pc-linux-gnu
arch i486
os linux-gnu
system i486, linux-gnu
status
major 2
minor 9.0
year 2009
month 04
day17
svn rev48333
language R
version.string R version 2.9.0 (2009-04-17)
Cheers,
Rainer
--
Rainer M. Krug, Centre of Excellence for Invasion Biology, Stellenbosch
University, South Africa
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__
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
--
Rainer M. Krug, Centre of Excellence for Invasion Biology, Stellenbosch
University, South Africa
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Coxph frailty model counting process error X matrix deemed singular
... Occasionally, (about 1 in every 100 simulations) I get the following warning: Error in coxph(Surv(start, end, censorind) ~ binary + uniform : X matrix deemed to be singular; variable 2 It is not uncommon for the X matrix in a Cox model to be close enough to singular that the program thinks it is singular, i.e., one of the pivot elements falls below the tolerance threshold while performing a Cholesky decomposition. To understand your case better you need to capture one of the offending data sets and look at the X matrix. I am not going to be able to guess. I normally ignore this (the indeterminate coefficient is set to NA) and go on. You however have chosen to override the default and set singular.ok=FALSE so that the program fails with an error message. You get what you asked for. Last, I am always surprised when people chose method='breslow' in coxph. Why substitute an inferior approximation for the better one? Of course with simulated data there are no ties, in which case all the methods are identical. Terry Therneau __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] variable driven summary of one column
On 6/25/2009 5:44 AM, Anne Skoeries wrote: Hello, how can I get a variable driven summary of one column of my data.frame? Usually I would do summary(data$columnname) to get a summary of column named columnname of my data.frame named data. In my case the columnname is not static but can be set dynamically. So I save the chosen columname in something like variable - columnname but how can I now get the summary of the specified column? summary(data$get(variable)) doesn't work. summary(paste(data$, variable, sep=) doesn't work either! and if I try summary(data[get(variable)] it gives me back a different result since the data isn't a factor anymore but a list. vname - Species summary(subset(iris, select=vname)) Species setosa:50 versicolor:50 virginica :50 vname - Sepal.Width summary(subset(iris, select=vname)) Sepal.Width Min. :2.000 1st Qu.:2.800 Median :3.000 Mean :3.057 3rd Qu.:3.300 Max. :4.400 Thanks for the help, Anne __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Chuck Cleland, Ph.D. NDRI, Inc. (www.ndri.org) 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] GeoXp package
Zeljko Vrba wrote: On Wed, Jun 10, 2009 at 08:21:06AM +0200, Poizot Emmanuel wrote: Error in fun(...) : GDAL Error 1: libgrass_I.so: Ne peut ouvrir le fichier d'objet partagé: Aucun fichier ou dossier de ce type (sorry for the french :) ) It would have been far more useful had you translated the error message to english than to have apologized. Try doing export LD_LIBRARY_PATH=/path/to/dir/where/so/is/located before starting R from the shell. __ 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. Hi, I tried this but with no success. Still enable to load the library rgdal and so GeoXp. This problem of for me now of high importance as I need to export interpolated data sets to grids. and rgdal allows that kind of operations. Regards -- View this message in context: http://www.nabble.com/GeoXp-package-tp23964536p24201943.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] stringsAsFactors has no impact in expand.grid()?
It works for me. Try a more recent version of R. a - c(PR, NC, A2, BS) b - c(1, 0.5, 0.25, 0.125, 0.0625, 0.03125) class(expand.grid(a, b, stringsAsFactors=FALSE)[[1]]) [1] character class(expand.grid(a, b, stringsAsFactors=TRUE)[[1]]) [1] factor R.version.string [1] R version 2.9.1 RC (2009-06-19 r48804) On Thu, Jun 25, 2009 at 7:21 AM, Rainer M Krugr.m.k...@gmail.com wrote: Hi I have the feeling, that the argument stringsAsFactors has no impact in the function expand.grid: a - c(PR, NC, A2, BS) b - c(1, 0.5, 0.25, 0.125, 0.0625, 0.03125) class(expand.grid(css, fscs, stringsAsFactors=FALSE)[[1]]) [1] factor class(expand.grid(css, fscs, stringsAsFactors=TRUE)[[1]]) [1] factor Also, when I look at the code of expand.grid, stringsAsFactors does not occur in the code. Am I missing something? version _ platform i486-pc-linux-gnu arch i486 os linux-gnu system i486, linux-gnu status major 2 minor 9.0 year 2009 month 04 day 17 svn rev 48333 language R version.string R version 2.9.0 (2009-04-17) Cheers, Rainer -- Rainer M. Krug, Centre of Excellence for Invasion Biology, Stellenbosch University, South Africa [[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] Using by() and stacking back sub-data frames to one data frame
Have a look at ddply from the plyr package, http://had.co.nz/plyr. It's made for exactly this type of operation. Hadley On Wed, Jun 24, 2009 at 10:34 PM, Stephan Lindnerlindn...@umich.edu wrote: Dear all, I have a code where I subset a data frame to match entries within levels of an factor (actually, the full script uses three difference factors do do that). I'm very happy with the precision with which I can work with R, but since I loop over factor levels, and the data frame is big, the process is slow. So I've been trying to speed up the process using by(), but I got stuck at the point where I want to stack back the sub- data frames, and I was wondering whether someone could help me out. Here is an example: -- y - data.frame(suid = c(rep(1074034,16),rep(1123003,4)), month = rep(c(12,1,2,3),5), esr = c(6,2,2,2,1,1,1,1,2,2,2,2,9,9,9,9,2,2,2,2)) by(y,y$month,function(x)return(x)) y$month: 1 suid month esr 2 1074034 1 2 6 1074034 1 1 10 1074034 1 2 14 1074034 1 9 18 1123003 1 2 y$month: 2 suid month esr 3 1074034 2 2 7 1074034 2 1 11 1074034 2 2 15 1074034 2 9 19 1123003 2 2 y$month: 3 suid month esr 4 1074034 3 2 8 1074034 3 1 12 1074034 3 2 16 1074034 3 9 20 1123003 3 2 y$month: 12 suid month esr 1 1074034 12 6 5 1074034 12 1 9 1074034 12 2 13 1074034 12 9 17 1123003 12 2 -- What I would like to do is stacking these four data frames back to one data frame, which in this simple example would just be y. I tried unlist(), unclass() and rbind(), but none of them would work. Thanks a lot, Stephan -- --- Stephan Lindner University of Michigan __ 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. -- 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] stringsAsFactors has no impact in expand.grid()?
On Thu, Jun 25, 2009 at 2:55 PM, Gabor Grothendieck ggrothendi...@gmail.com wrote: It works for me. Try a more recent version of R. Good to see that it is fixed in 2.9.1. I am using the latest release and am going to wait till 2.9.1 is released. Thanks, Rainer a - c(PR, NC, A2, BS) b - c(1, 0.5, 0.25, 0.125, 0.0625, 0.03125) class(expand.grid(a, b, stringsAsFactors=FALSE)[[1]]) [1] character class(expand.grid(a, b, stringsAsFactors=TRUE)[[1]]) [1] factor R.version.string [1] R version 2.9.1 RC (2009-06-19 r48804) On Thu, Jun 25, 2009 at 7:21 AM, Rainer M Krugr.m.k...@gmail.com wrote: Hi I have the feeling, that the argument stringsAsFactors has no impact in the function expand.grid: a - c(PR, NC, A2, BS) b - c(1, 0.5, 0.25, 0.125, 0.0625, 0.03125) class(expand.grid(css, fscs, stringsAsFactors=FALSE)[[1]]) [1] factor class(expand.grid(css, fscs, stringsAsFactors=TRUE)[[1]]) [1] factor Also, when I look at the code of expand.grid, stringsAsFactors does not occur in the code. Am I missing something? version _ platform i486-pc-linux-gnu arch i486 os linux-gnu system i486, linux-gnu status major 2 minor 9.0 year 2009 month 04 day17 svn rev48333 language R version.string R version 2.9.0 (2009-04-17) Cheers, Rainer -- Rainer M. Krug, Centre of Excellence for Invasion Biology, Stellenbosch University, South Africa [[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. -- Rainer M. Krug, Centre of Excellence for Invasion Biology, Stellenbosch University, South Africa [[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] Using by() and stacking back sub-data frames to one data frame
Your request for a more general approach is precisely the reason that Hadley Wickham wrote the plyr package. He describes a split-apply- combine strategy for a variety of data structures and tools to implement those strategies here: http://had.co.nz/plyr/plyr-intro-090510.pdf The argument to the by stp is a column name rather than a list or object as it would be in tapply or split. I is just the identity function which doubles for return(x) in your code. library(plyr) ddply(y, month, fun=I) suid month esr 1 1074034 1 2 2 1074034 1 1 3 1074034 1 2 4 1074034 1 9 5 1123003 1 2 6 1074034 2 2 7 1074034 2 1 8 1074034 2 2 9 1074034 2 9 10 1123003 2 2 11 1074034 3 2 12 1074034 3 1 13 1074034 3 2 14 1074034 3 9 15 1123003 3 2 16 107403412 6 17 107403412 1 18 107403412 2 19 107403412 9 20 112300312 2 On Jun 24, 2009, at 11:34 PM, Stephan Lindner wrote: Dear all, I have a code where I subset a data frame to match entries within levels of an factor (actually, the full script uses three difference factors do do that). I'm very happy with the precision with which I can work with R, but since I loop over factor levels, and the data frame is big, the process is slow. So I've been trying to speed up the process using by(), but I got stuck at the point where I want to stack back the sub- data frames, and I was wondering whether someone could help me out. Here is an example: -- y - data.frame(suid = c(rep(1074034,16),rep(1123003,4)), month = rep(c(12,1,2,3),5), esr = c(6,2,2,2,1,1,1,1,2,2,2,2,9,9,9,9,2,2,2,2)) by(y,y$month,function(x)return(x)) y$month: 1 suid month esr 2 1074034 1 2 6 1074034 1 1 10 1074034 1 2 14 1074034 1 9 18 1123003 1 2 y$month: 2 suid month esr 3 1074034 2 2 7 1074034 2 1 11 1074034 2 2 15 1074034 2 9 19 1123003 2 2 y$month: 3 suid month esr 4 1074034 3 2 8 1074034 3 1 12 1074034 3 2 16 1074034 3 9 20 1123003 3 2 y$month: 12 suid month esr 1 107403412 6 5 107403412 1 9 107403412 2 13 107403412 9 17 112300312 2 -- What I would like to do is stacking these four data frames back to one data frame, which in this simple example would just be y. I tried unlist(), unclass() and rbind(), but none of them would work. Thanks a lot, Stephan -- --- Stephan Lindner University of Michigan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Heritage Laboratories West Hartford, CT __ 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] get a dataframe subset based on time interval
Hi, I have a big dataframe with a POSIXct column and I'd like to extract a subset contained in a given time interval, from Date 1 to Date 2. Paulo E. Cardoso __ 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] ANOVA with means and SDs as input
Look at the anova.mean function in the HH package. It does what you are asking, although limited to one-way ANOVA. __ 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] Efficient lookup on a two-dimensional table
Dear R-Users,
I need to lookup values from a 2-d table using the row names and column
names as indices. I was wondering if there's a way to do this without an
explicit loop.
Example:
#x is the 2-d table that holds the values
x - matrix(rnorm(26*12),nrow=26)
rownames(x) - letters
colnames(x) - month.name
#y is a data frame that has the keys I want to use as indices into x
y - data.frame(ltrs=sample(letters,5),mnths=sample(month.name,5),values=0)
#I want to fill in the values column using the ltrs and mnths columns
as keys to look up
# the associated value from x
#One way to do this is with a FOR loop
for (i in 1:nrow(y)) {y$val[i] - x[y$ltrs[i],y$mnths[i]]}
My question: Is there a more efficient way (e.g., one without using an
explicit loop) to do this?
Thanks in advance!
-Rama Ramakrishnan
[[alternative HTML version deleted]]
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[R] Fwd: Efficient lookup on a two-dimensional table
Resending after fixing a mistake in the earlier email ... sorry for the
confusion.
**
Dear R-Users,
I need to lookup values from a 2-d table using the row names and column
names as indices. I was wondering if there's a way to do this without an
explicit loop.
Example:
#x is the 2-d table that holds the values
x - matrix(rnorm(26*12),nrow=26)
rownames(x) - letters
colnames(x) - month.name
#y is a data frame that has the keys I want to use as indices into x
y - data.frame(ltrs=sample(letters,5),mnths=sample(month.name,5),values=0)
#I want to fill in the values column using the ltrs and mnths columns
as keys to look up
# the associated value from x
#One way to do this is with a FOR loop
for (i in 1:nrow(y)) {y$values[i] - x[as.character(y$ltrs[i]),as.character(
y$mnths[i])]}
My question: Is there a more efficient way (e.g., one without using an
explicit loop) to do this?
Thanks in advance!
-Rama Ramakrishnan
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Efficient lookup on a two-dimensional table
That works!! Very nice way to do it! Thank you, Henrique!
Rama Ramakrishnan
On Thu, Jun 25, 2009 at 10:11 AM, Henrique Dallazuanna www...@gmail.comwrote:
Try this:
y$values - diag(x[y$ltrs, y$mnths])
On Thu, Jun 25, 2009 at 11:02 AM, Rama Ramakrishnan r...@alum.mit.eduwrote:
Dear R-Users,
I need to lookup values from a 2-d table using the row names and column
names as indices. I was wondering if there's a way to do this without an
explicit loop.
Example:
#x is the 2-d table that holds the values
x - matrix(rnorm(26*12),nrow=26)
rownames(x) - letters
colnames(x) - month.name
#y is a data frame that has the keys I want to use as indices into x
y - data.frame(ltrs=sample(letters,5),mnths=sample(month.name
,5),values=0)
#I want to fill in the values column using the ltrs and mnths
columns
as keys to look up
# the associated value from x
#One way to do this is with a FOR loop
for (i in 1:nrow(y)) {y$val[i] - x[y$ltrs[i],y$mnths[i]]}
My question: Is there a more efficient way (e.g., one without using an
explicit loop) to do this?
Thanks in advance!
-Rama Ramakrishnan
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Efficient lookup on a two-dimensional table
Try this:
y$values - diag(x[y$ltrs, y$mnths])
On Thu, Jun 25, 2009 at 11:02 AM, Rama Ramakrishnan r...@alum.mit.eduwrote:
Dear R-Users,
I need to lookup values from a 2-d table using the row names and column
names as indices. I was wondering if there's a way to do this without an
explicit loop.
Example:
#x is the 2-d table that holds the values
x - matrix(rnorm(26*12),nrow=26)
rownames(x) - letters
colnames(x) - month.name
#y is a data frame that has the keys I want to use as indices into x
y - data.frame(ltrs=sample(letters,5),mnths=sample(month.name
,5),values=0)
#I want to fill in the values column using the ltrs and mnths columns
as keys to look up
# the associated value from x
#One way to do this is with a FOR loop
for (i in 1:nrow(y)) {y$val[i] - x[y$ltrs[i],y$mnths[i]]}
My question: Is there a more efficient way (e.g., one without using an
explicit loop) to do this?
Thanks in advance!
-Rama Ramakrishnan
[[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.
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
[[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.
[R] [e1071] Inconsistent results when using matrix.csr for svm() - possibly scaling problem
Dear all,
I'm training an SVM with default settings on a matrix csr (SparseM
package). I realized that if I train
the SVM with the (hopefully) equivalent matrix (Matrix package)
representation, the returned models and predictions
sometimes differ. I expected both representations of the same data
to lead to the same results though.
It could be that it is a scaling problem, because unscaled results are
equal(see below).
I'm using the SparseM_0.80 and e1071_1.5-19.
This is what I do (details can be found below):
#run model on csr matrixes
model - svm(matrixcsrTraining,classfactorTraining)
predict(model, matrixcsrTest)
1
0.944625
#run model on matrix representation (Matrix) of the csr matrix from above
model - svm(as.matrix(matrixcsrTraining),classfactorTraining)
predict(model, as.matrix(matrixcsrTest))
1
0.8325838
Possibly this is a scaling problem with sparse matrices, because
results are equal,
if scaling is disabled.
#run model on csr matrixes without scaling
model - svm(matrixcsrTraining,classfactorTraining, scale = FALSE)
predict(model, matrixcsrTest)
1
0.944625
#run model on normal matrixes without scaling
model - svm(as.matrix(matrixcsrTraining),classfactorTraining, scale = FALSE)
predict(model, as.matrix(matrixcsrTest))
1
0.944625
Is scaling different for both formats? Or is there no scaling for SparseM?
Thank you very much for your help,
Eva
CS bachelor student
---
---
Details:
Code below, files attached
#read in data
coordinates - read.csv('vector.data',head=TRUE)
j - subset(coordinates,select=c(j))
ja - as.integer(j[1:dim(j)[1],])
i-subset(coordinates,select=c(i))
ia - as.integer(i[1:dim(i)[1],])
classes - read.csv(classes.data,head=TRUE)
classfactorTraining - classes[1:(max(ia)),]
#build matrixcoo first, then matrixcsr for training
dim - as.integer(c(max(ia),max(ja)))
matrixcoo = new(matrix.coo,ra=rep(1,dim(j)[1]),ja=ja,ia=ia,dim=dim)
matrixcsrTraining = as.matrix.csr(matrixcoo)
#build a simple matrix for testing
matrixcoo =
new(matrix.coo,ra=rep(1,1),ja=as.integer(c(13)),ia=as.integer(c(1)),dim=as.integer(c(1,max(ja
matrixcsrTest = as.matrix.csr(matrixcoo)
#run model on csr matrixes
model - svm(matrixcsrTraining,classfactorTraining, scale = FALSE)
predict(model, matrixcsrTest)
#run model on normal matrixes
model - svm(as.matrix(matrixcsrTraining),classfactorTraining, scale = FALSE)
predict(model, as.matrix(matrixcsrTest))
--
Masked Methods:
The following object(s) are masked from package:stats :
model.response
The following object(s) are masked from package:base :
backsolve,
chol
-
Session info:
sessionInfo()
R version 2.9.0 (2009-04-17)
x86_64-pc-linux-gnu
locale:
LC_CTYPE=en_US.UTF-8;LC_NUMERIC=C;LC_TIME=en_US.UTF-8;LC_COLLATE=en_US.UTF-8;LC_MONETARY=C;LC_MESSAGES=en_US.UTF-8;LC_PAPER=en_US.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_US.UTF-8;LC_IDENTIFICATION=C
attached base packages:
[1] methods stats graphics grDevices utils datasets base
other attached packages:
[1] SparseM_0.80 e1071_1.5-19 class_7.2-47
--
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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] Efficient lookup on a two-dimensional table
Follow-on question: is there a way to do this for higher-dimensional (i.e.
more than 2 dimensions) arrays?
On Thu, Jun 25, 2009 at 10:17 AM, Rama Ramakrishnan r...@alum.mit.eduwrote:
That works!! Very nice way to do it! Thank you, Henrique!
Rama Ramakrishnan
On Thu, Jun 25, 2009 at 10:11 AM, Henrique Dallazuanna
www...@gmail.comwrote:
Try this:
y$values - diag(x[y$ltrs, y$mnths])
On Thu, Jun 25, 2009 at 11:02 AM, Rama Ramakrishnan r...@alum.mit.eduwrote:
Dear R-Users,
I need to lookup values from a 2-d table using the row names and column
names as indices. I was wondering if there's a way to do this without an
explicit loop.
Example:
#x is the 2-d table that holds the values
x - matrix(rnorm(26*12),nrow=26)
rownames(x) - letters
colnames(x) - month.name
#y is a data frame that has the keys I want to use as indices into x
y - data.frame(ltrs=sample(letters,5),mnths=sample(month.name
,5),values=0)
#I want to fill in the values column using the ltrs and mnths
columns
as keys to look up
# the associated value from x
#One way to do this is with a FOR loop
for (i in 1:nrow(y)) {y$val[i] - x[y$ltrs[i],y$mnths[i]]}
My question: Is there a more efficient way (e.g., one without using an
explicit loop) to do this?
Thanks in advance!
-Rama Ramakrishnan
[[alternative HTML version deleted]]
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Efficient lookup on a two-dimensional table
On Jun 25, 2009, at 10:02 AM, Rama Ramakrishnan wrote:
Dear R-Users,
I need to lookup values from a 2-d table using the row names and
column
names as indices. I was wondering if there's a way to do this
without an
explicit loop.
Example:
#x is the 2-d table that holds the values
x - matrix(rnorm(26*12),nrow=26)
rownames(x) - letters
colnames(x) - month.name
#y is a data frame that has the keys I want to use as indices into x
y - data.frame(ltrs=sample(letters,5),mnths=sample(month.name,
5),values=0)
#I want to fill in the values column using the ltrs and mnths
columns
as keys to look up
# the associated value from x
?apply
x$value - apply(y, 1, function(z) x[ z['ltrs'], z['mnths'] ])
#One way to do this is with a FOR loop
for (i in 1:nrow(y)) {y$val[i] - x[y$ltrs[i],y$mnths[i]]}
My question: Is there a more efficient way (e.g., one without using an
explicit loop) to do this?
Thanks in advance!
-Rama Ramakrishnan
[[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.
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
__
R-help@r-project.org mailing list
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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] RODBC: Trying to read an Excel file
On 6/25/2009 7:20 AM, John Sorkin wrote:
R 2.8
Windows XP
Excel 2003
I am trying to read an Excel spread sheet. I have looked at the RODBC help
pages and am having trouble setting up code that will work. My code and the
results are pasted below:
jo-odbcConnectExcel(i:\\all\\sorkinjohn\\stats\\silvermannatalie\\NEMOcombined06-24-09.xls,readOnly
= TRUE)
mo-sqlGetResults(jo)
You set up a connection, but didn't submit a query, so there are no
results to get.
You need to do something like sqlQuery(select * from tablename), where
tablename is a table within your spreadsheet. The ?odbcConnectExcel
page links to
http://support.microsoft.com/default.aspx?scid=KB;EN-US;Q195951 which
describes how to specify one.
Duncan Murdoch
mo
[1] -1
I have tried to set up the connection string using odbcConnectExcel and then
have tried to read the file using sqlGetResults. Can someone tell me what I am
doing wrong?
Thanks,
John
John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
Confidentiality Statement:
This email message, including any attachments, is for th...{{dropped:6}}
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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.
__
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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] Efficient lookup on a two-dimensional table
On Jun 25, 2009, at 10:24 AM, Rama Ramakrishnan wrote: Follow-on question: is there a way to do this for higher-dimensional (i.e. more than 2 dimensions) arrays? The apply method I just posted generalizes to higher dimensional arrays. -- DW On Thu, Jun 25, 2009 at 10:17 AM, Rama Ramakrishnan r...@alum.mit.eduwrote: That works!! Very nice way to do it! Thank you, Henrique! Rama Ramakrishnan On Thu, Jun 25, 2009 at 10:11 AM, Henrique Dallazuanna www...@gmail.com wrote: Try this: y$values - diag(x[y$ltrs, y$mnths]) On Thu, Jun 25, 2009 at 11:02 AM, Rama Ramakrishnan r...@alum.mit.edu wrote: Dear R-Users, I need to lookup values from a 2-d table using the row names and column names as indices. I was wondering if there's a way to do this without an explicit loop. Example: #x is the 2-d table that holds the values x - matrix(rnorm(26*12),nrow=26) rownames(x) - letters colnames(x) - month.name #y is a data frame that has the keys I want to use as indices into x y - data.frame(ltrs=sample(letters,5),mnths=sample(month.name ,5),values=0) #I want to fill in the values column using the ltrs and mnths columns as keys to look up snip [[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. David Winsemius, MD Heritage Laboratories West Hartford, CT __ 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] Make functions aware of theyr own slots.
Hello, Is there a way to access function's slots from inside the function? I want to make functions slot dependent without recurring to generic function mechanism. Probably this goes a bit against R philosophy, but otherwise I don't really see the use of extending functions in R. Would be nice to have something like: setClass(myFunc,list(type=numeric),contains=function) foo-new(myFunc, function(x) m...@type^x,type=0) Many thanks, Vitalie. __ 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] Efficient lookup on a two-dimensional table
Try this (shown for stated problem but generalizes by just adding
additional arguments):
mapply([, list(x), ltrs, mnths)
On Thu, Jun 25, 2009 at 10:24 AM, Rama Ramakrishnanr...@alum.mit.edu wrote:
Follow-on question: is there a way to do this for higher-dimensional (i.e.
more than 2 dimensions) arrays?
On Thu, Jun 25, 2009 at 10:17 AM, Rama Ramakrishnan r...@alum.mit.eduwrote:
That works!! Very nice way to do it! Thank you, Henrique!
Rama Ramakrishnan
On Thu, Jun 25, 2009 at 10:11 AM, Henrique Dallazuanna
www...@gmail.comwrote:
Try this:
y$values - diag(x[y$ltrs, y$mnths])
On Thu, Jun 25, 2009 at 11:02 AM, Rama Ramakrishnan
r...@alum.mit.eduwrote:
Dear R-Users,
I need to lookup values from a 2-d table using the row names and column
names as indices. I was wondering if there's a way to do this without an
explicit loop.
Example:
#x is the 2-d table that holds the values
x - matrix(rnorm(26*12),nrow=26)
rownames(x) - letters
colnames(x) - month.name
#y is a data frame that has the keys I want to use as indices into x
y - data.frame(ltrs=sample(letters,5),mnths=sample(month.name
,5),values=0)
#I want to fill in the values column using the ltrs and mnths
columns
as keys to look up
# the associated value from x
#One way to do this is with a FOR loop
for (i in 1:nrow(y)) {y$val[i] - x[y$ltrs[i],y$mnths[i]]}
My question: Is there a more efficient way (e.g., one without using an
explicit loop) to do this?
Thanks in advance!
-Rama Ramakrishnan
[[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.
--
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.
__
R-help@r-project.org mailing list
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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] exportation of xml file from R
Hello everybody, I have programed a function to translate the R object structure into a XML data structure but I haven't found a function in the XML package to export this data in an XML file! my data look like that: FactorAssessor:codec/Factor Sum Sq 33.98159/Sum Sq Df 28/Df F value 6.510894/F value Pr(F) 1.725149e-22/Pr(F) /line line 6 FactorAssessor:Sample/Factor Sum Sq 33.82264/Sum Sq Df 126/Df F value 1.440098/F value Pr(F) 1.752742e-03/Pr(F) /line line 7 Factorcodec:Sample/Factor Sum Sq253.04055/Sum Sq Df 18/Df F value 75.417587/F value Pr(F)4.055009e-180/Pr(F) /line line 8 FactorResiduals/Factor Sum Sq214.73234/Sum Sq Df1152/Df F valueNA/F value Pr(F)NA/Pr(F) /line /anova:2 /anova for data I basically used the function xmlNode,addChildren,xmlchildren. I don't know if it's sufficient to recognise a xml file So thanks for your help! Regards, Guillaume [[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: system is computationally singular: reciprocal condition number
I get this error while computing partial correlation.
*Error in solve.default(Szz) :
system is computationally singular: reciprocal condition number =
4.90109e-18*
Why is it?Can anyone give me some idea ,how do i get rid it it?
This is the function i use for calculating partial correlation.
pcor.mat - function(x,y,z,method=p,na.rm=T){
x - c(x)
y - c(y)
z - as.data.frame(z)
if(dim(z)[2] == 0){
stop(There should be given data\n)
}
data - data.frame(x,y,z)
if(na.rm == T){
data = na.omit(data)
}
xdata - na.omit(data.frame(data[,c(1,2)]))
Sxx - cov(xdata,xdata,m=method)
xzdata - na.omit(data)
xdata - data.frame(xzdata[,c(1,2)])
zdata - data.frame(xzdata[,-c(1,2)])
Sxz - cov(xdata,zdata,m=method)
zdata - na.omit(data.frame(data[,-c(1,2)]))
Szz - cov(zdata,zdata,m=method)
# is Szz positive definite?
zz.ev - eigen(Szz)$values
if(min(zz.ev)[1]0){
stop(\'Szz\' is not positive definite!\n)
}
# partial correlation
Sxx.z - Sxx - Sxz %*% solve(Szz) %*% t(Sxz)
print(Sxx.z) # this gets printed
rxx.z - cov2cor(Sxx.z)[1,2] #some problem in this function
function (V)
{
print(cov2cor)
p - (d - dim(V))[1]
if (!is.numeric(V) || length(d) != 2L || p != d[2L])
stop('V' is not a square numeric matrix)
Is - sqrt(1/diag(V))
if (any(!is.finite(Is)))
warning(diag(.) had 0 or NA entries; non-finite result is
doubtful)
r - V
r[] - Is * V * rep(Is, each = p)
r[cbind(1L:p, 1L:p)] - 1
r
}
return(rxx.z)
}
--
Thanks
Moumita
[[alternative HTML version deleted]]
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Re: [R] exportation of xml file from R
See the StatDataML package. On Thu, Jun 25, 2009 at 10:54 AM, guillaume Le Rayleray.guilla...@gmail.com wrote: Hello everybody, I have programed a function to translate the R object structure into a XML data structure but I haven't found a function in the XML package to export this data in an XML file! my data look like that: FactorAssessor:codec/Factor Sum Sq 33.98159/Sum Sq Df 28/Df F value 6.510894/F value Pr(F) 1.725149e-22/Pr(F) /line line 6 FactorAssessor:Sample/Factor Sum Sq 33.82264/Sum Sq Df 126/Df F value 1.440098/F value Pr(F) 1.752742e-03/Pr(F) /line line 7 Factorcodec:Sample/Factor Sum Sq253.04055/Sum Sq Df 18/Df F value 75.417587/F value Pr(F)4.055009e-180/Pr(F) /line line 8 FactorResiduals/Factor Sum Sq214.73234/Sum Sq Df1152/Df F valueNA/F value Pr(F)NA/Pr(F) /line /anova:2 /anova for data I basically used the function xmlNode,addChildren,xmlchildren. I don't know if it's sufficient to recognise a xml file So thanks for your help! Regards, Guillaume [[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] exportation of xml file from R
Hi Guillaume Once you have created an XML representation using any of the various ways to represent XML in R via the XML package (e.g. internal nodes, lists of lists, hash trees), you can use the saveXML() function: saveXML(myXML, fileName.xml) Personally, I use newXMLNode() and friends and build the tree with internal nodes. D. guillaume Le Ray wrote: Hello everybody, I have programed a function to translate the R object structure into a XML data structure but I haven't found a function in the XML package to export this data in an XML file! my data look like that: FactorAssessor:codec/Factor Sum Sq 33.98159/Sum Sq Df 28/Df F value 6.510894/F value Pr(F) 1.725149e-22/Pr(F) /line line 6 FactorAssessor:Sample/Factor Sum Sq 33.82264/Sum Sq Df 126/Df F value 1.440098/F value Pr(F) 1.752742e-03/Pr(F) /line line 7 Factorcodec:Sample/Factor Sum Sq253.04055/Sum Sq Df 18/Df F value 75.417587/F value Pr(F)4.055009e-180/Pr(F) /line line 8 FactorResiduals/Factor Sum Sq214.73234/Sum Sq Df1152/Df F valueNA/F value Pr(F)NA/Pr(F) /line /anova:2 /anova for data I basically used the function xmlNode,addChildren,xmlchildren. I don't know if it's sufficient to recognise a xml file So thanks for your help! Regards, Guillaume [[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] Efficient lookup on a two-dimensional table
Thanks, David, that works too! On Thu, Jun 25, 2009 at 10:30 AM, David Winsemius dwinsem...@comcast.netwrote: On Jun 25, 2009, at 10:24 AM, Rama Ramakrishnan wrote: Follow-on question: is there a way to do this for higher-dimensional (i.e. more than 2 dimensions) arrays? The apply method I just posted generalizes to higher dimensional arrays. -- DW On Thu, Jun 25, 2009 at 10:17 AM, Rama Ramakrishnan r...@alum.mit.edu wrote: That works!! Very nice way to do it! Thank you, Henrique! Rama Ramakrishnan On Thu, Jun 25, 2009 at 10:11 AM, Henrique Dallazuanna www...@gmail.com wrote: Try this: y$values - diag(x[y$ltrs, y$mnths]) On Thu, Jun 25, 2009 at 11:02 AM, Rama Ramakrishnan r...@alum.mit.edu wrote: Dear R-Users, I need to lookup values from a 2-d table using the row names and column names as indices. I was wondering if there's a way to do this without an explicit loop. Example: #x is the 2-d table that holds the values x - matrix(rnorm(26*12),nrow=26) rownames(x) - letters colnames(x) - month.name #y is a data frame that has the keys I want to use as indices into x y - data.frame(ltrs=sample(letters,5),mnths=sample(month.name ,5),values=0) #I want to fill in the values column using the ltrs and mnths columns as keys to look up snip [[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. David Winsemius, MD Heritage Laboratories West Hartford, CT [[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-pkgs] WriteXLS - New Version 1.8.1
The updated package has been submitted to CRAN and will propagate to mirrors over the next day or so. It is maintained on R-Forge at http://r-forge.r-project.org/projects/writexls , where downloads will be available as well. There is a transient problem at the moment with R-Forge and the build system, so there may be delays using R-Forge to get the updated package builds. Package: WriteXLS Version: 1.8.1 Description: Cross-platform perl based R function to create Excel (XLS) files from one or more data frames. Each data frame will be written to a separate named worksheet in the Excel spreadsheet. The worksheet name will be the name of the data frame it contains or can be specified by the user. Author(s): Marc Schwartz marc_schwa...@me.com Maintainer: Marc Schwartz marc_schwa...@me.com License: GPL (=2) Changes since version 1.8.0: The data frames will be written to the Excel file in the order that they appear in 'x'. Note that this is a change in the default behavior, where due to Perl's file globbing, they would previously have been written in alpha sort order. The prior default behavior could result in a mis-match between the data frames written to the Excel worksheets and the user specified worksheet names as indicated in the 'SheetNames' argument (added recently in version 1.7.1), if present. As a result of the above change, File::Glob is no longer used and has been removed as a required Perl module for the package. References to File::Glob have been removed from the testPerl() function and all supporting files. The INSTALL file has been copied to the 'inst' package folder, so that it is installed in the main package directory. It will therefore be available both on the CRAN mirror page for the package and on the user's system as referenced in the help files. ___ R-packages mailing list r-packa...@r-project.org 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] exportation of xml file from R
BTW, the XML you show is not legal XML. For instance, the element Pr(F)NA/Pr(F) is not a legal XML element name. Similarly /anova:2 is not a legal name since it is 2. And anova:2 means an element with name 2 with (XML) name space prefix anova (which must be declared previously) As Gabor mentioned, StatDataML is one approach to generating an XML representation of an R object. Since the objects look related to fitted models, you might also want to look at PMML (Predictive Model Markup Language) as that is a more widely used standard. D. guillaume Le Ray wrote: Hello everybody, I have programed a function to translate the R object structure into a XML data structure but I haven't found a function in the XML package to export this data in an XML file! my data look like that: FactorAssessor:codec/Factor Sum Sq 33.98159/Sum Sq Df 28/Df F value 6.510894/F value Pr(F) 1.725149e-22/Pr(F) /line line 6 FactorAssessor:Sample/Factor Sum Sq 33.82264/Sum Sq Df 126/Df F value 1.440098/F value Pr(F) 1.752742e-03/Pr(F) /line line 7 Factorcodec:Sample/Factor Sum Sq253.04055/Sum Sq Df 18/Df F value 75.417587/F value Pr(F)4.055009e-180/Pr(F) /line line 8 FactorResiduals/Factor Sum Sq214.73234/Sum Sq Df1152/Df F valueNA/F value Pr(F)NA/Pr(F) /line /anova:2 /anova for data I basically used the function xmlNode,addChildren,xmlchildren. I don't know if it's sufficient to recognise a xml file So thanks for your help! Regards, Guillaume [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How do I define the method for gcheckboxgroup in gWidgets?
Hi All...
I¹m trying to build a small demo using gWidgets which permits interactive
scaling and selection among different things to plot. I can get the widgets
for scaling to work just fine. I am using gcheckboxgroup to make the
(possibly multiple) selections. However, I can¹t seem to figure out how to
properly define the gcheckboxgroup; I can draw the widget properly, I think
my handler would use the svalue right if it actually received it. Part of
the problem is using the index of the possible values rather than the values
themselves, but I'm pretty sure this is not all of the problem. I've been
unable to find an example like this in any of the various resources I've
come across.
BTW, report.which is really only there for troubleshooting. It works to
return the values, I can't get it to return the indices, which are probably
what I need in this case.
A demo script is at the bottom and the error is just below.
tmp - gcheckboxgroup(stuff, handler = report.which, index = TRUE,
+ checked = c(TRUE, FALSE, FALSE, FALSE, FALSE), container = leftPanel)
add(tmp, value = 1, expand = TRUE)
Error in function (classes, fdef, mtable) :
unable to find an inherited method for function .add, for signature
gCheckboxgroupRGtk, guiWidgetsToolkitRGtk2, numeric
This error suggests that I don't have a method - I agree, but I don't know
what goes into the method for gcheckboxgroup.
For the sliders, it's clear to me how the actions and drawing of the widgets
differ, but not so for gcheckboxgroup.
A big TIA, Bryan
*
Bryan Hanson
Professor of Chemistry Biochemistry
DePauw University, Greencastle IN USA
Full Script:
x - 1:10
y1 - x
y2 - x^2
y3 - x^0.5
y4 - y^3
df - as.data.frame(cbind(x, y1, y2, y3, y4))
stuff - c(y = x, y = x^2, y = x^0.5, y = x^3)
which.y - 2 # inital value, to be changed later by the widget
# Define a function for the widget handlers
update.Plot - function(h,...) {
plot(df[,1], df[,svalue(which.y)], type = l,
ylim = c(0, svalue(yrange)), main = Interactive Selection Scaling,
xlab = x values, ylab = y values)
}
report.which - function(h, ...) { print(svalue(h$obj), index = TRUE) }
# Define the actions type of widget, along with returned values.
# Must be done before packing widgets.
yrange - gslider(from = 0, to = max(y), by = 1.0,
value = max(y), handler = update.Plot)
which.y - gcheckboxgroup(stuff, handler = report.which, index = TRUE,
checked = c(TRUE, FALSE, FALSE, FALSE, FALSE))
# Assemble the graphics window groups of containers
mainWin - gwindow(Interactive Plotting)
bigGroup - ggroup(cont = mainWin)
leftPanel - ggroup(horizontal = FALSE, container = bigGroup)
# Format and pack the widgets, link to their actions/type
tmp - gframe(y range, container = leftPanel)
add(tmp, yrange, expand = TRUE)
tmp - gcheckboxgroup(stuff, handler = report.which, index = TRUE,
checked = c(TRUE, FALSE, FALSE, FALSE, FALSE), container = leftPanel)
add(tmp, value = 1, expand = TRUE)
# Put it all together
add(mainWin, ggraphics()) # puts the active graphic window w/i mainWin
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Re: [R] Efficient lookup on a two-dimensional table
Thanks, Gabor. Works great!
On Thu, Jun 25, 2009 at 10:38 AM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
Try this (shown for stated problem but generalizes by just adding
additional arguments):
mapply([, list(x), ltrs, mnths)
On Thu, Jun 25, 2009 at 10:24 AM, Rama Ramakrishnanr...@alum.mit.edu
wrote:
Follow-on question: is there a way to do this for higher-dimensional
(i.e.
more than 2 dimensions) arrays?
On Thu, Jun 25, 2009 at 10:17 AM, Rama Ramakrishnan r...@alum.mit.edu
wrote:
That works!! Very nice way to do it! Thank you, Henrique!
Rama Ramakrishnan
On Thu, Jun 25, 2009 at 10:11 AM, Henrique Dallazuanna
www...@gmail.comwrote:
Try this:
y$values - diag(x[y$ltrs, y$mnths])
On Thu, Jun 25, 2009 at 11:02 AM, Rama Ramakrishnan r...@alum.mit.edu
wrote:
Dear R-Users,
I need to lookup values from a 2-d table using the row names and
column
names as indices. I was wondering if there's a way to do this without
an
explicit loop.
Example:
#x is the 2-d table that holds the values
x - matrix(rnorm(26*12),nrow=26)
rownames(x) - letters
colnames(x) - month.name
#y is a data frame that has the keys I want to use as indices into x
y - data.frame(ltrs=sample(letters,5),mnths=sample(month.name
,5),values=0)
#I want to fill in the values column using the ltrs and mnths
columns
as keys to look up
# the associated value from x
#One way to do this is with a FOR loop
for (i in 1:nrow(y)) {y$val[i] - x[y$ltrs[i],y$mnths[i]]}
My question: Is there a more efficient way (e.g., one without using an
explicit loop) to do this?
Thanks in advance!
-Rama Ramakrishnan
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[R] Problems with subsets in NLME
I am trying to estimate models with subsets using the NLME package. However, I am getting an error in the case below (among others): subset - c(rep(TRUE, 107), FALSE) fm2 - lme(distance ~ age + Sex, data = Orthodont, random = ~ 1, subset=subset) Error in xj[i] : invalid subscript type 'closure' fm2 - lme(distance ~ age + Sex, data = Orthodont, random = ~ 1, subset=1:107) The second estimation works. Does anyone know if there is another work-around? (I have also e-mailed the package maintainers, but one of the e-mails bounced, so I am trying this list as well.) Thank you! Rebecca -- Rebecca Sela IOMS/Statistics Group Stern School of Business New York University __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] crr - computationally singular
Dear R-help, I'm very sorry to ask 2 questions in a week. I am using the package 'crr' and it does exactly what I need it to when I use the dataset a. However, when I use dataset b I get the following error message: Error in drop(.Call(La_dgesv, a, as.matrix(b), tol, PACKAGE = base)) : system is computationally singular: reciprocal condition number = 1.28654e-24 This is obviously as a result of a problem with the data but apart from dataset a having 1674 rows and dataset b having 701 rows there is really no difference between them. The code I am using is as follows where covaea and covaeb are matrices of covarites, all coded as binary variables. In case a: covaea - cbind(sexa,fsha,fdra,nsigna,eega,th1a,th2a,stype1a,stype2a,stype3a,pgu1a,pgu2a,log(agea),firstinta/1000,totsezbasea) fita - crr(snearma$with.Withtime,csaea,covaea,failcode=2,cencode=0) and in case b: covaeb - cbind(sexb,fshb,fdrb,nsignb,eegb,th1b,th2b,stype1b,stype2b,stype3b,stype4b,stype5b,pgu1b,pgu2b,(ageb/10)^(-1),firstintb,log(totsezbaseb)) fitb - crr(snearmb$with.Withtime,csaeb,covaeb,failcode=2,cencode=0) csaea and csaeb are the censoring indicators for a and b respectively which equal 1 for the event of interest, 2 for the competing risks event and 0 otherwise. Can anyone suggest a reason for the error message? I've tried running fitb with variants of covaeb and irrespective of the order of the covariates in the matrix, the code runs fine with 16 of the 17 covariates included but then produces an error message when the 17th is added. Thank you for your help, Laura __ 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] Problems with subsets in NLME
On Jun 25, 2009, at 11:35 AM, Rebecca Sela wrote: I am trying to estimate models with subsets using the NLME package. However, I am getting an error in the case below (among others): subset - c(rep(TRUE, 107), FALSE) fm2 - lme(distance ~ age + Sex, data = Orthodont, random = ~ 1, subset=subset) Error in xj[i] : invalid subscript type 'closure' fm2 - lme(distance ~ age + Sex, data = Orthodont, random = ~ 1, subset=1:107) The second estimation works. Does anyone know if there is another work-around? (I have also e- mailed the package maintainers, but one of the e-mails bounced, so I am trying this list as well.) Hard to tell. Closure is a type of function. It appears that R might be mistaking its own function subset for the object that you intended to pass. Perhaps if you stopped calling your dogs, Dog? library(fortunes) fortune(dog) Thank you! Rebecca -- Rebecca Sela IOMS/Statistics Group Stern School of Business New York University __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Heritage Laboratories West Hartford, CT __ 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] Problems with subsets in NLME
Rebecca, I think the problem is that subset is a nume of an R function. If you do something like subs - c(rep(TRUE, 107), FALSE) fm2 - lme(distance ~ age + Sex, data = Orthodont, random = ~ 1, subset=subs) everything works fine. Hope this helps, Andy __ Andy Jaworski 518-1-01 Process Laboratory 3M Corporate Research Laboratory - E-mail: apjawor...@mmm.com Tel: (651) 733-6092 Fax: (651) 736-3122 Rebecca Sela rs...@stern.nyu.edu Sent by: r-help-boun...@r-project.org 06/25/2009 10:37 AM To r-help r-help@r-project.org cc Subject [R] Problems with subsets in NLME I am trying to estimate models with subsets using the NLME package. However, I am getting an error in the case below (among others): subset - c(rep(TRUE, 107), FALSE) fm2 - lme(distance ~ age + Sex, data = Orthodont, random = ~ 1, subset=subset) Error in xj[i] : invalid subscript type 'closure' fm2 - lme(distance ~ age + Sex, data = Orthodont, random = ~ 1, subset=1:107) The second estimation works. Does anyone know if there is another work-around? (I have also e-mailed the package maintainers, but one of the e-mails bounced, so I am trying this list as well.) Thank you! Rebecca -- Rebecca Sela IOMS/Statistics Group Stern School of Business New York University __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[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] crr - computationally singular
This means that your design matrix or model matrix is rank deficient, i.e it does not have linearly independent columns. Your predictors are collinear! Just take your design matrices covaea or covaeb with 17 predcitors and compute their rank or try to invert them. You will see the problem. 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: rvarad...@jhmi.edu Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty_personal_pages/Varadhan.h tml -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Laura Bonnett Sent: Thursday, June 25, 2009 11:39 AM To: r-help@r-project.org Subject: [R] crr - computationally singular Dear R-help, I'm very sorry to ask 2 questions in a week. I am using the package 'crr' and it does exactly what I need it to when I use the dataset a. However, when I use dataset b I get the following error message: Error in drop(.Call(La_dgesv, a, as.matrix(b), tol, PACKAGE = base)) : system is computationally singular: reciprocal condition number = 1.28654e-24 This is obviously as a result of a problem with the data but apart from dataset a having 1674 rows and dataset b having 701 rows there is really no difference between them. The code I am using is as follows where covaea and covaeb are matrices of covarites, all coded as binary variables. In case a: covaea - cbind(sexa,fsha,fdra,nsigna,eega,th1a,th2a,stype1a,stype2a,stype3a,pgu 1a,pgu2a,log(agea),firstinta/1000,totsezbasea) fita - crr(snearma$with.Withtime,csaea,covaea,failcode=2,cencode=0) and in case b: covaeb - cbind(sexb,fshb,fdrb,nsignb,eegb,th1b,th2b,stype1b,stype2b,stype3b,sty pe4b,stype5b,pgu1b,pgu2b,(ageb/10)^(-1),firstintb,log(totsezbaseb)) fitb - crr(snearmb$with.Withtime,csaeb,covaeb,failcode=2,cencode=0) csaea and csaeb are the censoring indicators for a and b respectively which equal 1 for the event of interest, 2 for the competing risks event and 0 otherwise. Can anyone suggest a reason for the error message? I've tried running fitb with variants of covaeb and irrespective of the order of the covariates in the matrix, the code runs fine with 16 of the 17 covariates included but then produces an error message when the 17th is added. Thank you for your help, Laura __ 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] How to avoid ifelse statement converting factor to character
On Wed, Jun 24, 2009 at 9:04 PM, Rolf Turnerr.tur...@auckland.ac.nz wrote:
Do not get your knickers in a twist. R works simply and straightforwardly
in simple straightforward situations.
Though I find R an incredibly useful tool, alas, it is simply not true
that R works simply and straightforwardly in simple straightforward
situations. No doubt this is for understandable historical reasons
and backwards compatibility, but there it is.
Some examples of simple straightforward situations:
I think it is reasonable to expect that appending a list/vector of
class X to another list/vector of class X would result in a
list/vector of class X. Similarly for the union of a list/vector of
class X. But in fact, not only is this not true for some of R's
important classes (factors, date/time, and delta-date/time), but the
result class is inconsistent by function and by class:
ff - factor(b)
c(ff,ff)= 1 1# class integer
union(ff,ff) = b# class character
tt - as.POSIXct('2009-01-01')
c(tt,tt) = 2009-01-01 EST 2009-01-01 EST # class POSIXt/POSIXct
union(tt,tt) 1230786000# class numeric
dt - tt - tt # class difftime
c(dt,dt) = 0 0 # class numeric
union(dt,dt) = 0 # class numeric
Similarly, the simplest, most straightforward situation I can think of
for ifelse is when the yes and no arguments are identical, and in that
case, I would (I think reasonably) expect that the result is of the
same class as the arguments, but it is not:
ifelse(TRUE,factor(b),factor(b)) = 1 (integer)
ifelse(TRUE,dd,dd) = 1230786000 (class numeric)
I hope you will agree that all of these are very simple and
straightforward situations, and that R is not working simply and
straightforwardly in them.
The less simple and less straightforward situations are of course more
complicated.
In respect of the current discussion of ifelse() --- the original problem
arose
because the values of ``yes'' and ``no'' were of different modes. It is
obvious
that in such instances a decision will have to be made about the mode
of the result. The appropriateness of the designers' decision may be
disputed,
Indeed.
If you don't understand what's going on, then just stick to using
ifelse() only when ``yes'' and ``no'' have the same mode.
That's not enough. They have to be of a basic class as well. See above.
Bottom line: R is easy to use at any level, but in order to use it a
``high'' level you need to understand the high level. Don't attempt
to run before you can crawl.
Bottom line: Some very basic things in R violate users' reasonable
expectations and moreover are internally inconsistent. You have to be
careful about this whenever you work in R, even at an elementary
level.
-s
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Re: [R] Error: system is computationally singular: reciprocal conditionnumber
Your covariance matrix Szz is not positive definite. It is singular. The
following test that you are doing is neither necessary nor useful:
zz.ev - eigen(Szz)$values
if(min(zz.ev)[1]0){
stop(\'Szz\' is not positive definite!\n)
}
You may want to use Moore-Penrose inverse, also known as generalized inverse
or pseudoinverse to overcome this problem. This approach uses
singular-value decomposition (SVD). Take a look at the corpcor package.
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: rvarad...@jhmi.edu
Webpage:
http://www.jhsph.edu/agingandhealth/People/Faculty_personal_pages/Varadhan.h
tml
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Moumita Das
Sent: Thursday, June 25, 2009 10:59 AM
To: r-help@r-project.org
Subject: [R] Error: system is computationally singular: reciprocal
conditionnumber
I get this error while computing partial correlation.
*Error in solve.default(Szz) :
system is computationally singular: reciprocal condition number =
4.90109e-18*
Why is it?Can anyone give me some idea ,how do i get rid it it?
This is the function i use for calculating partial correlation.
pcor.mat - function(x,y,z,method=p,na.rm=T){
x - c(x)
y - c(y)
z - as.data.frame(z)
if(dim(z)[2] == 0){
stop(There should be given data\n)
}
data - data.frame(x,y,z)
if(na.rm == T){
data = na.omit(data)
}
xdata - na.omit(data.frame(data[,c(1,2)]))
Sxx - cov(xdata,xdata,m=method)
xzdata - na.omit(data)
xdata - data.frame(xzdata[,c(1,2)])
zdata - data.frame(xzdata[,-c(1,2)])
Sxz - cov(xdata,zdata,m=method)
zdata - na.omit(data.frame(data[,-c(1,2)]))
Szz - cov(zdata,zdata,m=method)
# is Szz positive definite?
zz.ev - eigen(Szz)$values
if(min(zz.ev)[1]0){
stop(\'Szz\' is not positive definite!\n)
}
# partial correlation
Sxx.z - Sxx - Sxz %*% solve(Szz) %*% t(Sxz)
print(Sxx.z) # this gets printed
rxx.z - cov2cor(Sxx.z)[1,2] #some problem in this function function (V)
{
print(cov2cor)
p - (d - dim(V))[1]
if (!is.numeric(V) || length(d) != 2L || p != d[2L])
stop('V' is not a square numeric matrix)
Is - sqrt(1/diag(V))
if (any(!is.finite(Is)))
warning(diag(.) had 0 or NA entries; non-finite result is
doubtful)
r - V
r[] - Is * V * rep(Is, each = p)
r[cbind(1L:p, 1L:p)] - 1
r
}
return(rxx.z)
}
--
Thanks
Moumita
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Re: [R] How to avoid ifelse statement converting factor to character
Erratum: ifelse(TRUE,dd,dd) = 1230786000 (class numeric) should be ifelse(TRUE,tt,tt) = 1230786000 (class numeric) __ 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] Software lifecycle for R releases (aka practical limits of support for older versions)
Hello useRs:
Does anyone have thoughts on the lifecycle of older releases of R? I
know that currently the 2.8.x and 2.9.x releases seem to be actively
supported on the mailing lists, but what about older releases, say
2.4.x? Curious to hear when people think older versions of R become
obsolete and unsupportable on the lists (or other venues).
Regards,
Brian
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[R] Long memory residuals
Hi, How can I obtain the residuals of my long memory model: x.fd = fracdiff(dif, nar=1, nma=2, M=30) There is no function as acf() as for arima or garch models... Many thanks Ana [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to avoid ifelse statement converting factor to character
Dear Rolf,
Rolf Turner wrote:
On 25/06/2009, at 12:27 PM, Craig P. Pyrame wrote:
Dear Stavros,
What you discuss below is somewhat scary to me as an R newbie. Is this
just an incident, a bug perhaps, or rather the way things typically go
in R, as your Welcome to R! seems to suggest? I have just started to
learn R, and my initial euphoria of the I can do anything with it!
sort is gradually turning into an I can't get why it doesn't work and
I can't get how to make this work depression. I would be happy to
blame this on my incompetence and incapability, but would also like to
hear if it is not R itself that causes me to fail.
Do not get your knickers in a twist. R works simply and
straightforwardly
in simple straightforward situations. In less simple and less
straightforward
situations life gets more complicated. Don't dive into such
situations
without making sure you understand them. Check your results to
make sure
you have not overlooked a subtlety.
Yes, sure, but this sounds to me almost like what some others seem to
have already suggested on this list - use R for tasks where you can
simply apply any of the numerous excellent package functions (for
example, to fit a model), but avoid using it for programming, because it
inevitably leads to problems - is this what you are saying?
In respect of the current discussion of ifelse() --- the original
problem arose
because the values of ``yes'' and ``no'' were of different modes.
It is obvious
that in such instances a decision will have to be made about the
mode of the
result. The appropriateness of the designers' decision may be
disputed, but
you have to admit that some decision had to be made. Recognize
that and all
the mystery goes away.
I wasn't really updated on the original example, but yes, creating a
programming language involves numerous difficult decisions that have to
be made, often in a relatively ad hoc fashion. But once you point me to
the original example, I have looked into what Stavros wrote, and the
following strikes me as one of those cases where I would feel that
either me or R is wrong:
ifelse(TRUE,factor('x'),factor('x')) = 1 (integer)
The man page Stavros quotes states that the class attribute of the
result is taken from 'test', which clearly is not the case:
class(TRUE) = logical
class(ifelse(TRUE,factor('x'),factor('x'))) = integer
And also, I find myself incapable of making sense of the may in the
mode of the result may depend on the value of 'test' - may in what
sense? Will in 2.9 but not in 2.8? Will if test is logical but not if
it is raw? I am not particularly into programming languages, and
especially their semantics, so it may be my fault that I don't know what
such may may mean, but I am not quite sure. Could you help me?
If you don't understand what's going on, then just stick to using
ifelse() only
when ``yes'' and ``no'' have the same mode.
Yes, in my poor R programming I make efforts to avoid involving myself
in tricky situations, but it is not always helpful enough, I assure you.
Using things like as.raw() or taking one of ``yes'' and ``no'' to
be a list is getting
into territory where you need to be quite sophisticated and quite
careful. Unless
you are both, don't go there. I consider myself to be both (in
respect of R at
least) and I *still* would be very reluctant to go there.
I am not sure, is this supposed to be an argument in favor of using R?
Bottom line: R is easy to use at any level, but in order to use
it a ``high'' level
you need to understand the high level. Don't attempt to run
before you can crawl.
I am a bit reluctant to tell you this, Rolf, as there is likely much my
own fault involved, but I attempt to crawl, not to run, and this is
where I get into pains with R. Examples such as those of Stavros have
no place in my code, and yet I find it surprisingly difficult to
understand why large parts of my code don't work as expected - having
read the man pages carefully enough, I'd think.
Best regards,
Craig
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Re: [R] Software lifecycle for R releases (aka practical limits of support for older versions)
On Thu, 25 Jun 2009, Rowe, Brian Lee Yung (Portfolio Analytics) wrote: Does anyone have thoughts on the lifecycle of older releases of R? I know that currently the 2.8.x and 2.9.x releases seem to be actively supported on the mailing lists, but what about older releases, say 2.4.x? Curious to hear when people think older versions of R become obsolete and unsupportable on the lists (or other venues). Opinions vary, but: - reporting bugs (or asking if something is a bug) based on any older version of R than 2.9.x would likely get you flamed. - if your problem could be solved by updating to the current version, I think you would be expected to do so. My personal feeling is that you can just about get away with updating R only annually. Since you can easily keep an archive of previous versions available, there's no need to avoid updating on that account. Based just on R itself a longer update delay might be ok, but CRAN doesn't supply binaries of new or updated packages for old versions of R. Many packages will become seriously outdated much faster than base R. -thomas Thomas Lumley Assoc. Professor, Biostatistics tlum...@u.washington.eduUniversity of Washington, Seattle __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Software lifecycle for R releases (aka practical limits of support for older versions)
On Thu, Jun 25, 2009 at 9:38 AM, Rowe, Brian Lee Yung (Portfolio Analytics)b_r...@ml.com wrote: Hello useRs: Does anyone have thoughts on the lifecycle of older releases of R? I know that currently the 2.8.x and 2.9.x releases seem to be actively supported on the mailing lists, but what about older releases, say 2.4.x? Curious to hear when people think older versions of R become obsolete and unsupportable on the lists (or other venues). Regards, Brian This is actually a fairly common question from R users at commercial institutions, where for various reasons upgrading to the latest release of R isn't always possible. This might be for regulatory reasons (only a certain distribution of R has been validated), because of IT policies, of because R is incorporated into a production application where the risk of breaking the application outweighs the potential benefits of an upgrade. This is actually one of the main reasons why the release cycle for REvolution R isn't as frequent as that for CRAN R. The beneficial side effect is that we can therefore provide support for older versions of R in our distributions. We support R 2.7.2 through our distribution of REvolution R Enterprise, for example. More info here: http://www.revolution-computing.com/products/revolution-enterprise.php # David Smith -- David M Smith da...@revolution-computing.com Director of Community, REvolution Computing www.revolution-computing.com Tel: +1 (206) 577-4778 x3203 (San Francisco, USA) Check out our upcoming events schedule at www.revolution-computing.com/events __ 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] Software lifecycle for R releases (aka practical limits of support for older versions)
There is an archive for all packages for older versions of R, but if
you want up-to-date functionality of packages then you need the
newest versions.
my 2 cents
stephen
On Thu, Jun 25, 2009 at 12:38 PM, Rowe, Brian Lee Yung (Portfolio
Analytics)b_r...@ml.com wrote:
Hello useRs:
Does anyone have thoughts on the lifecycle of older releases of R? I
know that currently the 2.8.x and 2.9.x releases seem to be actively
supported on the mailing lists, but what about older releases, say
2.4.x? Curious to hear when people think older versions of R become
obsolete and unsupportable on the lists (or other venues).
Regards,
Brian
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[R] Problems with help
Hi, I was trying to read some help of functions and in all functions I try to see is giving me this error: ?write.table Erro em print.help_files_with_topic(C:/ARQUIV~1/R/R-29~1.0/library/utils/chm/write.table) : CHM file could not be displayed Anybody knows what is happening? Atenciosamente, Leandro Lins Marino Centro de Avaliação Fundação CESGRANRIO Rua Santa Alexandrina, 1011 - 2º andar Rio de Janeiro, RJ - CEP: 20261-903 R (21) 2103-9600 R.:236 ( (21) 8777-7907 ( lean...@cesgranrio.org.br Aquele que suporta o peso da sociedade é precisamente aquele que obtém as menores vantagens. (SMITH, Adam) Antes de imprimir pense em sua responsabilidade e compromisso com o MEIO AMBIENTE Esta mensagem, incluindo seus anexos, pode conter informacoes privilegiadas e/ou de carater confidencial, nao podendo ser retransmitida sem autorizacao do remetente. Se voce nao e o destinatario ou pessoa autorizada a recebe-la, informamos que o seu uso, divulgacao, copia ou arquivamento sao proibidos. Portanto, se você recebeu esta mensagem por engano, por favor, nos informe respondendo imediatamente a este e-mail e em seguida apague-a. __ 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] Software lifecycle for R releases (aka practical limits of support for older versions)
Good to know. I know that other software projects (whether languages, OSes, applications) tend to keep recent versions in maintenance mode for a certain period of time prior to retiring them. I wonder if that would happen with R, either by design or out of necessity of an increasing user base. Brian -Original Message- From: Thomas Lumley [mailto:tlum...@u.washington.edu] Sent: Thursday, June 25, 2009 12:50 PM To: Rowe, Brian Lee Yung (Portfolio Analytics) Cc: r-h...@stat.math.ethz.ch Subject: Re: [R] Software lifecycle for R releases (aka practical limits of support for older versions) On Thu, 25 Jun 2009, Rowe, Brian Lee Yung (Portfolio Analytics) wrote: Does anyone have thoughts on the lifecycle of older releases of R? I know that currently the 2.8.x and 2.9.x releases seem to be actively supported on the mailing lists, but what about older releases, say 2.4.x? Curious to hear when people think older versions of R become obsolete and unsupportable on the lists (or other venues). Opinions vary, but: - reporting bugs (or asking if something is a bug) based on any older version of R than 2.9.x would likely get you flamed. - if your problem could be solved by updating to the current version, I think you would be expected to do so. My personal feeling is that you can just about get away with updating R only annually. Since you can easily keep an archive of previous versions available, there's no need to avoid updating on that account. Based just on R itself a longer update delay might be ok, but CRAN doesn't supply binaries of new or updated packages for old versions of R. Many packages will become seriously outdated much faster than base R. -thomas Thomas Lumley Assoc. Professor, Biostatistics tlum...@u.washington.eduUniversity of Washington, Seattle -- This message w/attachments (message) may be privileged, confidential or proprietary, and if you are not an intended recipient, please notify the sender, do not use or share it and delete it. Unless specifically indicated, this message is not an offer to sell or a solicitation of any investment products or other financial product or service, an official confirmation of any transaction, or an official statement of Merrill Lynch. Subject to applicable law, Merrill Lynch may monitor, review and retain e-communications (EC) traveling through its networks/systems. The laws of the country of each sender/recipient may impact the handling of EC, and EC may be archived, supervised and produced in countries other than the country in which you are located. This message cannot be guaranteed to be secure or error-free. References to Merrill Lynch are references to any company in the Merrill Lynch Co., Inc. group of companies, which are wholly-owned by Bank of America Corporation. Secu! rities and Insurance Products: * Are Not FDIC Insured * Are Not Bank Guaranteed * May Lose Value * Are Not a Bank Deposit * Are Not a Condition to Any Banking Service or Activity * Are Not Insured by Any Federal Government Agency. Attachments that are part of this E-communication may have additional important disclosures and disclaimers, which you should read. This message is subject to terms available at the following link: http://www.ml.com/e-communications_terms/. By messaging with Merrill Lynch you consent to the foregoing. -- __ 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] Software lifecycle for R releases (aka practical limits of support for older versions)
On Jun 25, 2009, at 11:38 AM, Rowe, Brian Lee Yung (Portfolio Analytics) wrote: Hello useRs: Does anyone have thoughts on the lifecycle of older releases of R? I know that currently the 2.8.x and 2.9.x releases seem to be actively supported on the mailing lists, but what about older releases, say 2.4.x? Curious to hear when people think older versions of R become obsolete and unsupportable on the lists (or other venues). Regards, Brian For a description of R's formal SDLC, read: http://www.r-project.org/doc/R-FDA.pdf While that document is targeted to R users in the domain of regulated clinical trials, much of the content is relevant to other general use domains. From the perspective of getting community support on R-Help, if you are using version 2.4.0 and you post a version independent query to the list, you will get a helpful reply, especially if you don't include in the post that you are running version 2.4.0. However, it is possible that in the replies, there may be references to functions, function arguments or packages that are part of or are designed for newer versions of R. Upon reading that reply, you may end up scratching your head, wondering why you cannot find them in your version, which may prompt you to reply requesting clarification. That may lead you down the path to the next scenario... If you include in your post (or a follow up) that you are actually using version 2.4.0, you will get a series of rather curt recommendations to update to the current release version of R included in any responses to your query. However, if you post a query pertaining to what you perceive as a bug in 2.4.0 or a more recent version (possibly even 2.9.0 with 2.9.1 imminent), you will get a pretty rapid stream of replies, with a level of hostility (flaming) included. Those replies will tell you in no uncertain terms, that you better upgrade to the most recent version of R (which may include a patched version) before reporting bugs against versions that from a development standpoint, are no longer supported. You would be expected to check the most recent version that you can install to see if the behavior that you perceive as a bug is still present. The worst case scenario perhaps, in terms of being on the receiving end of flames, would be to actually submit a formal bug report on an older version of R, as that requires a **volunteer** member of R Core to have to stop what they are doing and spend time manually administering that report. Finally, a good reference to go along with this general discussion, is the Posting Guide, listed at the bottom of all e-mails coming from the list: http://www.r-project.org/posting-guide.html 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.
[R] interaction terms formula error
Hi,
I'm trying to fit a binary logistic regression model, and would like to
consider certain characteristics B and C only for people with variable A=1 and
not for those with variable A=0, so I'm trying to do the following:
model- lrm(formula= y ~ A: (B+C) + D + E +...)
I've had no problem with adding interaction terms using * , but every time I
try to use : , as in my example, I get an error message like:
Error in if (!length(fname) || !any(fname == zname)) { :
Missing value where TRUE/FALSE needed
I would really appreciate any advice on how to solve this, I don't know what in
the notation should be different for different operators.
Thanks in advance, kind regards,
Alejandra Solís
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Re: [R] interaction terms formula error
Alejandra Solis Herrera wrote:
Hi,
I'm trying to fit a binary logistic regression model, and would like to
consider certain characteristics B and C only for people with variable A=1 and
not for those with variable A=0, so I'm trying to do the following:
model- lrm(formula= y ~ A: (B+C) + D + E +...)
I've had no problem with adding interaction terms using * , but every time I try to
use : , as in my example, I get an error message like:
The Design package doesn't support that. It uses the (highly
recommended) hierarchy principle so it assumes you will use * for
interactions.
Frank
Error in if (!length(fname) || !any(fname == zname)) { :
Missing value where TRUE/FALSE needed
I would really appreciate any advice on how to solve this, I don't know what in
the notation should be different for different operators.
Thanks in advance, kind regards,
Alejandra Solís
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[R] How to draw a line in plot when I know the start point(x1, y1) and end point(x2, y2)?
Hello all, How to draw a line in plot when I know the start point(x1,y1) and end point(x2,y2)? I need make this as additional information in the graph: plot(wl2[[1]],wl2[[2]]) I think that is possible make this with the function abline(), is possible? I looked the function lines() too, but don't understand as make. Thanks! Lesandro Veja quais são os assuntos do momento no Yahoo! +Buscados [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to draw a line in plot when I know the start point(x1, y1) and end point(x2, y2)?
Dear Lesandro, Take a look at ?segments HTH, Jorge On Thu, Jun 25, 2009 at 2:30 PM, Lesandro lesand...@yahoo.com.br wrote: Hello all, How to draw a line in plot when I know the start point(x1,y1) and end point(x2,y2)? I need make this as additional information in the graph: plot(wl2[[1]],wl2[[2]]) I think that is possible make this with the function abline(), is possible? I looked the function lines() too, but don't understand as make. Thanks! Lesandro Veja quais são os assuntos do momento no Yahoo! +Buscados [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to draw a line in plot when I know the start point(x1, y1) and end point(x2, y2)?
On Jun 25, 2009, at 1:30 PM, Lesandro wrote: Hello all, How to draw a line in plot when I know the start point(x1,y1) and end point(x2,y2)? I need make this as additional information in the graph: plot(wl2[[1]],wl2[[2]]) I think that is possible make this with the function abline(), is possible? I looked the function lines() too, but don't understand as make. Thanks! Lesandro See ?segments which does just what you are looking for. lines() is more designed for a series of connected lines (eg. a polygon) rather than a single line segment. abline() can draw a straight line, at a given vertical or horizontal position, or if given a linear model object, the fitted line. 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.
Re: [R] How to draw a line in plot when I know the start point(x
On 25-Jun-09 18:38:37, Marc Schwartz wrote: On Jun 25, 2009, at 1:30 PM, Lesandro wrote: Hello all, How to draw a line in plot when I know the start point(x1,y1) and end point(x2,y2)? I need make this as additional information in the graph: plot(wl2[[1]],wl2[[2]]) I think that is possible make this with the function abline(), is possible? I looked the function lines() too, but don't understand as make. Thanks! Lesandro See ?segments which does just what you are looking for. lines() is more designed for a series of connected lines (eg. a polygon) rather than a single line segment. abline() can draw a straight line, at a given vertical or horizontal position, or if given a linear model object, the fitted line. HTH, Marc Schwartz Hmm ... for this particular purpose I don't see what is wrong with plot(wl2[[1]],wl2[[2]]) lines(c(x1,x2),c(y1,y2)) along with any additional paramaters to lines() for line-type, colour, etc. -- I do this all the time ... Ted. E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk Fax-to-email: +44 (0)870 094 0861 Date: 25-Jun-09 Time: 19:51:20 -- 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.
[R] LaTeX references inside R code using SweaveListingUtils
I have sent a note to Peter Ruckdeschel who wrote the excellent
SweaveListingUtils package but find myself up against a deadline in
preparing a handout for useR! 2009.
The following is supposed to work:
=
plot(x, y) # Figure `\ref{myfig}`
@
Where the back tick ` is an escape character and \ref{myfig} will be
replaced by LaTeX with the correct figure number that was created by a
\label{myfig} command. I have not been able to get this to work. I
just get `\ref{myfig}` verbatim on the R output.
Does anyone know of a way to do this?
Thanks
Frank
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Re: [R] How to draw a line in plot when I know the start point(x
On Jun 25, 2009, at 1:51 PM, Ted Harding wrote: On 25-Jun-09 18:38:37, Marc Schwartz wrote: On Jun 25, 2009, at 1:30 PM, Lesandro wrote: Hello all, How to draw a line in plot when I know the start point(x1,y1) and end point(x2,y2)? I need make this as additional information in the graph: plot(wl2[[1]],wl2[[2]]) I think that is possible make this with the function abline(), is possible? I looked the function lines() too, but don't understand as make. Thanks! Lesandro See ?segments which does just what you are looking for. lines() is more designed for a series of connected lines (eg. a polygon) rather than a single line segment. abline() can draw a straight line, at a given vertical or horizontal position, or if given a linear model object, the fitted line. HTH, Marc Schwartz Hmm ... for this particular purpose I don't see what is wrong with plot(wl2[[1]],wl2[[2]]) lines(c(x1,x2),c(y1,y2)) along with any additional paramaters to lines() for line-type, colour, etc. -- I do this all the time ... Ted. Nothing wrong at all Ted. It will of course work. For example: plot(1:10) lines(c(2, 4), c(6, 8)) That will give you the same result as: plot(1:10) segments(2, 6, 4, 8) In this case, it may be a matter of choice. For single lines, I tend to use segments(). As is frequently the case with R, there is more than one way to skin the feline... Regards, Marc __ 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] (Resolved) How to draw a line in plot when I know the start point(x1, y1) and end point(x2, y2)?
Thanks Jorge and Marc, I drew the line using the function: segments(x0, y0, x1, y1) Lesandro --- Em qui, 25/6/09, Marc Schwartz marc_schwa...@me.com escreveu: De: Marc Schwartz marc_schwa...@me.com Assunto: Re: [R] How to draw a line in plot when I know the start point(x1, y1) and end point(x2, y2)? Cc: r-help@r-project.org Data: Quinta-feira, 25 de Junho de 2009, 15:38 On Jun 25, 2009, at 1:30 PM, Lesandro wrote: Hello all, How to draw a line in plot when I know the start point(x1,y1) and end point(x2,y2)? I need make this as additional information in the graph: plot(wl2[[1]],wl2[[2]]) I think that is possible make this with the function abline(), is possible? I looked the function lines() too, but don't understand as make. Thanks! Lesandro See ?segments which does just what you are looking for. lines() is more designed for a series of connected lines (eg. a polygon) rather than a single line segment. abline() can draw a straight line, at a given vertical or horizontal position, or if given a linear model object, the fitted line. HTH, Marc Schwartz [[elided Yahoo spam]] [[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] what happened to the xlsReadWrite package
A naive question: what happened to the xlsReadWrite package? http://cran.r-project.org/web/packages/xlsReadWrite/ It says that it was removed from the CRAN repository. Are there any plans for it be available again? Thanks, Andrew [[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] grid.polygon() + color gradient
Glad if it helps.
check out this page of examples for tikz,
http://www.texample.net/tikz/examples/feature/shadings/
If you do choose this route, you could perhaps read the new wiki page on
importing graphics in a R plot,
http://wiki.r-project.org/rwiki/doku.php?id=tips:graphics-misc:display-images
(you'll want to convert the result of Tikz into a bitmap format first, with
imagemagick for example).
As for the pure R graphics solution, you'll have to play with Grid viewports
(and perhaps the gridBase package if your Figure is using base graphics).
Can you post a minimal, self-contained, reproducible example of what you're
trying to achieve to the list?
baptiste
2009/6/25 Kexin Ji kex...@gmail.com
Hi,
The triangle looks great!!
And thanks for mentioning the TeX package Tikz! Maybe I'll check it out
later.
The only problem is that I need to append this color-gradient triangle into
a another Figure I'm working on. But when I try to do that, this wonderful
triangle overwrites the other one. Have tried to append it with not much
luck..
Much appreciation to your help though!!!
Kexin
On 6/25/09, baptiste auguie baptiste.aug...@gmail.com wrote:
Hi,
I don't think the fill parameter can be a colour gradient. You'll need to
create small polygons, each with its own fill (200, say). Try this,
x= c(0, 0.5, 1)
y= c(0.5, 1, 0.5)
grid.polygon(x=x, y=y, gp=gpar(fill=grey90, col=grey90))
xx - seq(range(x)[1],range(x)[2], length=100)
yy - rep(max(y), length(xx))
cols - colorRampPalette(c(green, lightgray))(length(xx))
for(ii in seq_along(xx[-length(xx)])) {
grid.clip(x=xx[ii], y=0.5,
width= xx[ii+1],
height=1,
just=bottom)
grid.polygon(x=c(0, 0.5, 1), y=c(0.5, 1, 0.5), gp=gpar(fill=cols[ii],
col=NA))
}
Note that the situation would become rather more complicated for a
gradient at some angle (see ?grobX if you need to).
If you're free to choose an external tool to produce this, the TeX package
Tikz has good support for gradients and clipping.
HTH,
baptiste
Kexin Ji wrote:
Hi,
I wonder whether there is a way to generate a polygon (a triangle in my
case) with color gradient using grid.polygon() in package grid?
I tried something like
library(grid)
grid.polygon(x=c(0, 0.5, 1), y=c(0.5, 1, 0.5), gp=gpar(col=NA,
fill=colorRampPalette(c(green, lightgray),
space=Lab)(200)))
But am only getting a triangle filled with color green, whereas the aim
is a triangle of color gradient from green to lightgray.
Can grid.polygon() generate a color gradient, or am I being mistaken?
Best to my knowledge, is it true that R currently doesn't contain any
other function that might generate a polygon with color gradient?
Thank you!
Kexin
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_
Baptiste Auguié
School of Physics
University of Exeter
Stocker Road,
Exeter, Devon,
EX4 4QL, UK
Phone: +44 1392 264187
http://newton.ex.ac.uk/research/emag
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--
_
Baptiste Auguié
School of Physics
University of Exeter
Stocker Road,
Exeter, Devon,
EX4 4QL, UK
Phone: +44 1392 264187
http://newton.ex.ac.uk/research/emag
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] change the height or scale of the y axis
Hi, Jim, Thank you for your reply. I just want to increase the height of y axis in the second plot in order to show all the indices (x1, x2, ...). Can you help me? Thank you again. Legen Jim Lemon-2 wrote: legen wrote: Hallo, All, I have a question about changing the height or scale of the y axis. When I use following two R codes, I can get two plots. Please look at the y axes, the number of indices (x1, x2, …) on the y axis in the first plot is smaller than that in the second plot, and hence the space between any two indices in the first plot is wider than that in the second plot. As the number of indices increases, the space will vanish and the indices will overlap. I want to display all the indices on the y axis in the second plot, just look like that in the first plot. How to separate the indices on the y axis in the second plot? I guess maybe changing the height or scale of y axis is a way to solve my problem, but I failed to do it after several trails. Anybody can help me? Thank you in advance. Legen The first R code: x-c(x1,x2,x3,x4,x5,x6,x7,x8,x9,x10) y-sample(0:100,10,replace=F) d-data.frame(y) rownames(d)-x r-nrow(d) i-1:r plot(d$y,i,xlab=,ylab=,axes=F) axis(1) axis(2,at=i,labels=x) The second R code: x-c(x1,x2,x3,x4,x5,x6,x7,x8,x9,x10, x11,x12,x13,x14,x15,x16,x17,x18,x19,x20, x21,x22,x23,x24,x25,x26,x27,x28,x29,x30, x31,x32,x33,x34,x35,x36,x37,x38,x39,x40, x41,x42,x43,x44,x45,x46,x47,x48,x49,x50, x51,x52,x53,x54,x55,x56,x57,x58,x59,x60, x61,x62,x63,x64,x65,x66,x67,x68,x69,x70, x71,x72,x73,x74,x75,x76,x77,x78,x79,x80, x81,x82,x83,x84,x85,x86,x87,x88,x89,x90, x91,x92,x93,x94,x95,x96,x97,x98,x99,x100, x101,x102,x103,x104,x105,x106,x107,x108,x109,x110, x111,x112,x113,x114,x115,x116,x117,x118,x119,x120, x121,x122,x123,x124,x125,x126,x127,x128,x129,x130, x131,x132,x133,x134,x135,x136,x137,x138,x139,x140, x141,x142,x143,x144,x145,x146,x147,x148,x149,x150, x151,x152,x153,x154,x155,x156,x157,x158,x159,x160, x161,x162,x163,x164,x165,x166,x167,x168,x169,x170, x171,x172,x173,x174,x175,x176,x177,x178,x179,x180, x181,x182,x183,x184,x185,x186,x187,x188,x189,x190, x191,x192,x193,x194,x195,x196,x197,x198,x199,x200) y-sample(0:300,200,replace=F) d-data.frame(y) rownames(d)-x r-nrow(d) i-1:r plot(d$y,i,xlab=,ylab=,axes=F) axis(1) axis(2,at=i,labels=x) Hi legen, Your problem centers about the fact that you have about 3 times the vertical scale on the second plot and 20 times the number of ticks and labels. You could make the second plot 3 times the height of the first one, and you would get approximately the same spacing of equal intervals, but the intervals on the second plot are likely to be about 1.5 times those of the first. I can't work out why you are trying to do what you say you want to do. Maybe some more explanation would help. Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/change-the-height-or-scale-of-the-y-axis-tp24187351p24206583.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] change the height or scale of the y axis
Hi, Dieter Menne, Thank you for your help. I tried par(las=1,cex=0.5), but it changed only the size of indices on the y axis in the second plot relative to the default. I really want to increase the height of y axis in order to show all the indices (x1, x2, ...). In the genetic study, we aften draw plots of chromosomes with lots of genes (x1, x2, ...) and their positons on the chromosome (numerical values). Legen Dieter Menne wrote: legen wrote: I have a question about changing the height or scale of the y axis. When I use following two R codes, I can get two plots. Please look at the y axes, the number of indices (x1, x2, …) on the y axis in the first plot is smaller than that in the second plot, and hence the space between any two indices in the first plot is wider than that in the second plot. As the number of indices increases, the space will vanish and the indices will overlap. I want to display all the indices on the y axis in the second plot, just look like that in the first plot. How to separate the indices on the y axis in the second plot? I guess maybe changing the height or scale of y axis is a way to solve my problem, but I failed to do it after several trails. Anybody can help me? Thank you in advance. You really put a lot of typing work into your example code, but it work, so it's fine. As a first attempt, put the following before the second plot(). par(las=1,cex=0.5) And think about the paper size. I suggest to have a look at lattice graphics for this type of work as a more elegant alternative. And Hadley Wickham can certainly tell you how to do this in ggplot47 -- View this message in context: http://www.nabble.com/change-the-height-or-scale-of-the-y-axis-tp24187351p24206684.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] Calculating home ranges using mcp in adehabitat
Hello, I've been trying to calculate home range sizes (for Icelandic geese!) using minimum convex polygons with the adehabitat package. I've tried to use the R code shown by demo(homerange) in adehabitat and when that didn't work I've fiddled around with it but to no avail...Below is the output of the demo that I've attempted to follow, followed by a subset of my data (for one individual) and the code I've been trying. Can anyone tell me where I may be going wrong??? #Demo output... data(puechabon) xy-puechabon$locs[,c(X,Y)] id-puechabon$locs$Name ## The data are: xy[1:4,] ## relocations coordinates X Y 1 699889 3161559 2 700046 3161541 3 698840 3161033 4 699809 3161496 id[1:4] ## ID [1] Brock Brock Brock Brock Levels: Brock Calou Chou Jean ### ### ### ### Home ranges ## MCP hr-mcp(xy, id) ## home range estimation # My attempts... mydata Name XY 1 XDRY -21.98389 64.06457 2 XDRY -21.99759 64.08291 3 XDRY -21.98784 64.06467 4 XDRY -21.98333 64.06058 5 XDRY -21.97889 64.06257 6 XDRY -21.98284 64.06044 7 XDRY -21.97886 64.06358 8 XDRY -21.99741 64.08124 9 XDRY -21.99715 64.08330 10 XDRY -22.00397 64.09331 11 XDRY -21.99811 64.08239 12 XDRY -22.00453 64.09337 13 XDRY -21.99713 64.08299 14 XDRY -21.99608 64.08307 15 XDRY -21.99646 64.08352 16 XDRY -21.99326 64.08361 17 XDRY -22.00770 64.09090 18 XDRY -21.98160 64.06400 19 XDRY -21.97966 64.06367 xy-mydata$locs[,c(X,Y)] id-mydata$locs$Name xy[1:19,] NULL id[1:19,] NULL mcp(xy,id,percent=95) Error in if (length(id) != nrow(xy)) stop(xy and id should be of the same length) : argument is of length zero #I've also tried things like the code below. But didn't know how to incorporate 'locs' into this... xy-cbind(X,Y) id-Name[1:19] xy[1:19,] XY [1,] -21.98389 64.06457 [2,] -21.99759 64.08291 [3,] -21.98784 64.06467 [4,] -21.98333 64.06058 [5,] -21.97889 64.06257 [6,] -21.98284 64.06044 [7,] -21.97886 64.06358 [8,] -21.99741 64.08124 [9,] -21.99715 64.08330 [10,] -22.00397 64.09331 [11,] -21.99811 64.08239 [12,] -22.00453 64.09337 [13,] -21.99713 64.08299 [14,] -21.99608 64.08307 [15,] -21.99646 64.08352 [16,] -21.99326 64.08361 [17,] -22.00770 64.09090 [18,] -21.98160 64.06400 [19,] -21.97966 64.06367 id-Name[1:19] id[1:19] [1] XDRY XDRY XDRY XDRY XDRY XDRY XDRY XDRY XDRY XDRY XDRY [12] XDRY XDRY XDRY XDRY XDRY XDRY XDRY XDRY Levels: XDRY mcp(xy,id,percent=95) Error in apply(xy, 2, mean) : dim(X) must have a positive length xy1-xy[1:19,] id1-id[1:19] mcp(xy1,id1,percent=95) Error in apply(xy, 2, mean) : dim(X) must have a positive length Would be most grateful for any suggestions to where I'm going wrong! Tom Mason, University of Exeter, UK. -- View this message in context: http://www.nabble.com/Calculating-home-ranges-using-mcp-in-adehabitat-tp24207532p24207532.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] Converting S-plus project folders to R
I have many S-plus project folders that I need to convert to R workspaces. For the smaller project folders ( 200MB), using data.dump with oldStyle = T and data.restore (in the foreign package) within R seems to work fine. However, I have several project folders that are quite large (~ 4GB). When I use this procedure to try to convert these project folders, R always crashes when I perform the data.restore command. Does anyone have any suggestions? Thanks, Dave Bosley -- View this message in context: http://www.nabble.com/Converting-S-plus-project-folders-to-R-tp24209758p24209758.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.
