Re: [R] ReadWrite.xls problem
On Dec 8, 2010, at 15:52 , tkdweber wrote: Dear community, I have now taken my R-file from lectures and intend to use it at home, but have a problem reading the Data from the file. I have installed and loaded the Package xlsReadWrite so far. xlsReadWrite is a contributed package with a complicated installation procedure. It has a maintainer and a website. Offhand I would guess that you haven't downloaded the proprietary DLL that goes with it, and maybe that the placeholder DLL has not been updated when the number of arguments was changed. Is there any good reason not to just use the save/load mechanism? I have also Changed directory. This is what I have entered daten=read.xls(Daten A2) This is my Error-Message in its German original: Fehler in .Call(ReadXls, file, colNames, sheet, type, from, rowNames, : Falsche Anzahl von Argumenten (11), erwarte 10 für ReadXls This is my Error-Message in its english translation: Error in .Call(ReadXls, file, colNames, sheet, type, from, rowNames, : Wrong Number of Arguments (11), expect 10 for ReadXls Please help me in solving this problem. Without being able to read data, the programme renders pointless for me :-( Chers, Toby -- View this message in context: http://r.789695.n4.nabble.com/ReadWrite-xls-problem-tp3078348p3078348.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. -- Peter Dalgaard Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.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] Sweave: Setting options with SweaveOpts{} when using driver=RweaveHTML
When using Sweave in connection with the driver RweaveLatex(), global options
can be set with \SweaveOpts{}, e.g.
\SweaveOpts{keep.source=T}.
Does anybody know if it is possible to set global options in the same way when
using Sweave with the driver RweaveHTML().
Regards
Søren
[[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] importing date vector with read.table
Hi, I understand this should be an easy task but I am still struggling a bit to read a .txt file with a date vector. My code is as below: data-read.table(file.choose(),header=TRUE,sep=\t,dec=,,colClasses=c(Date,numeric,numeric,numeric)) But I am getting an error: Error in charToDate(x) : character string is not in a standard unambiguous format because my date vector is in a format below: format=%d.%m.%Y But how can i define the date format for date vector in read.table function? Br, Jack [[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] .por files
hi, i have a spss portable file which i cannot open in R. i tried library(Hmisc) File.spss - spss.get(File.por, use.value.labels=T) and received: Fehler in read.spss(File.por, : Datei »File.por« ist nicht in einem unterstützten SPSS-Format (in engl.: error, not supported spss format) then i tried library(memisc) File - spss.portable.file(File.por) and got: WARNING: slash not found ans = l832112+JGJJs6tiskYVsGfoVisDRTZekfjY462+52+7,+5+41KLD5+8+2+5+8+2+3K+5ZeXReXjUlRkFehler in readStringPorStream(stream) : kann Speicherblock der Größe 3.3 GB nicht zuteilen the file is 1023 kb i can open the file in spss - so it is not damaged or anything. any ideas? sessionInfo() R version 2.12.0 (2010-10-15) Platform: i486-pc-linux-gnu (32-bit) locale: [1] LC_CTYPE=de_CH.UTF-8 LC_NUMERIC=C [3] LC_TIME=de_CH.UTF-8 LC_COLLATE=de_CH.UTF-8 [5] LC_MONETARY=de_CH.UTF-8 LC_MESSAGES=de_CH.UTF-8 [7] LC_PAPER=de_CH.UTF-8 LC_NAME=de_CH.UTF-8 [9] LC_ADDRESS=de_CH.UTF-8LC_TELEPHONE=de_CH.UTF-8 [11] LC_MEASUREMENT=de_CH.UTF-8LC_IDENTIFICATION=de_CH.UTF-8 attached base packages: [1] splines grid stats graphics grDevices utils datasets [8] methods base other attached packages: [1] memisc_0.95-31 Hmisc_3.8-3 Deducer_0.4-2foreign_0.8-41 [5] effects_2.0-10 colorspace_1.0-1 lattice_0.19-13 multcomp_1.2-4 [9] mvtnorm_0.9-95 car_2.0-6survival_2.36-2 nnet_7.3-1 [13] MASS_7.3-5 ggplot2_0.8.8proto_0.3-8 reshape_0.8.3 [17] plyr_1.2.1 JGR_1.7-4iplots_1.1-3 JavaGD_0.5-3 [21] rJava_0.8-8 loaded via a namespace (and not attached): [1] cluster_1.13.2 tools_2.12.0 kat __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] importing date vector with read.table
On Fri, 10 Dec 2010, Jack Johnson wrote: Hi, I understand this should be an easy task but I am still struggling a bit to read a .txt file with a date vector. My code is as below: data-read.table(file.choose(),header=TRUE,sep=\t,dec=,,colClasses=c(Date,numeric,numeric,numeric)) But I am getting an error: Error in charToDate(x) : character string is not in a standard unambiguous format because my date vector is in a format below: format=%d.%m.%Y But how can i define the date format for date vector in read.table function? It is easiest to simply read the column as character, and convert later by as.Date(x, format=%d.%m.%Y) You might want to consider using read.delim2 to get all the usual defaults for a tab-delimited file in a locale using comma as decimal separator. Br, Jack [[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. -- 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] Projecting data on a world map using long/lat
On Fri, Dec 10, 2010 at 2:21 AM, mathijsdevaan mathijsdev...@gmail.com wrote: Hi, I have a dataset (CSV) with some counts of firms located around the globe. Each count is assigned to the longitude and latitude of the specific location. Now I want to plot these counts on a world map using dots (size of dots represent the count). I have been unable to find any info on whether this is possible and if so, how? Can you please help me? Thanks! Plotting points is trivial - plot(data$x,data$y,pch=19,cex=data$size) will do for a start. i'm guessing your real problem is when you say 'on a world map'. How detailed a world map do you need? There's an outline one in the 'maps' package, or you should be able to find a shapefile of the world on the web somewhere and use that via the rgdal package. Other options include making a KML file of your points and overlaying on google earth. Or getting google map tiles and overlaying on that Or exporting your data to a GIS format and doing the pretty map in something like Quantum GIS. What are you trying to do exactly? also, you might want to post to r-sig-geo Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Minimization of the distance
On Thu, Dec 09, 2010 at 07:43:52PM -0800, bluesky wrote: I just contect R,and still learn how to write the code. I have a problem with argmin sum d(pi,p)/n for example I have 3 points (a1,b1)(a2,b2)(a3,b3) ,then I want to find p(x,y) make sure that (sqrt((a1-x)^2+(b1-y)^2)+sqrt((a2-x)^2+(b2-y)^2)+sqrt((a3-x)^2+(b3-y)^2))/3 is the minimum. The following code solves the example as i understand it. # rows of matrix a are three points in the plane a - rbind( c(1, 1), c(2.3, 1), c(3, 3)) d - function(x, a) mean(sqrt(rowSums((a - rep(x, each=nrow(a)))^2))) xinit - colMeans(a) x - optim(xinit, d, a=a)$par plot(a) points(rbind(x), col=2) Is this, what you mean? Function optim() has further parameters, which influence efficiency and accuracy, and there are also other optimization functions. Petr Savicky. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] RES: Barplot with Independent Lines Y axis
OK that's it. Working nicely. I sent the final graph with the note of the
twoord.plot help attached to it. And some personal comments
Thank you very much for all the help and remarks.
Rodrigo.
2010/12/10 Dennis Murphy djmu...@gmail.com
Hi:
Like Peter Ehlers, I'm not a big fan of multiple response variables on a
page with different y-axis scaling, but if you have to do it, try not to let
one graphical metaphor interfere with/obscure/dominate the other (e.g., bars
and lines). This is my attempt, both with connecting lines and points, but
am holding my nose in the process:
library(plotrix)
dd - read.table(textConnection(
Point RainSalt
Fev/03 365.6 13
Mar/03 235 18
Abr/03 115.1 18
Mai/03 47.418.75
Jun/03 112 15
Jul/03 156.8 17
Ago/03 66.115
Set/03 149.8 14
Out/03 167.1 11.5
Nov/03 269.3 17.5
Dez/03 283.7 NA
Jan/04 415 1.3
Fev/04 322 8.5
Mar/04 258.7 10.5), header = TRUE, stringsAsFactors = FALSE)
closeAllConnections()
dd$mo - seq(nrow(dd))
with(dd,
twoord.plot(lx = mo, ly = Rain, rx = mo, ry = Salt,
lylim = c(0, 450), rylim = c(0, 20),
lcol = 'red', rcol = 'blue',
lpch = 1, rpch = 16, type = c('b', 'b'),
xtickpos = mo, xticklab = Point)
)
legend('bottomleft', leg = c('Rain (left axis)', 'Salt (right axis)'),
text.col = c('red', 'blue'), col = c('red', 'blue'), lty = c(1, 1)
)
I understand why these types of plots exist in general, and I can see why
you might want to compare two variables that are temporally related but have
different units of measurement, but by doing so, you are increasing the
cognitive task of the average viewer. As Greg Snow mentioned, read the
section of ?twoord,plot headlined 'Note' and observe that this plot has
crossing profiles (points connected by lines). And remember that it's your
responsibility to properly convey the message of the plot to the viewer...
HTH,
Dennis
On Thu, Dec 9, 2010 at 2:22 PM, Rodrigo Aluizio r.alui...@gmail.comwrote:
Oh sorry. An example say lots more than words. The data below, when
submitted to twoord.plot return the mentioned error. Rain are bars and
Salt
lines, the bars appear and the error occurs with the salt data.
Point RainSalt
Fev/03 365.6 13
Mar/03 235 18
Abr/03 115.1 18
Mai/03 47.418.75
Jun/03 112 15
Jul/03 156.8 17
Ago/03 66.115
Set/03 149.8 14
Out/03 167,1 11.5
Nov/03 269.3 17.5
Dez/03 283.7
Jan/04 415 1,3
Fev/04 322 8,5
Mar/04 258.7 10.5
-Mensagem original-
De: Greg Snow [mailto:greg.s...@imail.org]
Enviada em: quinta-feira, 9 de dezembro de 2010 17:57
Para: Rodrigo Aluizio; 'R Help'
Assunto: RE: [R] Barplot with Independent Lines Y axis
Without seeing a reproducible example we cannot be sure, but my guess is
that you are letting twoord.plot set the limits and the function does not
remove missing values, if you specify rylim and or lylim arguments
specifically, then it should not run into the problem you are seeing (I
hope).
If that does not work, then send a small reproducible example (the dput
command is great for the data part) which will help us find the problem.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From: Rodrigo Aluizio [mailto:r.alui...@gmail.com]
Sent: Thursday, December 09, 2010 12:27 PM
To: Greg Snow; 'R Help'
Subject: RES: [R] Barplot with Independent Lines Y axis
Thank you for the function suggestion, works nicely for complete data
vectors. Just another question. When using the twoord.plot I reached
another
issue, it seems that this function can't handle NAs in one of the
variables
(the data of one month for another variable is missing). The lines and
barplot functions just interrupt the plot and continue after the NA but
twoord.plot gives me an error: Error at plot.window(...) : finite
values are
necessary for 'ylim'
Is there a way to work around this error?
P.S.: I'm aware of the problems regarding this type of graph, but this
time
it's not a choice of mine (unfortunately). Thank you anyway for the
highlight, maybe it will help me arguing.
Regards
Rodrigo.
-Mensagem original-
De: Greg Snow [mailto:greg.s...@imail.org]
Enviada em: quinta-feira, 9 de dezembro de 2010 15:56
Para: Rodrigo Aluizio; R Help
Assunto: RE: [R] Barplot with Independent Lines Y axis
Look at the twoord.plot function in the plotrix package, but be sure to
read
the note on the help page, then reread it and take its advice if you
decide
to stick with this type of plot.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From: r-help-boun...@r-project.org
Re: [R] RES: Barplot with Independent Lines Y axis
On 12/10/2010 08:48 PM, Rodrigo Aluizio wrote: OK that's it. Working nicely. I sent the final graph with the note of the twoord.plot help attached to it. And some personal comments Hi Rodrigo, Although I came rather late to this, I might as well show how I did it: # first I corrected the commas as Peter pointed out library(plotrix) par(las=3) twoord.plot(ly=pluv_sal$Rain,lx=pluv_sal$Salt,mar=c(6,4,4,2), xlab=,ylab=Rainfall (mm),rylab=Salinity,type=c(bar,b), main=Rainfall and salinity,lcol=NA,rcol=2, xticklab=pluv_sal$Point,xaxt=n) par(las=0) mtext(Month,side=1,at=7,line=3) You did find a bug in the twoord.plot function, and the next version of plotrix will contain one more bug fix, with thanks to you. Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Compare one level of a factor with *all* other non-missing levels
Dear list, I try to compare the mean of a variable given a value of a factor with the mean of the same variable for all K-1 other non-missing values of this factor. This procedure I want to repeat for each level of the factor. Having read the recommendations of this list I want to avoid creating K-1 dummy variables and searched for options of the pairwise.t.test. But couldn't find a solution. Anyone with a suggestion how to do the comparisions? Cheers, Derik -- View this message in context: http://r.789695.n4.nabble.com/Compare-one-level-of-a-factor-with-all-other-non-missing-levels-tp3081777p3081777.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] Need help on nnet
Hi, Am working on neural network. Below is the coding and the output library (nnet) uplift.nn-nnet (PVU~ConsumerValue+Duration+PromoVolShare,y,size=3) # weights: 16 initial value 4068.052704 final value 3434.194253 converged summary (uplift.nn) a 3-3-1 network with 16 weights options were - b-h1 i1-h1 i2-h1 i3-h1 16.646.62 149.932.24 b-h2 i1-h2 i2-h2 i3-h2 -42.79 -17.40 -507.50 -5.14 b-h3 i1-h3 i2-h3 i3-h3 3.451.87 18.890.61 b-o h1-o h2-o h3-o 402.81 41.29 236.766.06 I have few questions, please i need help Q1: How to interpret the above output Q2: My objective is to know the contribution of each independent variable. Q3: Which package of neural network provides the AIC or BIC values Regards jothy -- View this message in context: http://r.789695.n4.nabble.com/Need-help-on-nnet-tp3081744p3081744.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] subset with two factors
Dear all, I have a dataframe of the following strucutre numacc_b coverage_b Geschlecht GG 10 1 W A 20 1 M A 30 1 M B 40 1 M B 50 1 W A 60 1 M B I would like to form a subset consisting of all entries with Geschlecht=M and GG=A. Using T1 - subset(daten1, Geschlecht==M, GG==A) delievers data frame with 0 columns and 6 rows T1 - subset(daten1, Geschlecht==M) delievers numacc_b coverage_b Geschlecht GG 2 0 1 M A 3 0 1 M B 4 0 1 M B 6 0 1 M B 9 0 1 M B 100 1 M B But I want to select the dataframe according to both factos. What can I do? Thank you answers in advance! Best, Martin -- GMX DSL Doppel-Flat ab 19,99 euro;/mtl.! Jetzt auch mit gratis Notebook-Flat! http://portal.gmx.net/de/go/dsl __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 enable Arial font for postcript/pdf figure on Windows?
Hi Agnes, I converted the Arial font files from ttf to afm using ttf2afm from MikTex complete installation. When used in R with the line recommended by Plos, they seem to give correct Arial font graphics: I checked by opening the ps file with a viewer (gsview), a text editor (notepad++) and Adobe illustrator. However I did not try (but hopefully will do soon) the ultimate test: submission to Plos. Here is a way to do it: (Mind the trick at step 5) 1) Download ttf2afm.exe (available for ex in the install directory of MikTex complete installation) 2) Fetch the arial ttf files in C:\Windows\Fonts 3) Place ttf2afm.exe and the ttf files in a directory (eg C:/ttf2afm/) 4) Open a DOS window (using cmd). Place yourself into the created directory (using cd). Then type ttf2afm.exe arial.ttf arial.afm ttf2afm.exe arialbd.ttf arial-Bold.afm ttf2afm.exe ariali.ttf arial-Oblique.afm ttf2afm.exe arialbi.ttf arial-BoldOblique.afm 5) Now, if you used ttf2afm.exe from MikTex you should open the created afm files with a text editor (ex Notepad++) and correct the following things: • Remove the copyright line (or make it start zith comment and get rid of the (c) copyright symbol ) • At the beginning of 4 of the first lines, the variable name is missing, so add it. Eg in arial.afm: o FontName ArialMT o FullName Arial o FamilyName Arial o Weight Normal Now you can use these fonts with the postscript function in R with the following line: postscript(file=try.ps, horizontal=F, onefile=F, width=4, height=4, pointsize=12, family=c ( “C:/ttf2afm/arial.afm, “C:/ttf2afm/arial-Bold.afm , “C:/ttf2afm/arial-Oblique.afm , “C:/ttf2afm/arial-BoldOblique.afm ) ) hist( rnorm(100) ) dev.off() Cheers, Camille -- View this message in context: http://r.789695.n4.nabble.com/How-to-enable-Arial-font-for-postcript-pdf-figure-on-Windows-tp3017809p3081774.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] [lattice xyplot] Help needed in help in customizing the panel.abline() function
Hi Girish,
Try this:
disc - xyplot(cnt_gt50pct_disc ~ week_num|sku_num, data=DF,type =
h,lwd=2,panel = function(x, y, ...) {
panel.abline(v = x[which.max(y)], lty = 2)
panel.xyplot(x, y, ...)
})
-Felix
On 9 December 2010 17:35, Girish A.R. garam...@gmail.com wrote:
Hi folks,
I need some help in customizing the abline() function to be used in a
lattice plot. I have attached a reproducible example below.
I need help in the following snippet:
disc - xyplot(cnt_gt50pct_disc ~ week_num|sku_num, data=DF,type =
h,lwd=2,panel = function(...) {
panel.abline(v = 8, lty = 2)
panel.xyplot(...)
})
Is there a way I can give panel.abline() input from a which.max() function?
Essentially I need the vertical line to be drawn at the week_num
corresponding to the max (cnt_gt50pct_disc).
Thanks in advance,
-Girish
===
Lines - sku_num week_num pct_inv_left cnt_gt50pct_disc
1 1 99.88 47
1 2 99.54 109
1 3 98.7 260
1 4 97.83 202
1 5 96.53 389
1 6 94.11 450
1 7 90.42 459
1 8 86.63 448
1 9 83.39 411
1 10 77 478
1 11 71.65 476
1 12 67.3 463
1 13 62.45 472
1 14 52.47 488
1 15 40.86 486
1 16 31.34 484
1 17 23.2 472
1 18 17 458
1 19 12.66 423
1 20 10.18 364
1 21 7.6 343
1 22 3.09 343
1 23 1.05 211
2 1 99.94 30
2 2 99.4 151
2 3 98.85 146
2 4 97.92 274
2 5 97.03 204
2 6 95.59 378
2 7 92.81 452
2 8 89.07 470
2 9 85.11 454
2 10 81.68 421
2 11 75.34 479
2 12 70.05 476
2 13 66.11 456
2 14 61.85 465
2 15 53.2 485
2 16 42.75 486
2 17 33.58 481
2 18 25 477
2 19 18.13 450
2 20 12.97 416
2 21 10.03 343
2 22 7.03 293
2 23 2.33 283
2 24 0.77 116
DF - read.table(con- textConnection(Lines), skip = 1);
names(DF) - scan(textConnection(Lines), what = , nlines = 1) ;
close(con);
require(latticeExtra)
DF$sku_num - as.factor(DF$sku_num)
disc - xyplot(cnt_gt50pct_disc ~ week_num|sku_num, data=DF,type =
h,lwd=2,panel = function(...) {
panel.abline(v = 8, lty = 2)
panel.xyplot(...)
})
sales - xyplot(pct_inv_left ~ week_num|sku_num, data=swtop16,type =
l,lwd=2,panel = function(...) {
panel.abline(h = 75, lty = 2)
panel.xyplot(...)
})
doubleYScale(disc, sales, style1 = 0, style2 = 2, add.ylab2 = TRUE,text =
c(# stores with gt 50pct disc, % Unsold))
--
View this message in context:
http://r.789695.n4.nabble.com/lattice-xyplot-Help-needed-in-help-in-customizing-the-panel-abline-function-tp3079656p3079656.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.
--
Felix Andrews / 安福立
http://www.neurofractal.org/felix/
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] pgfSweave 1.1.1 Released
The next release of pgfSweave is now on CRAN! pgfSweave has seen some
significant changes in the past couple of months.
The main new features are:
- Automatic code highlighting via the highlight package. This can be
turned off with the new `highlight` option.
- Tidying of source code output via the tidy option.
- Access to tikzDevice sanitization through a code chunk option `sanitize`
- Automatic addition of the \pgfrealjobname command if it does not
exist similarly to the addition of the \usepackage{Sweave} line.
- Setting tex.driver=latex will now (in addition to working) generate
an eps file
And of course bug fixes:
- Fixes for bunches of issues related to the changes in Sweave in R
2.12. I think these issues are now resolved (fingers crossed)
- keep.source actually works now.
See the NEWS file for the complete list of changes and the vignette
for information on now to use the new options.
Cheers!
-Cameron
___
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] subset with two factors
Hello Martin, You were almost there :) T1 - subset(daten1, Geschlecht==M GG==A) Hope this helps. Michael On 10 December 2010 22:25, Martin Spindler martin.spind...@gmx.de wrote: Dear all, I have a dataframe of the following strucutre numacc_b coverage_b Geschlecht GG 1 0 1 W A 2 0 1 M A 3 0 1 M B 4 0 1 M B 5 0 1 W A 6 0 1 M B I would like to form a subset consisting of all entries with Geschlecht=M and GG=A. Using T1 - subset(daten1, Geschlecht==M, GG==A) delievers data frame with 0 columns and 6 rows T1 - subset(daten1, Geschlecht==M) delievers numacc_b coverage_b Geschlecht GG 2 0 1 M A 3 0 1 M B 4 0 1 M B 6 0 1 M B 9 0 1 M B 10 0 1 M B But I want to select the dataframe according to both factos. What can I do? Thank you answers in advance! Best, Martin -- GMX DSL Doppel-Flat ab 19,99 euro;/mtl.! Jetzt auch mit gratis Notebook-Flat! http://portal.gmx.net/de/go/dsl __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] [lattice xyplot] Help needed in help in customizing the panel.abline() function
Thanks, Felix! That works. best, -Girish -- View this message in context: http://r.789695.n4.nabble.com/lattice-xyplot-Help-needed-in-help-in-customizing-the-panel-abline-function-tp3079656p3081792.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] New Installs, Same Trouble Loading doBy and coin Packages
I tried Tal's suggestion of deleting the doBy and coin packages and then reinstalling them from a different mirror. The first install was from the Harvard mirror and the second was from the Case Western Univ. mirror. The new packages generate the same errors when I call them using the library() command. Also, I tried to load these packages using R and its script editor thinking that the problem may have something to do with Tinn-R, but the same errors are generated on the R terminal when I use the library() function. Any help would be appreciated. Again, the errors for these two packages: Error in length(label) : could not find function .extendsForS3 Error: package/namespace load failed for 'doBy' library(coin) Loading required package: mvtnorm Loading required package: modeltools Loading required package: stats4 #This is odd. I cannot find any reference for this package. AC Error in length(sig) : could not find function .extendsForS3 Error: package 'stats4' could not be loaded - Forwarded Message From: Adam Carr adamlc...@yahoo.com To: Tal Galili tal.gal...@gmail.com Cc: r-help@r-project.org Sent: Thu, December 9, 2010 1:12:21 PM Subject: Re: [R] Trouble Loading doBy and coin Packages Hi Tal: No I have not tried this. I will do it this evening and we'll see what happens. Thanks for the suggestion. Adam From: Tal Galili tal.gal...@gmail.com Cc: r-help@r-project.org Sent: Thu, December 9, 2010 12:29:20 PM Subject: Re: [R] Trouble Loading doBy and coin Packages I Adam, Have you tried deleting the package files and then reinstalling them from a different CRAN mirror? Tal Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- Good Evening R-Help Community: I have attached a file that contains the output from sessionInfo() and a summary of my Win XP system. I am running R 2.12.0 and using Tinn-R 2.3.6.2 as my interface. When I attempt to call either the doBy or coin packages R generates an error that I do not understand and have so far not been able to resolve by searching R resources. I exchanged a couple of emails with Soren Hojsgaard who does not think the doBy error is directly related to the package itself, and he suggested that I post this problem for input from others. When the doBy package is loaded, the following error appears in the Tinn-R log: Error in length(label) : could not find function .extendsForS3 Error: package/namespace load failed for 'doBy' When the coin package is called, this error appears in the Tinn-R log: Error in length(sig) : could not find function .extendsForS3 Error: package 'stats4' could not be loaded No functions in either package work, and when I attempt to call them the same errors are generated in the log. Any help or direction would be appreciated. Thanks very much, Adam __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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..Neural Network
Hi all, I am trying to develop a neural network with single target variable and 5 input variables to predict the importance of input variables using R. I used the packages nnet and RSNNS. But unfortunately I could not interpret the out put properly and the documentation of that packages also not giving proper direction. Please help me to find a good package with a proper documentation for neural network. Advance thanks s.sadanand -- View this message in context: http://r.789695.n4.nabble.com/Help-Neural-Network-tp3081800p3081800.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] subset with two factors
Hey Michael, Thank you very much. It works! Best, Martin Original-Nachricht Datum: Fri, 10 Dec 2010 22:35:56 +1100 Von: Michael Bedward michael.bedw...@gmail.com An: Martin Spindler martin.spind...@gmx.de CC: r-help@r-project.org Betreff: Re: [R] subset with two factors Hello Martin, You were almost there :) T1 - subset(daten1, Geschlecht==M GG==A) Hope this helps. Michael On 10 December 2010 22:25, Martin Spindler martin.spind...@gmx.de wrote: Dear all, I have a dataframe of the following strucutre numacc_b coverage_b Geschlecht GG 1 0 1 W A 2 0 1 M A 3 0 1 M B 4 0 1 M B 5 0 1 W A 6 0 1 M B I would like to form a subset consisting of all entries with Geschlecht=M and GG=A. Using T1 - subset(daten1, Geschlecht==M, GG==A) delievers data frame with 0 columns and 6 rows T1 - subset(daten1, Geschlecht==M) delievers numacc_b coverage_b Geschlecht GG 2 0 1 M A 3 0 1 M B 4 0 1 M B 6 0 1 M B 9 0 1 M B 10 0 1 M B But I want to select the dataframe according to both factos. What can I do? Thank you answers in advance! Best, Martin -- GMX DSL Doppel-Flat ab 19,99 euro;/mtl.! Jetzt auch mit gratis Notebook-Flat! http://portal.gmx.net/de/go/dsl __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- GRATIS! Movie-FLAT mit über 300 Videos. -- GMX DSL Doppel-Flat ab 19,99 euro;/mtl.! Jetzt auch mit gratis Notebook-Flat! http://portal.gmx.net/de/go/dsl __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] ReadWrite.xls problem
Toby, haruo0409, 2010/12/8 tkdweber tkd.we...@gmail.com: This is my Error-Message in its German original: Fehler in .Call(ReadXls, file, colNames, sheet, type, from, rowNames, : Falsche Anzahl von Argumenten (11), erwarte 10 für ReadXls There was a wrong DLL for a short while in the old 1.5.2 version (I fixed a R2.12.0 related issue and unfortunately introduced this error). If you delete the old xlsReadWrite package and re-install the package (either from CRAN or see www.swissr.org/download) it really should work. 2010/12/10 haruo0409 eixcx...@bca.bai.ne.jp: I'm also annoyed at same problem. I installed xlsReadWriter today and entered x - read.xls(data.xls,sheet=1) But I got Error Message: 以下にエラー .Call(ReadXls, file, colNames, sheet, type, from, rowNames, : 引数の個数(11)が不正です。10 個が ReadXls に対しては必要です (It's Japanese.Its English translation is the same as yours) What's the 'library(xlsReadWrite)' startup message? For the current version it should be: 'xlsReadWrite version 1.5.3 (0b78c1)'. Could you please give more details about 'I installed xlsReadWriter today' (which CRAN mirror, 'R.version' and '.Platform' output, is there only one 'xlsReadWrite.dll' file on your system). It is supposed to work and I am unable to find any problem. Thanks a lot! 2010/12/8 tkdweber tkd.we...@gmail.com: Without being able to read data, the programme renders pointless for me :-( There are many ways to read/write data in R: * Other than load/save you could use read.table/write.table (see R Data Import/Export). * Using Excel files is not the recommended way. However when you want/need it, there are several options (see http://rwiki.sciviews.org/doku.php?id=tips:data-io:ms_windows) Cheers, Hans-Peter PS. not that I mind to discuss things here, but as these are package specific problems I'd suggest to switch to the xlsReadWrite forum (http://dev.swissr.org/projects/xlsreadwrite/boards). You also can create an issue (http://dev.swissr.org/projects/xlsreadwrite/issues/new) or just send an email to 'support' at 'swissr.org'). __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] New Installs, Same Trouble Loading doBy and coin Packages
On 2010-12-10 03:43, Adam Carr wrote: I tried Tal's suggestion of deleting the doBy and coin packages and then reinstalling them from a different mirror. The first install was from the Harvard mirror and the second was from the Case Western Univ. mirror. The new packages generate the same errors when I call them using the library() command. Also, I tried to load these packages using R and its script editor thinking that the problem may have something to do with Tinn-R, but the same errors are generated on the R terminal when I use the library() function. Any help would be appreciated. Again, the errors for these two packages: Error in length(label) : could not find function .extendsForS3 Error: package/namespace load failed for 'doBy' library(coin) Loading required package: mvtnorm Loading required package: modeltools Loading required package: stats4 #This is odd. I cannot find any reference for this package. AC Error in length(sig) : could not find function .extendsForS3 Error: package 'stats4' could not be loaded I would remove and re-install R. 'stats4' is a base package and if that can't be loaded, your installation may be broken. Try require(stats4) or help(package=stats4) Peter Ehlers - Forwarded Message From: Adam Carradamlc...@yahoo.com To: Tal Galilital.gal...@gmail.com Cc: r-help@r-project.org Sent: Thu, December 9, 2010 1:12:21 PM Subject: Re: [R] Trouble Loading doBy and coin Packages Hi Tal: No I have not tried this. I will do it this evening and we'll see what happens. Thanks for the suggestion. Adam From: Tal Galilital.gal...@gmail.com Cc: r-help@r-project.org Sent: Thu, December 9, 2010 12:29:20 PM Subject: Re: [R] Trouble Loading doBy and coin Packages I Adam, Have you tried deleting the package files and then reinstalling them from a different CRAN mirror? Tal Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- Good Evening R-Help Community: I have attached a file that contains the output from sessionInfo() and a summary of my Win XP system. I am running R 2.12.0 and using Tinn-R 2.3.6.2 as my interface. When I attempt to call either the doBy or coin packages R generates an error that I do not understand and have so far not been able to resolve by searching R resources. I exchanged a couple of emails with Soren Hojsgaard who does not think the doBy error is directly related to the package itself, and he suggested that I post this problem for input from others. When the doBy package is loaded, the following error appears in the Tinn-R log: Error in length(label) : could not find function .extendsForS3 Error: package/namespace load failed for 'doBy' When the coin package is called, this error appears in the Tinn-R log: Error in length(sig) : could not find function .extendsForS3 Error: package 'stats4' could not be loaded No functions in either package work, and when I attempt to call them the same errors are generated in the log. Any help or direction would be appreciated. Thanks very much, Adam __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Using Lagsarlm
This has been answered offlist (the poster also wrote directly to me as package maintainer, but did not post on the R-sig-geo list, as would have seemed natural). The resolution was to read ?formula, and to use either errorsarlm() or lagsarlm() in spdep with formula=y ~ 1. Apparently an insurance analyst in a hurry ... Roger Saswati Neogi wrote: I'm trying to use the spdep package to calculate this: y = rho W y + e I don't want to use explanatory variables, just the lag from the dependent variable. How would I code this? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - Roger Bivand Economic Geography Section Department of Economics Norwegian School of Economics and Business Administration Helleveien 30 N-5045 Bergen, Norway -- View this message in context: http://r.789695.n4.nabble.com/Using-Lagsarlm-tp3080368p3081833.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] Projecting data on a world map using long/lat
Oh, whoops I was looking for the vote up button and accidentally hit Reply All. On Fri, Dec 10, 2010 at 8:25 PM, Barry Rowlingson b.rowling...@lancaster.ac.uk wrote: On Fri, Dec 10, 2010 at 2:21 AM, mathijsdevaan mathijsdev...@gmail.com wrote: Hi, I have a dataset (CSV) with some counts of firms located around the globe. Each count is assigned to the longitude and latitude of the specific location. Now I want to plot these counts on a world map using dots (size of dots represent the count). I have been unable to find any info on whether this is possible and if so, how? Can you please help me? Thanks! Plotting points is trivial - plot(data$x,data$y,pch=19,cex=data$size) will do for a start. i'm guessing your real problem is when you say 'on a world map'. How detailed a world map do you need? There's an outline one in the 'maps' package, or you should be able to find a shapefile of the world on the web somewhere and use that via the rgdal package. Other options include making a KML file of your points and overlaying on google earth. Or getting google map tiles and overlaying on that Or exporting your data to a GIS format and doing the pretty map in something like Quantum GIS. What are you trying to do exactly? also, you might want to post to r-sig-geo Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Michael Sumner Institute for Marine and Antarctic Studies, University of Tasmania Hobart, Australia e-mail: mdsum...@gmail.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] Adding numbers in Outputs
two OutputsHello! I am Amelia from Auckland and work for a bank. I am new to R and I have started my venture with R just a couple of weeks back and this is my first mail to R-forum. I need following assistance Suppose my R code generates following outputs as X [[1]] [1] 40 [[2]] [1] 80 160 [[3]] [1] 160 80 400 Y [[1]] [1] 10 [[2]] [1] 10 30 [[3]] [1] 5 18 20 and suppose Z = c(1, 2, 3) I need to perform the calculation where I will be multiplying corresponding terms of X and Y individually and multiplying their sum by Z and store these results in a dataframe. I.e. I need to calculate (40*10) * 1 # (first element of X + First element of Y) * Z[1] = 400 ((80*10)+(160*30)) * 2 # 2 row of X and 2nd row of Y = 11200 ((160*5)+(80*18)+(400*20)) * 3 # 3rd row of X and 3 row of Y and Z[3] = 30720 So the final output should be 400 11200 30720 One way of doing it is write R code for individual rows and arrive at the result e.g. ([[X]][1]*[[Y]][1])*1 will result in 400. However, I was just trying to know some smart way of doing it as there could be number of rows and writing code for each row will be a cumbersome job. So is there any better way to do it? Please guide me. I thank you in advance. Thanking all Amelia [[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] Compare one level of a factor with *all* other non-missing levels
On 2010-12-10 03:24, deriK2000 wrote: Dear list, I try to compare the mean of a variable given a value of a factor with the mean of the same variable for all K-1 other non-missing values of this factor. This procedure I want to repeat for each level of the factor. Having read the recommendations of this list I want to avoid creating K-1 dummy variables and searched for options of the pairwise.t.test. But couldn't find a solution. Anyone with a suggestion how to do the comparisions? Sounds like you want the Dunnett test procedure which seems to be implemented in a number of packages: multcomp, asd, MCPAN and others. It would probably be a good idea to install package 'sos' and learn how to search with it. Peter Ehlers Cheers, Derik __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sweave: Setting options with SweaveOpts{} when using driver=RweaveHTML
On 10/12/2010 3:40 AM, Søren Højsgaard wrote:
When using Sweave in connection with the driver RweaveLatex(), global options
can be set with \SweaveOpts{}, e.g.
\SweaveOpts{keep.source=T}.
Does anybody know if it is possible to set global options in the same way when
using Sweave with the driver RweaveHTML().
I haven't used R2HTML, but it looks from the SweaveSyntaxHTML variable that the
same syntax should work there.
Regards
Søren
[[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] Adding numbers in Outputs
try this:
x - list(40, c(80,160), c(160,80,400))
y - list(10, c(10,30), c(5,18,20))
z - c(1,2,3)
mapply(function(a1,a2,a3){
+ a3 * sum(a1 * a2)
+ }
+ , x
+ , y
+ , z
+ )
[1] 400 11200 30720
On Fri, Dec 10, 2010 at 5:41 AM, Amelia Vettori
amelia_vett...@yahoo.co.nz wrote:
two OutputsHello!
I am Amelia from Auckland and work for a bank. I am new to R and I have
started my venture with R just a couple of weeks back and this is my first
mail to R-forum. I need following assistance
Suppose my R code generates following outputs as
X
[[1]]
[1] 40
[[2]]
[1] 80 160
[[3]]
[1] 160 80 400
Y
[[1]]
[1] 10
[[2]]
[1] 10 30
[[3]]
[1] 5 18 20
and suppose
Z = c(1, 2, 3)
I need to perform the calculation where I will be multiplying corresponding
terms of X and Y individually and multiplying their sum by Z and store these
results in a dataframe.
I.e. I need to calculate
(40*10) * 1 # (first element of X + First
element of Y) * Z[1] = 400
((80*10)+(160*30)) * 2 # 2 row of X and 2nd row of Y = 11200
((160*5)+(80*18)+(400*20)) * 3 # 3rd row of X and 3 row of Y and Z[3] =
30720
So the final output should be
400
11200
30720
One way of doing it is write R code for individual rows and
arrive at the result e.g.
([[X]][1]*[[Y]][1])*1 will result in 400. However, I was just trying to know
some smart way of doing it as there could be number of rows and writing code
for each row will be a cumbersome job. So is there any better way to do it?
Please guide me.
I thank you in advance.
Thanking
all
Amelia
[[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.
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Adding numbers in Outputs
On 2010-12-10 02:41, Amelia Vettori wrote:
two OutputsHello!
I am Amelia from Auckland and work for a bank. I am new to R and I have started
my venture with R just a couple of weeks back and this is my first mail to
R-forum. I need following assistance
Suppose my R code generates following outputs as
X
[[1]]
[1] 40
[[2]]
[1] 80160
[[3]]
[1] 160 80 400
Y
[[1]]
[1] 10
[[2]]
[1] 1030
[[3]]
[1] 5 18 20
and suppose
Z = c(1, 2, 3)
I need to perform the calculation where I will be multiplying corresponding
terms of X and Y individually and multiplying their sum by Z and store these
results in a dataframe.
I.e. I need to calculate
(40*10) * 1 # (first element of X + First
element of Y) * Z[1] = 400
((80*10)+(160*30)) * 2 # 2 row of X and 2nd row of Y = 11200
((160*5)+(80*18)+(400*20)) * 3 # 3rd row of X and 3 row of Y and Z[3] =
30720
So the final output should be
400
11200
30720
One way of doing it is write R code for individual rows and
arrive at the result e.g.
([[X]][1]*[[Y]][1])*1 will result in 400. However, I was just trying to know
some smart way of doing it as there could be number of rows and writing code
for each row will be a cumbersome job. So is there any better way to do it?
Please guide me.
Why not just write a function to do what
you've done by hand:
f - function(x, y, z){
len - length(x)
res - rep(NA, len)
for(i in 1:len){
res[i] - sum(x[[i]] * y[[i]]) * z[i]
}
res
}
f(X, Y, Z)
Peter Ehlers
I thank you in advance.
Thanking
all
Amelia
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Adding numbers in Outputs
X-list(40,c(80,160),c(160,80,400)) Y-list(10,c(10,30),c(5,18,20)) Z-c(1,2,3) as.data.frame(do.call(rbind,X))-x as.data.frame(do.call(rbind,Y))-y x*y*Z-r r[upper.tri(r)] - 0 rowSums(r) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Adding numbers in Outputs
Dear Mr Holtman Sir,
Thanks a lot for your great solution. This certainly is helping me achieve what
I need to get. However, I shall be hugely thankful to you if you can guide me
in one respect.
Sir, you have used following commands to assign values to x and y.
x - list(40, c(80,160), c(160,80,400))
y - list(10, c(10,30), c(5,18,20))
z - c(1,2,3)
But Sir, the problem is these values are basically outputs of some other
process which I am running and chances are these will vary. Sir, it will be a
great help if you can guide me to convert the output (which I am getting)
X
[[1]]
[1] 40
[[2]]
[1] 80 160
[[3]]
[1] 160 80 400
to what you have suggested
x - list(40, c(80,160), c(160,80,400))
So, in that case once I get output in my format, I will convert that output as
provided by you.
I apologize for taking the liberty of writing to you, but I shall be really
grateful to you, as I have just started getting the feel of 'R' and I know I
need to take lots of efforts to begin with.
Thanks and eagerly waiting for your guidance.
Amelia Vettori
--- On Fri, 10/12/10, jim holtman jholt...@gmail.com wrote:
From: jim holtman jholt...@gmail.com
Subject: Re: [R] Adding numbers in Outputs
To: Amelia Vettori amelia_vett...@yahoo.co.nz
Cc: r-help@r-project.org
Received: Friday, 10 December, 2010, 1:43 PM
try this:
x - list(40, c(80,160), c(160,80,400))
y - list(10, c(10,30), c(5,18,20))
z - c(1,2,3)
mapply(function(a1,a2,a3){
+ a3 * sum(a1 * a2)
+ }
+ , x
+ , y
+ , z
+ )
[1] 400 11200 30720
On Fri, Dec 10, 2010 at 5:41 AM, Amelia Vettori
amelia_vett...@yahoo.co.nz wrote:
two OutputsHello!
I am Amelia from Auckland and work for a bank. I am new to R and I have
started my venture with R just a couple of weeks back and this is my first
mail to R-forum. I need following assistance
Suppose my R code generates following outputs as
X
[[1]]
[1] 40
[[2]]
[1] 80 160
[[3]]
[1] 160 80 400
Y
[[1]]
[1] 10
[[2]]
[1] 10 30
[[3]]
[1] 5 18 20
and suppose
Z = c(1, 2, 3)
I need to perform the calculation where I will be multiplying corresponding
terms of X and Y individually and multiplying their sum by Z and store these
results in a dataframe.
I.e. I need to calculate
(40*10) * 1 # (first element of X + First
element of Y) * Z[1] = 400
((80*10)+(160*30)) * 2 # 2 row of X and 2nd row of Y = 11200
((160*5)+(80*18)+(400*20)) * 3 # 3rd row of X and 3 row of Y and Z[3] =
30720
So the final output should be
400
11200
30720
One way of doing it is write R code for individual rows and
arrive at the result e.g.
([[X]][1]*[[Y]][1])*1 will result in 400. However, I was just trying to know
some smart way of doing it as there could be number of rows and writing code
for each row will be a cumbersome job. So is there any better way to do it?
Please guide me.
I thank you in advance.
Thanking
all
Amelia
[[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.
--
Jim Holtman
Data Munger Guru
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] Compare one level of a factor with *all* other non-missing levels
Peter Ehlers wrote: Sounds like you want the Dunnett test procedure which seems to be implemented in a number of packages: multcomp, asd, MCPAN and others. It would probably be a good idea to install package 'sos' and learn how to search with it. Peter Ehlers Thanks for the hints! Unfortunately, Dunnett compares the mean(x) for a factor level with the means(x) of all single K-1 other levels resulting in K-1 comparisions for each level (printed in a lower triangular matrix for the results). Instead, I just want to compare this one mean(x) with one other mean(x) of all the K-1 other levels (printed in a vector of length K for the results). Concerning sos: sounds like a good idea! Cheers, Derik -- View this message in context: http://r.789695.n4.nabble.com/Compare-one-level-of-a-factor-with-all-other-non-missing-levels-tp3081777p3082005.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] (no subject)
Hi R-help, I am trying to find a way to select five highest values in data frame according some variable. I will demonstrate: c X1 X2 1 1 1 2 1 2 3 1 3 4 1 4 5 1 5 6 1 6 7 1 7 8 1 8 9 1 9 10 1 10 11 2 11 12 2 12 13 2 13 14 2 14 15 2 15 16 2 16 17 2 17 18 2 18 19 2 19 20 2 20 21 2 21 22 2 22 23 2 23 24 2 24 25 2 25 So I would like to select a rows with higest values of X2 inside X1. Expected result should be: X1 X2 1 10 1 9 1 8 1 7 1 6 2 25 2 24 2 23 2 22 2 21 I first oreded the data frame using c=c[with(c,order(X1,-X2)),] but I need a help to select highes five. It is easy to select when I have just 2 unique values of X1 but what is if I have 500 unique values in X1? Thanks Andrija [[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] (no subject)
try this:
do.call(rbind, lapply(split(x, x$X1), function(.grp){
+ .ord - .grp[order(.grp$X2, decreasing = TRUE),]
+ .ord[seq(min(5, nrow(.grp))),]
+ }))
X1 X2
1.10 1 10
1.9 1 9
1.8 1 8
1.7 1 7
1.6 1 6
2.25 2 25
2.24 2 24
2.23 2 23
2.22 2 22
2.21 2 21
On Fri, Dec 10, 2010 at 9:18 AM, andrija djurovic djandr...@gmail.com wrote:
Hi R-help,
I am trying to find a way to select five highest values in data frame
according some variable. I will demonstrate:
c
X1 X2
1 1 1
2 1 2
3 1 3
4 1 4
5 1 5
6 1 6
7 1 7
8 1 8
9 1 9
10 1 10
11 2 11
12 2 12
13 2 13
14 2 14
15 2 15
16 2 16
17 2 17
18 2 18
19 2 19
20 2 20
21 2 21
22 2 22
23 2 23
24 2 24
25 2 25
So I would like to select a rows with higest values of X2 inside X1.
Expected result should be:
X1 X2
1 10
1 9
1 8
1 7
1 6
2 25
2 24
2 23
2 22
2 21
I first oreded the data frame using
c=c[with(c,order(X1,-X2)),]
but I need a help to select highes five. It is easy to select when I have
just 2 unique values of X1 but what is if I have 500 unique values in X1?
Thanks
Andrija
[[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.
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] melt causes errors when characters and values are used
Hello,
I am finding that the melt function from the reshape library causes
errors when applied to a data.frame that contains numeric and character
columns. For example,
melt(id.vars=ID,data.frame(ID=1:3,date=c(a,b,c),value=c(1,4,5)))
ID variable value
1 1 date a
2 2 date b
3 3 date c
4 1value NA
5 2value NA
6 3value NA
Warning message:
In `[-.factor`(`*tmp*`, ri, value = c(1, 4, 5)) :
invalid factor level, NAs generated
It would be useful in this situation that the numerical column got
converted to a character column in this situation. Any ways round this?
In actual fact I have got a situation where it is more like this
ID Date_1 Value_1 Date_2 Value_2 ...
and I would like to convert it to a data.frame of ID, Date Value but I
thought the above would be an appropriate middle step.
Thanks
Dan
--
**
Daniel Brewer, Ph.D.
Institute of Cancer Research
Molecular Carcinogenesis
Email: daniel.bre...@icr.ac.uk
**
The Institute of Cancer Research: Royal Cancer Hospital, a charitable Company
Limited by Guarantee, Registered in England under Company No. 534147 with its
Registered Office at 123 Old Brompton Road, London SW7 3RP.
This e-mail message is confidential and for use by the a...{{dropped:2}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Adding numbers in Outputs
You should be able to use whatever values you are getting from your script
right now. I just did the assignment to match what you were showing on the
output. The easiest thing to do is to do 'str(X)' from your data and
compare it to the 'x' I created -- str(x).
Here is what 'str(x)' gives:
x - list(40, c(80,160), c(160,80,400))
str(x)
List of 3
$ : num 40
$ : num [1:2] 80 160
$ : num [1:3] 160 80 400
When providing sample data, it is probably best to use 'dput' to it can be
reconstructed by the reader:
dput(x)
list(40, c(80, 160), c(160, 80, 400))
This gives back what I was using and hopefully it compares with your
current output.
On Fri, Dec 10, 2010 at 9:00 AM, Amelia Vettori
amelia_vett...@yahoo.co.nzwrote:
Dear Mr Holtman Sir,
Thanks a lot for your great solution. This certainly is helping me achieve
what I need to get. However, I shall be hugely thankful to you if you can
guide me in one respect.
Sir, you have used following commands to assign values to x and y.
x - list(40, c(80,160), c(160,80,400))
y - list(10, c(10,30), c(5,18,20))
z - c(1,2,3)
But Sir, the problem is these values are basically outputs of some other
process which I am running and chances are these will vary. Sir, it will be
a great help if you can guide me to convert the output (which I am getting)
X
[[1]]
[1] 40
[[2]]
[1] 80160
[[3]]
[1] 160 80 400
to what you have suggested
x - list(40, c(80,160), c(160,80,400))
So, in that case once I get output in my format, I will convert that output
as provided by you.
I apologize for taking the liberty of writing to you, but I shall be really
grateful to you, as I have just started getting the feel of 'R' and I know I
need to take lots of efforts to begin with.
Thanks and eagerly waiting for your guidance.
Amelia Vettori
--- On *Fri, 10/12/10, jim holtman jholt...@gmail.com* wrote:
From: jim holtman jholt...@gmail.com
Subject: Re: [R] Adding numbers in Outputs
To: Amelia Vettori amelia_vett...@yahoo.co.nz
Cc: r-help@r-project.org
Received: Friday, 10 December, 2010, 1:43 PM
try this:
x - list(40, c(80,160), c(160,80,400))
y - list(10, c(10,30), c(5,18,20))
z - c(1,2,3)
mapply(function(a1,a2,a3){
+ a3 * sum(a1 * a2)
+ }
+ , x
+ , y
+ , z
+ )
[1] 400 11200 30720
On Fri, Dec 10, 2010 at 5:41 AM, Amelia Vettori
amelia_vett...@yahoo.co.nzhttp://mc/compose?to=amelia_vett...@yahoo.co.nz
wrote:
two OutputsHello!
I am Amelia from Auckland and work for a bank. I am new to R and I have
started my venture with R just a couple of weeks back and this is my first
mail to R-forum. I need following assistance
Suppose my R code generates following outputs as
X
[[1]]
[1] 40
[[2]]
[1] 80160
[[3]]
[1] 160 80 400
Y
[[1]]
[1] 10
[[2]]
[1] 1030
[[3]]
[1] 5 18 20
and suppose
Z = c(1, 2, 3)
I need to perform the calculation where I will be multiplying
corresponding terms of X and Y individually and multiplying their sum by Z
and store these results in a dataframe.
I.e. I need to calculate
(40*10) * 1 # (first element of X + First
element of Y) * Z[1] = 400
((80*10)+(160*30)) * 2 # 2 row of X and 2nd row of Y =
11200
((160*5)+(80*18)+(400*20)) * 3 # 3rd row of X and 3 row of Y and Z[3]
= 30720
So the final output should be
400
11200
30720
One way of doing it is write R code for individual rows and
arrive at the result e.g.
([[X]][1]*[[Y]][1])*1 will result in 400. However, I was just trying to
know some smart way of doing it as there could be number of rows and writing
code for each row will be a cumbersome job. So is there any better way to do
it?
Please guide me.
I thank you in advance.
Thanking
all
Amelia
[[alternative HTML version deleted]]
__
R-help@r-project.org http://mc/compose?to=r-h...@r-project.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
Data Munger Guru
What is the problem that you are trying to solve?
--
Jim Holtman
Data Munger Guru
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] ReadWrite.xls problem
Hans-Peter I have checked the 'library(xlsReadWrite)' startup message. I found that I just failed to 'xls.getshlib()'. Entering 'xls.getshlib()', read.xls() works regularly. Thank you. -- View this message in context: http://r.789695.n4.nabble.com/ReadWrite-xls-problem-tp3078348p3082108.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] Minimization of the distance
its really help,thanks a lot -- View this message in context: http://r.789695.n4.nabble.com/Minimization-of-the-distance-tp3081345p3081900.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] Textwrangler Languages Folder
Dear R Community, I recently switched to a Mac (10.6.5), and have installed Textwrangler to run code to R. However, I can't install the syntax highlighting file because I can't find the directory: ~Users/username/Library/Application Support/TextWrangler/Language Modules/. Is there a different location I can place the syntax highlighting file? Scott [[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] overlap different line in a xyplot (lattice)
dear [R] users, is there a way to plot different data (but with the same x-variables) in the same xyplot window? There are already a similar question, but the answer is not enought explanatory... Thanks a lot, Francesco [[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] (no subject)
Andrija, You should be able to extract the data that you want using a call like this (AD substituted for your c) with(AD, tapply(X2, X1, function(x) sort(x, dec=T)[1:5])) That returns a list like this: $`1` [1] 10 9 8 7 6 $`2` [1] 25 24 23 22 21 Just package it the way that you want. Dave From: andrija djurovic djandr...@gmail.com To: r-help@r-project.org Date: 12/10/2010 08:21 AM Subject: [R] (no subject) Sent by: r-help-boun...@r-project.org Hi R-help, I am trying to find a way to select five highest values in data frame according some variable. I will demonstrate: c X1 X2 1 1 1 2 1 2 3 1 3 4 1 4 5 1 5 6 1 6 7 1 7 8 1 8 9 1 9 10 1 10 11 2 11 12 2 12 13 2 13 14 2 14 15 2 15 16 2 16 17 2 17 18 2 18 19 2 19 20 2 20 21 2 21 22 2 22 23 2 23 24 2 24 25 2 25 So I would like to select a rows with higest values of X2 inside X1. Expected result should be: X1 X2 1 10 1 9 1 8 1 7 1 6 2 25 2 24 2 23 2 22 2 21 I first oreded the data frame using c=c[with(c,order(X1,-X2)),] but I need a help to select highes five. It is easy to select when I have just 2 unique values of X1 but what is if I have 500 unique values in X1? Thanks Andrija [[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] survival: ridge log-likelihood workaround
-- begin inclusion - Dear all, I need to calculate likelihood ratio test for ridge regression. In February I have reported a bug where coxph returns unpenalized log-likelihood for final beta estimates for ridge coxph regression. In high-dimensional settings ridge regression models usually fail for lower values of lambda. As the result of it, in such settings the ridge regressions have higher values of lambda (e.g. over 100) which means that the difference between unpenalized log-likelihood and penalized log-likelihood is not insignificant. I would be grateful if someone can confirm that the below code is correct workaround. --- end included message First, the bug you report is not a bug. The log partial likelihood from a Cox model LPL(beta) is well defined for any vector of coefficients beta, whether they are result of a maximization or taken from your daily horoscope. The loglik component of coxph is the LPL for the reported coefficients. For a ridge regression the coxph function maximizes LPL(beta) - penalty(beta) = penalized partial likelihood = PPL(beta). You have correctly recreated the PPL. Second: how do you do formal tests on such a model? This is hard. The difference LPL1- LPL2 is a chi-square when each is the result of maximizing the Cox LPL over a set of coefficients; when using a PPL we are maximizing over something else. The distribution of the difference of constrained LPL values can be argued to be a weighed sum of squared normals where the weights are in (0,1), which is something more complex than a chisq distribution. In a world with infinite free time I'd have pursued this, worked it all out, and added appropriate code to coxph. What about the difference in PPL values, which is the test you propose? I'm not aware of any theory showing that these have any relation to a chi-square distribution. (Said theory may well exist, and I'd be happy for pointers.) 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] Textwrangler Languages Folder
Hi, On Dec 10, 2010, at 9:27 AM, Scott Chamberlain wrote: Dear R Community, I recently switched to a Mac (10.6.5), and have installed Textwrangler to run code to R. However, I can't install the syntax highlighting file because I can't find the directory: ~Users/username/Library/Application Support/TextWrangler/Language Modules/. Is there a different location I can place the syntax highlighting file? I don't have OSX 10.6 and I happen to use SubEthaEdit, but I suspect that you are really looking for something like the following: /Users/ben/Library/Application Support/TextWrangler/Language Modules/ Note that you would place your user name in the place where mine (ben) appears. If the TextWrangler/Language Modules directory doesn't exist in the Application Support directory you can create it. Cheers, Ben Ben Tupper Bigelow Laboratory for Ocean Sciences 180 McKown Point Rd. P.O. Box 475 West Boothbay Harbor, Maine 04575-0475 http://www.bigelow.org/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Remove 100 years from a date object
Hello,
I have some data that has dates in the form 27.02.37. I convert them to
a date object as follows:
as.Date(data$date,format=%d.%m.%y)
But this gives me years such as 2037 when I would like them to be 1937.
I thought of trying to take off some time i.e.
as.Date(camCD$DoB,format=%d.%m.%y) - 100*365
But that doesn't seem to work out correctly. Any ideas how to do this?
Thanks
Dan
--
**
Daniel Brewer, Ph.D.
Institute of Cancer Research
Molecular Carcinogenesis
Email: daniel.bre...@icr.ac.uk
**
The Institute of Cancer Research: Royal Cancer Hospital, a charitable Company
Limited by Guarantee, Registered in England under Company No. 534147 with its
Registered Office at 123 Old Brompton Road, London SW7 3RP.
This e-mail message is confidential and for use by the a...{{dropped:2}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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 requested
HI friends, I have very lengthy graph data in edge list format. I want to convert it into node list format. example: EDGE LIST FORMAT 1 2 1 3 1 4 1 5 2 3 2 4 3 2 4 1 4 3 4 5 5 2 5 4 ITS NODE LIST FORMAT SHOULD BE LIKE: 1 2 3 4 5 2 3 4 3 2 4 1 3 5 2 4 Kindly suggest me which package in R provides the support to do my task. Thank u friends in advance. -- View this message in context: http://r.789695.n4.nabble.com/help-requested-tp3082147p3082147.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] 45 Degree labels on barplot? Help understanding code previously posted.
Dear colleagues, i found a line or two of code in the help archives from Uwe Ligges about creating slanted x-labels for a barplot and it works well for my purposes (code below). However, I was hoping someone could explain to me precisely what the code is doing. I'm aware it's invoking the text command, and I know the first ttwo arguments to text are x and y co-ordinates. I'm also aware that par(usr)[3] is grabbing the third element of the vector of plotting co-ordinates. But I tried replacing par(usr)[3] with just 0 and that didn't work; all the labels got bunched up on the left. Is it necessary to create a new object via barplot and then quote that in the x,y coordinates of text? Like I said, the code works great, but I'm trying to actually understand the rationale behind the elements so I can apply it in future. Yours, Simon Kiss #Reproducible Code mydat-data.frame(countries=c(Canada, Denmark, Framce, United Kingdom, Germany, Australia, New Zealand, Switzerland, Belgium, Netherlands), stories_total=c(429, 25, 239, 99, 100, 96, 18, 21, 0, 6), avg=c(4.165048544, 6.25, 6.459459459, 0.908256881, 1.923076923, 1.103448276, 1.058823529, 1.615384615, 0, 0.107142857), steps=c(2, 2, 2, 0,1, 1, 1, 0,0,0), newspapers=c(103, 4, 37, 109, 52, 87, 17, 13, 10, 56)) mydat.sort1-mydat[order(-mydat$avg), ] myplot-barplot(mydat.sort1$avg, col=c(black, black, black, grey, white, grey, grey, white, white, white), ylim=c(0,7), main=Regulatory Action On Bisphenol A By Newspaper Coverage) col.vec=c(black, grey, white) legend(topright, col=col.vec, fill=c(black, grey, white), legend=c(Meaningful Ban, Recommendations To Withdraw, No Legislative Action)) labels=mydat.sort1$countries #These lines create the labels text(myplot, par(usr)[3], labels=labels, srt=35, offset=1, adj=1, xpd=TRUE) axis(2) par(usr)[3] * Simon J. Kiss, PhD Assistant Professor, Wilfrid Laurier University 73 George Street Brantford, Ontario, Canada N3T 2C9 Cell: +1 519 761 7606 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Remove 100 years from a date object
On Fri, Dec 10, 2010 at 3:27 PM, Daniel Brewer daniel.bre...@icr.ac.uk wrote: Hello, I have some data that has dates in the form 27.02.37. I convert them to a date object as follows: as.Date(data$date,format=%d.%m.%y) But this gives me years such as 2037 when I would like them to be 1937. I thought of trying to take off some time i.e. as.Date(camCD$DoB,format=%d.%m.%y) - 100*365 But that doesn't seem to work out correctly. Any ideas how to do this? Normally to adjust dates you can use as.difftime() and do arithmetic, but a year is a variable thing (can be 365 or 366 days) so you cant make a difftime of years. Days are variable things if you worry about leap seconds... Also, you could end up with an invalid date if you have 29-Feb-2000 and 29-Feb-1900. One wasn't a leap year... A solution minus those caveats is to convert to POSIXlt and adjust the $year element: dob=27.02.37 as.Date(dob,format=%d.%m.%y) [1] 2037-02-27 dobp = as.POSIXlt(as.Date(dob,format=%d.%m.%y)) dobp$year = dobp$year - 100 dobp [1] 1937-02-27 UTC as.Date(dobp) [1] 1937-02-27 although it might be easier to paste a '19' into your character variable paste(substr(dob,1,6),19,substr(dob,7,9),sep=) [1] 27.02.1937 and then do it the way you started. Assumes you have leading zeroes on all fields though. Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] help requested
awk '{arr[$1]=arr[$1] $2}END{for( i in arr){print i,arr[i]}}'
edgelist.txt | sort -k1
On Fri, Dec 10, 2010 at 4:20 PM, profaar prof...@live.com wrote:
1 2
1 3
1 4
1 5
2 3
2 4
3 2
4 1
4 3
4 5
5 2
5 4
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] help requested
Try this: DF V1 V2 1 1 2 2 1 3 3 1 4 4 1 5 5 2 3 6 2 4 7 3 2 8 4 1 9 4 3 10 4 5 11 5 2 12 5 4 aggregate(V2 ~ V1, DF, paste, collapse = ' ') V1 V2 1 1 2 3 4 5 2 2 3 4 3 3 2 4 4 1 3 5 5 5 2 4 On Fri, Dec 10, 2010 at 1:20 PM, profaar prof...@live.com wrote: HI friends, I have very lengthy graph data in edge list format. I want to convert it into node list format. example: EDGE LIST FORMAT 1 2 1 3 1 4 1 5 2 3 2 4 3 2 4 1 4 3 4 5 5 2 5 4 ITS NODE LIST FORMAT SHOULD BE LIKE: 1 2 3 4 5 2 3 4 3 2 4 1 3 5 2 4 Kindly suggest me which package in R provides the support to do my task. Thank u friends in advance. -- View this message in context: http://r.789695.n4.nabble.com/help-requested-tp3082147p3082147.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. -- 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] survreg vs. aftreg (eha) - the relationship between fitted coefficients?
Dear R-users,
I need to use the aftreg function in package 'eha' to estimate failure times
for left truncated survival data. Apparently, survreg still cannot fit such
models. Both functions should be fitting the accelerated failure time (Weibull)
model. However, as Göran Broström points out in the help file for aftreg, the
parameterisation is different giving rise to different coefficients. The betas
for adjusted covariates are opposite in sign but otherwise identical, whereas
the intercept is quite different in a non-obvious way. The log-likelihoods are
similar also, but not identical. I would like to find out how I can convert one
set of coefficients to the other so as to obtain the same linear predictors
using either model. Any ideas???
#the example below uses right-censored data for simplicity (the principle
should be the same with left truncation I hope)
library(survival)
library(eha)
# COMPARE coefs between survreg ('survival' pkg) and aftreg ('eha' pkg)
#Fitting NULL models (no covariates) results in (approximately) the same coefs
(which is good!)
m1_NULL=survreg(Surv(futime/365, status==1) ~ 1, data=pbcseq)
m2_NULL=aftreg(Surv(futime/365, status==1) ~ 1, data=pbcseq)
c(m1_NULL$coef, 1/m1_NULL$scale) #-- intercept= 3.878656 , shape = 1.478177
c(m2_NULL$coef[1], exp(m2_NULL$coef[2])) #-- intercept= 3.878859 ,
shape=1.478150
# NOW I adjust for covariates
m1=survreg(Surv(futime/365, status==1) ~ chol+stage, data=pbcseq)
m2= aftreg(Surv(futime/365, status==1) ~ chol+stage, data=pbcseq)
### m2 ###
#Coefficients:
# (Intercept) cholstage
# 5.944641913 -0.001692574 -0.470861324
#Scale= 0.6416744
#Loglik(model)= -483.9 Loglik(intercept only)= -506.8
#Chisq= 45.91 on 2 degrees of freedom, p= 1.1e-10
#n=1124 (821 observations deleted due to missingness)
### m2 ###
#Covariate W.mean Coef Exp(Coef) se(Coef)Wald p
#chol 303.777 0.002 1.002 0.000 0.000
#stage 3.298 0.460 1.584 0.119 0.000
#
#log(scale)5.029 152.807 0.477 0.000
#log(shape)0.467 1.595 0.095 0.000
#
#Events92
#Total time at risk 9017
#Max. log. likelihood -484.31
#LR test statistic 45.0
#Degrees of freedom2
#Overall p-value 1.64669e-10
Many thanks for any help you may be able to provide.
Eleni Rapsomaniki
Research Associate
University of Cambridge
Institute of Primary and Public Health
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Remove 100 years from a date object
There still may be a problem if the dates go back far enough, e.g., 1909. Is '09' 1909 or 2009? No matter what, you have to decide which values need 1900 added and which need 2000. I'd split the date on the delimiter '.', decide whether to add 1900 or 2000, and then paste them together and then as.Date(). Clint -- Clint BowmanINTERNET: cl...@ecy.wa.gov Air Quality Modeler INTERNET: cl...@math.utah.edu Department of Ecology VOICE: (360) 407-6815 PO Box 47600FAX:(360) 407-7534 Olympia, WA 98504-7600 USPS: PO Box 47600, Olympia, WA 98504-7600 Parcels:300 Desmond Drive, Lacey, WA 98503-1274 On Fri, 10 Dec 2010, Barry Rowlingson wrote: On Fri, Dec 10, 2010 at 3:27 PM, Daniel Brewer daniel.bre...@icr.ac.uk wrote: Hello, I have some data that has dates in the form 27.02.37. I convert them to a date object as follows: as.Date(data$date,format=%d.%m.%y) But this gives me years such as 2037 when I would like them to be 1937. I thought of trying to take off some time i.e. as.Date(camCD$DoB,format=%d.%m.%y) - 100*365 But that doesn't seem to work out correctly. Any ideas how to do this? Normally to adjust dates you can use as.difftime() and do arithmetic, but a year is a variable thing (can be 365 or 366 days) so you cant make a difftime of years. Days are variable things if you worry about leap seconds... Also, you could end up with an invalid date if you have 29-Feb-2000 and 29-Feb-1900. One wasn't a leap year... A solution minus those caveats is to convert to POSIXlt and adjust the $year element: dob=27.02.37 as.Date(dob,format=%d.%m.%y) [1] 2037-02-27 dobp = as.POSIXlt(as.Date(dob,format=%d.%m.%y)) dobp$year = dobp$year - 100 dobp [1] 1937-02-27 UTC as.Date(dobp) [1] 1937-02-27 although it might be easier to paste a '19' into your character variable paste(substr(dob,1,6),19,substr(dob,7,9),sep=) [1] 27.02.1937 and then do it the way you started. Assumes you have leading zeroes on all fields though. Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Remove 100 years from a date object
On Fri, Dec 10, 2010 at 10:27 AM, Daniel Brewer daniel.bre...@icr.ac.uk wrote: Hello, I have some data that has dates in the form 27.02.37. I convert them to a date object as follows: as.Date(data$date,format=%d.%m.%y) But this gives me years such as 2037 when I would like them to be 1937. I thought of trying to take off some time i.e. as.Date(camCD$DoB,format=%d.%m.%y) - 100*365 But that doesn't seem to work out correctly. Any ideas how to do this? The easiest is just to use chron dates since it uses a cut.off of 30 by default. That is, if yy is less than that then 2000+yy is used and if greater than that then 1900+yy is used. Thus try this: library(chron) d - 27.02.37 as.Date(dates(d, format = d.m.y)) # 1937-02-27 as.Date(d, format = %d.%m.%y) # 2037-02-27 Also if that is not good enough and you want a different value for the cut.off then note that the default in chron is to use the year.expand function to expand two digit dates but you can change that via something like this: options(chron.year.expand = function(..., cut.off = 25) year.expand(..., cut.off = cut.off)) -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.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] Stricter read.table?
read.table gives idiosyncratic results when the input is formatted strangely, for example: read.table(textConnection(a'b\nc'd\n),header=FALSE,fill=TRUE,sep=,quote=') = c'd a'b c'd read.table(textConnection(a'b\nc'd\nf'\n'\n),header=FALSE,fill=TRUE,sep=,quote=') = f' \na b c'd f' \n Though read.table doesn't specify the syntax of its input precisely, these results don't seem particularly useful or consistent. Is there a stricter version of read.table (perhaps in a package) that gives errors or warnings if it finds quotation marks in the middle of fields or encounters other such peculiar situations? Thanks, -s [[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] Remove 100 years from a date object
On 10/12/2010 4:17 PM, Barry Rowlingson wrote:
On Fri, Dec 10, 2010 at 3:27 PM, Daniel Brewer daniel.bre...@icr.ac.uk
wrote:
Hello,
I have some data that has dates in the form 27.02.37. I convert them to
a date object as follows:
as.Date(data$date,format=%d.%m.%y)
But this gives me years such as 2037 when I would like them to be 1937.
I thought of trying to take off some time i.e.
as.Date(camCD$DoB,format=%d.%m.%y) - 100*365
But that doesn't seem to work out correctly. Any ideas how to do this?
Normally to adjust dates you can use as.difftime() and do arithmetic,
but a year is a variable thing (can be 365 or 366 days) so you cant
make a difftime of years. Days are variable things if you worry about
leap seconds...
Also, you could end up with an invalid date if you have 29-Feb-2000
and 29-Feb-1900. One wasn't a leap year...
A solution minus those caveats is to convert to POSIXlt and adjust the
$year element:
dob=27.02.37
as.Date(dob,format=%d.%m.%y)
[1] 2037-02-27
dobp = as.POSIXlt(as.Date(dob,format=%d.%m.%y))
dobp$year = dobp$year - 100
dobp
[1] 1937-02-27 UTC
as.Date(dobp)
[1] 1937-02-27
although it might be easier to paste a '19' into your character variable
paste(substr(dob,1,6),19,substr(dob,7,9),sep=)
[1] 27.02.1937
and then do it the way you started. Assumes you have leading zeroes on
all fields though.
Barry
Many thanks. Thats great
Dan
--
**
Daniel Brewer, Ph.D.
Institute of Cancer Research
Molecular Carcinogenesis
Email: daniel.bre...@icr.ac.uk
**
The Institute of Cancer Research: Royal Cancer Hospital, a charitable Company
Limited by Guarantee, Registered in England under Company No. 534147 with its
Registered Office at 123 Old Brompton Road, London SW7 3RP.
This e-mail message is confidential and for use by the a...{{dropped:2}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Perl cut equivalent in R
SL == Steve Lianoglou mailinglist.honey...@gmail.com
on Mon, 6 Dec 2010 14:21:59 -0500 writes:
if(FALSE) { stuff your don't want executed }
Switching a block of code off/on with editing a single
character may be done using 0/1 instead of FALSE/TRUE.
SL Or even F/T
Bad Idea:
F - 1
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Projecting data on a world map using long/lat
Thanks for the suggestions, but I am not there yet (I'm a real novice). In the code provided by Patrick (see below), I changed the shape input (from sids to world) which I downloaded here: http://thematicmapping.org/downloads/world_borders.php. As a result I also need to change the CNTY_ID and id in the code, but I have no idea what to put there. Could you please help me? Thanks! Mathijs library(maptools) library(ggplot2) gpclibPermit() myshp- readShapeSpatial(system.file(shapes/sids.shp, package=maptools)) ## see licence, not GPL myshp.points- fortify.SpatialPolygonsDataFrame(myshp, region=CNTY_ID) shpm- merge(myshp.points, myshp, by.x=id, by.y=CNTY_ID) head(shpm) p - ggplot(shpm, aes(long, lat, group=group, fill=NWBIR74)) p - p + geom_polygon() + geom_path(color=white) + coord_equal() ## Add some locations cities - read.table(textConnection( longlat city val -78.644722 35.818889 Raleigh 323 -80.84 35.226944 Charlotte 510 -82.555833 35.58 Asheville400), header = TRUE) p - p + geom_point(aes( fill=NULL, group = NULL, size=val), data = cities, color= 'black') p -- View this message in context: http://r.789695.n4.nabble.com/Projecting-data-on-a-world-map-using-long-lat-tp3081298p3082305.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] Perl cut equivalent in R
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Martin Maechler
Sent: Friday, December 10, 2010 8:54 AM
To: Steve Lianoglou
Cc: r-help@r-project.org
Subject: Re: [R] Perl cut equivalent in R
SL == Steve Lianoglou mailinglist.honey...@gmail.com
on Mon, 6 Dec 2010 14:21:59 -0500 writes:
if(FALSE) { stuff your don't want executed }
Switching a block of code off/on with editing a single
character may be done using 0/1 instead of FALSE/TRUE.
SL Or even F/T
Bad Idea:
F - 1
Another approach is to write the following function
dontRun - function(expr) {}
and replace that
if (FALSE) { ... questionable code ... }
with
dontRun( {... questionable code ...} )
If you do want the questionable code to run,
redefine dontRun to be
dontRun - function(expr) { expr }
You can use this approach to put assertion tests
into your code that only get run when the assertion
function is defined to do something.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] new edition of R Companion to Applied Regression
Dear all, Sandy Weisberg and I would like to announce the publication of the second edition of An R Companion to Applied Regression (Sage, 2011). As is immediately clear, the book now has two authors and S-PLUS is gone from the title (and the book). The R Companion has also been thoroughly rewritten, covering developments in the nearly 10 years since the first edition was written and expanding coverage of topics such as R graphics and R programming. As before, however, the R Companion provides a general introduction to R in the context of applied regression analysis, broadly construed. It is available from the publisher at http://www.sagepub.com/books/Book233899? (US) or http://www.uk.sagepub.com/books/Book233899? (UK), and from Amazon at http://www.amazon.ca/R-Companion-Applied-Regression/dp/141297514X/ref=sr_1_ 3?s=booksie=UTF8qid=1291995545sr=1-3. The book is augmented by a web site http://socserv.mcmaster.ca/jfox/Books/Companion/ with data sets, appendices on a variety of topics, and more, and it associated with the car package on CRAN, which has recently undergone an overhaul. Regards, John and Sandy John Fox Senator William McMaster Professor of Social Statistics Department of Sociology McMaster University Hamilton, Ontario, Canada web: socserv.mcmaster.ca/jfox __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] help requested
From: jinyan...@gmail.com
Date: Fri, 10 Dec 2010 17:20:00 +0100
To: prof...@live.com
CC: r-help@r-project.org
Subject: Re: [R] help requested
awk '{arr[$1]=arr[$1] $2}END{for( i in arr){print i,arr[i]}}'
edgelist.txt | sort -k1
My first thought PERL hash but I guess my answer would
still be to consider any R hash-like structures. I guess any array
that accepts arbitrary subscripts amounts to a hash.
On Fri, Dec 10, 2010 at 4:20 PM, profaar wrote:
1 2
1 3
1 4
1 5
2 3
2 4
3 2
4 1
4 3
4 5
5 2
5 4
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Perl cut equivalent in R
On 10/12/2010 12:05 PM, William Dunlap wrote:
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Martin Maechler
Sent: Friday, December 10, 2010 8:54 AM
To: Steve Lianoglou
Cc: r-help@r-project.org
Subject: Re: [R] Perl cut equivalent in R
SL == Steve Lianogloumailinglist.honey...@gmail.com
on Mon, 6 Dec 2010 14:21:59 -0500 writes:
if(FALSE) { stuff your don't want executed }
Switching a block of code off/on with editing a single
character may be done using 0/1 instead of FALSE/TRUE.
SL Or even F/T
Bad Idea:
F- 1
Another approach is to write the following function
dontRun- function(expr) {}
and replace that
if (FALSE) { ... questionable code ... }
with
dontRun( {... questionable code ...} )
If you do want the questionable code to run,
redefine dontRun to be
dontRun- function(expr) { expr }
You can use this approach to put assertion tests
into your code that only get run when the assertion
function is defined to do something.
That's a nice idea!
Duncan Murdoch
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] [r] overlap different line in a xyplot (lattice)
On 2010-12-10 07:04, Francesco Nutini wrote: dear [R] users, is there a way to plot different data (but with the same x-variables) in the same xyplot window? There are already a similar question, but the answer is not enought explanatory... Something like this? x - rep(1:10, 2) y1 - rnorm(10); y2 - rnorm(10) + 2 y - c(y1, y2) g - gl(2, 10) xyplot( y ~ x, groups = g, type = 'b') Peter Ehlers Thanks a lot, Francesco __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Could concurrent R sessions mix up variables?
Hi, I'm working in R 2.11.1 x64 on Windows x86_64-pc-mingw32. I'm experiencing a strange problem in R that I'm not even sure how to begin to fix. I've got a huge (forty-pages printed) simulation written in R that I'd like to run multiple times. When I open up R and run it on its own, it works fine. At the beginning of the program, there's a variable X that I set to 1, 5, 10, 20, depending on how sensitive I want the simulation to be to a certain parameter. When I just run one instance of R, the X variable stays the same throughout the program. I have a quad-core machine, so I'd like to take advantage of all four processors. If I open up four sessions and set X to 1, 5, 10, and 20 in those different sessions, then run all four simulations all the way through (about eighteen hours of processing time) at the same time, the variable X ends up being 20 at the end of all four sessions. It's as if R mixed up the variable setting between the four concurrent sessions. I can't figure out why else my variable X would ever get changed to 20 in the three simulations that I set it to 1, 5, and 10, respeectively (it doesn't get updated anywhere during the simulation). When I have all four of these simulations running concurrently, I am absolutely maxing out my computer. All four processors are at 100%, and my Windows Task Manager says I'm using almost 100% of my 16 GB of RAM. Is it possible that intense resource use would cause a variable conflict like this? I have no idea where to start troubleshooting this error, so any advice would be appreciated. Thanks! Anthony Damico Kaiser Family Foundation __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] help with RSQLite adding a new column
I'm new to using sql so I'm having difficulties (and worries) in adding a new column of data to a table I have. Its a very large file (around 5 Gb) which is why I'm having to use SQL I have a table with variables ID, IDrec and IDdes and the variables IDrec and IDdes give a mapping of some other values but the other values are associated with the ID variable (think of IDrec and IDdes being character strings and ID being numeric) (Imagine the transposed) Table1: ID: 1,2,3,4,... IDrec: A,B,C,D... IDdes: B,C,A,E... So I've created a table with the final form I need it to be in dbGetQuery(db, CREATE TABLE Map (ID int, IDrec int, IDrec1 int, IDdes int, IDdes1 int)) And the finished table would look something like: Map: ID: 1, 2, 3, 4,... IDrec: 1, 2, 3, 4,... IDrec1: A, B, C, D,... IDdes: 2, 3, 1, 5, IDdes1: B, C, A, E,... So I copy in the first set of values easily: dbGetQuery(db, INSERT INTO Map(ID, IDrec, IDrec1, IDdes1) SELECT ID, ID, IDrec, IDdes FROM Ntemp) Giving me a table that looks like: Map: ID: 1, 2, 3, 4,... IDrec: 1, 2, 3, 4,... IDrec1: A, B, C, D,... IDdes: NA,NA,NA,NA,... IDdes1: B, C, A, E,... Then I create a new table with just the IDdes values I need: dbGetQuery(db, Create table temp2 as SELECT temp.ID FROM Ntemp, temp WHERE Ntemp.IDdes1 = temp.IDrec1) Giving me temp2 (not sure what the variable name is) V1: 2, 3, 1, 5,... But when I try to copy in the new data: dbGetQuery(db, INSERT INTO Map(IDdes) SELECT * FROM temp2) My map table isn't updated: Map: ID: 1, 2, 3, 4,... IDrec: 1, 2, 3, 4,... IDrec1: A, B, C, D,... IDdes: NA,NA,NA,NA,... IDdes1: B, C, A, E,... Is there something I'm missing? Or am I just going about inserting the IDdes variables the wrong way? Thanks for the help. Michael [[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] help requested
On Fri, Dec 10, 2010 at 07:20:55AM -0800, profaar wrote: HI friends, I have very lengthy graph data in edge list format. I want to convert it into node list format. example: EDGE LIST FORMAT 1 2 1 3 1 4 1 5 2 3 2 4 3 2 4 1 4 3 4 5 5 2 5 4 ITS NODE LIST FORMAT SHOULD BE LIKE: 1 2 3 4 5 2 3 4 3 2 4 1 3 5 2 4 Kindly suggest me which package in R provides the support to do my task. How long the list of egdes is? For not too large lists, consider also library(graph) G - new(graphNEL, edgemode=directed) G - addNode(as.character(1:5), G) edges - read.table(file=stdin(), colClasses=character) 1 2 1 3 1 4 1 5 2 3 2 4 3 2 4 1 4 3 4 5 5 2 5 4 G - addEdge(from=edges[, 1], to=edges[, 2], G) edgeL(G) $`1` $`1`$edges [1] 2 3 4 5 $`2` $`2`$edges [1] 3 4 $`3` $`3`$edges [1] 2 $`4` $`4`$edges [1] 1 3 5 $`5` $`5`$edges [1] 2 4 Very large lists can be handled by unix/linux sort command (if not sorted already) and by extracting the end-nodes of the edges starting in each node. In a sorted file, they form blocks of consecutive lines, so a simple text processing with perl is sufficient. Petr Savicky. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] 45 Degree labels on barplot? Help understanding code previously posted.
On Dec 10, 2010, at 10:25 AM, Simon Kiss wrote: Dear colleagues, i found a line or two of code in the help archives from Uwe Ligges about creating slanted x-labels for a barplot and it works well for my purposes (code below). However, I was hoping someone could explain to me precisely what the code is doing. I'm aware it's invoking the text command, and I know the first ttwo arguments to text are x and y co-ordinates. I'm also aware that par(usr)[3] is grabbing the third element of the vector of plotting co-ordinates. More accurately the limits of the plot area in plot dimensions. But I tried replacing par(usr)[3] with just 0 and that didn't work; all the labels got bunched up on the left. That was the y argument, not the rotation argument. (Which means I am surprised that it bunched things to the side ... and for me it did nothing at all... same graphic.) It is the srt argument that controls the angle. Is it necessary to create a new object via barplot That gives you appropriate positions for the labels in plot coordinate terms and the xpd argument allows these locations to be used outside the plot area. and then quote that in the x,y coordinates of text? What do you mean by then quote it in the x,y,coordinates? I don't see any quotes. You could of course just look at the plot area and supply your own locations. You would need to figure out what the unlabeled x-axis scale really was, but that too is documented. -- David. Like I said, the code works great, but I'm trying to actually understand the rationale behind the elements so I can apply it in future. Yours, Simon Kiss #Reproducible Code mydat-data.frame(countries=c(Canada, Denmark, Framce, United Kingdom, Germany, Australia, New Zealand, Switzerland, Belgium, Netherlands), stories_total=c(429, 25, 239, 99, 100, 96, 18, 21, 0, 6), avg=c(4.165048544, 6.25, 6.459459459, 0.908256881, 1.923076923, 1.103448276, 1.058823529, 1.615384615, 0, 0.107142857), steps=c(2, 2, 2, 0,1, 1, 1, 0,0,0), newspapers=c(103, 4, 37, 109, 52, 87, 17, 13, 10, 56)) mydat.sort1-mydat[order(-mydat$avg), ] myplot-barplot(mydat.sort1$avg, col=c(black, black, black, grey, white, grey, grey, white, white, white), ylim=c(0,7), main=Regulatory Action On Bisphenol A By Newspaper Coverage) col.vec=c(black, grey, white) legend(topright, col=col.vec, fill=c(black, grey, white), legend=c(Meaningful Ban, Recommendations To Withdraw, No Legislative Action)) labels=mydat.sort1$countries #These lines create the labels text(myplot, par(usr)[3], labels=labels, srt=35, offset=1, adj=1, xpd=TRUE) axis(2) par(usr)[3] * Simon J. Kiss, PhD Assistant Professor, Wilfrid Laurier University 73 George Street Brantford, Ontario, Canada N3T 2C9 Cell: +1 519 761 7606 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Could concurrent R sessions mix up variables?
On 10/12/2010 1:13 PM, Anthony Damico wrote: Hi, I'm working in R 2.11.1 x64 on Windows x86_64-pc-mingw32. I'm experiencing a strange problem in R that I'm not even sure how to begin to fix. I've got a huge (forty-pages printed) simulation written in R that I'd like to run multiple times. When I open up R and run it on its own, it works fine. At the beginning of the program, there's a variable X that I set to 1, 5, 10, 20, depending on how sensitive I want the simulation to be to a certain parameter. When I just run one instance of R, the X variable stays the same throughout the program. I have a quad-core machine, so I'd like to take advantage of all four processors. If I open up four sessions and set X to 1, 5, 10, and 20 in those different sessions, then run all four simulations all the way through (about eighteen hours of processing time) at the same time, the variable X ends up being 20 at the end of all four sessions. It's as if R mixed up the variable setting between the four concurrent sessions. I can't figure out why else my variable X would ever get changed to 20 in the three simulations that I set it to 1, 5, and 10, respeectively (it doesn't get updated anywhere during the simulation). When I have all four of these simulations running concurrently, I am absolutely maxing out my computer. All four processors are at 100%, and my Windows Task Manager says I'm using almost 100% of my 16 GB of RAM. Is it possible that intense resource use would cause a variable conflict like this? I have no idea where to start troubleshooting this error, so any advice would be appreciated. If you are running something that takes 18 hours to complete, a common practice is to save intermediate results to disk occasionally. Have you (or whoever wrote the simulation) done this and forgotten about it? If all 4 processes are saving to the same place, then reading results back, you'd see something like you describe. If all calculations are held in memory, you shouldn't. A simple approach that might debug this is to create a new variables initX, set equal to X. Then sprinkle statements stopifnot(X == initX) through your simulation code. That should quit when the change happens, and you can try to figure out why it happened. Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Could concurrent R sessions mix up variables?
Anthony - I would advise you to use the multicore or snowfall packages to utilize multiple CPUs. As an example using multicore: library(multicore) sim = function(mu)max(replicate(10,max(rnorm(100,mu library(multicore) unlist(mclapply(c(1,5,10,20),sim)) [1] 6.569332 10.268091 15.335847 25.291502 Using snowfall: library(snowfall) sim = function(mu)max(replicate(10,max(rnorm(100,mu sfInit(cpus=4,type='SOCK',parallel=TRUE) sfSapply(c(1,5,10,20),sim) [1] 6.200161 10.307807 15.271581 25.055950 Hope this helps. - Phil Spector Statistical Computing Facility Department of Statistics UC Berkeley spec...@stat.berkeley.edu On Fri, 10 Dec 2010, Anthony Damico wrote: Hi, I'm working in R 2.11.1 x64 on Windows x86_64-pc-mingw32. I'm experiencing a strange problem in R that I'm not even sure how to begin to fix. I've got a huge (forty-pages printed) simulation written in R that I'd like to run multiple times. When I open up R and run it on its own, it works fine. At the beginning of the program, there's a variable X that I set to 1, 5, 10, 20, depending on how sensitive I want the simulation to be to a certain parameter. When I just run one instance of R, the X variable stays the same throughout the program. I have a quad-core machine, so I'd like to take advantage of all four processors. If I open up four sessions and set X to 1, 5, 10, and 20 in those different sessions, then run all four simulations all the way through (about eighteen hours of processing time) at the same time, the variable X ends up being 20 at the end of all four sessions. It's as if R mixed up the variable setting between the four concurrent sessions. I can't figure out why else my variable X would ever get changed to 20 in the three simulations that I set it to 1, 5, and 10, respeectively (it doesn't get updated anywhere during the simulation). When I have all four of these simulations running concurrently, I am absolutely maxing out my computer. All four processors are at 100%, and my Windows Task Manager says I'm using almost 100% of my 16 GB of RAM. Is it possible that intense resource use would cause a variable conflict like this? I have no idea where to start troubleshooting this error, so any advice would be appreciated. Thanks! Anthony Damico Kaiser Family Foundation __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] survival package - calculating probability to survive a given time
Dear R users, i try to calculate the probabilty to survive a given time by using the estimated survival curve by kaplan meier. What is the right way to do that? as far as is see i cannot use the predict-methods from the survival package? library(survival) set.seed(1) time - cumsum(rexp(1000)/10) status - rbinom(1000, 1, 0.5) ## kaplan meier estimates fit - survfit(Surv(time, status) ~ 1) s - summary(fit) ## 1. possibility to get the probability for surviving 20 units of time ind - findInterval(20, s$time) cbind(s$surv[ind], s$time[ind]) ## 2. possibility to get the probability for surviving 20 units of time ind - s$time = 20 sum(ind) / length(ind) Thanks and best regards Andreas __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Compare one level of a factor with *all* other non-missing levels
On 2010-12-10 05:58, deriK2000 wrote:
Peter Ehlers wrote:
Sounds like you want the Dunnett test procedure which seems
to be implemented in a number of packages: multcomp, asd, MCPAN
and others.
It would probably be a good idea to install package 'sos' and
learn how to search with it.
Peter Ehlers
Thanks for the hints!
Unfortunately, Dunnett compares the mean(x) for a factor level with the
means(x) of all single K-1 other levels resulting in K-1 comparisions for
each level (printed in a lower triangular matrix for the results). Instead,
I just want to compare this one mean(x) with one other mean(x) of all the
K-1 other levels (printed in a vector of length K for the results).
Okay, I misunderstood; should have read more carefully.
I would just use a loop (I'm not as loop-averse as
some R users).
x - rnorm(20)
f - gl(4, 5, lab = letters[1:4])
lev - levels(f)
len - length(lev)
pv - numeric(len)
for(i in 1:len){
pv[i] - t.test(x[f == lev[i]], x[f != lev[i]])$p.value
}
pv
For pvalue adjustment (if you think that's needed),
see ?p.adjust.
Concerning sos: sounds like a good idea!
Yes, it's an excellent tool.
Peter Ehlers
Cheers,
Derik
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] survival package - calculating probability to survive a given time
On Dec 10, 2010, at 2:07 PM, Andreas Wittmann wrote: Dear R users, i try to calculate the probabilty to survive a given time by using the estimated survival curve by kaplan meier. What is the right way to do that? as far as is see i cannot use the predict-methods from the survival package? library(survival) set.seed(1) time - cumsum(rexp(1000)/10) status - rbinom(1000, 1, 0.5) ## kaplan meier estimates fit - survfit(Surv(time, status) ~ 1) s - summary(fit) ## 1. possibility to get the probability for surviving 20 units of time ind - findInterval(20, s$time) cbind(s$surv[ind], s$time[ind]) See if this helps: head(which(s$surv 0.5)) [1] 368 369 370 371 372 373 plot(fit) abline(h=0.5) abline(v=s$time[368]) ## 2. possibility to get the probability for surviving 20 units of time ind - s$time = 20 sum(ind) / length(ind) -- David Winsemius, MD 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] Reorder factor and address embedded escapes
I am trying to reorder a factor variable that has embedded escape characters. The data begins as a csv file with a factor that includes embedded new line characters. By the time read.table has rendered it into a data frame, the variable now has an extra backslash. e.g. This\nLabel in the csv becomes This\\nLabel in the data frame. So, I am trying to reorder the factor and deal with the introduction of a secondary \ . Here's a small example: A = c(A\\nB, C\\nD) test -data.frame(A) str(test) test$reorderA -factor(test$A, c(C\\nD, A\\nB)) str(test) test$reorderB -sub(n, \n, test$reorderA) str(test) When sub is applied to the now-correctly ordered factor, it returns to the default ordering. Suggestions? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Time Series Row Label
Simple question... I know that when referencing data in a multivariate time series matrix, I can use the variable name instead of the column number (such as budget.ts[4,incometax]). Is there a way I can use the time unit (say, the year in an annual time series) instead of the row number? -- View this message in context: http://r.789695.n4.nabble.com/Time-Series-Row-Label-tp3082639p3082639.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] Reorder factor and address embedded escapes
Does the following help? A = c(A\\nB, C\\nD) test -data.frame(A) #access levels directly to change names levels(test$A) - sub(n, \n, levels(test$A)) #re-order levels of the factor test$A - relevel(test$A, C\nD) Rob James wrote: I am trying to reorder a factor variable that has embedded escape characters. The data begins as a csv file with a factor that includes embedded new line characters. By the time read.table has rendered it into a data frame, the variable now has an extra backslash. e.g. This\nLabel in the csv becomes This\\nLabel in the data frame. So, I am trying to reorder the factor and deal with the introduction of a secondary \ . Here's a small example: A = c(A\\nB, C\\nD) test -data.frame(A) str(test) test$reorderA -factor(test$A, c(C\\nD, A\\nB)) str(test) test$reorderB -sub(n, \n, test$reorderA) str(test) When sub is applied to the now-correctly ordered factor, it returns to the default ordering. Suggestions? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Reorder factor and address embedded escapes
On Dec 10, 2010, at 3:14 PM, Rob James wrote: I am trying to reorder a factor variable that has embedded escape characters. The data begins as a csv file with a factor that includes embedded new line characters. By the time read.table has rendered it into a data frame, the variable now has an extra backslash. e.g. This\nLabel in the csv becomes This\\nLabel in the data frame. So, I am trying to reorder the factor and deal with the introduction of a secondary \ . Here's a small example: A = c(A\\nB, C\\nD) test -data.frame(A) str(test) test$reorderA -factor(test$A, c(C\\nD, A\\nB)) str(test) test$reorderB -sub(n, \n, test$reorderA) str(test) When sub is applied to the now-correctly ordered factor, it returns to the default ordering. Actually it returns a character vector rather than a factor. Suggestions? Perhaps you should ask a question. -- David Winsemius, MD 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] Encoding problem - I fails to read Hebrew text from online
Hi Matt and everyone else,
Thanks for the help so far.
I ended up using the tips provided to create a dirty hack based on a
translation table between the code and the Hebrew letters.
For the future (and for any suggestions), I am attaching this code bellow:
Best,
Tal
# the translation table:
translation.table.Hebrew - structure(list(V1 = structure(1:27, .Label =
c(05D0, 05D1,
05D2, 05D3, 05D4, 05D5, 05D6, 05D7, 05D8,
05D9, 05DA, 05DB, 05DC, 05DD, 05DE, 05DF,
05E0, 05E1, 05E2, 05E3, 05E4, 05E5, 05E6,
05E7, 05E8, 05E9, 05EA), class = factor), V2 = structure(1:27,
.Label = c(×,
×, ×, ×, ×, ×, ×, ×, ×, ×, ×, ×, ×, ×,
×, ×, × , ס, ×¢, ×£, פ, ×¥, צ, ק, ר, ש, ת
), class = factor)), .Names = c(CODE, HEBREW), class = data.frame,
row.names = c(NA,
-27L))
# translation.table
# STRING = inp
turn_nohash - function(STRING)
{
require(stringr)
nohash - str_replace(STRING, #, 0) # cvrt # to 0
nohash - str_replace(nohash, ;, ) # cvrt # to 0
nohash - str_replace(nohash, , ) # cvrt # to 0
nohash - str_replace(nohash, x, ) # cvrt # to 0
return(nohash)
}
translate.all.chars - function(STRING, TABLE = translation.table.Hebrew)
{
# TABLE is of the form:
# CODE HEBREW
# 1 05D0 ×
# 2 05D1 ×
# 3 05D2 ×
require(stringr)
i.chars.to.check - seq_len(dim(TABLE)[1])
for(i in i.chars.to.check)
{
STRING - str_replace(STRING, as.character(TABLE[i,1]),
as.character(TABLE[i,2]))
}
return(STRING)
}
HTML_heb_decode - function(STRING, TABLE = translation.table.Hebrew)
{
STRING - turn_nohash(STRING)
STRING - translate.all.chars(STRING, TABLE)
return(STRING)
}
# example of use:
inp - #x5E9;#x5DC;#x5D5;#x5DD;
HTML_heb_decode(inp)
inp - #x5E9;#x5DC;#x5D5;#x5DD; #x5D7;#x5E0;#x5D5;#x5DA;\
HTML_heb_decode(inp)
ourput:
HTML_heb_decode(inp)
Loading required package: stringr
Loading required package: plyr
[1] ש×××
inp - #x5E9;#x5DC;#x5D5;#x5DD; #x5D7;#x5E0;#x5D5;#x5DA;\
HTML_heb_decode(inp)
[1] ש××× ×× ××
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
--
On Fri, Dec 10, 2010 at 12:00 AM, Matt Shotwell shotw...@musc.edu wrote:
Tal,
OK, let me clarify my understanding. The original and decoded file are
text, encoded by UTF-8. In the original file, there are HTML `entities'
that represent UTF-8 Hebrew characters. In the decoded file, the
entities are converted to UTF-8 characters. The question is how to
convert these entities within R. It's not the same as converting between
character encodings, otherwise iconv() might offer a solution.
I'll have a look around to find a solution, and I hope others will too.
My first idea is to check RCurl, XML, and the related utils::URLdecode.
If there really is no existing solution, I think it might be worthwhile
to look at how PHP and Python do it (and maybe borrow some code :) ).
-Matt
On Thu, 2010-12-09 at 14:27 -0500, Tal Galili wrote:
Hi Matt,
Thanks for having a look at this.
I just spent some time looking around and couldn't find any R function
to decode decimal HTML code.
Do you (or someone else on the list) knows how to program this sort of
thing? (is there a formula for the translation?
p.s:
For it to work on my end I added the encoding parameter:
readLines(http://biostatmatt.com/temp/Hebrew-decoded;, warn=FALSE,
encoding= UTF-8)
p.p.s: The Hebrew word I used means peace
Cheers,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew)
| www.r-statistics.com (English)
--
On Thu, Dec 9, 2010 at 8:38 PM, Matt Shotwell shotw...@musc.edu
wrote:
Tal,
It looks like the data you received has HTML special hex
characters.
That is, '#x5E9;' is just an ASCII HTML representation of a
hex
character. It's not encoded in a special manner.
The trick is to substitute the HTML encoded hex character for
its binary
representation, or decode the character. I don't know of any
R
function that does this, but there are web services, for
example:
http://www.hashemian.com/tools/html-url-encode-decode.php
I decoded your file using this service and posted it on my
website. You
can see the difference by running:
readLines(http://biostatmatt.com/temp/Hebrew-original;,
warn=FALSE)
[R] spatial clusters
Dear all, I am looking for a clustering method usefull to classify the countries in some clusters taking account of: a) the geographical distance (in km) between countries and b) of some macroeconomic indicators (gdp, life expectancy...). Are there some packages in R usefull for this? Thanks a lot for your help, Dorina __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Time Series Row Label
On Fri, Dec 10, 2010 at 3:16 PM, dreadgazebo ernie.tedes...@gmail.com wrote: Simple question... I know that when referencing data in a multivariate time series matrix, I can use the variable name instead of the column number (such as budget.ts[4,incometax]). Is there a way I can use the time unit (say, the year in an annual time series) instead of the row number? You can use window(), e.g. tt - ts(1:4, start = 2000) window(tt, start = 2001, end = 2002) -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.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] WriteXLS error:Error in get(x, envir = envir) : variable names are limited to 256 bytes
Hello all, I don't understand why this won't work. I have entered: WriteXLS(alldata,'test.xls') and I get this error message: Error in get(x, envir = envir) : variable names are limited to 256 bytes. My variable names are not very long, and are accepted by write.csv. alldata is a list containing 4 dataframes, with each dataframe having the the same variable names, which are: names(avg8302) [1] IDcluster rec.unit int.hib yr.hibyr0309.hibint.hib.seyr.hib.se yr0309.hib.se int.cl [11] yr.cl yr0309.cl int.cl.se yr.cl.se yr0309.cl.se int.ruyr.ru yr0309.ru int.ru.se yr.ru.se [21] yr0309.ru.se int.spyr.sp yr0309.sp int.sp.se yr.sp.se yr0309.sp.se Does anybody know how I can fix this? Or another way to write a multi-sheet xls? Thank you. [[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] How to print colorful R output??
Hi All, I wonder if there is a way to print the R output with COLOR? Not the color plots, but the outputs in the console. Thank. casper -- View this message in context: http://r.789695.n4.nabble.com/How-to-print-colorful-R-output-tp3082750p3082750.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] WriteXLS error:Error in get(x, envir = envir) : variable names are limited to 256 bytes
On Dec 10, 2010, at 4:39 PM, Patrick McKann wrote:
Hello all,
I don't understand why this won't work. I have entered:
WriteXLS(alldata,'test.xls')
I have gotten tripped up by the argument syntax in WriteXLS myself,
many times. Please check the help page for argument names and use
them, especially paying attention to the fact that the first argument
needs to be a character _vector_ (and I suspect that passing it a list
may not qualify) and I always use the name for the Excel file
argument. I suspect that this may work:
WriteXLS('alldata','test.xls')
--
David.
and I get this error message:
Error in get(x, envir = envir) : variable names are limited to 256
bytes.
My variable names are not very long, and are accepted by write.csv.
alldata is a list containing 4 dataframes, with each dataframe
having the
the same variable names, which are:
names(avg8302)
[1] IDcluster rec.unit int.hib
yr.hibyr0309.hibint.hib.seyr.hib.se
yr0309.hib.se int.cl
[11] yr.cl yr0309.cl int.cl.se yr.cl.se
yr0309.cl.se int.ruyr.ru yr0309.ru
int.ru.se
yr.ru.se
[21] yr0309.ru.se int.spyr.sp yr0309.sp
int.sp.se yr.sp.se yr0309.sp.se
Does anybody know how I can fix this? Or another way to write a
multi-sheet
xls?
Thank you.
[[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
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] WriteXLS error:Error in get(x, envir = envir) : variable names are limited to 256 bytes
On Dec 10, 2010, at 5:02 PM, David Winsemius wrote:
On Dec 10, 2010, at 4:39 PM, Patrick McKann wrote:
Hello all,
I don't understand why this won't work. I have entered:
WriteXLS(alldata,'test.xls')
I have gotten tripped up by the argument syntax in WriteXLS myself,
many times. Please check the help page for argument names and use
them, especially paying attention to the fact that the first
argument needs to be a character _vector_ (and I suspect that
passing it a list may not qualify) and I always use the name for the
Excel file argument. I suspect that this may work:
WriteXLS('alldata','test.xls')
OOOPs. I wrote that before I noted that you said you were using a
list, and I forgot to go back and fix it, so that would NOT work.
--
David.
and I get this error message:
Error in get(x, envir = envir) : variable names are limited to 256
bytes.
My variable names are not very long, and are accepted by write.csv.
alldata is a list containing 4 dataframes, with each dataframe
having the
the same variable names, which are:
names(avg8302)
[1] IDcluster rec.unit int.hib
yr.hibyr0309.hibint.hib.seyr.hib.se
yr0309.hib.se int.cl
[11] yr.cl yr0309.cl int.cl.se
yr.cl.se
yr0309.cl.se int.ruyr.ru yr0309.ru
int.ru.se
yr.ru.se
[21] yr0309.ru.se int.spyr.sp
yr0309.sp
int.sp.se yr.sp.se yr0309.sp.se
Does anybody know how I can fix this? Or another way to write a
multi-sheet
xls?
Thank you.
[[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
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.
David Winsemius, MD
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] Stricter read.table?
Stavros Macrakis macrakis at alum.mit.edu writes: read.table gives idiosyncratic results when the input is formatted strangely, for example: read.table(textConnection( a'b\nc'd\n),header=FALSE, fill=TRUE,sep=,quote=') = c'd a'b c'd read.table(textConnection( a'b\nc'd\nf'\n'\n), header=FALSE,fill=TRUE sep=,quote=') = f' \na b c'd f' \n Though read.table doesn't specify the syntax of its input precisely, these results don't seem particularly useful or consistent. Is there a stricter version of read.table (perhaps in a package) that gives errors or warnings if it finds quotation marks in the middle of fields or encounters other such peculiar situations? I dissected this behavior a bit more here https://stat.ethz.ch/pipermail/r-devel/2010-November/059016.html (it is due to an inconsistency between the way that scan() and readLines() handle lines with unterminated quotes, IIRC) and Martin Maechler said https://stat.ethz.ch/pipermail/r-devel/2010-November/059107.html I think it can be defended to file as a bug, but it is tricky to pinpoint exactly what the issue is. I don't know of a stricter version of read.table(), but if you had the time and inclination to pick through the code and (i) provide a careful definition of desired behavior and (ii) supply patches, you could do your little bit to make R better. (If I posted a bug report would you annotate it with a discussion of desired behavior?) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Quantile with discrete types
I don't understand why 'quantile' works in this case:
tt - rep(c('a','b'),c(10,3))
sapply(0:6/6,function(q) quantile(tt,probs=q,type=1))
0% 16.7% 33.3% 50% 66.7% 83.3% 100%
a a a a a b b
and also
quantile(tt,0:5/5,type=1)
0% 20% 40% 60% 80% 100%
a a a a b b
but gives an error in this, which I would have thought equivalent to the
first case above:
quantile(tt,0:6/6,type=1)
Error in sort.int(x, na.last = na.last, decreasing = decreasing, ...) :
argument is not a numeric vector
I could of course write something like
sort(tt)[seq(1,length(tt),length.out=7)] -- but I'm wondering why quantile
fails in this case.
Thanks,
-s
[[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] [r] overlap different line in a xyplot (lattice)
Peter Ehlers ehlers at ucalgary.ca writes: On 2010-12-10 07:04, Francesco Nutini wrote: dear [R] users, is there a way to plot different data (but with the same x-variables) in the same xyplot window? There are already a similar question, but the answer is not enought explanatory... Something like this? [snip] Also possibly the layer() command in the latticeExtra package. If there is an answer that doesn't make sense to you it might be most efficient to post an edited version of that question/answer, attempting to clarify which parts of the answer you do and don't understand ... A reproducible example would be nice too. Ben Bolker __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] locfit weights not working as expected
Hello! I am having a problem understanding what the weights option in the locfit command of the locfit package is doing. I have written a sample program which illustrates the issue (below). The example involves using bootstrap however, that is not my main goal but it illustrates where my problem lies. As you know, to compute a bootstrap estimate of a particular quantity using a sample size of size N, I would sample a multinomial vector which should be of size N with possible values 1:N. I should get the exact same answer whether I a) estimate the quantity using this multinomial vector as the indices i.e. literally make a new sample by choosing each element according to the number of time in the corresponding index or b) estimate the quantity using the original data and specify this multinomial vector as the weights. In all commands that I have tried, the weights functions works such that this is always true. This is not true for the locfit command. The models are different, the coefficients are different, the predictions are different. I have tried several combinations of things including specifying different options for the evaluation points, different kernels, different datasets to predict and I cannot find any way to make these equal. For my particular simulation, I need to understand exactly what the weights option is doing since it is clearly not doing what I expected it to. I would gratefully appreciate any advice or help that anyone can give on this issue and I appreciate your time very much. library(MASS) library(splines) library(sm) library(quantreg) library(locfit) set.seed(20) Zi = rnorm(5000,0, 2) ei = mvrnorm(5000, c(.6,2), matrix(c(.7,0,0,.9),2,2)) X1i = exp(-.5*Zi + ei[,1]) X2i = exp(-.5*(Zi*log(T1i) -log(T1i) + Zi) + ei[,2]) X1i = log(X1i) boot.weight = rmultinom(1,5000, prob=rep(1/5000, 5000)) boot.subset = rep(1:5000, boot.weight) loc.model.1 = locfit(1*(X2i[boot.subset] 10) ~ lp(X1i[boot.subset], Zi[boot.subset], style = c (n,cpar), h=0.4, deg = 1, nn=0), family = binomial, link = logit) loc.model.2 = locfit(1*(X2i 10) ~ lp(X1i, Zi, style = c(n,cpar), h=0.4, deg = 1,nn=0), family = binomial, link = logit, weights=boot.weight) print(loc.model.1) print(loc.model.2) t.vector = log(seq(0.5,10, length=300)) p.0.1 = predict(loc.model.1, newdata=cbind(t.vector, rep(0,length(t.vector p.0.2 = predict(loc.model.2, newdata=cbind(t.vector, rep(0,length(t.vector #quantity below should be vector of zeros but it is not. p.0.1-p.0.2 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 question: memory usage
Hi I have a large R progress that that is constently running into memory issues, I am trying to rewrite some of the code to make more efficeint However, one thing I have found is that the memory usage shown by gc() is very different from what i see from unix termilnal Garbage collection 2047 = 1398+243+406 (level 2) ... 12.4 Mbytes of cons cells used (3%) 998.6 Mbytes of vectors used (23%) used (Mb) gc trigger (Mb) max used (Mb) Ncells231884 12.48140556 434.8 12719620 679.4 Vcells 130881306 998.6 572307468 4366.4 713657158 5444.8 vs PID USER PR NI VIRT RES SHR S %CPU %MEMTIME+ COMMAND 26291 s_edge25 0 2883m 8.2g 4276 R 99.7 48.0 10:35.31 R Am i missing something here? Thanks a lot for the help Simon [[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] randomForest: help with combine() function
I've built two RF objects (RF1 and RF2) and have tried to combine them, but I get the following error: Error in rf$votes + ifelse(is.na(rflist[[i]]$votes), 0, rflist[[i]]$votes) : non-conformable arrays In addition: Warning message: In rf$oob.times + rflist[[i]]$oob.times : longer object length is not a multiple of shorter object length Both RF models use the same variables, although the NAs in both models likely differ (using na.roughfix in both models). I assume this is part of the reason that my arrays are non-conformable. If so, does anyone have any suggestions on how to combine in such a situation? How similar do RFs have to be in order to combine? Cheers __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 print colorful R output??
Yes (depending on what you mean by output), but maybe there is a long way to go. The vignette of the Rd2roxygen package is an example: http://cran.r-project.org/web/packages/Rd2roxygen/vignettes/Rd2roxygen.pdf It makes use of the highlight package and Sweave to output colored code. If the style of the above vignette is what you want, more details are here: http://yihui.name/en/2010/10/how-to-start-using-pgfsweave-in-lyx-in-one-minute/ Regards, Yihui -- Yihui Xie xieyi...@gmail.com Phone: 515-294-2465 Web: http://yihui.name Department of Statistics, Iowa State University 2215 Snedecor Hall, Ames, IA On Fri, Dec 10, 2010 at 3:53 PM, casperyc caspe...@hotmail.co.uk wrote: Hi All, I wonder if there is a way to print the R output with COLOR? Not the color plots, but the outputs in the console. Thank. casper -- View this message in context: http://r.789695.n4.nabble.com/How-to-print-colorful-R-output-tp3082750p3082750.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] WriteXLS error:Error in get(x, envir = envir) : variable names are limited to 256 bytes
On Dec 10, 2010, at 4:14 PM, David Winsemius wrote:
On Dec 10, 2010, at 5:02 PM, David Winsemius wrote:
On Dec 10, 2010, at 4:39 PM, Patrick McKann wrote:
Hello all,
I don't understand why this won't work. I have entered:
WriteXLS(alldata,'test.xls')
I have gotten tripped up by the argument syntax in WriteXLS myself, many
times. Please check the help page for argument names and use them,
especially paying attention to the fact that the first argument needs to be
a character _vector_ (and I suspect that passing it a list may not qualify)
and I always use the name for the Excel file argument. I suspect that this
may work:
WriteXLS('alldata','test.xls')
OOOPs. I wrote that before I noted that you said you were using a list, and I
forgot to go back and fix it, so that would NOT work.
--
David.
and I get this error message:
Error in get(x, envir = envir) : variable names are limited to 256 bytes.
My variable names are not very long, and are accepted by write.csv.
alldata is a list containing 4 dataframes, with each dataframe having the
the same variable names, which are:
names(avg8302)
[1] IDcluster rec.unit int.hib
yr.hibyr0309.hibint.hib.seyr.hib.se
yr0309.hib.se int.cl
[11] yr.cl yr0309.cl int.cl.se yr.cl.se
yr0309.cl.se int.ruyr.ru yr0309.ru int.ru.se
yr.ru.se
[21] yr0309.ru.se int.spyr.sp yr0309.sp
int.sp.se yr.sp.se yr0309.sp.se
Does anybody know how I can fix this? Or another way to write a multi-sheet
xls?
Thank you.
Hi David and Patrick,
Apologies for the delay in my reply as I am away on vacation at the moment.
As David surmised initially, the name of the object(s) to be exported, needs to
be passed as a character vector. The vector can either contain the names of one
or more data frames, or can be the single name of a list of data frames. The
latter option was added in a September update. See the help page for an example
of use.
Thus:
WriteXLS(alldata, test.xls)
should work.
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.
