Re: [R] [R-sig-Geo] Using RGDAL to copy header information...
On Wed, 5 Nov 2008, Jonathan Greenberg wrote:
R-geographers...
I'm trying to solve a problem to implement a line-by-line tiled processing
using RGDAL (read 1 line of an image, process the one line, write one line of
the image to a binary file). Everything except for the final step I'm able
to do using a combination of RGDAL and r-base commands. Below is the basic
structure, the input file elev can be any image file GDAL supports -- the
code below just copies the image one line at a time:
###
library(rgdal)
infile='elev'
outfile_base='testout'
outfile_ext='.bil'
outfile=paste(outfile_base,outfile_ext,sep='')
outcon - file(outfile, wb)
infile_info=GDALinfo(infile)
nl=infile_info[[1]]
ns=infile_info[[2]]
for (row in 1:nl) {
templine - readGDAL(infile,region.dim=c(1,ns),offset=c(row-1,0))
writeBin(templine[[1]], outcon,size=4)
}
close(outcon)
# Below doesn't work
# writeGDAL(templine,outfile_base,drivername='EHdr',type=Float32)
###
The issue is this: I need to be able to effectively copy the
geographic/header information from the input file (the last line almost
works, but as you can see it sets the line #s to 1, not ns), and parse it
over to the output file (which is some form of a flat binary file -- in this
case, an Arc Binary Raster, but an ENVI binary output would also be good) --
ideally I'd like to be able to modify this header, particularly the number
type (e.g. if the input file is an integer, and I'm writing out a floating
point binary, I need that reflected in the output header). I can do this by
*manually* creating the output header via a series of ascii-writes, but I was
wondering if there is a more effective way of doing this using RGDAL that
might apply generically to the header of any binary image file I might write?
Thanks!
It is the internals of writeGDAL() and create2GDAL() without the bands:
data(meuse.grid)
gridded(meuse.grid) - ~x+y
d.drv - new(GDALDriver, EHdr)
gp - gridparameters(meuse.grid)
cellsize - gp$cellsize
offset - gp$cellcentre.offset
dims - gp$cells.dim
nbands - 1
tds.out - new(GDALTransientDataset, driver = d.drv, rows = dims[2],
cols = dims[1], bands = nbands, type = Float32)
gt - c(offset[1] - 0.5 * cellsize[1], cellsize[1], 0.0,
offset[2] + (dims[2] -0.5) * cellsize[2], 0.0, -cellsize[2])
.Call(RGDAL_SetGeoTransform, tds.out, gt, PACKAGE = rgdal)
list.files(tempdir())
fn - tempfile()
saveDataset(tds.out, fn)
list.files(tempdir())
GDAL.close(tds.out)
list.files(tempdir())
However, the driver does commit the data file too, so you'd need to run
the above code to produce the header so as not to overwrite your output
data. The key thing is to get the input grid parameters right, but you've
got them anyway. If you want the projection information out too, look in
create2GDAL for the .Call() you need. Note that passing unchecked
variables through .Call may crash your session - going through
GridTopology should make sure that they are stored correctly.
Hope this helps,
Roger
--j
--
Roger Bivand
Economic Geography Section, Department of Economics, Norwegian School of
Economics and Business Administration, Helleveien 30, N-5045 Bergen,
Norway. voice: +47 55 95 93 55; fax +47 55 95 95 43
e-mail: [EMAIL PROTECTED]
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[R] Mean Computation for Data from External File
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[R] Data manipulation question
Dear R-listers, I am a relatively inexperienced R-user currently migrating from Stata. I am deeply frustrated by this data manipulation question: I know how I could do it in Stata, but I cannot make it work in R. I have a data frame of hospitalization data where each row represents an admission. I need to know when patients were first discharged, but the problem is that patients were sometimes transferred between hospital departments. In my data a transfer looks like a new admission, except that it has a 'start' date equal to the previous admission's 'stop' date. Here is an example: id - c(rep(a,4),rep(b,2), rep(c,5), rep(d,1)) start - c(c(0,6,17,20),c(0,1),c(0,5,10,11,50),c(0)) stop - c(c(6,12,20,30),c(1,10),c(3,10,11,30,55),c(6)) data - as.data.frame(cbind(id,start,stop)) data #id start stop # 1 a 06 # 2 a 6 12 # 3 a17 20 # 4 a20 30 # 5 b 01 # 6 b 1 10 # 7 c 03 # 8 c 5 10 # 9 c10 11 # 10 c11 30 # 11 c50 55 # 12 d 06 So, what I want to end up with is this: id start stop a 0 12 # This patient was transferred at time 6 and discharged at time 12. The admission starting at time 17 is therefore irrelevant. b 0 10 c 0 3 d 0 6 I have tried tons of variations over lapply, sapply, split, for etc., all to no avail. Thank you in advance for any assistance. Best regards, Peter Jepsen, MD. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Reshape a matrix
Dear Dinesh, Have a look at ?cast for the reshape package dataset - data.frame(Col1 = LETTERS[1:3], Col2 = sort(rev(LETTERS)[1:3]), value = 1:3) dataset Col1 Col2 value 1AX 1 2BY 2 3CZ 3 library(reshape) cast(Col1 ~ Col2, data = dataset, fill = 0) Col1 X Y Z 1A 1 0 0 2B 0 2 0 3C 0 0 3 HTH, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology and quality assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 [EMAIL PROTECTED] www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey -Oorspronkelijk bericht- Van: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Namens dinesh kumar Verzonden: donderdag 6 november 2008 1:53 Aan: r-help@r-project.org Onderwerp: [R] Reshape a matrix Dear R users, I have a matrix like A X1 B Y2 C Z3 I want to reshape this matrix into this format X Y Z A 1 B 2 C 3 Thanks in advance for your help. Dinesh -- Dinesh Kumar Barupal Junior Specialist Metabolomics Fiehn Lab UCD Genome Center 451 East Health Science Drive GBSF Builidng University of California DAVIS 95616 http://fiehnlab.ucdavis.edu/staff/kumar [[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. Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Memory limits for large data sets
Alan Cohen wrote: Hello, I have several very large data sets (1-7 million observations, sometimes hundreds of variables) that I'm trying to work with in R, and memory seems to be a big issue. I'm currently using a 2 GB Windows setup, but might have the option to run R on a server remotely. Windows R seems basically limited to 2 GB memory if I'm right; is there the possibility to go much beyond that with server-based R? In other words, am I limited by R or by my hardware, and how much might R be able to handle if I get the hardware necessary? Hi, R under Windows can handle up to 2GB, and under certain conditions up to 3GB. Note that no 64-bit version for Windows is available yet. Anyway, under 64-bit Linux their are almost no limits other than hardware restrictions. Best wishes, Uwe Also, any possibility of using web-based R for this kind of thing? Cheers, Alan Cohen Alan Cohen Post-doctoral Fellow Centre for Global Health Research 70 Richmond St. East, Suite 202A Toronto, ON M5C 1N8 Canada (416) 854-3121 (cell) (416) 864-6060 ext. 3156 (0ffice) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Methods dispatch and inheritance R.oo
Thanks for reply Henrik, seems obvious now.
Can child class (B) access argument of the parent class, i.e. can i rewrite
definition of the class B as
setConstructorS3(ClassB, function() {
extend(ClassA(), ClassB,
.size2 = A
);
})
it didn't work for me, so guess i'm doing smth wrong and i couldn't find any
reference on inheritance in the help files.
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[R] lars
Hi, Can anyone tell me how we can have the optimal value of lambda from lars package? I want to extract the estimates against this optimal value of lambda. I have tried the predict.lars and coef.lars but these provide entire sequence of coefficients. I don't know how to have the optimal one. Cheers, -- Sohail Chand [[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] splinefun and interpSpline
On the splinefun help page it refers to the interpSpline function from the splines package. What are the differences in the results, other than splinefun returns a function and with interpSpline you use predict. The only difference from the help pages seems to be the deriv option of splinefun, are there other differences? -- Rune __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] unlist dataframes
I have forgotten to say that the elements in the list are not necessarily equal... meaning that x1 and x2 are from different dimensions so I can't use data.frame(l) More suitable example: x1=data.frame(Name=LETTERS[1:2],Age=1:2) x2=data.frame(Name=LETTERS[3:5],Age=3:5) l=list(x1,x2) l l [[1]] Name Age 1A 1 2B 2 [[2]] Name Age 1C 3 2D 4 3E 5 Naira Naira wrote: Dear all, I would like to know whether it is possible to unlist elements and keep the original format of the data. To make it more clear, let me give an exemple: I have a list l of dataframes that I created with apply but which looks like this: x1=data.frame(Name=LETTERS[1:2],Age=1:2) x2=data.frame(Name=LETTERS[3:4],Age=3:4) l=list(x1,x2) l [[1]] Name Age 1A 1 2B 2 [[2]] Name Age 1C 3 2D 4 I would like to unlist l to create a dataframe with 2 columns and 4 rows but keeping the format of Name (character) and Age (numeric). Now when I unlist l, I obtain : unlist(l) Name1 Name2 Age1 Age2 Name1 Name2 Age1 Age2 1 2 1 2 1 2 3 4 Is there a way to at least obtain something like A 1 B 2 C 3 D 4 as result from the unlist?? Thanks a lot for your replies Naira -- View this message in context: http://www.nabble.com/unlist---dataframes-tp20358993p20359097.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] R Mixed Anova
Hi Rodrigo, Here are two options; for each type, the second version gives 2nd order interactions ## aov T.aovmod - aov(response ~ Season + Beach + Line + Error(Block/Strata)) T.aovmod - aov(response ~ (Season + Beach + Line)^2 + Error(Block/Strata)) ## lme library(nlme) T.lmemod - lme(response ~ Season + Beach + Line, random = ~ 1 | Block/Strata) T.lmemod - lme(response ~ (Season + Beach + Line)^2, random = ~ 1 | Block/Strata) anova(T.lmemod) ?anova.lme ## Use package multcomp for doing post-hoc analysis. Regards, Mark. Rodrigo Aluizio wrote: Hi list, I was searching how to properly write a command line for a mixed ANOVA. Well honestly, there are so many material on the older post of the list that just confused me. I have five factors. Season (fixed) Beach (fixed) Line (fixed) Block (random) Strata (random) nested in Block And for each of the tree strata per block I got 3 replicates. I saw lots of things about different linear models commands lm, lme On R help I found aov and anova Any one knows what should be more adequate and how I write and appropriate command line? I need an ANOVA table, a p value, and be able to run post hoc tests. Thanks for your attention and patience. ___ MSc. mailto:[EMAIL PROTECTED] Rodrigo Aluizio Centro de Estudos do Mar/UFPR Laboratório de Micropaleontologia Avenida Beira Mar s/n - CEP 83255-000 Pontal do Paraná - PR - BRASIL Fone: (0**41) 3455-1496 ramal 217 Fax: (0**41) 3455-1105 [[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. -- View this message in context: http://www.nabble.com/R-Mixed-Anova-tp20358811p20359309.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] unlist dataframes
Thanks a lot :) It is doing exactly what I want Naira David Freedman wrote: try newdata=do.call(rbind,l) David Freedman, Atlanta Naira wrote: Dear all, I would like to know whether it is possible to unlist elements and keep the original format of the data. To make it more clear, let me give an exemple: I have a list l of dataframes that I created with apply but which looks like this: x1=data.frame(Name=LETTERS[1:2],Age=1:2) x2=data.frame(Name=LETTERS[3:4],Age=3:4) l=list(x1,x2) l [[1]] Name Age 1A 1 2B 2 [[2]] Name Age 1C 3 2D 4 I would like to unlist l to create a dataframe with 2 columns and 4 rows but keeping the format of Name (character) and Age (numeric). Now when I unlist l, I obtain : unlist(l) Name1 Name2 Age1 Age2 Name1 Name2 Age1 Age2 1 2 1 2 1 2 3 4 Is there a way to at least obtain something like A 1 B 2 C 3 D 4 as result from the unlist?? Thanks a lot for your replies Naira -- View this message in context: http://www.nabble.com/unlist---dataframes-tp20358993p20359182.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to make a multiple plot
perhaps you could also look into ggplot2 or lattice package to display several plots on the same page. They take care of important but annoying details such as scaling, layout, limits, legend, ... Admittedly, there is a learning curve when you're used to base graphics, but in the long term it is a more efficient, robust and elegant approach. Baptiste On 6 Nov 2008, at 07:15, Dieter Menne wrote: amy rheanita amee_86 at yahoo.com writes: I'm a student. now, I'm studying Nelson-Siegel Extended, a term structure model. I can analyze - estimate parameters and make aplot - manually, from data bond in a day. I can analyze bond data in a month, like make multiple plot for different bont date and make a multiple plot to compare parameters in different date. Try par(mfrow=c(2,2)) before creating the plot. Dieter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. _ Baptiste Auguié School of Physics University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] unlist dataframes
Try this: unlist(lapply(l, t)) On Thu, Nov 6, 2008 at 9:05 AM, Naira [EMAIL PROTECTED] wrote: I have forgotten to say that the elements in the list are not necessarily equal... meaning that x1 and x2 are from different dimensions so I can't use data.frame(l) More suitable example: x1=data.frame(Name=LETTERS[1:2],Age=1:2) x2=data.frame(Name=LETTERS[3:5],Age=3:5) l=list(x1,x2) l l [[1]] Name Age 1A 1 2B 2 [[2]] Name Age 1C 3 2D 4 3E 5 Naira Naira wrote: Dear all, I would like to know whether it is possible to unlist elements and keep the original format of the data. To make it more clear, let me give an exemple: I have a list l of dataframes that I created with apply but which looks like this: x1=data.frame(Name=LETTERS[1:2],Age=1:2) x2=data.frame(Name=LETTERS[3:4],Age=3:4) l=list(x1,x2) l [[1]] Name Age 1A 1 2B 2 [[2]] Name Age 1C 3 2D 4 I would like to unlist l to create a dataframe with 2 columns and 4 rows but keeping the format of Name (character) and Age (numeric). Now when I unlist l, I obtain : unlist(l) Name1 Name2 Age1 Age2 Name1 Name2 Age1 Age2 1 2 1 2 1 2 3 4 Is there a way to at least obtain something like A 1 B 2 C 3 D 4 as result from the unlist?? Thanks a lot for your replies Naira -- View this message in context: http://www.nabble.com/unlist---dataframes-tp20358993p20359097.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.
Re: [R] r script
Alfredo Alessandrini wrote: A simple question :-) I'm writing a R scritp(#!/usr/bin/Rscript) How can execute a bash command inside the script, like the command for change the directory (cd)? Thanks in advance, Alfredo __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. ?system -- Drs. Paul Hiemstra Department of Physical Geography Faculty of Geosciences University of Utrecht Heidelberglaan 2 P.O. Box 80.115 3508 TC Utrecht Phone: +31302535773 Fax:+31302531145 http://intamap.geo.uu.nl/~paul __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] r script
A simple question :-) I'm writing a R scritp(#!/usr/bin/Rscript) How can execute a bash command inside the script, like the command for change the directory (cd)? Thanks in advance, Alfredo __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] unlist dataframes
try newdata=do.call(rbind,l) David Freedman, Atlanta Naira wrote: Dear all, I would like to know whether it is possible to unlist elements and keep the original format of the data. To make it more clear, let me give an exemple: I have a list l of dataframes that I created with apply but which looks like this: x1=data.frame(Name=LETTERS[1:2],Age=1:2) x2=data.frame(Name=LETTERS[3:4],Age=3:4) l=list(x1,x2) l [[1]] Name Age 1A 1 2B 2 [[2]] Name Age 1C 3 2D 4 I would like to unlist l to create a dataframe with 2 columns and 4 rows but keeping the format of Name (character) and Age (numeric). Now when I unlist l, I obtain : unlist(l) Name1 Name2 Age1 Age2 Name1 Name2 Age1 Age2 1 2 1 2 1 2 3 4 Is there a way to at least obtain something like A 1 B 2 C 3 D 4 as result from the unlist?? Thanks a lot for your replies Naira - David Freedman Atlanta -- View this message in context: http://www.nabble.com/unlist---dataframes-tp20358993p20359063.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] gamlss.dist
Hi, I'm not sure how use curve(dexGAUS( None of the following four works: rt- rexGAUS(100, mu=300, nu=100, sigma=35) m1-gamlss(rt~1, family=exGAUS) curve(dexGAUS(rt=x, mu=300 ,sigma=35,nu=100), 100, 600, main = The ex- GAUS density mu=300 ,sigma=35,nu=100) curve(dexGAUS(x=rt, mu=300 ,sigma=35,nu=100), 100, 600, main = The ex- GAUS density mu=300 ,sigma=35,nu=100) curve(dexGAUS(y=rt, mu=300 ,sigma=35,nu=100), 100, 600, main = The ex- GAUS density mu=300 ,sigma=35,nu=100) curve(dexGAUS(rt=y, mu=300 ,sigma=35,nu=100), 100, 600, main = The ex- GAUS density mu=300 ,sigma=35,nu=100) But this does: y- rexGAUS(100, mu=300, nu=100, sigma=35) m1-gamlss(y~1, family=exGAUS) curve(dexGAUS(y=x, mu=300 ,sigma=35,nu=100), 100, 600, main = The ex- GAUS density mu=300 ,sigma=35,nu=100) _ Professor Michael Kubovy University of Virginia Department of Psychology USPS: P.O.Box 400400Charlottesville, VA 22904-4400 Parcels:Room 102Gilmer Hall McCormick RoadCharlottesville, VA 22903 Office:B011+1-434-982-4729 Lab:B019+1-434-982-4751 Fax:+1-434-982-4766 WWW:http://www.people.virginia.edu/~mk9y/ [[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] read Idrisi file
Abdou Ali a.ali at agrhymet.ne writes: Dear Sirs, Is there a R function to read an Idrisi image (*.rdc *.rst) Yes, please see the rgdal package, which can read this format. The Spatial Task View (CRAN) might have got you the information you need without posting. Depending on the size of your data, used the GDAL bindings directly, or use sp classes if the raster isn't huge. Roger Bivand Thanks __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Inference and confidence interval for a restricted cubic spline function in a hurdle model
Dear list, I'm currently analyzing some count data using a hurdle model. I've used the rcspline.eval function in the Hmisc-library to contruct the spline terms for the regression model, and what I want in the end is the ability to compute coefficients and confidence intervals for different changes in the smooth function as well as plotting the smooth function along with the confidence interval at the values of the x-variable. An example using the zero-part of the hurdle model: library(Hmisc) library(pscl) # Simulate some data set.seed(1) y-c(rep(0,50),rnbinom(50,0.9,0.2)) x-sin(y)+rnorm(100) # Set up the spline terms ssp-rcspline.eval(x,inclx=T) # Fit the model and construct the smooth function f-hurdle(y~ssp) knots-attr(ssp,knots) coef-f$coefficients$zero w-rcspline.restate(knots,coef) fun-eval(attr(w,function)) The coefficient for a change in x from -0.1 to 0.1 is fun(0.1)-fun(-0.1). My question is therefore how do I compute the confidence interval for this change? This is easy to do with the Design-library for the zero-part but as far as I know, zero-truncated negative binomial data can't be fitted using Design's functions as they are. Does someone know any neat tricks that would make this possible? Any help on this would be greatly appreciated. Thanks for your time. /Erik Lampa, Statistician, Uppsala University Hospital, Sweden [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to get the length of an UTF-8 string
On Thu, 6 Nov 2008, Fán Lóng wrote: Hi there, I am intending to get the length of an UTF-8 string which contains some Japanese characters (let's say, rstr) in R language. I try to use the nchar(rstr) to get its length, however, it returns the NA for it contains some multi-byte characters. Is there any alternatives to return the length of this rstr? Use a UTF-8 locale, then nchar() will get it right. Or convert it to the Japanese locale on your system, and use nchar in that locale. Any suggestion is appreciated. Long __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Please do: we are missing the 'at a minimum' information needed to answer this question. Locales do matter. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] mean computation for external data
christabel_jane prudencio wrote: I have an external data (.txt) for annual peak flood. The first column is the year, second column is the observation date, and the last is the observed discharge. My task is to calculate the mean, skewness and kurtosis of the said data. I was advised to use read.table() to read the entire data. Please help me on how to perform the required computation. I am obviously a new user of this statistical software. Thank you so much for helping. Then please read the posting guide of this list and learn that we will not solve your homeworks. Uwe Ligges Christabel Jane P. Rubio M.S. Student Water Resources Engineering Dept. of Construction Environmental Engineering Kongju National University (Cheonan Campus) 102th Office,The 5th Engineering Building, 275, Budae-dong, Cheonan-si, Chungnam-do, 330-717, Korea TEL : +82-41-521-9316 FAX : +82-41-568-0287 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Simple rep() question duplicating times and dates.
Hi John, I had the same problem yesterday and solved it like following: seq(ISOdate(Year,Month,Day,Hour), by=hour, length=24*365) for example: seq(ISOdate(2005,1,1,0), by=hour, length=8760) Regards, Solveig EIFER Europäisches Institut für Energieforschung Institut européen de recherche sur l'énergie European Institute for Energy Research Solveig Mimler Sociologist M.A. Emmy-Noether-Straße 11 76131 Karlsruhe Tel: +49 721 6105 1351 Fax: +49 721 6105 1332 mailto: [EMAIL PROTECTED] _ EIFER Europäisches Institut für Energieforschung, Electricité de France / Universität Karlsruhe (TH) EWIV Sitz der europäischen wirtschaftlichen Interessenvereinigung: Karlsruhe Handelsregister: Amtsgericht Karlsruhe HRA 4823 Vorsitzender des Aufsichtsrats: Prof. Dr.-Ing. RAINER REIMERT Geschäftsführer: Dr. FREDERIC BARON (Institutsleiter) _ This e-mail and any attachment is for authorized use by the intended recipient(s) only. It may contain proprietary material, confidential information and/or be subject to legal privilege. It should not be copied, disclosed to, retained or used by any other party. If you are not an intended recipient then please promptly delete this e-mail and any attachment and all copies and inform the sender. _ Message: 27 Date: Wed, 5 Nov 2008 16:14:19 +0100 From: Gustaf Rydevik [EMAIL PROTECTED] Subject: Re: [R] Simple rep() question duplicating times and dates. To: [EMAIL PROTECTED] Cc: R R-help [EMAIL PROTECTED] Message-ID: [EMAIL PROTECTED] Content-Type: text/plain; charset=ISO-8859-1 On Wed, Nov 5, 2008 at 4:02 PM, John Kane [EMAIL PROTECTED] wrote: I want to create a data.frame of time and date for a year. I started with the idea of simply producing two vectors (time and date) The first part ( time) is easy. rep(1:24, 365) But how do I get a series of 24 dates for O1 January 2005 and repeat this to 31 December 2005. It should be easy but I don't see it. Thanks Hi John, ?Date leads you to (among other things) ?seq.Date. Something like this should work: time-rep(1:24, 365) dates-seq(as.Date(01012005,format=%d%m%Y),as.Date(31122005,format=%d%m%Y),by=1) TimeFrame-data.frame(time) TimeFrame$dates-rep(dates,each=24) Regards, Gustaf -- Gustaf Rydevik, M.Sci. tel: +46(0)703 051 451 address:Essingetorget 40,112 66 Stockholm, SE skype:gustaf_rydevik -- Message: 28 Date: Wed, 5 Nov 2008 07:30:31 -0800 (PST) From: John Kane [EMAIL PROTECTED] Subject: Re: [R] Simple rep() question duplicating times and dates. To: Gustaf Rydevik [EMAIL PROTECTED] Cc: R R-help [EMAIL PROTECTED] Message-ID: [EMAIL PROTECTED] Content-Type: text/plain; charset=us-ascii I don't have R on this machine to try it but it looks good to me. I had even got as far as seq() but completely missed the use of each. Thanks very much. --- On Wed, 11/5/08, Gustaf Rydevik [EMAIL PROTECTED] wrote: From: Gustaf Rydevik [EMAIL PROTECTED] Subject: Re: [R] Simple rep() question duplicating times and dates. Cc: R R-help [EMAIL PROTECTED] Received: Wednesday, November 5, 2008, 10:14 AM On Wed, Nov 5, 2008 at 4:02 PM, John Kane I want to create a data.frame of time and date for a year. I started with the idea of simply producing two vectors (time and date) The first part ( time) is easy. rep(1:24, 365) But how do I get a series of 24 dates for O1 January 2005 and repeat this to 31 December 2005. It should be easy but I don't see it. Thanks Hi John, ?Date leads you to (among other things) ?seq.Date. Something like this should work: time-rep(1:24, 365) dates-seq(as.Date(01012005,format=%d%m%Y),as.Date(31122005,format=%d%m%Y),by=1) TimeFrame-data.frame(time) TimeFrame$dates-rep(dates,each=24) Regards, Gustaf -- Gustaf Rydevik, M.Sci. tel: +46(0)703 051 451 address:Essingetorget 40,112 66 Stockholm, SE skype:gustaf_rydevik [[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] mean computation for external data
I have an external data (.txt) for annual peak flood. The first column is the year, second column is the observation date, and the last is the observed discharge. My task is to calculate the mean, skewness and kurtosis of the said data. I was advised to use read.table() to read the entire data. Please help me on how to perform the required computation. I am obviously a new user of this statistical software. Thank you so much for helping. Christabel Jane P. Rubio M.S. Student Water Resources Engineering Dept. of Construction Environmental Engineering Kongju National University (Cheonan Campus) 102th Office,The 5th Engineering Building, 275, Budae-dong, Cheonan-si, Chungnam-do, 330-717, Korea TEL : +82-41-521-9316 FAX : +82-41-568-0287 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] RES: R Mixed Anova
Thanks Mark, it was exactly what I was looking for. -Mensagem original- De: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Em nome de Mark Difford Enviada em: quinta-feira, 6 de novembro de 2008 09:22 Para: r-help@r-project.org Assunto: Re: [R] R Mixed Anova Hi Rodrigo, Here are two options; for each type, the second version gives 2nd order interactions ## aov T.aovmod - aov(response ~ Season + Beach + Line + Error(Block/Strata)) T.aovmod - aov(response ~ (Season + Beach + Line)^2 + Error(Block/Strata)) ## lme library(nlme) T.lmemod - lme(response ~ Season + Beach + Line, random = ~ 1 | Block/Strata) T.lmemod - lme(response ~ (Season + Beach + Line)^2, random = ~ 1 | Block/Strata) anova(T.lmemod) ?anova.lme ## Use package multcomp for doing post-hoc analysis. Regards, Mark. Rodrigo Aluizio wrote: Hi list, I was searching how to properly write a command line for a mixed ANOVA. Well honestly, there are so many material on the older post of the list that just confused me. I have five factors. Season (fixed) Beach (fixed) Line (fixed) Block (random) Strata (random) nested in Block And for each of the tree strata per block I got 3 replicates. I saw lots of things about “different” linear models commands… lm, lme… On R help I found aov and anova… Any one knows what should be more adequate and how I write and appropriate command line? I need an ANOVA table, a p value, and be able to run post hoc tests. Thanks for your attention and patience. ___ MSc. mailto:[EMAIL PROTECTED] Rodrigo Aluizio Centro de Estudos do Mar/UFPR Laboratório de Micropaleontologia Avenida Beira Mar s/n - CEP 83255-000 Pontal do Paraná - PR - BRASIL Fone: (0**41) 3455-1496 ramal 217 Fax: (0**41) 3455-1105 [[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. -- View this message in context: http://www.nabble.com/R-Mixed-Anova-tp20358811p20359309.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] replicate and as.matrix: different behaviour between batch and non-batch mode
Hello,
for a simulation I tried the following:
=
sampmeanvec - function (from, n, repititions)
{
print( paste(samplesize n:, n, repititions:, repititions) )
samples.mat - as.matrix( replicate( repititions, sample(from, n) ) )
# would that case-check be necessary?
#if( n == 1 )
#{
# samples.mat - t (samples.mat)
#}
print( Dim of matrix:)
print( dim(samples.mat) )
meanvec - apply(samples.mat, 2, mean)
return(meanvec)
}
gleichsamp - runif(1)
for( sampsize in c(1,2,4,8,16,32,64,128) )
{
sampmeanvec(gleichsamp, sampsize, 20)
}
=
The following result:
--
source(central_limit_theorem.R)
[1] samplesize n: 1 repititions: 20
[1] Dim of matrix:
[1] 20 1
[1] samplesize n: 2 repititions: 20
[1] Dim of matrix:
[1] 2 20
[1] samplesize n: 4 repititions: 20
[1] Dim of matrix:
[1] 4 20
[1] samplesize n: 8 repititions: 20
[1] Dim of matrix:
[1] 8 20
[1] samplesize n: 16 repititions: 20
[1] Dim of matrix:
[1] 16 20
[1] samplesize n: 32 repititions: 20
[1] Dim of matrix:
[1] 32 20
[1] samplesize n: 64 repititions: 20
[1] Dim of matrix:
[1] 64 20
[1] samplesize n: 128 repititions: 20
[1] Dim of matrix:
[1] 128 20
Look at the first dimension: there the cols and rows are
changed.
I tried directly in the R-shell:
x - 1:20
dim( as.matrix(replicate(1, sample(x, length(x)) ) ))
[1] 20 1
dim( as.matrix(replicate(2, sample(x, length(x)) ) ))
[1] 20 2
dim( as.matrix(replicate(3, sample(x, length(x)) ) ))
[1] 20 3
dim( as.matrix(replicate(4, sample(x, length(x)) ) ))
[1] 20 4
This looks good (and correct to me).
Can you locate my problem here?
Why is the cols and rows dimensions be changed?
Without using as.matrix the result of replicte is just a vector.
So I have to use it.
But only in the script/batch-mode.
Typed in directly (see above), it works as expected.
Any ideas on this behaviour?
Ciao,
Oliver
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Re: [R] r script
2008/11/6 Paul Hiemstra [EMAIL PROTECTED]: How can execute a bash command inside the script, like the command for change the directory (cd)? ?system The system function won't work for changing the working directory of R though: getwd() [1] /home/rowlings system(cd /) getwd() [1] /home/rowlings For that you need R the function setwd(newWorkingDirectory). 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] RE : Text function in 3D graph
I produced the 3D plot with cloud function (lattice package), but rgl package works well: I can easily plot in 3D and annotate. Thank you very much for your help. Sebastian. - Merci de répondre à cette adresse e-mail et à [EMAIL PROTECTED] Please reply both to this email address and to [EMAIL PROTECTED] -Message d'origine- De : Duncan Murdoch [mailto:[EMAIL PROTECTED] Envoyé : mercredi 5 novembre 2008 19:02 À : EVANS David-William Cc : r-help@r-project.org Objet : Re: [R] Text function in 3D graph On 11/5/2008 10:59 AM, EVANS David-William wrote: Fellow R users, I often use « text » to annotate plots in 2 dimensions. Having grown up to 3 dimensions recently, I am looking for a similar function. Could anyone provide code for how to annotate points in a 3-d plot? You need to describe how you are producing the 3d plot. The rgl package has text3d, but it's different in the scatterplot3d package or lattice. Duncan Murdoch Your help is greatly appreciated. For Sébastien Velazquez. - Merci de répondre à cette adresse e-mail et à [EMAIL PROTECTED] mailto:[EMAIL PROTECTED] . Please reply both to this email address and to [EMAIL PROTECTED] mailto:[EMAIL PROTECTED] . [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 Mixed Anova
Hi list, I was searching how to properly write a command line for a mixed ANOVA. Well honestly, there are so many material on the older post of the list that just confused me. I have five factors. Season (fixed) Beach (fixed) Line (fixed) Block (random) Strata (random) nested in Block And for each of the tree strata per block I got 3 replicates. I saw lots of things about different linear models commands lm, lme On R help I found aov and anova Any one knows what should be more adequate and how I write and appropriate command line? I need an ANOVA table, a p value, and be able to run post hoc tests. Thanks for your attention and patience. ___ MSc. mailto:[EMAIL PROTECTED] Rodrigo Aluizio Centro de Estudos do Mar/UFPR Laboratório de Micropaleontologia Avenida Beira Mar s/n - CEP 83255-000 Pontal do Paraná - PR - BRASIL Fone: (0**41) 3455-1496 ramal 217 Fax: (0**41) 3455-1105 [[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] Queries about step() and stepAIC()
I have been learning how to use these functions and would like to know the following as I have so far been unable to find the answers in the documentation. 1) What stopping rules are used ? 2) Can the stopping rules be changed? 3) Can the results of each step be stored as objects in R and if so how ? 4) Is there a worked example somewhere of using the keep= argument ? Many thanks in anticipation, Philip This message should be regarded as confidential. If you have received this email in error please notify the sender and destroy it immediately. Statements of intent shall only become binding when confirmed in hard copy by an authorised signatory. The contents of this email may relate to dealings with other companies within the Detica Group plc group of companies. Detica Limited is registered in England under No: 1337451. Registered offices: Surrey Research Park, Guildford, Surrey, GU2 7YP, England. [[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 get the length of an UTF-8 string
Hi there, I am intending to get the length of an UTF-8 string which contains some Japanese characters (let's say, rstr) in R language. I try to use the nchar(rstr) to get its length, however, it returns the NA for it contains some multi-byte characters. Is there any alternatives to return the length of this rstr? Any suggestion is appreciated. Long __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] unlist dataframes
Dear all, I would like to know whether it is possible to unlist elements and keep the original format of the data. To make it more clear, let me give an exemple: I have a list l of dataframes that I created with apply but which looks like this: x1=data.frame(Name=LETTERS[1:2],Age=1:2) x2=data.frame(Name=LETTERS[3:4],Age=3:4) l=list(x1,x2) l [[1]] Name Age 1A 1 2B 2 [[2]] Name Age 1C 3 2D 4 I would like to unlist l to create a dataframe with 2 columns and 4 rows but keeping the format of Name (character) and Age (numeric). Now when I unlist l, I obtain : unlist(l) Name1 Name2 Age1 Age2 Name1 Name2 Age1 Age2 1 2 1 2 1 2 3 4 Is there a way to at least obtain something like A 1 B 2 C 3 D 4 as result from the unlist?? Thanks a lot for your replies Naira -- View this message in context: http://www.nabble.com/unlist---dataframes-tp20358993p20358993.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] Calling optim and .C
Thanks, your answer just helped me to find the error. Armin __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Data manipulation question
How about:
id - c(rep(a,4),rep(b,2), rep(c,5), rep(d,1))
start - c(c(0,6,17,20),c(0,1),c(0,5,10,11,50),c(0))
stop - c(c(6,12,20,30),c(1,10),c(3,10,11,30,55),c(6))
data - data.frame(id,start,stop)
f - function(data){
m - match(data$start,data$stop) + 1
if (length(m)==1 is.na(m)) m - 1
if (length(m) 1 is.na(m[2])) m - 1
data$stop[min(m,na.rm=T)]
}
by(data,data$id,f)
The if statements in the function are for some special cases, in all the
other cases the firs line will do the trick.
I would like to add that using data is a somewhat bad behavior, as this
overwrites the build in data function of R.
And I changed the way you made up the data.frame, as your method would
convert everything to factors.
Good luck
Bart
Peter Jepsen wrote:
Dear R-listers,
I am a relatively inexperienced R-user currently migrating from Stata. I
am deeply frustrated by this data manipulation question: I know how I
could do it in Stata, but I cannot make it work in R.
I have a data frame of hospitalization data where each row represents an
admission. I need to know when patients were first discharged, but the
problem is that patients were sometimes transferred between hospital
departments. In my data a transfer looks like a new admission, except
that it has a 'start' date equal to the previous admission's 'stop'
date.
Here is an example:
id - c(rep(a,4),rep(b,2), rep(c,5), rep(d,1))
start - c(c(0,6,17,20),c(0,1),c(0,5,10,11,50),c(0))
stop - c(c(6,12,20,30),c(1,10),c(3,10,11,30,55),c(6))
data - as.data.frame(cbind(id,start,stop))
data
#id start stop
# 1 a 06
# 2 a 6 12
# 3 a17 20
# 4 a20 30
# 5 b 01
# 6 b 1 10
# 7 c 03
# 8 c 5 10
# 9 c10 11
# 10 c11 30
# 11 c50 55
# 12 d 06
So, what I want to end up with is this:
id start stop
a 0 12 # This patient was transferred at time 6 and discharged at
time 12. The admission starting at time 17 is therefore irrelevant.
b 0 10
c 0 3
d 0 6
I have tried tons of variations over lapply, sapply, split, for etc.,
all to no avail.
Thank you in advance for any assistance.
Best regards,
Peter Jepsen, MD.
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View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
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Re: [R] Data manipulation question
On Thu, Nov 6, 2008 at 4:23 PM, Peter Jepsen [EMAIL PROTECTED] wrote:
Here is an example:
id - c(rep(a,4),rep(b,2), rep(c,5), rep(d,1))
start - c(c(0,6,17,20),c(0,1),c(0,5,10,11,50),c(0))
stop - c(c(6,12,20,30),c(1,10),c(3,10,11,30,55),c(6))
data - as.data.frame(cbind(id,start,stop))
data
#id start stop
# 1 a 06
# 2 a 6 12
# 3 a17 20
# 4 a20 30
# 5 b 01
# 6 b 1 10
# 7 c 03
# 8 c 5 10
# 9 c10 11
# 10 c11 30
# 11 c50 55
# 12 d 06
So, what I want to end up with is this:
id start stop
a 0 12 # This patient was transferred at time 6 and discharged at
time 12. The admission starting at time 17 is therefore irrelevant.
b 0 10
c 0 3
d 0 6
Try this:
result - list()
num - length(levels(factor(data$id)))
length(result) - 3*num
dim(result) - c(3,num)
result - data[data$start == 0,]
Y - as.integer(row.names(result))
for (i in 1:num) {
if (Y[i] == dim(data)[1]) (result[i,3] - data[dim(data)[1],3])
else (result[i,3] - data[Y[i]+1,3])
}
result
Sorry it is ugly cuz i am new too but hopefully it gives you some ideas.
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[R] mean computation for external data
__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Data manipulation question
Thank you for your prompt assistance, cruz and Bart.
Bart set me on the right track, and I modified his proposal to this:
f - function(data){
m - match(data$stop,data$start)
n - min(length(m),which(is.na(m)))
data$stop[n]
}
by(data,data$id,f)
It also handles some special cases outside my small example dataset.
Thank you again!
Peter.
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of bartjoosen
Sent: 6. november 2008 11:31
To: r-help@r-project.org
Subject: Re: [R] Data manipulation question
How about:
id - c(rep(a,4),rep(b,2), rep(c,5), rep(d,1))
start - c(c(0,6,17,20),c(0,1),c(0,5,10,11,50),c(0))
stop - c(c(6,12,20,30),c(1,10),c(3,10,11,30,55),c(6))
data - data.frame(id,start,stop)
f - function(data){
m - match(data$start,data$stop) + 1
if (length(m)==1 is.na(m)) m - 1
if (length(m) 1 is.na(m[2])) m - 1
data$stop[min(m,na.rm=T)]
}
by(data,data$id,f)
The if statements in the function are for some special cases, in all the
other cases the firs line will do the trick.
I would like to add that using data is a somewhat bad behavior, as this
overwrites the build in data function of R.
And I changed the way you made up the data.frame, as your method would
convert everything to factors.
Good luck
Bart
Peter Jepsen wrote:
Dear R-listers,
I am a relatively inexperienced R-user currently migrating from Stata.
I
am deeply frustrated by this data manipulation question: I know how I
could do it in Stata, but I cannot make it work in R.
I have a data frame of hospitalization data where each row represents
an
admission. I need to know when patients were first discharged, but the
problem is that patients were sometimes transferred between hospital
departments. In my data a transfer looks like a new admission, except
that it has a 'start' date equal to the previous admission's 'stop'
date.
Here is an example:
id - c(rep(a,4),rep(b,2), rep(c,5), rep(d,1))
start - c(c(0,6,17,20),c(0,1),c(0,5,10,11,50),c(0))
stop - c(c(6,12,20,30),c(1,10),c(3,10,11,30,55),c(6))
data - as.data.frame(cbind(id,start,stop))
data
#id start stop
# 1 a 06
# 2 a 6 12
# 3 a17 20
# 4 a20 30
# 5 b 01
# 6 b 1 10
# 7 c 03
# 8 c 5 10
# 9 c10 11
# 10 c11 30
# 11 c50 55
# 12 d 06
So, what I want to end up with is this:
id start stop
a 0 12 # This patient was transferred at time 6 and discharged
at
time 12. The admission starting at time 17 is therefore irrelevant.
b 0 10
c 0 3
d 0 6
I have tried tons of variations over lapply, sapply, split, for etc.,
all to no avail.
Thank you in advance for any assistance.
Best regards,
Peter Jepsen, MD.
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[R] need help in plotting barchart
Df contains Session_Setup DCT RevDataVols_bincounts comp 1Session_Setup RLL 1NA Session_Setup+RLL+1 2Session_Setup RLL 2NA Session_Setup+RLL+2 3Session_Setup RLL 3NA Session_Setup+RLL+3 4Session_Setup RLL 4NA Session_Setup+RLL+4 5Session_Setup RLL 5NA Session_Setup+RLL+5 6Session_Setup RLL 6NA Session_Setup+RLL+6 7Session_Setup RLL 7NA Session_Setup+RLL+7 8Session_Setup RLL 8NA Session_Setup+RLL+8 9Session_Setup RLL 9NA Session_Setup+RLL+9 10 Session_Setup RLL 10NA Session_Setup+RLL+10 11 Session_Setup RLL 11NA Session_Setup+RLL+11 12 Session_Setup RLL 12NA Session_Setup+RLL+12 13 Session_Setup RLL 13NA Session_Setup+RLL+13 14 Session_Setup RLL 14NA Session_Setup+RLL+14 15 Session_Setup RLL 15NA Session_Setup+RLL+15 16 Session_Setup RLL 16NA Session_Setup+RLL+16 17 Session_Setup RLL 17NA Session_Setup+RLL+17 18 Session_Setup RLL 18NA Session_Setup+RLL+18 19 Session_Setup RLL 19NA Session_Setup+RLL+19 20 Session_Setup RLL 20NA Session_Setup+RLL+20 21 Session_Setup RLL 21NA Session_Setup+RLL+21 22 Session_Setup RLL 22NA Session_Setup+RLL+22 23 Session_Setup RLL 23NA Session_Setup+RLL+23 24 Session_Setup RLL 24NA Session_Setup+RLL+24 25 Session_Setup RLL 25NA Session_Setup+RLL+25 26 Session_Setup RLL 26NA Session_Setup+RLL+26 27 Session_Setup RLL 27NA Session_Setup+RLL+27 28 Session_Setup RLL 28NA Session_Setup+RLL+28 29 Session_Setup RLL 29NA Session_Setup+RLL+29 30 Session_Setup RLL 30NA Session_Setup+RLL+30 31 User_Initiated RLL 1 8.487840 User_Initiated+RLL+1 32 User_Initiated RLL 2 6.066089 User_Initiated+RLL+2 33 User_Initiated RLL 3NA User_Initiated+RLL+3 34 User_Initiated RLL 4NA User_Initiated+RLL+4 35 User_Initiated RLL 5 5.906891 User_Initiated+RLL+5 36 User_Initiated RLL 6NA User_Initiated+RLL+6 37 User_Initiated RLL 7NA User_Initiated+RLL+7 38 User_Initiated RLL 8NA User_Initiated+RLL+8 39 User_Initiated RLL 9NA User_Initiated+RLL+9 40 User_Initiated RLL 10NA User_Initiated+RLL+10 41 User_Initiated RLL 11NA User_Initiated+RLL+11 42 User_Initiated RLL 12NA User_Initiated+RLL+12 43 User_Initiated RLL 13NA User_Initiated+RLL+13 44 User_Initiated RLL 14NA User_Initiated+RLL+14 45 User_Initiated RLL 15NA User_Initiated+RLL+15 46 User_Initiated RLL 16NA User_Initiated+RLL+16 47 User_Initiated RLL 17NA User_Initiated+RLL+17 48 User_Initiated RLL 18NA User_Initiated+RLL+18 49 User_Initiated RLL 19NA User_Initiated+RLL+19 50 User_Initiated RLL 20NA User_Initiated+RLL+20 51 User_Initiated RLL 21NA User_Initiated+RLL+21 52 User_Initiated RLL 22NA User_Initiated+RLL+22 53 User_Initiated RLL 23NA User_Initiated+RLL+23 54 User_Initiated RLL 24NA User_Initiated+RLL+24 55 User_Initiated RLL 25NA User_Initiated+RLL+25 56 User_Initiated RLL 26NA User_Initiated+RLL+26 57 User_Initiated RLL 27NA User_Initiated+RLL+27 58 User_Initiated RLL 28NA User_Initiated+RLL+28 59 User_Initiated RLL 29NA User_Initiated+RLL+29 60 User_Initiated RLL 30NA User_Initiated+RLL+30 61 Session_Setup NoRLL 1 8.906891 Session_Setup+NoRLL+1 62 Session_Setup NoRLL 2NA Session_Setup+NoRLL+2 63 Session_Setup NoRLL 3NA Session_Setup+NoRLL+3 64 Session_Setup NoRLL 4NA Session_Setup+NoRLL+4 65 Session_Setup NoRLL 5NA Session_Setup+NoRLL+5 66 Session_Setup NoRLL 6NA Session_Setup+NoRLL+6 67 Session_Setup NoRLL 7NA Session_Setup+NoRLL+7 68
[R] Bio7 1.3 for Linux released!
Release notes: http://www.nabble.com/Bio7-1.3-Linux-released!-td20360723.html#a20360723 http://www.nabble.com/Bio7-1.3-Linux-released!-td20360723.html#a20360723 With kind regards M.Austenfeld -- View this message in context: http://www.nabble.com/New-version-of-ecological-modeling-software-using-R-tp20318525p20362952.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] Including graphics files in MS office / open office
Hi all, I'm trying to write up some recommendations for what graphics formats are most useful for inclusion into ms office and openoffice. There have been a few discussions on the list in the past, but I haven't seen a summary. These are the options I've seen so far, along with there costs and benefits: * high-resolution (600-dpi) png output (or tiff or jpg or other raster format). The main disadvantage is that it's a raster format, so you need to know the eventual output size. * windows metafile: works well in MS office, but does not support transparency. Can only be produced on windows, and openoffice support isn't great * encapsulated postscript: supported by both MS office and open office, but won't display a preview image unless you add one with an external program. Prints fine. * svg: R output devices still experimental and open office import still experimental. No support in ms office. Have I missed anything? Is the information correct? Hadley -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Including graphics files in MS office / open office
Hello Hadley, I have started this: http://wiki.r-project.org/rwiki/doku.php?id=tips:graphics-misc:export. One solution that works not too bad in OpenOffice is to output the graph in XFig format, and then use fig2dev from transfig to get an EMF file. That one is rather well readable by OpenOffice. There are some limitations of XFig with R graphs, see ?xfig. Best, Philippe ..°})) ) ) ) ) ) ( ( ( ( (Prof. Philippe Grosjean ) ) ) ) ) ( ( ( ( (Numerical Ecology of Aquatic Systems ) ) ) ) ) Mons-Hainaut University, Belgium ( ( ( ( ( .. hadley wickham wrote: Hi all, I'm trying to write up some recommendations for what graphics formats are most useful for inclusion into ms office and openoffice. There have been a few discussions on the list in the past, but I haven't seen a summary. These are the options I've seen so far, along with there costs and benefits: * high-resolution (600-dpi) png output (or tiff or jpg or other raster format). The main disadvantage is that it's a raster format, so you need to know the eventual output size. * windows metafile: works well in MS office, but does not support transparency. Can only be produced on windows, and openoffice support isn't great * encapsulated postscript: supported by both MS office and open office, but won't display a preview image unless you add one with an external program. Prints fine. * svg: R output devices still experimental and open office import still experimental. No support in ms office. Have I missed anything? Is the information correct? Hadley __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] wireframe
I've been using lattice/wireframe succesfully to visualize some data. I have one question. I want to be able to change the viewpoint ( i.e. rotate the plotted figure a bit left or right or up and down ) Is there a way to do that? Or is there some other package around that could help? Thanks. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Re : Dataframe help
Hi Ramya, Sorry if I missed something, but unless I have problems reading your message it seems that there is no line in your example matching both your test conditions... Hope this helps. Olivier Hi there, I have a dataframe length.unique.info length.unique.info abc 12 345 def 16 550 lmn 6 600 I want those names that fall under the condition (length.unique.info[,2][i] =5 length.unique.info[,3][i] =500) abcder-length.unique.info[which(length.unique.info[,2][i] =5 length.unique.info[,3][i] = 500),1] will look for both the condition.It isnt returning names is there anything i am missing.Kindly suggest me the way to do it. Regards Ramya -- Olivier ETERRADOSSI Maître-Assistant CMGD / Equipe Propriétés Psycho-Sensorielles des Matériaux Ecole des Mines d'Alès Hélioparc, 2 av. P. Angot, F-64053 PAU CEDEX 9 tel std: +33 (0)5.59.30.54.25 tel direct: +33 (0)5.59.30.90.35 fax: +33 (0)5.59.30.63.68 http://www.ema.fr __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] replacing characters in formulae / models
Dear all, How can I replace text in objects that are of class formula? y=a * x + b class(y)=formula grep(x,y) y[1] Suppose I would like to replace the x by w in the formula object y. How can this be done? Somehow, the methods that can be used in character objects do not work 1:1 in formula objects... Many thanks and best wishes Christoph -- Dr. rer.nat. Christoph Scherber University of Goettingen DNPW, Agroecology Waldweg 26 D-37073 Goettingen Germany phone +49 (0)551 39 8807 fax +49 (0)551 39 8806 Homepage http://www.gwdg.de/~cscherb1 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] comparing matrices using max or min
Dear all, I have 3 matrices with the same dimension, A,B,C and I would like to produce a matrix D where in each position would retrieve the max(or min) value along A,B,C taken from the same position. I guess that apply functions should fit, but for matrices objects I am not getting it. thanks in advance, Diogo André Alagador [[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] wireframe
Am Donnerstag, den 06.11.2008, 10:38 -0500 schrieb Bill Szkotnicki: I've been using lattice/wireframe succesfully to visualize some data. I have one question. I want to be able to change the viewpoint ( i.e. rotate the plotted figure a bit left or right or up and down ) Is there a way to do that? Take a look at the parameter screen in ?wireframe: The viewing direction is given by a sequence of rotations specified by the 'screen' argument, starting from the positive Z-axis. e.g. screen = list(z = 20, x = -70, y = 3) from the example section. Kornelius. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] wireframe
Try the 'rgl' package. On Thu, Nov 6, 2008 at 10:38 AM, Bill Szkotnicki [EMAIL PROTECTED] wrote: I've been using lattice/wireframe succesfully to visualize some data. I have one question. I want to be able to change the viewpoint ( i.e. rotate the plotted figure a bit left or right or up and down ) Is there a way to do that? Or is there some other package around that could help? Thanks. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 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] Including graphics files in MS office / open office
* svg: R output devices still experimental I've been using the svg device in the Cairo package for a while now. I've never had any issues with it and wouldn't characterize it as experimental (of course, others may have had issues). I have had problems generating svg using some of the non-Cairo packages, but this was some time ago. Until OO can import them without going to great lengths, I guess it is a moot point. -- Max __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] replicate and as.matrix: different behaviour between batch and non-batch mode
On Thu, 6 Nov 2008, Oliver Bandel wrote:
Hello,
for a simulation I tried the following:
=
sampmeanvec - function (from, n, repititions)
{
print( paste(samplesize n:, n, repititions:, repititions) )
samples.mat - as.matrix( replicate( repititions, sample(from, n) ) )
# would that case-check be necessary?
#if( n == 1 )
#{
# samples.mat - t (samples.mat)
#}
print( Dim of matrix:)
print( dim(samples.mat) )
meanvec - apply(samples.mat, 2, mean)
return(meanvec)
}
gleichsamp - runif(1)
for( sampsize in c(1,2,4,8,16,32,64,128) )
{
sampmeanvec(gleichsamp, sampsize, 20)
}
=
The following result:
--
source(central_limit_theorem.R)
[1] samplesize n: 1 repititions: 20
[1] Dim of matrix:
[1] 20 1
[1] samplesize n: 2 repititions: 20
[1] Dim of matrix:
[1] 2 20
[1] samplesize n: 4 repititions: 20
[1] Dim of matrix:
[1] 4 20
[1] samplesize n: 8 repititions: 20
[1] Dim of matrix:
[1] 8 20
[1] samplesize n: 16 repititions: 20
[1] Dim of matrix:
[1] 16 20
[1] samplesize n: 32 repititions: 20
[1] Dim of matrix:
[1] 32 20
[1] samplesize n: 64 repititions: 20
[1] Dim of matrix:
[1] 64 20
[1] samplesize n: 128 repititions: 20
[1] Dim of matrix:
[1] 128 20
Look at the first dimension: there the cols and rows are
changed.
I tried directly in the R-shell:
x - 1:20
dim( as.matrix(replicate(1, sample(x, length(x)) ) ))
[1] 20 1
dim( as.matrix(replicate(2, sample(x, length(x)) ) ))
[1] 20 2
dim( as.matrix(replicate(3, sample(x, length(x)) ) ))
[1] 20 3
dim( as.matrix(replicate(4, sample(x, length(x)) ) ))
[1] 20 4
This looks good (and correct to me).
Look again .
It is not the same as what you have above.
If studying your code does not reveal your error, you might put a
browser() call in your function after the samples.mat - ... line to
get a sense of what is happening. See
?browser
Can you locate my problem here?
Why is the cols and rows dimensions be changed?
Without using as.matrix the result of replicte is just a vector.
Not!
is.matrix( replicate(4, sample(x, length(x)) ) )
[1] TRUE
So I have to use it.
But only in the script/batch-mode.
Typed in directly (see above), it works as expected.
Any ideas on this behaviour?
Get some rest. Then make a big pot of coffee. Stuff like this happens to
many of us. You need to slow down and work thru your code before
jumping to conclusions about odd behavior of well tested, well documented
functions.
HTH,
Chuck
Ciao,
Oliver
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Charles C. Berry(858) 534-2098
Dept of Family/Preventive Medicine
E mailto:[EMAIL PROTECTED] UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901
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Re: [R] Including graphics files in MS office / open office
On Thu, Nov 6, 2008 at 10:05 AM, Max Kuhn [EMAIL PROTECTED] wrote: * svg: R output devices still experimental I've been using the svg device in the Cairo package for a while now. I've never had any issues with it and wouldn't characterize it as experimental (of course, others may have had issues). Ah, ok. There's a bewildering array of options for producing svg (rsvgdevice, svg with grDevices, svg with Cairo, svg with cairoDevice, gridSVG) and it's hard to figure out which is the most mature. Thanks for the info. Hadley -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] New to R - Errors in plotting
I am new to R and am running into trouble with the function plot. When I enter in the simple code: x-1:4 y-5:8 plot(x,y) I get a scatter plot with 4 points as expected. However, with my own data, A and B are both vectors of length ~85, each entry a decimal in [0,1]. Using the same plot(A,B) with this data, the plot function no longer gives me a simple plot with 85 points. Instead there are many points, and what looks to be several boxwhisker plots also included on the plot. This is a link to the actual output. http://www.nabble.com/file/p20364310/plotAB.bmp plotAB.bmp Why is the plot function doing this? How can I get it to simply give me a scatterplot? (From there I want to do a lsline, etc..) Any help is greatly appreciated... Thanks! -- View this message in context: http://www.nabble.com/New-to-R---Errors-in-plotting-tp20364310p20364310.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] Including graphics files in MS office / open office
Hi Phillippe, Thanks for the pointer. It looks like a nice resource. Hadley On Thu, Nov 6, 2008 at 9:34 AM, Philippe Grosjean [EMAIL PROTECTED] wrote: Hello Hadley, I have started this: http://wiki.r-project.org/rwiki/doku.php?id=tips:graphics-misc:export. One solution that works not too bad in OpenOffice is to output the graph in XFig format, and then use fig2dev from transfig to get an EMF file. That one is rather well readable by OpenOffice. There are some limitations of XFig with R graphs, see ?xfig. Best, Philippe ..°})) ) ) ) ) ) ( ( ( ( (Prof. Philippe Grosjean ) ) ) ) ) ( ( ( ( (Numerical Ecology of Aquatic Systems ) ) ) ) ) Mons-Hainaut University, Belgium ( ( ( ( ( .. hadley wickham wrote: Hi all, I'm trying to write up some recommendations for what graphics formats are most useful for inclusion into ms office and openoffice. There have been a few discussions on the list in the past, but I haven't seen a summary. These are the options I've seen so far, along with there costs and benefits: * high-resolution (600-dpi) png output (or tiff or jpg or other raster format). The main disadvantage is that it's a raster format, so you need to know the eventual output size. * windows metafile: works well in MS office, but does not support transparency. Can only be produced on windows, and openoffice support isn't great * encapsulated postscript: supported by both MS office and open office, but won't display a preview image unless you add one with an external program. Prints fine. * svg: R output devices still experimental and open office import still experimental. No support in ms office. Have I missed anything? Is the information correct? Hadley -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Reshape a matrix
Assuming you mean data frame and not matrix and depending on whether the empty spaces are intended to represent NA or 0 we have: DF - data.frame(V1 = LETTERS[1:3], V2 = LETTERS[24:26], V3 = 1:3) tapply(DF[[3]], DF[1:2], c) V2 V1 X Y Z A 1 NA NA B NA 2 NA C NA NA 3 xtabs(V3 ~ V1 + V2, DF) V2 V1 X Y Z A 1 0 0 B 0 2 0 C 0 0 3 On Wed, Nov 5, 2008 at 7:53 PM, dinesh kumar [EMAIL PROTECTED] wrote: Dear R users, I have a matrix like A X1 B Y2 C Z3 I want to reshape this matrix into this format X Y Z A 1 B 2 C 3 Thanks in advance for your help. Dinesh -- Dinesh Kumar Barupal Junior Specialist Metabolomics Fiehn Lab UCD Genome Center 451 East Health Science Drive GBSF Builidng University of California DAVIS 95616 http://fiehnlab.ucdavis.edu/staff/kumar [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to avoid $ operator is invalid for atomic vectors
Hi, I am writing this in a wrong way, can someone please correct me? A - matrix() length(A) - 6 dim(A) - c(3,2) colnames(A) - c(X,Y) A X Y [1,] NA NA [2,] NA NA [3,] NA NA A$X Error in A$X : $ operator is invalid for atomic vectors Thanks, cruz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 to R - Errors in plotting
You need to provide more information. PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Show at least an 'str' of your data if you can not include it and the commands that you were using. On Thu, Nov 6, 2008 at 11:18 AM, BKMooney [EMAIL PROTECTED] wrote: I am new to R and am running into trouble with the function plot. When I enter in the simple code: x-1:4 y-5:8 plot(x,y) I get a scatter plot with 4 points as expected. However, with my own data, A and B are both vectors of length ~85, each entry a decimal in [0,1]. Using the same plot(A,B) with this data, the plot function no longer gives me a simple plot with 85 points. Instead there are many points, and what looks to be several boxwhisker plots also included on the plot. This is a link to the actual output. http://www.nabble.com/file/p20364310/plotAB.bmp plotAB.bmp Why is the plot function doing this? How can I get it to simply give me a scatterplot? (From there I want to do a lsline, etc..) Any help is greatly appreciated... Thanks! -- View this message in context: http://www.nabble.com/New-to-R---Errors-in-plotting-tp20364310p20364310.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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 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] replacing characters in formulae / models
On Thu, 6 Nov 2008, Christoph Scherber wrote: Dear all, How can I replace text in objects that are of class formula? y=a * x + b class(y)=formula grep(x,y) y[1] What exactly are you trying to accomplish?? And why did you assign 'formula' as the class of a character string? 'y' is not a valid formula object: lm(y) Error in terms.formula(formula, data = data) : argument is not a valid model = Perhaps, you need to review ?formula and 11 Statistical models in R from Introduction to R. Oh, yes. There is the matter of reviewing the _posting guide_ before posting, too. HTH, Chuck Suppose I would like to replace the x by w in the formula object y. How can this be done? Somehow, the methods that can be used in character objects do not work 1:1 in formula objects... Many thanks and best wishes Christoph -- Dr. rer.nat. Christoph Scherber University of Goettingen DNPW, Agroecology Waldweg 26 D-37073 Goettingen Germany phone +49 (0)551 39 8807 fax +49 (0)551 39 8806 Homepage http://www.gwdg.de/~cscherb1 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Charles C. Berry(858) 534-2098 Dept of Family/Preventive Medicine E mailto:[EMAIL PROTECTED] UC San Diego http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] New to R - Errors in plotting
Hello - In you example, what are the classes of x and y? x-1:4 y-5:8 plot(x,y) class(x) class(y) In your 'real' data, what are the classes of A and B class(A) class(B) One may be a factor? How are you reading your data into R, read.table? Make sure your data are numeric, then plot them, and it should do what you'd like. Since you're new to R, one of my tips would be to learn the class() and str() functions. Many functions, such as plot, operate differently depending on the class of data given to them, therefore it's very important to know the classes of your data objects. BKMooney wrote: I am new to R and am running into trouble with the function plot. When I enter in the simple code: x-1:4 y-5:8 plot(x,y) I get a scatter plot with 4 points as expected. However, with my own data, A and B are both vectors of length ~85, each entry a decimal in [0,1]. Using the same plot(A,B) with this data, the plot function no longer gives me a simple plot with 85 points. Instead there are many points, and what looks to be several boxwhisker plots also included on the plot. This is a link to the actual output. http://www.nabble.com/file/p20364310/plotAB.bmp plotAB.bmp Why is the plot function doing this? How can I get it to simply give me a scatterplot? (From there I want to do a lsline, etc..) Any help is greatly appreciated... Thanks! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] New to R - Errors in plotting
On 06/11/2008 11:18 AM, BKMooney wrote: I am new to R and am running into trouble with the function plot. When I enter in the simple code: x-1:4 y-5:8 plot(x,y) I get a scatter plot with 4 points as expected. However, with my own data, A and B are both vectors of length ~85, each entry a decimal in [0,1]. Using the same plot(A,B) with this data, the plot function no longer gives me a simple plot with 85 points. Instead there are many points, and what looks to be several boxwhisker plots also included on the plot. This is a link to the actual output. http://www.nabble.com/file/p20364310/plotAB.bmp plotAB.bmp Why is the plot function doing this? How can I get it to simply give me a scatterplot? (From there I want to do a lsline, etc..) Any help is greatly appreciated... Most likely your A vector is not numeric: it's being read as strings like 0.4688, etc. Take a look at str(A) and str(B) to see their structure. (I can't tell from the plot, but I'd guess it's being read as a factor, because some of the values are not recognized as numeric.) 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] How to avoid $ operator is invalid for atomic vectors
Hello - cruz wrote: Hi, I am writing this in a wrong way, can someone please correct me? A - matrix() length(A) - 6 dim(A) - c(3,2) colnames(A) - c(X,Y) A X Y [1,] NA NA [2,] NA NA [3,] NA NA A$X Error in A$X : $ operator is invalid for atomic vectors A[, X] may be what you want? See the details of ?Extract about how '$' only works on recursive (list-like) objects, of which your matrix A is not one of. Thanks, cruz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] replicate and as.matrix: different behaviour between batch and non-batch mode
Hello Charles, thank you for the hint. Zitat von Charles C. Berry [EMAIL PROTECTED]: [...] This looks good (and correct to me). Look again . It is not the same as what you have above. [...] OK, yes, you are right! I mixed the two parameters... Now I get the same problem also at the shell directly typed in: = as.matrix(replicate(10, sample(x, 3) ) ) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] 12 153 12744 14620 [2,]6 11146 19 17 11 1911 [3,] 20 17683 14 19 10214 as.matrix(replicate(10, sample(x, 2) ) ) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,]4 17 18 146 115 17 1313 [2,] 143 16 12 177 20 16217 as.matrix(replicate(10, sample(x, 1) ) ) [,1] [1,] 19 [2,] 17 [3,] 12 [4,] 14 [5,] 12 [6,] 13 [7,] 18 [8,] 12 [9,]5 [10,] 20 dim( as.matrix(replicate(10, sample(x, 3) ) ) ) [1] 3 10 dim( as.matrix(replicate(10, sample(x, 2) ) ) ) [1] 2 10 dim( as.matrix(replicate(10, sample(x, 1) ) ) ) [1] 10 1 = So, the behaviour is the same... ...but is not really that fine. :( ...how could I avoid the necessity of the transposition of the matrix in the case of only one sample? I mean, it's not extremely a problem, because there are not many loops aroud it, but it looks somehow ugly. :( Ciao, Oliver __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] replacing characters in formulae / models
Dear all, Dear Keith,
Well, of course in fact the problem is more complicated than that. The example was just for
illustration.
I have several statistical models for which I want to retrieve predictions using delta.method (from
library(alr3)).
Now for this I need a character string such as e.g.
delta.method(model, a + exp(c * x)) # model could for example be an
exponential nls fit
where the x shall be replaced by a numeric value at which the predictions
plus s.e. shall be returned:
delta.method(model, a + exp(c * 10))
##
Because there are many models for which this shall be done, I would like to have a function that
does something like the following:
extract.estimates=function(model.list,xval)
for (i in 1:length(model.list)){
mod=model.list[i]
- extract the model formula using formula(mod)
- every time the variable x is present, it shall be replaced by xval
- create a pasted character string pastestring to be used by delta.method
result=list(delta.method(mod,pastestring))
return(result)
}
I hope this has helped illustrating my point.
All the best
Christoph
Keith Jewell schrieb:
Hi,
Firstly, I'd recommend using '-' for assignment, rather than '='; it saves
confusion
Secondly, I don't think you want 'a*x+b' as a formula, I think you want an
expression.
Thirdly, your 'y' has only one term, a 9 character constant = a * x + b
Consider instead,
y - expression(a*x+b)
a - 2
b - 3
x - 1:10
y
eval(y)
Now, how to replace 'x' by 'w'?
I'm not an expert, but this is the kind of thing I need to do, so I'd
welcome criticism of my approach.
I would view the expression as a list:
as.list(y)
as.list(y[[1]])
So y is an expression containing a sub-expression; that is y is '(a*x) + b'
You want to access the sub-expression 'a*x'
y[[1]][[2]]
as.list(y[[1]][[2]])
Now you want to replace the third item in that sub-expression with the name
(not the character) w
y[[1]][[2]][[3]] - as.name(w)
w - 11:20
y
eval(y)
hth
Keith J
P.S. Perhaps you really do want a formula; y ~ a*x+b ??
In that case I'd still probably manipulate it as a list.
---
Christoph Scherber [EMAIL PROTECTED] wrote in
message news:[EMAIL PROTECTED]
Dear all,
How can I replace text in objects that are of class formula?
y=a * x + b
class(y)=formula
grep(x,y)
y[1]
Suppose I would like to replace the x by w in the formula object y.
How can this be done? Somehow, the methods that can be used in character
objects do not work 1:1 in formula objects...
Many thanks and best wishes
Christoph
--
Dr. rer.nat. Christoph Scherber
University of Goettingen
DNPW, Agroecology
Waldweg 26
D-37073 Goettingen
Germany
phone +49 (0)551 39 8807
fax +49 (0)551 39 8806
Homepage http://www.gwdg.de/~cscherb1
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
.
--
Dr. rer.nat. Christoph Scherber
University of Goettingen
DNPW, Agroecology
Waldweg 26
D-37073 Goettingen
Germany
phone +49 (0)551 39 8807
fax +49 (0)551 39 8806
Homepage http://www.gwdg.de/~cscherb1
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Using a background photo in lattice--title and axes missing
I am trying to use a background photo in a lattice plot. I am using the rimage and TeachingDemos packages to plot the photo and translate from the photo coordinates in pixels to geographic coordinates, which is what I want to use for plotting contours, lines, etc. The (unrunable) code below does give me a plot showing the photo, color contours, contour lines, and colorkey, but not the plot title or axes. How can I get the title and axes to appear? Nevermind, I figured out that all I needed was to add the statement par(new=T) before plot(photo). Regards, Scott Waichler __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 with SIGNAL module
I sent this message to the maintainer's email address listed on the
signal package, but it bounced. Perhaps somebody on this list has more
insight into the signal package than I do (or knows the maintainer's new
address):
Subject:
question about buttord function in R signal module
From:
Michael Tiemann [EMAIL PROTECTED]
Date:
Thu, 06 Nov 2008 07:18:58 -0500
To:
[EMAIL PROTECTED]
Tom,
I was trying to compare lowpass and highpass filters and found that the
buttord example in the manual worked as expected, namely the $type was
low. But when I reversed the Wp and Ws parameters, it resulted in an
object with an empty $type instead of $type being high. I looked at
the source code and found an obvious reason why:
if (length(Wp) == 2) {
warning(buttord seems to overdesign bandpass and bandreject
filters)
if (any(stop))
type = stop
else type = pass
}
else {
## here we set to high or low, as expected
if (any(stop))
type = high
else type = low
}
## here we arbitrarily override the type if there is a stop band
defined...why?
if (any(stop))
type =
Thank you for any explanation you can provide.
M
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[R] How to manipulate the time data without the date?
Hi,all I only got the time data such as: tms-c(19:30:23,18:39:10.) I want to manipulate this time series data. For example, plus one second(or minute) or minus one second This data only has the time(h:m:s), without the date. I know that there are chron package, ISOPix class and the timeDate class, but all these class need the input of date. How can we manipulate the time data without the date? Thanks advance Ted -- View this message in context: http://www.nabble.com/How-to-manipulate-the-time-data-without-the-date--tp20365370p20365370.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] FW: [rkward-devel] questions on RKWard
Thought I should copy the list with Matthieu's response. H -Original Message- From: Matthieu Stigler [mailto:[EMAIL PROTECTED] Sent: Wednesday, November 05, 2008 8:29 PM To: Horace Tso; [EMAIL PROTECTED] Subject: Re: [rkward-devel] questions on RKWard some answer only for the third question: do you have the package r-doc-html installed? (on ubuntu check with dpkg -l r-*) this is maybe the solution Horace Tso a écrit : Folks, I'm making progress moving from Windows to Linux and have RKward up and running. I read somewhere that Rkward's supposed to be the Tinn-R for linux and Tinn-R has worked out great for me. So naturally I'd like to do the following, if possible, 1. How to ask Rkward not to load the last saved image. Right now whenever it starts, it loads an image from some obscure corner of my directory. I can't quite figure out where it gets that image from. On the command line, I can do --no-restore. But there seems to be no place to sneak in these command line options under Rkward. A startup config file hidden somewhere? 2. Customize CTRL-keys. Just out of the box, many menu options do not have a control key associated with them. Any way to pick my favorite key? 3. I'm used to typing ?command on the R-console and get a HTML help page pops up. But when I do that, a shockingly large window comes up and complains in many words about some error which I can't quite figure out what it's saying (sorry don't have linux in front of me right now). 4. I see there is an 'Output' tab by default. But command results are sent to R-console, and nothing seems to happen in 'Output'. I have R 2.7.1 running under KDE on openSUSE 10.3. TIA. Horace - This SF.Net email is sponsored by the Moblin Your Move Developer's challenge Build the coolest Linux based applications with Moblin SDK win great prizes Grand prize is a trip for two to an Open Source event anywhere in the world http://moblin-contest.org/redirect.php?banner_id=100url=/ ___ RKWard-devel mailing list [EMAIL PROTECTED] https://lists.sourceforge.net/lists/listinfo/rkward-devel __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 manipulate the time data without the date?
Why not give it an arbitrary date such 2008-01-01? On Thu, 6 Nov 2008, tedzzx wrote: Hi,all I only got the time data such as: tms-c(19:30:23,18:39:10.) I want to manipulate this time series data. For example, plus one second(or minute) or minus one second This data only has the time(h:m:s), without the date. I know that there are chron package, ISOPix class and the timeDate class, but all these class need the input of date. And in base R there is POSIXct. How can we manipulate the time data without the date? Not really, as DST changes mean which day does matter. But all other days will be the same. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to manipulate the time data without the date?
On Thu, Nov 6, 2008 at 12:10 PM, tedzzx [EMAIL PROTECTED] wrote: Hi,all I only got the time data such as: tms-c(19:30:23,18:39:10.) I want to manipulate this time series data. For example, plus one second(or minute) or minus one second This data only has the time(h:m:s), without the date. I know that there are chron package, ISOPix class and the timeDate class, but all these class need the input of date. The chron package's times class does not have that restriction: library(chron) times(10:34:21) [1] 10:34:21 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to avoid $ operator is invalid for atomic vectors
Hi Cruz you don't need to assign dimension or classes to your objects. It's easier if you do like this a=c(0,1,2,4,1,1) length(a) [1] 6 b=matrix(a,3,2,byrow=T) b [,1] [,2] [1,]01 [2,]24 [3,]11 of course you can change the colnames and assign what you prefer colnames(b)=c(x,y) but if you try to recall x with b$x is not going to work like that, you have two option: 1. switch form matrix to a dataframe: c=as.data.frame(b) c x y 1 0 1 2 2 4 3 1 1 c$x [1] 0 2 1 no problems. 2. Can get the column x on the matrix b as b[,1] [1] 0 2 1 just giving the position. Hope that this helps. Best Regards Anna Anna Freni Sterrantino Ph.D Student Department of Statistics University of Bologna, Italy via Belle Arti 41, 40124 BO. Da: cruz [EMAIL PROTECTED] A: r-help@r-project.org Inviato: Giovedì 6 novembre 2008, 17:22:42 Oggetto: [R] How to avoid $ operator is invalid for atomic vectors Hi, I am writing this in a wrong way, can someone please correct me? A - matrix() length(A) - 6 dim(A) - c(3,2) colnames(A) - c(X,Y) A X Y [1,] NA NA [2,] NA NA [3,] NA NA A$X Error in A$X : $ operator is invalid for atomic vectors Thanks, cruz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Unisciti alla community di Io fotografo e video, il nuovo corso di fotografia di Gazzetta dello sport: [[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] Strang line while plotting failure curves
Dear R helper, I encountered a problem when I tried to plot the cumulative failure rate (i.e. 1 - survival probability). I have used the following code to plot. The scenario is that patients are randomized to different treatment arm (rev in the code), the PCI revascularization was monitored over 5 years. #R code testfit - survfit(Surv(pcifu,pci)~rev,data=subproc) testfit$surv - 1 - testfit$surv testfail - plot(testfit, mark.time=FALSE,col=1:2, main='Failure Rate') #End of R code I arbitarily replaced testfit$surv by computing 1 minus the original survival rate. So far so good. However, when I plot the manipulated testfit, there is a vertical line plotted at x=0, y=0:1. I checked testfit$time and testfit$surv, nothing weird there. I am very confused where the vertical line at starting point of time 0 came from. How can I get rid of it? Would you pleae help me with this? Thanks a lot! Jiang [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to avoid $ operator is invalid for atomic vectors
-Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of anna freni sterrantino Sent: Thursday, November 06, 2008 10:00 AM To: cruz; r-help@r-project.org Subject: Re: [R] How to avoid $ operator is invalid for atomic vectors Hi Cruz you don't need to assign dimension or classes to your objects. It's easier if you do like this a=c(0,1,2,4,1,1) length(a) [1] 6 b=matrix(a,3,2,byrow=T) b [,1] [,2] [1,]01 [2,]24 [3,]11 of course you can change the colnames and assign what you prefer colnames(b)=c(x,y) but if you try to recall x with b$x is not going to work like that, you have two option: 1. switch form matrix to a dataframe: c=as.data.frame(b) c x y 1 0 1 2 2 4 3 1 1 c$x [1] 0 2 1 no problems. 2. Can get the column x on the matrix b as b[,1] [1] 0 2 1 just giving the position. Hope that this helps. Best Regards Anna Anna Freni Sterrantino Ph.D Student Department of Statistics University of Bologna, Italy via Belle Arti 41, 40124 BO. Da: cruz [EMAIL PROTECTED] A: r-help@r-project.org Inviato: Giovedì 6 novembre 2008, 17:22:42 Oggetto: [R] How to avoid $ operator is invalid for atomic vectors Hi, I am writing this in a wrong way, can someone please correct me? A - matrix() length(A) - 6 dim(A) - c(3,2) colnames(A) - c(X,Y) A X Y [1,] NA NA [2,] NA NA [3,] NA NA A$X Error in A$X : $ operator is invalid for atomic vectors Thanks, cruz You can also use the column name with the matrix A[,'X'] Hope this is helpful, Dan Daniel J. Nordlund Washington State Department of Social and Health Services Planning, Performance, and Accountability Research and Data Analysis Division Olympia, WA 98504-5204 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 manipulate the time data without the date?
Does this help? library(chron) tms-c(19:30:23,18:39:10) mytimes - times(tms) mytimes[1]-mytimes[2] --- On Thu, 11/6/08, tedzzx [EMAIL PROTECTED] wrote: From: tedzzx [EMAIL PROTECTED] Subject: [R] How to manipulate the time data without the date? To: r-help@r-project.org Received: Thursday, November 6, 2008, 12:10 PM Hi,all I only got the time data such as: tms-c(19:30:23,18:39:10.) I want to manipulate this time series data. For example, plus one second(or minute) or minus one second This data only has the time(h:m:s), without the date. I know that there are chron package, ISOPix class and the timeDate class, but all these class need the input of date. How can we manipulate the time data without the date? Thanks advance Ted -- __ [[elided Yahoo spam]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Simple rep() question duplicating times and dates.
Hi Solveig, Thanks very much, it's a nice solution and one I had not even begun to think about. I MUST learn more about dates! --- On Thu, 11/6/08, Solveig Mimler [EMAIL PROTECTED] wrote: From: Solveig Mimler [EMAIL PROTECTED] Subject: Re: Simple rep() question duplicating times and dates. To: [EMAIL PROTECTED] Cc: r-help@r-project.org Received: Thursday, November 6, 2008, 7:06 AM Hi John, I had the same problem yesterday and solved it like following: seq(ISOdate(Year,Month,Day,Hour), by=hour, length=24*365) for example: seq(ISOdate(2005,1,1,0), by=hour, length=8760) Regards, Solveig EIFER Europäisches Institut für Energieforschung Institut européen de recherche sur l'énergie European Institute for Energy Research Solveig Mimler Sociologist M.A. Emmy-Noether-Straße 11 76131 Karlsruhe Tel: +49 721 6105 1351 Fax: +49 721 6105 1332 mailto: [EMAIL PROTECTED] _ EIFER Europäisches Institut für Energieforschung, Electricité de France / Universität Karlsruhe (TH) EWIV Sitz der europäischen wirtschaftlichen Interessenvereinigung: Karlsruhe Handelsregister: Amtsgericht Karlsruhe HRA 4823 Vorsitzender des Aufsichtsrats: Prof. Dr.-Ing. RAINER REIMERT Geschäftsführer: Dr. FREDERIC BARON (Institutsleiter) _ This e-mail and any attachment is for authorized use by the intended recipient(s) only. It may contain proprietary material, confidential information and/or be subject to legal privilege. It should not be copied, disclosed to, retained or used by any other party. If you are not an intended recipient then please promptly delete this e-mail and any attachment and all copies and inform the sender. _ Message: 27 Date: Wed, 5 Nov 2008 16:14:19 +0100 From: Gustaf Rydevik [EMAIL PROTECTED] Subject: Re: [R] Simple rep() question duplicating times and dates. To: [EMAIL PROTECTED] Cc: R R-help [EMAIL PROTECTED] Message-ID: [EMAIL PROTECTED] Content-Type: text/plain; charset=ISO-8859-1 On Wed, Nov 5, 2008 at 4:02 PM, John Kane [EMAIL PROTECTED] wrote: I want to create a data.frame of time and date for a year. I started with the idea of simply producing two vectors (time and date) The first part ( time) is easy. rep(1:24, 365) But how do I get a series of 24 dates for O1 January 2005 and repeat this to 31 December 2005. It should be easy but I don't see it. Thanks Hi John, ?Date leads you to (among other things) ?seq.Date. Something like this should work: time-rep(1:24, 365) dates-seq(as.Date(01012005,format=%d%m%Y),as.Date(31122005,format=%d%m%Y),by=1) TimeFrame-data.frame(time) TimeFrame$dates-rep(dates,each=24) Regards, Gustaf -- Gustaf Rydevik, M.Sci. tel: +46(0)703 051 451 address:Essingetorget 40,112 66 Stockholm, SE skype:gustaf_rydevik -- Message: 28 Date: Wed, 5 Nov 2008 07:30:31 -0800 (PST) From: John Kane [EMAIL PROTECTED] Subject: Re: [R] Simple rep() question duplicating times and dates. To: Gustaf Rydevik [EMAIL PROTECTED] Cc: R R-help [EMAIL PROTECTED] Message-ID: [EMAIL PROTECTED] Content-Type: text/plain; charset=us-ascii I don't have R on this machine to try it but it looks good to me. I had even got as far as seq() but completely missed the use of each. Thanks very much. --- On Wed, 11/5/08, Gustaf Rydevik [EMAIL PROTECTED] wrote: From: Gustaf Rydevik [EMAIL PROTECTED] Subject: Re: [R] Simple rep() question duplicating times and dates. To: [EMAIL PROTECTED] Cc: R R-help [EMAIL PROTECTED] Received: Wednesday, November 5, 2008, 10:14 AM On Wed, Nov 5, 2008 at 4:02 PM, John Kane [EMAIL PROTECTED] wrote: I want to create a data.frame of time and date for a year. I started with the idea of simply producing two vectors (time and date) The first part ( time) is easy. rep(1:24, 365) But how do I get a series of 24 dates for O1 January 2005 and repeat this to 31 December 2005. It should be easy but I don't see it. Thanks Hi John, ?Date leads you to (among other things) ?seq.Date. Something like this should work: time-rep(1:24, 365) dates-seq(as.Date(01012005,format=%d%m%Y),as.Date(31122005,format=%d%m%Y),by=1) TimeFrame-data.frame(time) TimeFrame$dates-rep(dates,each=24) Regards, Gustaf -- Gustaf Rydevik, M.Sci. tel: +46(0)703 051 451 address:Essingetorget 40,112 66 Stockholm, SE skype:gustaf_rydevik __ Ask a question on any topic and get answers from real people. Go to Yahoo! Answers and share what you know at http://ca.answers.yahoo.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do
Re: [R] How to avoid $ operator is invalid for atomic vectors
Thanks for all the responses, they are all very helpful:) you don't need to assign dimension or classes to your objects. It's easier if you do like this this is something that really bothers me, when I need to define an object which i will later fill with data, the dimension of this object should not be fixed because it will grow... so in MATLAB, i.e. we define x = [ ], what about in R? Thanks, cruz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] mean computation for external data
?mean kurtosis http://finzi.psych.upenn.edu/R/Rhelp02a/archive/110186.html --- On Thu, 11/6/08, christabel_jane prudencio [EMAIL PROTECTED] wrote: From: christabel_jane prudencio [EMAIL PROTECTED] Subject: [R] mean computation for external data To: r-help@r-project.org Received: Thursday, November 6, 2008, 6:07 AM I have an external data (.txt) for annual peak flood. The first column is the year, second column is the observation date, and the last is the observed discharge. My task is to calculate the mean, skewness and kurtosis of the said data. I was advised to use read.table() to read the entire data. Please help me on how to perform the required computation. I am obviously a new user of this statistical software. Thank you so much for helping. Christabel Jane P. Rubio M.S. Student Water Resources Engineering Dept. of Construction Environmental Engineering Kongju National University (Cheonan Campus) 102th Office,The 5th Engineering Building, 275, Budae-dong, Cheonan-si, Chungnam-do, 330-717, Korea TEL : +82-41-521-9316 FAX : +82-41-568-0287 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ [[elided Yahoo spam]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Strang line while plotting failure curves
Lu, Jiang wrote: Dear R helper, I encountered a problem when I tried to plot the cumulative failure rate (i.e. 1 - survival probability). I have used the following code to plot. The scenario is that patients are randomized to different treatment arm (rev in the code), the PCI revascularization was monitored over 5 years. #R code testfit - survfit(Surv(pcifu,pci)~rev,data=subproc) testfit$surv - 1 - testfit$surv testfail - plot(testfit, mark.time=FALSE,col=1:2, main='Failure Rate') #End of R code I arbitarily replaced testfit$surv by computing 1 minus the original survival rate. So far so good. However, when I plot the manipulated testfit, there is a vertical line plotted at x=0, y=0:1. I checked testfit$time and testfit$surv, nothing weird there. I am very confused where the vertical line at starting point of time 0 came from. How can I get rid of it? Would you pleae help me with this? Thanks a lot! Jiang Also see the survplot.* functions in the Design package and their fun argument, e.g., fun=function(y)1-y Frank [[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. -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to avoid $ operator is invalid for atomic vectors
Does that answer your question? Thanks:) I received one from Erin: x - NULL __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to avoid $ operator is invalid for atomic vectors
cruz wrote on 11/06/2008 12:16 PM: Thanks for all the responses, they are all very helpful:) you don't need to assign dimension or classes to your objects. It's easier if you do like this this is something that really bothers me, when I need to define an object which i will later fill with data, the dimension of this object should not be fixed because it will grow... so in MATLAB, i.e. we define x = [ ], what about in R? ?numeric, ?character, others probably... x - numeric() x[10] - 42 print(x) [1] NA NA NA NA NA NA NA NA NA 42 Does that answer your question? Jeff Thanks, cruz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- http://biostat.mc.vanderbilt.edu/JeffreyHorner __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to avoid $ operator is invalid for atomic vectors
Does this help (mylist - list(NULL)) (mylist[[3]] - data.frame(a=1:4, b=letters[1:4])) mylist (mylist[[2]] - matrix(1:12, nrow=4)) mylist --- On Thu, 11/6/08, cruz [EMAIL PROTECTED] wrote: From: cruz [EMAIL PROTECTED] Subject: Re: [R] How to avoid $ operator is invalid for atomic vectors To: r-help@r-project.org Received: Thursday, November 6, 2008, 1:16 PM Thanks for all the responses, they are all very helpful:) you don't need to assign dimension or classes to your objects. It's easier if you do like this this is something that really bothers me, when I need to define an object which i will later fill with data, the dimension of this object should not be fixed because it will grow... so in MATLAB, i.e. we define x = [ ], what about in R? Thanks, cruz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ [[elided Yahoo spam]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] PCA
I need perform PCA analyst with a matrix with more variables than units. The princomp command don't match with this matrix. Anybody knows a good command to do it? -- Noela [[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] PCA
would you please provide a dummy example that explains your problem. Then maybe I can help you. thanks Stephen On Thu, Nov 6, 2008 at 1:42 PM, Noela Sánchez [EMAIL PROTECTED] wrote: I need perform PCA analyst with a matrix with more variables than units. The princomp command don't match with this matrix. Anybody knows a good command to do it? -- Noela [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] PCA
?princomp refers you to prcomp for that case: 'princomp' only handles so-called R-mode PCA, that is feature extraction of variables. If a data matrix is supplied (possibly via a formula) it is required that there are at least as many units as variables. For Q-mode PCA use 'prcomp'. On Thu, 6 Nov 2008, Noela Sánchez wrote: I need perform PCA analyst with a matrix with more variables than units. The princomp command don't match with this matrix. Anybody knows a good command to do it? -- Noela -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] PCA
Hi Noela, Take a loot at ?prcomp HTH, Jorge On Thu, Nov 6, 2008 at 1:42 PM, Noela Sánchez [EMAIL PROTECTED] wrote: I need perform PCA analyst with a matrix with more variables than units. The princomp command don't match with this matrix. Anybody knows a good command to do it? -- Noela [[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] mean computation for external data
Hi Christabel, Take a look at the function basicStats in the fBasics package. Here is an example: library(fBasics) set.seed(123) x=rnorm(20,24,2) basicStats(x) #x #nobs 20.00 #NAs 0.00 #Minimum 20.066766 #Maximum 27.573826 #1. Quartile 23.012892 #3. Quartile 25.097454 #Mean 24.283248 #Median 24.239970 #Sum 485.664952 #SE Mean 0.434989 #LCL Mean 23.372805 #UCL Mean 25.193690 #Variance 3.784311 #Stdev 1.945331 #Skewness -0.062495 #Kurtosis -0.547891 HTH, Jorge On Thu, Nov 6, 2008 at 6:07 AM, christabel_jane prudencio [EMAIL PROTECTED] wrote: I have an external data (.txt) for annual peak flood. The first column is the year, second column is the observation date, and the last is the observed discharge. My task is to calculate the mean, skewness and kurtosis of the said data. I was advised to use read.table() to read the entire data. Please help me on how to perform the required computation. I am obviously a new user of this statistical software. Thank you so much for helping. Christabel Jane P. Rubio M.S. Student Water Resources Engineering Dept. of Construction Environmental Engineering Kongju National University (Cheonan Campus) 102th Office,The 5th Engineering Building, 275, Budae-dong, Cheonan-si, Chungnam-do, 330-717, Korea TEL : +82-41-521-9316 FAX : +82-41-568-0287 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] PCA
My matrix have 436 registers and 518 variables. I need to do a PCA analyst. Usually I use princomp command to perform PCA analyst, but this time i can't because of my variables are more than my registers. 2008/11/6 stephen sefick [EMAIL PROTECTED] would you please provide a dummy example that explains your problem. Then maybe I can help you. thanks Stephen On Thu, Nov 6, 2008 at 1:42 PM, Noela Sánchez [EMAIL PROTECTED] wrote: I need perform PCA analyst with a matrix with more variables than units. The princomp command don't match with this matrix. Anybody knows a good command to do it? -- Noela [[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.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis -- Noela Grupo de Recursos Marinos y Pesquerías Universidad de A Coruña [[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] Methods dispatch and inheritance R.oo
Hi,
On Thu, Nov 6, 2008 at 2:12 AM, Yuri Volchik [EMAIL PROTECTED] wrote:
Thanks for reply Henrik, seems obvious now.
Can child class (B) access argument of the parent class, i.e. can i rewrite
definition of the class B as
setConstructorS3(ClassB, function() {
extend(ClassA(), ClassB,
.size2 = A
);
})
it didn't work for me, so guess i'm doing smth wrong and i couldn't find any
reference on inheritance in the help files.
Note that 'A' is only an argument (~= local variable) of the
constructor function ClassA(). There is no way the (constructor)
function ClassB() can know about that one; compare to:
foo - function(A=1) {
}
bar - function(B=2) {
# Inside here you have no access to 'A' in foo(), regardless if you
call foo() or not.
}
So, I'm not sure what you are trying to do in reality. Note, since
the value of 'A' is already assigned in ClassA() and part of its
returned object, then it will also be part of the returned object from
ClassB(), so you don't really have to assign 'A' again.
I think this is a better example. See if you want to achive what
ClassB or ClassC do:
setConstructorS3(ClassA, function(A=0) {
extend(Object(), ClassA,
A = A
);
})
objA - ClassA();
ll(objA)
member data.class dimension objectSize
1 Anumeric 1 32
objA$A
[1] 0
setConstructorS3(ClassB, function(A=0, B=1) {
extend(ClassA(A=A), ClassB,
B = B
);
})
objB - ClassB(A=1, B=2);
ll(objB);
member data.class dimension objectSize
1 Anumeric 1 32
2 Bnumeric 1 32
objB$A
[1] 1
objB$B
[1] 2
If you really want to access a field from a superclass inside the
constructor function, you have to first get the object returned by the
constructor of the superclass, and then from that object retrieve the
field:
setConstructorS3(ClassC, function(B=1) {
this - extend(ClassA(), ClassC,
B = B
);
this$copyOfA - this$A;
this;
})
objC - ClassC(B=3);
ll(objC);
member data.class dimension objectSize
1 Anumeric 1 32
2 Bnumeric 1 32
objC$A
[1] 0
objC$B
[1] 3
Note, there is nothing magic going on inside Object(), extend() or
any other R.oo methods. Most of the things you can think of as
regular lists and functions. The new think is that objects inheriting
from the Object class store their values inside an environment, which
is why the object behaves as it is passed by reference.
Hope this helps
/Henrik
--
View this message in context:
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to avoid $ operator is invalid for atomic vectors
cruz wrote: Thanks for all the responses, they are all very helpful:) you don't need to assign dimension or classes to your objects. It's easier if you do like this this is something that really bothers me, when I need to define an object which i will later fill with data, the dimension of this object should not be fixed because it will grow... so in MATLAB, i.e. we define x = [ ], what about in R? It is generally best in R to create an object as the size it will end up being and then make assignments into the object. But using the theory of giving you enough rope to hang yourself, you can do things like: x - numeric(0) or x - NULL Patrick Burns [EMAIL PROTECTED] +44 (0)20 8525 0696 http://www.burns-stat.com (home of S Poetry and A Guide for the Unwilling S User) Thanks, cruz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 plotrix and prettyR
Hi all, I'm noting the appearance of new versions of plotrix and prettyR as I found a bug in the brkdn function that messed up the order of value.labels if they had been imported from an SPSS data file. For anyone using brkdn, please upgrade to the new version of prettyR (1.3-5) if you are importing SPSS data files. If anyone discovers other problems, you know where to complain. Thanks. Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] wireframe
Look at the rotate.wireframe function in the TeachingDemos package. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] 801.408.8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] project.org] On Behalf Of Bill Szkotnicki Sent: Thursday, November 06, 2008 8:39 AM To: r-help@r-project.org Subject: [R] wireframe I've been using lattice/wireframe succesfully to visualize some data. I have one question. I want to be able to change the viewpoint ( i.e. rotate the plotted figure a bit left or right or up and down ) Is there a way to do that? Or is there some other package around that could help? Thanks. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] PCA
Neola I'm a bit rusty on this, but I believe you can conduct on singular-value decomposition on the 436 by 518 matrix. The squares of your singular values (max of 436, 518-436 will be zero) will be your eigenvalues, the same as in the PC analysis. The post-eigenvectors will be your components. As I have said, my knowledge is not completely trustworthy at this point, but SVD would be worth looking into. Joe -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Noela Sánchez Sent: Thursday, November 06, 2008 1:10 PM To: stephen sefick Cc: r-help@r-project.org Subject: Re: [R] PCA My matrix have 436 registers and 518 variables. I need to do a PCA analyst. Usually I use princomp command to perform PCA analyst, but this time i can't because of my variables are more than my registers. 2008/11/6 stephen sefick [EMAIL PROTECTED] would you please provide a dummy example that explains your problem. Then maybe I can help you. thanks Stephen On Thu, Nov 6, 2008 at 1:42 PM, Noela Sánchez [EMAIL PROTECTED] wrote: I need perform PCA analyst with a matrix with more variables than units. The princomp command don't match with this matrix. Anybody knows a good command to do it? -- Noela [[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.htmlhttp://www.r-project.org/p osting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis -- Noela Grupo de Recursos Marinos y Pesquerías Universidad de A Coruña [[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] Confidence limits for the parameter of the Poisson distribution
Hi all, So far I only know one way to get the confidence limit for the Poisson distribution is to use the look-up table given by the 2 parameter (the number of observation x and the confidence level, e.g. 95%) and the table is limit by the maximum number of observations (x = 50). I know the formula to compute the CI, however, mathematically it is not easy to do it. So, anyone know an R function to do this. Thanks Tuan. [[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] replicate and as.matrix: different behaviour between batch and non-batch mode
On Thu, 6 Nov 2008, Oliver Bandel wrote: Hello Charles, [snip] dim( as.matrix(replicate(10, sample(x, 3) ) ) ) [1] 3 10 dim( as.matrix(replicate(10, sample(x, 2) ) ) ) [1] 2 10 dim( as.matrix(replicate(10, sample(x, 1) ) ) ) [1] 10 1 = So, the behaviour is the same... ...but is not really that fine. :( ...how could I avoid the necessity of the transposition of the matrix in the case of only one sample? use matrix( your.result , nc = n.replicates ) or dim( your.result ) - c( n.samples, n.replicates ) instead of as.matrix( your.result ) HTH, Chuck I mean, it's not extremely a problem, because there are not many loops aroud it, but it looks somehow ugly. :( Ciao, Oliver __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Charles C. Berry(858) 534-2098 Dept of Family/Preventive Medicine E mailto:[EMAIL PROTECTED] UC San Diego http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Strang line while plotting failure curves
Thank you very much, Frank. I installed Design package and tried survplot().
#R code
survplot(testfit,time.inc=365.25,xaxt='n',xlim=c(0,1826.25),ylim=c(0,1),conf='none',
fun=function(y)1-y,label.curves=list(keys=c('Med','Rev')),
abbrev.label=TRUE,n.risk=TRUE)
# End of R code
I have achieved my goal with the 'fun' argument you advised. But I have a
difficult time to do the following fine tune.
1. The X axis scale was labeled as 'Days'. I would like something like
'xscale=365.25' in plot.survfit to put time into Years and label the ticks
from 0 to 5, instead of from 0 to 1826.25 by each 365.25 increment. I tried
xaxt='n' as you can see in the code above. Then I noticed that xaxt='n' only
work for plot(survfit), not in survplot(). Any advice how to change the tick
label using survplot()?
2. The n.risk was beautifully printed for each specified time point along
the x axis. However, since I am plotting the failure rate, the n.risk looks
busy with the failure rate curves. Is there a way to move the n.risk to the
top of the plot where there are lots of space?
3. I also tried label.curves=list(). It is very convenient. The curves are
labeled and the legend is created as well. Could I only keep the curve label
and get rid of the legend since the legend is not so necessary once the
curve is labeled. How do you think?
I really appreciate any help you give.
Best regards,
Jiang Lu
Statistician
Department of Epidemiology
University of Pittsburgh
On Thu, Nov 6, 2008 at 1:21 PM, Frank E Harrell Jr [EMAIL PROTECTED]
wrote:
Lu, Jiang wrote:
Dear R helper,
I encountered a problem when I tried to plot the cumulative failure rate
(i.e. 1 - survival probability). I have used the following code to plot.
The
scenario is that patients are randomized to different treatment arm (rev
in
the code), the PCI revascularization was monitored over 5 years.
#R code
testfit - survfit(Surv(pcifu,pci)~rev,data=subproc)
testfit$surv - 1 - testfit$surv
testfail - plot(testfit, mark.time=FALSE,col=1:2, main='Failure Rate')
#End of R code
I arbitarily replaced testfit$surv by computing 1 minus the original
survival rate. So far so good. However, when I plot the manipulated
testfit, there is a vertical line plotted at x=0, y=0:1. I checked
testfit$time and testfit$surv, nothing weird there. I am very confused
where
the vertical line at starting point of time 0 came from. How can I get rid
of it?
Would you pleae help me with this? Thanks a lot!
Jiang
Also see the survplot.* functions in the Design package and their fun
argument, e.g., fun=function(y)1-y
Frank
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
--
Frank E Harrell Jr Professor and Chair School of Medicine
Department of Biostatistics Vanderbilt University
[[alternative HTML version deleted]]
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[R] Umlaut read from csv-file
Dear All!
Reading character strings containing an umlaut
from a csv-file I find a (to me) surprising
behaviour in R 2.8.0, that I did not notice in R 2.7.2.
A comparison by == results in FALSE, while grep does find the aggreement.
See the example below.
The crucial line is x==div 1-2 Veränderungen,
with the result [1] FALSE in R 2.8.0 but
[1] TRUE in R 2.7.2.
Thank you in advance for your help
Heinz Tüchler
# in R 2.8.0 patched
x0 - div 1-2 Veränderungen # define a character string
write.csv(x0, 'chr.csv', row.names=FALSE) # write a csv-file with one line
rm(x0)
x - read.csv('chr.csv', skip=0, header=TRUE, as.is=TRUE)$x # read in csv-file
x
x==div 1-2 Veränderungen
[1] FALSE
grep(div 1-2 Veränderungen, x)
[1] 1
grep(div 1-2 Veränderungen, x, value=TRUE)
[1] div 1-2 Veränderungen
unlink('chr.csv') # delete file
Version:
platform = i386-pc-mingw32
arch = i386
os = mingw32
system = i386, mingw32
status = Patched
major = 2
minor = 8.0
year = 2008
month = 11
day = 04
svn rev = 46830
language = R
version.string = R version 2.8.0 Patched (2008-11-04 r46830)
Windows XP (build 2600) Service Pack 2
Locale:
LC_COLLATE=German_Austria.1252;LC_CTYPE=German_Austria.1252;LC_MONETARY=German_Austria.1252;LC_NUMERIC=C;LC_TIME=German_Austria.1252
Search Path:
.GlobalEnv, package:stats, package:graphics,
package:grDevices, package:utils,
package:datasets, package:methods, Autoloads, package:base
# in R 2.7.2 patched
x0 - div 1-2 Veränderungen # define a character string
write.csv(x0, 'chr.csv', row.names=FALSE) # write a csv-file with one line
rm(x0)
x - read.csv('chr.csv', skip=0, header=TRUE, as.is=TRUE)$x # read in csv-file
x
x==div 1-2 Veränderungen
[1] TRUE
grep(div 1-2 Veränderungen, x)
[1] 1
grep(div 1-2 Veränderungen, x, value=TRUE)
[1] div 1-2 Veränderungen
unlink('chr.csv') # delete file
Version:
platform = i386-pc-mingw32
arch = i386
os = mingw32
system = i386, mingw32
status = Patched
major = 2
minor = 7.2
year = 2008
month = 09
day = 02
svn rev = 46486
language = R
version.string = R version 2.7.2 Patched (2008-09-02 r46486)
Windows XP (build 2600) Service Pack 2
Locale:
LC_COLLATE=German_Austria.1252;LC_CTYPE=German_Austria.1252;LC_MONETARY=German_Austria.1252;LC_NUMERIC=C;LC_TIME=German_Austria.1252
Search Path:
.GlobalEnv, package:stats, package:graphics,
package:grDevices, package:utils,
package:datasets, package:methods, Autoloads, package:base
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Confidence limits for the parameter of the Poisson distribution
On Thu, 6 Nov 2008, Hoang Trong Minh Tuan wrote: Hi all, So far I only know one way to get the confidence limit for the Poisson distribution is to use the look-up table given by the 2 parameter (the number of observation x and the confidence level, e.g. 95%) and the table is limit by the maximum number of observations (x = 50). I know the formula to compute the CI, however, mathematically it is not easy to do it. So, anyone know an R function to do this. Thanks A 'marriage-of-convenience' between uniroot() and ppois() should do it. HTH, Chuck Tuan. [[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. Charles C. Berry(858) 534-2098 Dept of Family/Preventive Medicine E mailto:[EMAIL PROTECTED] UC San Diego http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Confidence limits for the parameter of the Poisson distribution
On 7/11/2008, at 9:26 AM, Hoang Trong Minh Tuan wrote:
Hi all,
So far I only know one way to get the confidence limit for the
Poisson
distribution is to use the look-up table given by the 2 parameter (the
number of observation x and the confidence level, e.g. 95%) and the
table is
limit by the maximum number of observations (x = 50).
I know the formula to compute the CI, however, mathematically it
is not
easy to do it. So, anyone know an R function to do this. Thanks
Tuan.
You may find the URL
http://www.math.mcmaster.ca/peter/s743/poissonalpha.html
useful.
cheers,
Rolf Turner
P.S. The foregoing URL discusses the issues in terms of a *single*
observation. Note that if X_1, ..., X_n are i.i.d. Poisson(lambda)
then Y = X_1 + ... + X_n is Poisson(n*lambda).
Note that the ``exact'' confidence limits are very conservative.
R. T.
##
Attention:\ This e-mail message is privileged and confid...{{dropped:9}}
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and provide commented, minimal, self-contained, reproducible code.
[R] nls: Fitting two models at once?
Hello,
I'm still a newbie user and struggling to automate some analyses from
SigmaPlot using R. R is a great help for me so far!
But the following problem makes me go nuts.
I have two spectra, both have to be fitted to reference data. Problem: the
both spectra are connected in some way: the stoichiometry of coefficients
cytf.v/cytb.v is 1/2.
{{In the SigmaPlot workflow one has to copy the two spectra into one column
beneath each other and the two spectra are somehow treated as one curve -
like in http://home.arcor.de/ballaschk/cytbf-help/sigmaplot%20formula.png}}
Can anybody help? :(
I tried to condense everything to the minimum R script below to give an
impression of what I'm talking about.
Martin
# Minimal R script reading remote data for convenience
### READ IN DATA
# first spectrum
asfe - read.table(http://home.arcor.de/ballaschk/cytbf-help/asfe.csv;)[, 1];
# second spectrum
dias - read.table(http://home.arcor.de/ballaschk/cytbf-help/dias.csv;)[, 1];
# reference data for fit, wavelength = wl
ref - read.table(http://home.arcor.de/ballaschk/cytbf-help/reference.csv;,
sep=\t, dec=., header=T);
attach(ref);
### FITTING, problem: 2*cytf.v == cytb.v
# fit first spectrum to two reference spectra
asfe.fit -
nls(
asfe ~ (cytf.v * 28) * CYTF + hp.v * B559HP,
start = list(cytf.v = 0.5, hp.v = 0.03) # arbitrary
);
# fit second spectrum to three reference spectra
dias.fit -
nls(
dias ~ (cytb.v * 24.6 * 2) * CYTB6 + lp.v * B559LP + c550.v * C550,
start = list(cytb.v = 1, lp.v = 1, c550.v = 1) # arbitrary
);
# draw stuff
plot(
1, 2,
type=n,
xlim = c(540, 575),
ylim=c(-0.002, 0.008),
);
# first spectrum and fit
lines(wl, asfe, type=b, pch=19); # solid circles
lines(wl, fitted(asfe.fit), col = red);
# second spectrum and fit
lines(wl, dias, type=b);
lines(wl, fitted(dias.fit), col = blue);
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Strang line while plotting failure curves
Lu, Jiang wrote:
Thank you very much, Frank. I installed Design package and tried
survplot().
#R code
survplot(testfit,time.inc=365.25,xaxt='n',xlim=c(0,1826.25),ylim=c(0,1),conf='none',
fun=function(y)1-y,label.curves=list(keys=c('Med','Rev')),
abbrev.label=TRUE,n.risk=TRUE)
# End of R code
I have achieved my goal with the 'fun' argument you advised. But I have
a difficult time to do the following fine tune.
1. The X axis scale was labeled as 'Days'. I would like something like
'xscale=365.25' in plot.survfit to put time into Years and label the
ticks from 0 to 5, instead of from 0 to 1826.25 by each 365.25
increment. I tried xaxt='n' as you can see in the code above. Then I
noticed that xaxt='n' only work for plot(survfit), not in survplot().
Any advice how to change the tick label using survplot()?
There are options to do all that, described in the documentation.
2. The n.risk was beautifully printed for each specified time point
along the x axis. However, since I am plotting the failure rate, the
n.risk looks busy with the failure rate curves. Is there a way to move
the n.risk to the top of the plot where there are lots of space?
May be best to move it to the bottom margin. See the help file.
3. I also tried label.curves=list(). It is very convenient. The curves
are labeled and the legend is created as well. Could I only keep the
curve label and get rid of the legend since the legend is not so
necessary once the curve is labeled. How do you think?
Should be options for that too.
Frank
I really appreciate any help you give.
Best regards,
Jiang Lu
Statistician
Department of Epidemiology
University of Pittsburgh
On Thu, Nov 6, 2008 at 1:21 PM, Frank E Harrell Jr
[EMAIL PROTECTED] mailto:[EMAIL PROTECTED] wrote:
Lu, Jiang wrote:
Dear R helper,
I encountered a problem when I tried to plot the cumulative
failure rate
(i.e. 1 - survival probability). I have used the following code
to plot. The
scenario is that patients are randomized to different treatment
arm (rev in
the code), the PCI revascularization was monitored over 5 years.
#R code
testfit - survfit(Surv(pcifu,pci)~rev,data=subproc)
testfit$surv - 1 - testfit$surv
testfail - plot(testfit, mark.time=FALSE,col=1:2,
main='Failure Rate')
#End of R code
I arbitarily replaced testfit$surv by computing 1 minus the original
survival rate. So far so good. However, when I plot the manipulated
testfit, there is a vertical line plotted at x=0, y=0:1. I checked
testfit$time and testfit$surv, nothing weird there. I am very
confused where
the vertical line at starting point of time 0 came from. How can
I get rid
of it?
Would you pleae help me with this? Thanks a lot!
Jiang
Also see the survplot.* functions in the Design package and their
fun argument, e.g., fun=function(y)1-y
Frank
[[alternative HTML version deleted]]
__
R-help@r-project.org mailto:R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
http://www.r-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Frank E Harrell Jr Professor and Chair School of Medicine
Department of Biostatistics Vanderbilt University
--
Frank E Harrell Jr Professor and Chair School of Medicine
Department of Biostatistics Vanderbilt University
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and provide commented, minimal, self-contained, reproducible code.
[R] Estimating mean and standard deviation of lognormal distribution between two points
This is more of a statistical question. Let's say I have two numbers. One is a lower bound and the other is a point estimate to the right of the lower bound. Now, let's say I want to be able to estimate the mean and standard deviation of a lognormal distribution, where 95% of the density falls within the range of the lower bound and the point estimate. Concomitantly, can an exponential distribution be described in the same fashion? Any help would be greatly appreciated. JB [[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] Looking for suggestions on how to debug pvm/snow proc's
Hi All - I'm running a faily long script that uses rpvm snowFT to spawn off multiple processes with the 'clusterApplyFT' function. Specifically, what happens is that the head node generates a number of seed clusters that are then spawned off to the pvm cluster (in this case, nodes on a 4 dual-core machine) using clusterApplyFT to apply a function to each seed cluster that optimizes each seed cluster and returns the new, optimized clusters. In some cases, however, clusterApplyFT will return try-error's for some of the clusters. Since it's a try-error, I know there's no way to get a traceback (I've tried using clusterApplyFT to call traceback with no luck), but even when I switch to snow instead of snowFT, I still can't figure out a way to get a traceback for the failed clusters. Any suggestions? Thanks! Peter [[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] PCA
try prcomp instead of princomp On Thu, Nov 6, 2008 at 3:20 PM, Lucke, Joseph F [EMAIL PROTECTED] wrote: Neola I'm a bit rusty on this, but I believe you can conduct on singular-value decomposition on the 436 by 518 matrix. The squares of your singular values (max of 436, 518-436 will be zero) will be your eigenvalues, the same as in the PC analysis. The post-eigenvectors will be your components. As I have said, my knowledge is not completely trustworthy at this point, but SVD would be worth looking into. Joe -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Noela Sánchez Sent: Thursday, November 06, 2008 1:10 PM To: stephen sefick Cc: r-help@r-project.org Subject: Re: [R] PCA My matrix have 436 registers and 518 variables. I need to do a PCA analyst. Usually I use princomp command to perform PCA analyst, but this time i can't because of my variables are more than my registers. 2008/11/6 stephen sefick [EMAIL PROTECTED] would you please provide a dummy example that explains your problem. Then maybe I can help you. thanks Stephen On Thu, Nov 6, 2008 at 1:42 PM, Noela Sánchez [EMAIL PROTECTED] wrote: I need perform PCA analyst with a matrix with more variables than units. The princomp command don't match with this matrix. Anybody knows a good command to do it? -- Noela [[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.htmlhttp://www.r-project.org/p osting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis -- Noela Grupo de Recursos Marinos y Pesquerías Universidad de A Coruña [[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] replacing values in a vector
Hello list.
I have a vector of values:
eg
head(diff_mirs_list)
[1] hsa-miR-26b hsa-miR-26b hsa-miR-23a hsa-miR-27b hsa-miR-29a
[6] hsa-miR-29b
and I would like to conditionally replace each value in this vector with a
number defined in a dataframe:
fc
Probe ave.fc
1 hsa-let-7a 1.28
2 hsa-miR-100 1.47
3 hsa-miR-125a-5p 1.31
4 hsa-miR-140-3p 1.28
5 hsa-miR-143 1.98
6 hsa-miR-193a-3p 1.37
7 hsa-miR-193b 1.48
8 hsa-miR-195 1.16
9 hsa-miR-214 1.22
10 hsa-miR-23a 1.21
11 hsa-miR-26b 1.13
12 hsa-miR-27b 1.37
13 hsa-miR-29a 1.24
14 hsa-miR-29b 1.69
15 hsa-miR-30b 1.16
16 hsa-miR-424 1.42
17 hsa-miR-768-3p 1.48
18 hsa-miR-886-3p 1.43
19 hsa-miR-933 1.23
ie every hsa-let-7a in the diff_mirs_list is replaced by 1.28, hsa-miR-100 by
1.47 etc etc
I have tried to make a loop to use gsub eg
for (i in 1:nrow(fc)){
+ test-gsub(fc[i,1], fc[i,2], diff_mirs_list)
}
but this obviously passes the unchanged vector to gsub each time and so I get
back my 'test' vector with only hsa-miR-933 changed. Could someone help me out
with this please.
Thanks
Iain
sessionInfo()
R version 2.8.0 (2008-10-20)
i486-pc-linux-gnu
locale:
LC_CTYPE=en_GB.UTF-8;LC_NUMERIC=C;LC_TIME=en_GB.UTF-8;LC_COLLATE=en_GB.UTF-8;LC_MONETARY=C;LC_MESSAGES=en_GB.UTF-8;LC_PAPER=en_GB.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_GB.UTF-8;LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
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