[R] Unable to load R
Dear all, I have recently installed R on a Red Hat Enterprise Linux (RHEL5) system. But I am unable to load R and it is giving the following error : */usr/local/lib/R/bin/exec/R: error while loading shared libraries: libreadline.so.5: cannot open shared object file: No such file or directory* I have checked for the presence of the above mention library and found that the library is present. I have run out of ideas. Can anyone help me out??? I will be greatly indebted. Cheers, Anupam Sinha, Graduate Student, Lab of Computational Biology, Centre for DNA Fingerprinting and Diagnostics, Hyderabad, India [[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] Unable to load R
On Thu, May 28, 2009 at 11:42:41AM +0530, anupam sinha wrote: I have checked for the presence of the above mention library and found that the library is present. I have run out of ideas. Can anyone help me out??? I will be greatly indebted. First: how did you check that the library is present? It might be that you have installed 32-bit R on 64-bit RHEL or vice-versa, so you need to install the requisite libraries in appropriate bitness. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Interaction plots as lines or bars?
Michael Kubovy wrote: An editor has suggested that I use bar plots to capture an interaction of two 2-level factors and an interaction of a 2 by 3 factorial experiment. (It would seem that there's a fear that someone might try to interpolate between, e.g., 'male' and 'female'.) In general it seems to me that an interaction plot with lines is easier to read, and not likely to mislead. Does anyone know if and where this has been discussed? Editors are paid by the ink. I have never seen anything else than lines for such a plot. However, I once was successful in using somewhat wider lines that looked barish. Which made me believe that the paid by the ink is not too bad an analogy: editors fear graphics that look too empty, so give them their black. I am sure Hadley will argue that this is the reason why ggplot2 by default has a gray background. Dieter -- View this message in context: http://www.nabble.com/Interaction-plots-as-lines-or-bars--tp23754973p23755909.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] question about using a remote system
On 28-May-09 00:58:17, Erin Hodgess wrote: Dear R People: I would like to set up a plug-in for Rcmdr to do the following: I would start on a Linux laptop. Then I would log into another outside system and run a some commands. Now, when I tried to do system(ssh e...@xxx.edu) password xx It goes to the remote system. how do I continue to issue commands from the Linux laptop please? (hope this makes sense) thanks, Erin Hi Erin, If I understand what you want, I think you may be asking a bit too much from R itself. When you, Erin, are interacting with your Linux laptop you can switch between two (or more) different X windows or consoles, on one of which you are logged in to the remote machine and on which you issue commands to the remote machine, and can read its output and, maybe, copy this to a terminal running commands on your Linux laptop. When you ask R to log in as you describe, you are tying that instance of R to the login. I don't know of any resources within R itself which can emulate your personal behaviour. Though maybe others do know ... However, there is in principle a work-round. You will need to get your toolbox out and get to grips with Unix plmubing; and, to make it accessible to R, you will need to create a Unix plumber. The basic component is the FIFO, or named pipe. To create one of these, use a Linux command like mkfifo myFIFO1 Then 'ls -l' in the directory you are working in will show this pipe as present with the name myFIFO1. For example, if in one terminal window [A] you start the command cat myFIFO1 then in another [B] you echo Some text to test my FIFO myFIFO1 you will find that this text will be output in the [A] window by the 'cat' command. Similarly, you could 'mkfifo myFIFO2' and write tc this while in the [A] window, and read from it while in the [B] window. So, if you can get R to write to myFIFO1 (which is being read from by another program which will send the output to a remote machine which that program is logged into), and read from myFIFO2 which is being written to by that other program (and that is easy to do in R, using connections), then you have the plumbing set up. But, to set it up, R needs to call in the plumber. In other words, R needs to execute a command like system(MakeMyFIFOs) where MakeMyFIFOs is a script that creates myFIFO1 and myFIFO2, logs in to the remote machine, watches myFIFO1 and sends anything it reads to the remote machine, watches for any feedback from the remote machine and then writes this into myFIFO2. Meanwhile, back in R, any time you want R to send something to the remote machine you write() it (or similar) to myFIFO1, and whenever you want to receive something from the remote machine you readLines() it (or similar) from myFIFO2. That's an outline of a possible work-round. Maybe other have solved your same problem in a better way (and maybe there's an R package which does just that ... ). If so, I hope they'll chip in! Ted. E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk Fax-to-email: +44 (0)870 094 0861 Date: 28-May-09 Time: 07:33:09 -- XFMail -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] alternative to built-in data editor
urlwolf wrote: I often have to peek at large data. While head and tail are convenient, at times I'd like some more comprehensive. I guess I debug better in a more visual way? I was wondering if there's a way to override the default data editor. I have never seen the data editor. The radical alternative I use is to do all data-peeking in an external database, e.g. Access or SQLServer Express. Writing using RODBC is so easy that write all intermediate results back to the database for inspection and re-use in a later step. Dieter -- View this message in context: http://www.nabble.com/alternative-to-built-in-data-editor-tp23747080p23756182.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] question about using a remote system
Hi. Do you need an interactive session at the remote machine, or are you simply wanting to run a pre-written script? If the latter, then you can ask ssh to execute a remote command, which conceivably could be R CMD x If you explain exactly why and what you are trying to do, then perhaps there's a better solution bw Mark 2009/5/28 Erin Hodgess erinm.hodg...@gmail.com: Dear R People: I would like to set up a plug-in for Rcmdr to do the following: I would start on a Linux laptop. Then I would log into another outside system and run a some commands. Now, when I tried to do system(ssh e...@xxx.edu) password xx It goes to the remote system. how do I continue to issue commands from the Linux laptop please? (hope this makes sense) thanks, Erin -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: erinm.hodg...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Dr. Mark Wardle Specialist registrar, Neurology Cardiff, UK __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How do I get removed from this mailing list?
Andrey Lyalko wrote: How do I get removed from this mailing list? [[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. By following the links in the footer! Specifically, bottom of page at https://stat.ethz.ch/mailman/listinfo/r-help -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - (p.dalga...@biostat.ku.dk) FAX: (+45) 35327907 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] 1L and 0L
Hi, Recently I come through those R-expressions and understood that 1L means 1 and 0L means 0. Why they are so? I mean, what the excess meanings they carry, instead writing simple 1 or 0? Is there any more this kind of expressions in R? Regards [[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] Object-oriented programming in R
Hi. I remember considering these options myself but concluded that for most analyses a strictly procedural approach was satisfactory. Although I may re-run multiple analyses, the data manipulation (and subsequent analysis - the former always more complex than the latter IMHO) is fairly project- and data-specific. As such, quick and specific code always seemed more appropriate than slow (to write) and generic. Obviously that doesn't apply to package creators and maintainers, creating something that is going to re-used in many different projects and can be made generic. That's the heuristic I now use - is this good enough for other people's consumption - if so, create a package and adopt some of the OO approaches seen in the base R packages. Otherwise, stick to bespoke and specific (procedural) functions. I've a small library of helper functions, but these don't use OO usually. They sometimes make assumptions about data passed in, don't have particularly robust error checking, but in general, work well. I suppose it depends on what you're trying to achieve and how much time you've got! Interested in other's thoughts too as clearly there's no right answer here. bw Mark 2009/5/27 Luc Villandre villa...@dms.umontreal.ca: Dear R-users, I have very recently started learning about object-oriented programming in R. I am far from being an expert in programming, although I do have an elementary C++ background. Please take a look at these lines of code. some.data = data.frame(V1 = 1:5, V2 = 6:10) ; p.plot = ggplot(data=some.data,aes(x=V1, y=V2)) ; class(p.plot) ; [1] ggplot My understanding is that the object p.plot belongs to the ggplot class. However, a new class definition like setClass(AClass, representation(mFirst = numeric, mSecond = ggplot)) ; yields the warning Warning message: In .completeClassSlots(ClassDef, where) : undefined slot classes in definition of AClass: mSecond(class ggplot) The ggplot object is also a list : is.list(p.plot) [1] TRUE So, I guess I could identify mSecond as being a list. However, I don't understand why ggplot is not considered a valid slot type. I thought setClass() was analogous to the class declaration in C++, but I guess I might be wrong. Would anyone care to provide additional explanations about this? I decided to explore object-oriented programming in R so that I could organize the output from my analysis in a more rigorous fashion and then define custom methods that would yield relevant output. However, I'm starting to wonder if this aspect is not better suited for package builders. R lists are already very powerful and convenient templates. Although it wouldn't be as elegant, I could define functions that would take lists outputted by the different steps of my analysis and do what I want with them. I'm wondering what the merits of both approaches in the context of R would be. If anyone has any thoughts about this, I'd be most glad to read them. Cheers, -- *Luc Villandré* /Biostatistician McGill University Health Center - Montreal Children's Hospital Research Institute/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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. Mark Wardle Specialist registrar, Neurology Cardiff, UK __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How do I get removed from this mailing list?
Duncan Murdoch wrote: On 27/05/2009 8:25 PM, Andrey Lyalko wrote: How do I get removed from this mailing list? Like most lists nowadays, it gives the instructions in each message header: List-Unsubscribe: https://stat.ethz.ch/mailman/options/r-help, mailto:r-help-requ...@r-project.org?subject=unsubscribe Duncan Murdoch Unfortunately, user-friendly mailers nowadays tends to hide such information from users. E.g. if I set headers to All in Thunderbird I get a window full of headers larger than the screen and no scrolling capacity. So it ends with X-Mailman something and the List-Unsubscribe: stuff is nowhere to be seen. -p -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - (p.dalga...@biostat.ku.dk) FAX: (+45) 35327907 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How do I get removed from this mailing list?
Peter Dalgaard wrote: Duncan Murdoch wrote: On 27/05/2009 8:25 PM, Andrey Lyalko wrote: How do I get removed from this mailing list? Like most lists nowadays, it gives the instructions in each message header: List-Unsubscribe: https://stat.ethz.ch/mailman/options/r-help, mailto:r-help-requ...@r-project.org?subject=unsubscribe Duncan Murdoch Unfortunately, user-friendly mailers nowadays tends to hide such information from users. E.g. if I set headers to All in Thunderbird I get a window full of headers larger than the screen and no scrolling capacity. So it ends with X-Mailman something and the List-Unsubscribe: stuff is nowhere to be seen. right, but there are (at least) three things you can do in thunderbird when all headers are on: - open a print preview -- there you have a scrollable content with all headers; - save the message and open it in a regular text editor; - install a suitable plugin (header scroll [1] works fine for me) the last one turns thunderbird a bit more user friendly. vQ [1] https://addons.mozilla.org/en-US/thunderbird/addon/1003 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ggplot2 adding vertical line at a certain date
check out geom_vline + geom_vline(xintercept=as.numeric(as.Date(2002-11-01))) [you may not need to convert the date to numeric in the most recent ggplot2 version] On 27 May 2009, at 20:31, stephen sefick wrote: library(ggplot2) melt.updn - (structure(list(date = structure(c(11808, 11869, 11961, 11992, 12084, 12173, 12265, 12418, 12600, 12631, 12753, 12996, 13057, 13149, 11808, 11869, 11961, 11992, 12084, 12173, 12265, 12418, 12600, 12631, 12753, 12996, 13057, 13149), class = Date), variable = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c(unrestored, restored), class = factor), value = c(1.1080259671261, 0.732188576856918, 0.410334408061265, 0.458980396410056, 0.429867902470711, 0.83126337241925, 0.602008712602784, 0.818751283264408, 1.12606382402475, 0.246174719479079, 0.941043753226865, 0.986511619794787, 0.291074883642735, 0.346361775752625, 1.36209038621623, 0.878561166753624, 0.525156715576168, 0.80305564765846, 1.08084449441812, 1.24906568558731, 0.970954515841768, 0.936838439269239, 1.26970090246036, 0.337831520417547, 0.909204325710795, 0.951009811036613, 0.290735620653709, 0.426683515714219)), .Names = c(date, variable, value), row.names = c(NA, -28L), class = data.frame)) qplot(date, value, data=melt.updn, shape=variable)+geom_smooth() #I would like to add a line at November 1, 2002 #thanks for the help -- Stephen Sefick Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] optima in unimode
Dear all i could not estimate the optima value or range value in unimodal plot in glm please help me out thanking you regard madan _ Hotmail® has a new way to see what's up with your friends. orial_WhatsNew1_052009 [[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] 1L and 0L
On Thu, 2009-05-28 at 12:50 +0530, bogaso.christofer wrote: Hi, Recently I come through those R-expressions and understood that 1L means 1 and 0L means 0. Why they are so? I mean, what the excess meanings they carry, instead writing simple 1 or 0? Is there any more this kind of expressions in R? typeof(1) [1] double typeof(1L) [1] integer So the L notation ensures that the number is stored as an integer not a double. HTH G -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How do I get removed from this mailing list?
On Thu, 2009-05-28 at 09:33 +0200, Wacek Kusnierczyk wrote: Peter Dalgaard wrote: Duncan Murdoch wrote: On 27/05/2009 8:25 PM, Andrey Lyalko wrote: How do I get removed from this mailing list? Like most lists nowadays, it gives the instructions in each message header: List-Unsubscribe: https://stat.ethz.ch/mailman/options/r-help, mailto:r-help-requ...@r-project.org?subject=unsubscribe Duncan Murdoch Unfortunately, user-friendly mailers nowadays tends to hide such information from users. E.g. if I set headers to All in Thunderbird I get a window full of headers larger than the screen and no scrolling capacity. So it ends with X-Mailman something and the List-Unsubscribe: stuff is nowhere to be seen. right, but there are (at least) three things you can do in thunderbird when all headers are on: The View message source would be a more direct way of viewing the full email, not just the bits TBird shows you in the preview pane. I forget how it is named exactly and under which menu it is found as it has been quite a while since I used TBird. G - open a print preview -- there you have a scrollable content with all headers; - save the message and open it in a regular text editor; - install a suitable plugin (header scroll [1] works fine for me) the last one turns thunderbird a bit more user friendly. vQ [1] https://addons.mozilla.org/en-US/thunderbird/addon/1003 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] optima in unimode
On Thu, 2009-05-28 at 07:50 +, ms.com wrote: Dear all i could not estimate the optima value or range value in unimodal plot in glm please help me out thanking you regard madan You're going to have to give us more to go on than that, but... Are you an ecologist and you want to estimate the optima and tolerance range from a fitted logit model? If so, here is an example: ## install.packages(analogue) if not installed require(analogue) data(ImbrieKipp, SumSST) ## fit a model mod - glm(G.pacR ~ SumSST + I(SumSST^2), data = ImbrieKipp/100, family = binomial) ## plot it pdat - data.frame(SumSST= seq(min(SumSST), max(SumSST), length = 100)) p2 - predict(mod, pdat, type = response) lines(p2 ~ SumSST, data = pdat, col = blue, lwd = 3) ## model ceofficients c.mod - coef(mod) ## optima -c.mod[2]/(2 * c.mod[3]) ## tolerance sqrt(-(1/(2 * c.mod[3]))) ## height == abundance at optimum exp(c.mod[1] - (c.mod[2]^2/(4 * c.mod[3]))) HTH G -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Unable to load R
Actually I did a search using locate but could not find the file. But when I do yum install readline (package containing the dependency libreadline.so.5) I get the following error: yum install readline Loading security plugin Loading rhnplugin plugin rhel-x86_64-client-5 100% |=| 1.3 kB 00:00 Setting up Install Process Parsing package install arguments Package readline - 5.1-1.1.x86_64 is already installed. Package readline - 5.1-1.1.i386 is already installed. Nothing to do Plzz help me out. Cheers, Anupam Sinha On Thu, May 28, 2009 at 11:54 AM, Zeljko Vrba zv...@ifi.uio.no wrote: On Thu, May 28, 2009 at 11:42:41AM +0530, anupam sinha wrote: I have checked for the presence of the above mention library and found that the library is present. I have run out of ideas. Can anyone help me out??? I will be greatly indebted. First: how did you check that the library is present? It might be that you have installed 32-bit R on 64-bit RHEL or vice-versa, so you need to install the requisite libraries in appropriate bitness. [[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] Unable to load R
On Thu, May 28, 2009 at 02:02:52PM +0530, anupam sinha wrote: Actually I did a search using locate but could not find the file. But when Locate reports useful results only if its database is up-to-date. Try running one of ldd /usr/lib/R/bin/exec/R ldd /usr/lib64/R/bin/exec/R and see what it reports. Maybe R is linked to another readline version than is installed on the system. I do yum install readline (package containing the dependency libreadline.so.5) I get the following error: yum install readline Loading security plugin Loading rhnplugin plugin rhel-x86_64-client-5 100% |=| 1.3 kB 00:00 Setting up Install Process Parsing package install arguments Package readline - 5.1-1.1.x86_64 is already installed. Package readline - 5.1-1.1.i386 is already installed. Nothing to do Otherwise, you could also ask on a mailing list related to your distribution. Plzz help me out. And learn how to spell please. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Delta Kronecker
Hi, Could some give some ideas on how to compute the spatiotemporal covariance matrix by a sum of Kronecker products in R. Is there any special function that can be used? Cheers. Firdaus [[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 help
I am interested in modeling hydrological extreme events. I found
MSClaio2008 very interesting function. In this function four
criterions for choosing distributions. Can we call these criterions
as model selection techniques or goodness of fit techniques or both?
Because goodness of fit techniques are usually performed after modle
selection.
What is MSClaio2008? I don't quite understand this bit. If you provide
code examples, then questions are often easier to answer.
Can I found chi-square, kolmogrov-sminov and cramer-von mises tests
for testing goodness of fit for proposed distributions?
Yes, you can do these things in R. The functions you want are:
chisq.test for the chi-square test
ks.test for the kolmogorov-smirnoff test.
You can view the help pages for these functions by typing a question mark
before the function name, e.g. ?chisq.test.
For the Cramer-von Mises tests, I didn't know, so I typed
RSiteSearch(cramer von mises test), from which there are several
suggestions. In particular, look at the CvM2SL1Test and CvM2SL2Test
packages.
Regards,
Richie.
Mathematical Sciences Unit
HSL
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[R] r-plot problem resolved
Thanks a lot for all the helpful comments! It finally works :handshake: (I settled with the code of Gavin) Best, Durden -- View this message in context: http://www.nabble.com/r-plot-tp23739356p23757458.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] R connectivity with Oracle DB
Hi Friends, I am working on R-2.9.0 and i want to connect oracle through R, but i could not find ROracle (zip file for windows) on CRAN Packages. Can anyone suggest me how to connect Oracle from R in windows platform? I would be grateful if u can provide me code aswell. Thanks Regards, Madan -- View this message in context: http://www.nabble.com/R-connectivity-with-Oracle-DB-tp23758768p23758768.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 do I get removed from this mailing list?
Gavin Simpson wrote: The View message source would be a more direct way of viewing the full email, not just the bits TBird shows you in the preview pane. I forget how it is named exactly and under which menu it is found as it has been quite a while since I used TBird. indeed; it's C-U, or under view message source vQ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] no internal function int.unzip in R 2.9.0 for Windows
The most recent version on CRAN is already fixed. We had a non-maintainer update end of March that fixed the R2HTML issues - the maintainer was unresponsive to our requests, unfortunately. Hence please update from CRAN. Best, Uwe Ligges Romain Francois wrote: Hi, I'll try to fix this soon. Could you log a bug request here: http://r-forge.r-project.org/tracker/?atid=1643group_id=405func=browse Regards, Romain Carson, John wrote: library(R2HTML) Loading required package: R2HTML Error in .Internal(int.unzip(zipname, NULL, dest)) : no internal function int.unzip Error : .onLoad failed in 'loadNamespace' for 'R2HTML' Error: package 'R2HTML' could not be loaded Version: R 2.9.0 for Windows __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 do I get removed from this mailing list?
On Thu, 28-May-2009 at 09:20AM +0200, Peter Dalgaard wrote:
Duncan Murdoch wrote:
On 27/05/2009 8:25 PM, Andrey Lyalko wrote:
How do I get removed from this mailing list?
Like most lists nowadays, it gives the instructions in each message header:
List-Unsubscribe: https://stat.ethz.ch/mailman/options/r-help,
mailto:r-help-requ...@r-project.org?subject=unsubscribe
Duncan Murdoch
Unfortunately, user-friendly mailers nowadays tends to hide such
information from users. E.g. if I set headers to All in Thunderbird I
get a window full of headers larger than the screen and no scrolling
capacity. So it ends with X-Mailman something and the List-Unsubscribe:
stuff is nowhere to be seen.
That's true, but because it takes up so much space, nobody would leave
it like that for long. Clicking on the '-' next to Subject
collapses headers to a single line and the rest of the message is
scrollable.
Even Outlook in HTML mode still doesn't hide the bottom, so I fail to
see why it becomes invisible. Perhaps such enquirers should be
directed to the footer, not the header.
--
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
___Patrick Connolly
{~._.~} Great minds discuss ideas
_( Y )_ Average minds discuss events
(:_~*~_:) Small minds discuss people
(_)-(_) . Eleanor Roosevelt
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
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Re: [R] no internal function int.unzip in R 2.9.0 for Windows
Hello Uwe, I have recently taken over maintenance of the package, which is now version controlled at r-forge (http://r-forge.r-project.org/projects/r2html/), and I am planning these changes: - rework the HTML function so that the html code is generated from a brew template, which in time will give more control to the user/package using R2HTML - enhance the sweave html driver so that it can take advantage of the source code highlighter I am currently writing. Do you know the extent of the modification. I'd like to propagate this to the r-forge version. Romain Uwe Ligges wrote: The most recent version on CRAN is already fixed. We had a non-maintainer update end of March that fixed the R2HTML issues - the maintainer was unresponsive to our requests, unfortunately. Hence please update from CRAN. Best, Uwe Ligges Romain Francois wrote: Hi, I'll try to fix this soon. Could you log a bug request here: http://r-forge.r-project.org/tracker/?atid=1643group_id=405func=browse Regards, Romain Carson, John wrote: library(R2HTML) Loading required package: R2HTML Error in .Internal(int.unzip(zipname, NULL, dest)) : no internal function int.unzip Error : .onLoad failed in 'loadNamespace' for 'R2HTML' Error: package 'R2HTML' could not be loaded Version: R 2.9.0 for Windows -- Romain Francois Independent R Consultant +33(0) 6 28 91 30 30 http://romainfrancois.blog.free.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] sample unique pairs from a matrix
Dear R users,
I have a matrix of both negative and positive values that I would like
to randomly sample with the following 2 conditions:
1. only sample positive values
2. once a cell in the matrix has been sampled the row and column of
that cell cannot be sampled from again.
#some dummy data
set.seed(101)
dataf - matrix(rnorm(36,1,2), nrow=6)
I can do this quite simply if all the values are positive by using the
sample function without replacement on the column and row indices.
samrow - sample(6,replace=F)
samcol - sample(6,replace=F)
values - numeric(6)
for(i in 1:6){
values[i] - dataf[samrow[i], samcol[i]]
}
However, I am not sure how to include the logical condition to only
include postitive values
Any help would be gratefully received.
Jos
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] How do I get removed from this mailing list?
On 28/05/2009 5:03 AM, Gavin Simpson wrote: On Thu, 2009-05-28 at 09:33 +0200, Wacek Kusnierczyk wrote: Peter Dalgaard wrote: Duncan Murdoch wrote: On 27/05/2009 8:25 PM, Andrey Lyalko wrote: How do I get removed from this mailing list? Like most lists nowadays, it gives the instructions in each message header: List-Unsubscribe: https://stat.ethz.ch/mailman/options/r-help, mailto:r-help-requ...@r-project.org?subject=unsubscribe Duncan Murdoch Unfortunately, user-friendly mailers nowadays tends to hide such information from users. E.g. if I set headers to All in Thunderbird I get a window full of headers larger than the screen and no scrolling capacity. So it ends with X-Mailman something and the List-Unsubscribe: stuff is nowhere to be seen. right, but there are (at least) three things you can do in thunderbird when all headers are on: The View message source would be a more direct way of viewing the full email, not just the bits TBird shows you in the preview pane. I forget how it is named exactly and under which menu it is found as it has been quite a while since I used TBird. That's what I use. It's Ctrl-U for the shortcut, View | Message source by menu. Duncan Murdoch G - open a print preview -- there you have a scrollable content with all headers; - save the message and open it in a regular text editor; - install a suitable plugin (header scroll [1] works fine for me) the last one turns thunderbird a bit more user friendly. vQ [1] https://addons.mozilla.org/en-US/thunderbird/addon/1003 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Object-oriented programming in R
p.plot is an S3 object and you are attempting to define an S4 class
so you if you really want to do that then you would need this to let
S4 know about the ggplot class:
# S4 class with S3 slot
setClass(ggplot) # make ggplot visible to S4
setClass(AClass, representation(mFirst = numeric, mSecond = ggplot))
a - new(AClass, mFirst = 1, mSecond = p.plot)
Probably you don't want to use S4 at all here but rather
define an S3 subclass, A of the ggplot class and
give it an attribute also called A.
# S3 subclass
AClass - function(mFirst, mSecond) {
attr(mSecond, A) - mFirst
class(mSecond) - c(A, setdiff(class(mSecond), A))
mSecond
}
a2 - AClass(1, p.plot)
or alternately represent A as an S3 list holding with the two objects
as components:
# S3 class but not subclass
AClass2 - function(mfirst, mSecond) {
out - list(mFirst = mFirst, mSecond = mSecond)
class(out) - A
out
}
a3 - AClass2(1, p.plot)
Also note that
1. ggplot uses S3 at the user level but uses the proto
package internally. Depending on what you want to do using proto
objects might be preferable. See home page at
http://r-proto.googlecode.com
2. R statements do not normally end in a semicolon.
On Wed, May 27, 2009 at 12:00 PM, Luc Villandre
villa...@dms.umontreal.ca wrote:
Dear R-users,
I have very recently started learning about object-oriented programming in
R. I am far from being an expert in programming, although I do have an
elementary C++ background.
Please take a look at these lines of code.
some.data = data.frame(V1 = 1:5, V2 = 6:10) ;
p.plot = ggplot(data=some.data,aes(x=V1, y=V2)) ;
class(p.plot) ;
[1] ggplot
My understanding is that the object p.plot belongs to the ggplot class.
However, a new class definition like
setClass(AClass, representation(mFirst = numeric, mSecond = ggplot))
;
yields the warning
Warning message:
In .completeClassSlots(ClassDef, where) :
undefined slot classes in definition of AClass: mSecond(class ggplot)
The ggplot object is also a list :
is.list(p.plot)
[1] TRUE
So, I guess I could identify mSecond as being a list.
However, I don't understand why ggplot is not considered a valid slot
type. I thought setClass() was analogous to the class declaration in C++,
but I guess I might be wrong. Would anyone care to provide additional
explanations about this?
I decided to explore object-oriented programming in R so that I could
organize the output from my analysis in a more rigorous fashion and then
define custom methods that would yield relevant output. However, I'm
starting to wonder if this aspect is not better suited for package builders.
R lists are already very powerful and convenient templates. Although it
wouldn't be as elegant, I could define functions that would take lists
outputted by the different steps of my analysis and do what I want with
them. I'm wondering what the merits of both approaches in the context of R
would be. If anyone has any thoughts about this, I'd be most glad to read
them.
Cheers,
--
*Luc Villandré*
/Biostatistician
McGill University Health Center -
Montreal Children's Hospital Research Institute/
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Re: [R] Labeling barplot bars by multiple factors
Thomas Levine wrote: I want to plot quantitative data as a function of three two-level factors. How do I group the bars on a barplot by level through labeling and spacing? Here http://www.thomaslevine.org/sample_multiple-factor_barplot.png's what I'm thinking of. Also, I'm pretty sure that I want a barplot, but there may be something better. Hi Tom, You may find that the hierobarp function in the plotrix package will do what you want. 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] R connectivity with Oracle DB
See http://cran.r-project.org/bin/windows/contrib/2.9/@ReadMe (ReadMe files are there to be read!) and consider the use of RODBC, which is used by lots of Oracle users as an R client on Windows. On Thu, 28 May 2009, Madan Mohan wrote: Hi Friends, I am working on R-2.9.0 and i want to connect oracle through R, but i could not find ROracle (zip file for windows) on CRAN Packages. Can anyone suggest me how to connect Oracle from R in windows platform? I would be grateful if u can provide me code aswell. Thanks Regards, Madan -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] 1L and 0L
On 28/05/2009 3:20 AM, bogaso.christofer wrote: Hi, Recently I come through those R-expressions and understood that 1L means 1 and 0L means 0. Why they are so? I mean, what the excess meanings they carry, instead writing simple 1 or 0? Is there any more this kind of expressions in R? Gavin explained the L marks the number as being stored as an integer. The i in 1i is somewhat similar: it marks the number as complex (but also changes the value, 1i is the imaginary value, not the real one). There's a discussion on this in section 3.1.1 Constants in the R Language Definition. Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R connectivity with Oracle DB
On 28/05/2009 6:03 AM, Madan Mohan wrote: Hi Friends, I am working on R-2.9.0 and i want to connect oracle through R, but i could not find ROracle (zip file for windows) on CRAN Packages. Can anyone suggest me how to connect Oracle from R in windows platform? I would be grateful if u can provide me code aswell. Get an ODBC driver from Oracle, and use RODBC. 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] sample unique pairs from a matrix
I have a matrix of both negative and positive values that I would like
to randomly sample with the following 2 conditions:
1. only sample positive values
2. once a cell in the matrix has been sampled the row and column of
that cell cannot be sampled from again.
#some dummy data
set.seed(101)
dataf - matrix(rnorm(36,1,2), nrow=6)
I can do this quite simply if all the values are positive by using the
sample function without replacement on the column and row indices.
samrow - sample(6,replace=F)
samcol - sample(6,replace=F)
values - numeric(6)
for(i in 1:6){
values[i] - dataf[samrow[i], samcol[i]]
}
However, I am not sure how to include the logical condition to only
include postitive values
You could create a new variable containing only the positive values of
dataf, and sample from that, e.g.
dataf - matrix(rnorm(36,1,2), nrow=6)
posdata - dataf[dataf 0]
sample(posdata, 6, replace=FALSE)
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] consultation
http://cran.r-project.org/web/views/Survival.html On May 11, 2009, at 1:46 PM, Nancy Tejerina wrote: *Dear R Users;* * * *I´m writing to ask you how can I do Survivals Curves using Time- dependent covariates? Which packages I need to Install?* *Thanks for your help, I look foward you reply.* ** *Sincerelly.* ** *Nancy Tejerina.* David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] sample unique pairs from a matrix
On May 28, 2009, at 6:33 AM, jos matejus wrote:
Dear R users,
I have a matrix of both negative and positive values that I would like
to randomly sample with the following 2 conditions:
1. only sample positive values
2. once a cell in the matrix has been sampled the row and column of
that cell cannot be sampled from again.
#some dummy data
set.seed(101)
dataf - matrix(rnorm(36,1,2), nrow=6)
I can do this quite simply if all the values are positive by using the
sample function without replacement on the column and row indices.
samrow - sample(6,replace=F)
samcol - sample(6,replace=F)
values - numeric(6)
for(i in 1:6){
values[i] - dataf[samrow[i], samcol[i]]
}
However, I am not sure how to include the logical condition to only
include postitive values
Any help would be gratefully received.
Jos
M - matrix(rnorm(36),nrow=6)
M[M[,]0]
[1] 1.65619781 0.56182830 0.23812890 0.81493915 1.01279243
1.29188874 0.64252343 0.53748655 0.31503112
[10] 0.37245358 0.07942883 0.56834586 0.62200056 0.39478167 0.02374574
0.04974857 0.56219171 0.52901658
sample(M[M[,]0],6,replace=F)
[1] 0.56834586 0.07942883 0.31503112 0.62200056 0.02374574 0.64252343
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
__
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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] no internal function int.unzip in R 2.9.0 for Windows
Will respond privately, since the thread becomes uninteresting for R-help, I guess. Uwe Romain Francois wrote: Hello Uwe, I have recently taken over maintenance of the package, which is now version controlled at r-forge (http://r-forge.r-project.org/projects/r2html/), and I am planning these changes: - rework the HTML function so that the html code is generated from a brew template, which in time will give more control to the user/package using R2HTML - enhance the sweave html driver so that it can take advantage of the source code highlighter I am currently writing. Do you know the extent of the modification. I'd like to propagate this to the r-forge version. Romain Uwe Ligges wrote: The most recent version on CRAN is already fixed. We had a non-maintainer update end of March that fixed the R2HTML issues - the maintainer was unresponsive to our requests, unfortunately. Hence please update from CRAN. Best, Uwe Ligges Romain Francois wrote: Hi, I'll try to fix this soon. Could you log a bug request here: http://r-forge.r-project.org/tracker/?atid=1643group_id=405func=browse Regards, Romain Carson, John wrote: library(R2HTML) Loading required package: R2HTML Error in .Internal(int.unzip(zipname, NULL, dest)) : no internal function int.unzip Error : .onLoad failed in 'loadNamespace' for 'R2HTML' Error: package 'R2HTML' could not be loaded Version: R 2.9.0 for Windows __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Read name multiple excel sheets using RODBC
I'd like to be able to read multiple sheets from an excel workbook and use the sheet name to name the resulting dataframe using RODBC. at the moment i've figured out how to do it the long way (see below) but feel sure that there is a speedier possibly automatic way to do it in R. i've tried to run a loop using sqlTables but it seemed to break the connection. unless i've missed something, i cant see a solution to this on the help list. grateful for any help or pointers simeon # long way library(RODBC) filepath - C:/Data/workbook.xlsx connect - odbcConnectExcel2007(filepath) tbls - sqlTables(connect) sheet1 -sqlFetch(channel=connect,sqtable='sheet1') sheet2 -sqlFetch(channel=connect,sqtable='sheet2') sheet3 -sqlFetch(channel=connect,sqtable='sheet3') .. etc close(connect) [[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] Full likelihood from survreg
How can I find out what survreg generates: the full likelihood or a likelihood with unnecessary constants dropped? Survreg returns a log-likelihood, not a deviance (LL with parameters dropped); and the result is labeled as such. It turns out that the deviance is not as easily defined for censored data. Terry Therneau __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How do I get removed from this mailing list?
Wacek Kusnierczyk wrote: Gavin Simpson wrote: The View message source would be a more direct way of viewing the full email, not just the bits TBird shows you in the preview pane. I forget how it is named exactly and under which menu it is found as it has been quite a while since I used TBird. indeed; it's C-U, or under view message source Nice, thanks. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - (p.dalga...@biostat.ku.dk) FAX: (+45) 35327907 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Labeling barplot bars by multiple factors
Both of those worked, but hierobarp looked a bit easier, so I used that. The one annoying thing is that it sorts alphabetically. Tom On Thu, May 28, 2009 at 6:46 AM, Jim Lemon j...@bitwrit.com.au wrote: Thomas Levine wrote: I want to plot quantitative data as a function of three two-level factors. How do I group the bars on a barplot by level through labeling and spacing? Here http://www.thomaslevine.org/sample_multiple-factor_barplot.png's what I'm thinking of. Also, I'm pretty sure that I want a barplot, but there may be something better. Hi Tom, You may find that the hierobarp function in the plotrix package will do what you want. Jim [[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] sample unique pairs from a matrix
Dear Ritchie and David,
Thanks very much for your advice. I had thought of this potential
solution, however it doesn't really fullfill my second criteria which
is that once a particular cell has been sampled, the row and column of
that cell can't be sampled from subsequently. In other words, the next
sample would be taken from a 5x5 matrix, and then a 4x4 matrix and so
on until I have my 6 values.
I will keep thinking!
Cheers
Jos
2009/5/28 David Winsemius dwinsem...@comcast.net:
On May 28, 2009, at 6:33 AM, jos matejus wrote:
Dear R users,
I have a matrix of both negative and positive values that I would like
to randomly sample with the following 2 conditions:
1. only sample positive values
2. once a cell in the matrix has been sampled the row and column of
that cell cannot be sampled from again.
#some dummy data
set.seed(101)
dataf - matrix(rnorm(36,1,2), nrow=6)
I can do this quite simply if all the values are positive by using the
sample function without replacement on the column and row indices.
samrow - sample(6,replace=F)
samcol - sample(6,replace=F)
values - numeric(6)
for(i in 1:6){
values[i] - dataf[samrow[i], samcol[i]]
}
However, I am not sure how to include the logical condition to only
include postitive values
Any help would be gratefully received.
Jos
M - matrix(rnorm(36),nrow=6)
M[M[,]0]
[1] 1.65619781 0.56182830 0.23812890 0.81493915 1.01279243 1.29188874
0.64252343 0.53748655 0.31503112
[10] 0.37245358 0.07942883 0.56834586 0.62200056 0.39478167 0.02374574
0.04974857 0.56219171 0.52901658
sample(M[M[,]0],6,replace=F)
[1] 0.56834586 0.07942883 0.31503112 0.62200056 0.02374574 0.64252343
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] sample unique pairs from a matrix
This is one way of sampling all the values in the matrix that meet your
criteria:
set.seed(1)
(x - matrix(rnorm(100), 10))
[,1][,2][,3][,4] [,5]
[,6][,7] [,8] [,9] [,10]
[1,] -0.6264538 1.51178117 0.91897737 1.35867955 -0.1645236 0.3981059
2.40161776 0.475509529 -0.5686687 -0.5425200
[2,] 0.1836433 0.38984324 0.78213630 -0.10278773 -0.2533617 -0.6120264
-0.03924000 -0.709946431 -0.1351786 1.2078678
[3,] -0.8356286 -0.62124058 0.07456498 0.38767161 0.6969634 0.3411197
0.68973936 0.610726353 1.1780870 1.1604026
[4,] 1.5952808 -2.21469989 -1.98935170 -0.05380504 0.5566632 -1.1293631
0.02800216 -0.934097632 -1.5235668 0.7002136
[5,] 0.3295078 1.12493092 0.61982575 -1.37705956 -0.6887557 1.4330237
-0.74327321 -1.253633400 0.5939462 1.5868335
[6,] -0.8204684 -0.04493361 -0.05612874 -0.41499456 -0.7074952 1.9803999
0.18879230 0.291446236 0.3329504 0.5584864
[7,] 0.4874291 -0.01619026 -0.15579551 -0.39428995 0.3645820 -0.3672215
-1.80495863 -0.443291873 1.0630998 -1.2765922
[8,] 0.7383247 0.94383621 -1.47075238 -0.05931340 0.7685329 -1.0441346
1.46555486 0.001105352 -0.3041839 -0.5732654
[9,] 0.5757814 0.82122120 -0.47815006 1.10002537 -0.1123462 0.5697196
0.15325334 0.074341324 0.3700188 -1.2246126
[10,] -0.3053884 0.59390132 0.41794156 0.76317575 0.8811077 -0.1350546
2.17261167 -0.589520946 0.2670988 -0.4734006
# create new matrix with R/C indicator
x.ind - cbind(R=as.vector(row(x)), C=as.vector(col(x)), val=as.vector(x))
# only keep positive values
x.ind - x.ind[x.ind[, 'val'] = 0,, drop=FALSE]
# now loop and select a value then delete others in the same R/C
while (nrow(x.ind) 0){
+ i - sample(nrow(x.ind), 1)
+ cat(sample:, x.ind[i,], \n)
+ # remove matching R/C
+ x.ind - x.ind[!((x.ind[, R] == x.ind[i, R]) | (x.ind[, C] ==
x.ind[i, C])),, drop=FALSE]
+ }
sample: 3 3 0.07456498
sample: 8 2 0.9438362
sample: 4 7 0.02800216
sample: 10 4 0.7631757
sample: 9 1 0.5757814
sample: 5 9 0.5939462
sample: 1 8 0.4755095
sample: 7 5 0.3645820
sample: 6 6 1.9804
sample: 2 10 1.207868
On Thu, May 28, 2009 at 8:21 AM, jos matejus matejus...@googlemail.comwrote:
Dear Ritchie and David,
Thanks very much for your advice. I had thought of this potential
solution, however it doesn't really fullfill my second criteria which
is that once a particular cell has been sampled, the row and column of
that cell can't be sampled from subsequently. In other words, the next
sample would be taken from a 5x5 matrix, and then a 4x4 matrix and so
on until I have my 6 values.
I will keep thinking!
Cheers
Jos
2009/5/28 David Winsemius dwinsem...@comcast.net:
On May 28, 2009, at 6:33 AM, jos matejus wrote:
Dear R users,
I have a matrix of both negative and positive values that I would like
to randomly sample with the following 2 conditions:
1. only sample positive values
2. once a cell in the matrix has been sampled the row and column of
that cell cannot be sampled from again.
#some dummy data
set.seed(101)
dataf - matrix(rnorm(36,1,2), nrow=6)
I can do this quite simply if all the values are positive by using the
sample function without replacement on the column and row indices.
samrow - sample(6,replace=F)
samcol - sample(6,replace=F)
values - numeric(6)
for(i in 1:6){
values[i] - dataf[samrow[i], samcol[i]]
}
However, I am not sure how to include the logical condition to only
include postitive values
Any help would be gratefully received.
Jos
M - matrix(rnorm(36),nrow=6)
M[M[,]0]
[1] 1.65619781 0.56182830 0.23812890 0.81493915 1.01279243 1.29188874
0.64252343 0.53748655 0.31503112
[10] 0.37245358 0.07942883 0.56834586 0.62200056 0.39478167 0.02374574
0.04974857 0.56219171 0.52901658
sample(M[M[,]0],6,replace=F)
[1] 0.56834586 0.07942883 0.31503112 0.62200056 0.02374574 0.64252343
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
[[alternative HTML version deleted]]
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Re: [R] split strings
(diverted to r-devel, a source code patch attached)
Wacek Kusnierczyk wrote:
Allan Engelhardt wrote:
Immaterial, yes, but it is always good to test :) and your solution
*is* faster and it is even faster if you can assume byte strings:
:)
indeed; though if the speed is immaterial (and in this case it
supposedly was), it's probably not worth risking fixed=TRUE removing
'.tif' from the middle of the name, however unlikely this might be (cf
murphy's laws).
but if you can assume that each string ends with a '.tif' (or any other
\..{3} substring), then substr is marginally faster than sub, even as a
three-pass approach, while avoiding the risk of removing '.tif' from the
middle:
strings = sprintf('f:/foo/bar//%s.tif', replicate(1000,
paste(sample(letters, 10), collapse='')))
library(rbenchmark)
benchmark(columns=c('test', 'elapsed'), replications=1000, order=NULL,
substr={basenames=basename(strings); substr(basenames, 1,
nchar(basenames)-4)},
sub=sub('.tif', '', basename(strings), fixed=TRUE, useBytes=TRUE))
# test elapsed
# 1 substr 3.176
# 2sub 3.296
btw., i wonder why negative indices default to 1 in substr:
substr('foobar', -5, 5)
# fooba
# substr('foobar', 1, 5)
substr('foobar', 2, -2)
#
# substr('foobar', 2, 1)
this does not seem to be documented in ?substr. there are ways to make
negative indices meaningful, e.g., by taking them as indexing from
behind (as in, e.g., perl):
# hypothetical
substr('foobar', -5, 5)
# ooba
# substr('foobar', 6-5+1, 5)
substr('foobar', 2, -2)
# ooba
# substr('foobar', 2, 6-2+1)
there is a trivial fix to src/main/character.c that gives substr the
extended functionality -- see the attached patch. the patch has been
created and tested as follows:
svn co https://svn.r-project.org/R/trunk r-devel
cd r-devel
# modifications made to src/main/character.c
svn diff character.c.diff
svn revert -R .
patch -p0 character.c.diff
./configure
make
make check-all
# no problems reported
with the patched substr, the original problem can now be solved more
concisely, using a two-pass approach, with performance still better than
the sub/fixed/bytes one, as follows:
strings = sprintf('f:/foo/bar//%s.tif', replicate(1000,
paste(sample(letters, 10), collapse='')))
library(rbenchmark)
benchmark(columns=c('test', 'elapsed'), replications=1000, order=NULL,
substr=substr(basename(strings), 1, -5),
'substr-nchar'={
basenames=basename(strings)
substr(basenames, 1, nchar(basenames)-4) },
sub=sub('.tif', '', basename(strings), fixed=TRUE, useBytes=TRUE))
# test elapsed
# 1 substr 2.981
# 2 substr-nchar 3.206
# 3 sub 3.273
if this sounds interesting, i can update the docs accordingly.
vQ
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Re: [R] Re ad name multiple excel sheets using RODBC
simeon duckworth wrote: I'd like to be able to read multiple sheets from an excel workbook and use the sheet name to name the resulting dataframe using RODBC. In Microsoft theory, something like the below should be ok (note the $), but never managed to get this to work. The same method works perfectly when you have name ranges within a spreadsheet. If you only have the sheet names, you should use package xlsReadWrite which is rather fast, but has some limitations in the non-commercial version. Dieter library(RODBC) filepath - workbook.xlsx tabls = list() channel - odbcConnectExcel2007(filepath) for (i in 1:2) tabls[[i]] = sqlQuery(channel, paste(SELECT * from [Tabelle,i,$],sep=)) odbcClose(channel) # I prefer an explicit odbcClose, but generic close.RODBC should work. -- View this message in context: http://www.nabble.com/Read---name-multiple-excel-sheets-using-RODBC-tp23760020p23760912.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] Still can't find missing data
Farley, Robert wrote: I can't get the syntax that will allow me to show NA values (rows) in the xtabs. lengthy non-reproducible example removed If you want a reproducible answer, prepare a reproducible result. And check that the syntax is na.action=na.pass Dieter -- View this message in context: http://www.nabble.com/Still-can%27t-find-missing-data-tp23730627p23761006.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] question about using a remote system
My goal is for a user to sit down at a Linux laptop, get to an Rcmdr type screen, submit jobs on a remote system and then get the results back in R. We will assume that the user is naive, and the only thing he/she can do is get to the Rcmdr screen. The Rcmdr plugin will have a submit jobs menu. The user presses that option, and gets results, without having to know the underside. thanks, Erin On Thu, May 28, 2009 at 1:54 AM, Mark Wardle m...@wardle.org wrote: Hi. Do you need an interactive session at the remote machine, or are you simply wanting to run a pre-written script? If the latter, then you can ask ssh to execute a remote command, which conceivably could be R CMD x If you explain exactly why and what you are trying to do, then perhaps there's a better solution bw Mark 2009/5/28 Erin Hodgess erinm.hodg...@gmail.com: Dear R People: I would like to set up a plug-in for Rcmdr to do the following: I would start on a Linux laptop. Then I would log into another outside system and run a some commands. Now, when I tried to do system(ssh e...@xxx.edu) password xx It goes to the remote system. how do I continue to issue commands from the Linux laptop please? (hope this makes sense) thanks, Erin -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: erinm.hodg...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Dr. Mark Wardle Specialist registrar, Neurology Cardiff, UK -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: erinm.hodg...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How could I terminate a script (without leaving R)?
Good morning!
Which is the preferred method to leave a sourced script and returning back to
the '' prompt?
For example I search for certain files to be processed, but nothing should be
done if they
are not present. Normally I do in my script:
f-dir(pattern=qq)
if(length(f) 0) {
...process file list...
}
# script end
But I don't like those {} encompassing the whole script so I'm searching for
something to
put in place of the '???' here:
f-dir(pattern=qq)
if(length(f) == 0) '???'
...process file list...
# script end
Using stop(No file found, call.=F) works, but the Error message is ugly in
this case.
Any suggestion?
Thanks for your help!
mario
--
Ing. Mario Valle
Data Analysis and Visualization Group| http://www.cscs.ch/~mvalle
Swiss National Supercomputing Centre (CSCS) | Tel: +41 (91) 610.82.60
v. Cantonale Galleria 2, 6928 Manno, Switzerland | Fax: +41 (91) 610.82.82
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Re: [R] R-help Digest, Vol 75, Issue 28
From: Jeff Newmiller jdnew...@dcn.davis.ca.us To: R Heberto Ghezzo, Dr heberto.ghe...@mcgill.ca Date: Wed, 27 May 2009 09:00:44 -0700 Subject: Re: [R] R in Ubunto R Heberto Ghezzo, Dr wrote: Hello , I do not know anything abount Ubunto, but I found a Portable Ubunto for Windows and since so many people prefer Linux to Windows I decided to give it a try. It runs very nicely, so I tried to load R, following Instructions in CRAN I added the line deb http://probability.ca/cran/bin/linux/ubuntu hardy/ to /etc/apt/sources.list and then from a console I did sudo apt-get update sudo apt-get install r-base a lot of printout and when it inishes I typed R in the console and surprise! I got R 2.6.2!! in Windows I have R 2.9.0?? Did I do something wrong or there is another way to get the latest version of R? From: stephen sefick ssef...@gmail.com To: R Heberto Ghezzo, Dr heberto.ghe...@mcgill.ca Date: Wed, 27 May 2009 11:51:25 -0400 Subject: Re: [R] R in Ubunto I don't remember what the version of R in deb repositories is, but 2.6.2 is probably about right. One of the things the Debian project is focused on is the stability of the operating system, so they do not update packages as readily as some other distributions. I had this with Debian 5.0 and just decided to compile R from source after getting the R development package and some x11 development libraries. sudo apt-get install r-base-dev I can help you through this process if you like, or there are good instructions for this process at the R website. FAQ 2.5.1 How can R be installed (Unix) On the web page http://probability.ca/cran/bin/linux/ubuntu/ it presents instructions for activating this repository. Special instructions are included for hardy regarding activating backports also. I'm running the latest Ubuntu version, and I get the latest version of R no problem. I see that http://probability.ca/cran/bin/linux/ubuntu/hardy/ does list R 2.9.0, so it seems to me that if the repository was correctly added you should get it. Were there any errors when you ran sudo apt-get update? Also, I second the idea of activating the backports repository. -Ista __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] can you help me please :)
hi there :) i want to use barplot with if else but i dont know how to do it ? i tried this but it is not working with me SNP - read.table(my.txt) SNP[,2] [1] 1175 483 240 170 99 79 76 45 38 35 21 16 14 19 16 [16]333 1021686820511 [31]10620 13050501000 [46]055010000000000 [61]000000000000007 [76]000030000000000 [91]000000000000030 [106]000000000000000 [121]000000000000000 [136]100011000000000 [151]000000000000000 [166]002000000100000 [181]000000000000001 [196]000000000000000 [211]000000001000000 [226]1 SNP[,2] == 0 [1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE [13] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE [25] FALSE FALSE TRUE FALSE FALSE FALSE FALSE TRUE FALSE FALSE TRUE FALSE [37] TRUE FALSE TRUE FALSE TRUE FALSE TRUE TRUE TRUE TRUE FALSE FALSE [49] TRUE FALSE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [61] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [73] TRUE TRUE FALSE TRUE TRUE TRUE TRUE FALSE TRUE TRUE TRUE TRUE [85] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [97] TRUE TRUE TRUE TRUE TRUE TRUE TRUE FALSE TRUE TRUE TRUE TRUE [109] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [121] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [133] TRUE TRUE TRUE FALSE TRUE TRUE TRUE FALSE FALSE TRUE TRUE TRUE [145] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [157] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE FALSE [169] TRUE TRUE TRUE TRUE TRUE TRUE FALSE TRUE TRUE TRUE TRUE TRUE [181] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [193] TRUE TRUE FALSE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [205] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [217] TRUE TRUE FALSE TRUE TRUE TRUE TRUE TRUE TRUE FALSE SNP[,2][SNP[,2] == 0] [1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [38] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [75] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [112] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [149] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 SNP[,2][! SNP[,2] == 0] [1] 1175 483 240 170 99 79 76 45 38 35 21 16 14 19 16 [16]333 1021686825111 [31]62 13551551733111 [46]2111 SNP[,1] [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 [19] 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 [37] 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 [55] 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 [73] 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 [91] 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 [109] 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 [127] 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 [145] 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 [163] 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 [181] 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 [199] 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 [217] 217 218 219 220 221 222 223 224 225 226 zz - SNP[,2][! SNP[,2] == 0] zz [1] 1175 483 240 170 99 79 76 45 38 35 21 16 14 19 16 [16]333 1021686825111 [31]62 13551551733111 [46]21111 [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 [20] 20 21 22 23 24 25 26 28 29 30
Re: [R] How could I terminate a script (without leaving R)?
On 5/28/2009 8:58 AM, Mario Valle wrote:
Good morning!
Which is the preferred method to leave a sourced script and returning back to the
'' prompt?
For example I search for certain files to be processed, but nothing should be
done if they
are not present. Normally I do in my script:
f-dir(pattern=qq)
if(length(f) 0) {
...process file list...
}
# script end
But I don't like those {} encompassing the whole script so I'm searching for
something to
put in place of the '???' here:
f-dir(pattern=qq)
if(length(f) == 0) '???'
...process file list...
# script end
Using stop(No file found, call.=F) works, but the Error message is ugly in
this case.
Any suggestion?
Put the code in a function, call the function as the last line of the
script, and return from the function when you want to exit the script.
Duncan Murdoch
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Re: [R] Re ad name multiple excel sheets using RODBC
2009/5/28 Dieter Menne dieter.me...@menne-biomed.de: If you only have the sheet names, you should use package xlsReadWrite which is rather fast, but has some limitations in the non-commercial version. what limitations, i.e. features do you miss? Cheers, Hans-Peter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] [Rd] split strings
Bill Dunlap
TIBCO Software Inc - Spotfire Division
wdunlap tibco.com
-Original Message-
From: r-devel-boun...@r-project.org
[mailto:r-devel-boun...@r-project.org] On Behalf Of Wacek Kusnierczyk
Sent: Thursday, May 28, 2009 5:30 AM
Cc: R help project; r-de...@r-project.org; Allan Engelhardt
Subject: Re: [Rd] [R] split strings
(diverted to r-devel, a source code patch attached)
Wacek Kusnierczyk wrote:
Allan Engelhardt wrote:
Immaterial, yes, but it is always good to test :) and your solution
*is* faster and it is even faster if you can assume byte strings:
:)
indeed; though if the speed is immaterial (and in this case it
supposedly was), it's probably not worth risking fixed=TRUE removing
'.tif' from the middle of the name, however unlikely this
might be (cf
murphy's laws).
but if you can assume that each string ends with a '.tif'
(or any other
\..{3} substring), then substr is marginally faster than
sub, even as a
three-pass approach, while avoiding the risk of removing
'.tif' from the
middle:
strings = sprintf('f:/foo/bar//%s.tif', replicate(1000,
paste(sample(letters, 10), collapse='')))
library(rbenchmark)
benchmark(columns=c('test', 'elapsed'),
replications=1000, order=NULL,
substr={basenames=basename(strings); substr(basenames, 1,
nchar(basenames)-4)},
sub=sub('.tif', '', basename(strings), fixed=TRUE,
useBytes=TRUE))
# test elapsed
# 1 substr 3.176
# 2sub 3.296
btw., i wonder why negative indices default to 1 in substr:
substr('foobar', -5, 5)
# fooba
# substr('foobar', 1, 5)
substr('foobar', 2, -2)
#
# substr('foobar', 2, 1)
this does not seem to be documented in ?substr.
Would your patched code affect the following
use of regexpr's output as input to substr, to
pull out the matched text from the string?
x-c(ooo,good food,bad)
r-regexpr(o+, x)
substring(x,r,attr(r,match.length)+r-1)
[1] ooo oo
substr(x,r,attr(r,match.length)+r-1)
[1] ooo oo
r
[1] 1 2 -1
attr(,match.length)
[1] 3 2 -1
attr(r,match.length)+r-1
[1] 3 3 -3
attr(,match.length)
[1] 3 2 -1
there are
ways to make
negative indices meaningful, e.g., by taking them as indexing from
behind (as in, e.g., perl):
# hypothetical
substr('foobar', -5, 5)
# ooba
# substr('foobar', 6-5+1, 5)
substr('foobar', 2, -2)
# ooba
# substr('foobar', 2, 6-2+1)
there is a trivial fix to src/main/character.c that gives substr the
extended functionality -- see the attached patch. the patch has been
created and tested as follows:
svn co https://svn.r-project.org/R/trunk r-devel
cd r-devel
# modifications made to src/main/character.c
svn diff character.c.diff
svn revert -R .
patch -p0 character.c.diff
./configure
make
make check-all
# no problems reported
with the patched substr, the original problem can now be solved more
concisely, using a two-pass approach, with performance still
better than
the sub/fixed/bytes one, as follows:
strings = sprintf('f:/foo/bar//%s.tif', replicate(1000,
paste(sample(letters, 10), collapse='')))
library(rbenchmark)
benchmark(columns=c('test', 'elapsed'),
replications=1000, order=NULL,
substr=substr(basename(strings), 1, -5),
'substr-nchar'={
basenames=basename(strings)
substr(basenames, 1, nchar(basenames)-4) },
sub=sub('.tif', '', basename(strings), fixed=TRUE,
useBytes=TRUE))
# test elapsed
# 1 substr 2.981
# 2 substr-nchar 3.206
# 3 sub 3.273
if this sounds interesting, i can update the docs accordingly.
vQ
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Re ad name multiple excel sheets using RODBC
Hans-Peter Suter wrote: If you only have the sheet names, you should use package xlsReadWrite which is rather fast, but has some limitations in the non-commercial version. what limitations, i.e. features do you miss? Reading of named ranges. Dieter -- View this message in context: http://www.nabble.com/Read---name-multiple-excel-sheets-using-RODBC-tp23760020p23761745.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] Re ad name multiple excel sheets using RODBC
Using the package rcom (by Thomas Baier) you can do everything you want since you have the full power of COM at your disposal. Dieter Menne wrote: Hans-Peter Suter wrote: If you only have the sheet names, you should use package xlsReadWrite which is rather fast, but has some limitations in the non-commercial version. what limitations, i.e. features do you miss? Reading of named ranges. Dieter -- Erich Neuwirth, University of Vienna Faculty of Computer Science Computer Supported Didactics Working Group Visit our SunSITE at http://sunsite.univie.ac.at Phone: +43-1-4277-39464 Fax: +43-1-4277-39459 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 gularly spaced points on multi-segement line
Dear all ... I need your help for the following purpose : I would like to create regularly spaced points on a multi-segment line (a 'psp class' object). A function of the spatstat library( = pointsOnLines() ) is dedicated to that objective but the problem is that the probability of falling on a particular segment is proportional to the length of each segment of the multi-segment line. What about a solution using the whole multi-segment line and not individual segments for points distribution ? Many thanks in advance for your help ! Plaff -- View this message in context: http://www.nabble.com/regularly-spaced-points-on-multi-segement-line-tp23761782p23761782.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] Changing data point size in quilt.plot
Hi, I am new to R and am using quilt.plot (from fields package) to plot gas concentrations in Europe. I am using the following code to plot my data library(maps) library(fields) test = read.csv(%change 1996_2005.txt, sep=\t) colnames(test) = c(Station, Measurement_BaseO3, Model_BaseO3, Measuerment_PeakO3, Model_PeakO3, Longitude, Latitude) quilt.plot(test$Longitude, test$Latitude, test$Model_BaseO3, breaks = c(-50, -40, -30, -20, -10, 0, 10, 20, 30, 40, 50), col = tim.colors(10), xlim=c(-11,25), ylim=c(40,60), zlim=c(-50, 50), xlab=Longitude, ylab=Latitude, asp=1, main=Modelled Base Ozone) map(add=TRUE) It all plots beautifully, except the size of the plotted data points is far too small. I have tried using cex = xx into quilt.plot() to change the size of the data points but that doesn't seem to work. Maybe I am just using this wrong?! I was wondering if someone could help me to alter the data point size (increase it to x 3 the current size). Thanks, Becca [[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] Urgent Statistician Position near Philadelphia
Dear R-list, I know is going to be weird, but I need to do something about and I know the R-list is big group of good people. Im moving to Philadelphia in August 2009 (my fiancé is there) and I need to find a job. I know to do so many things, but my skill is in science and research. I would like to know if you know a possible position near Philadelphia, New Jersey or New York. I was working in so many projects, doing statistical analysis, research in parasitology, cancer research and public health in dentistry problems. I would love to work doing cancer research or public health. I was working with cancer epidemiology using survival analysis and modeling. I know how work and programing with many softwares, R, SAS, S-plus, MathLab. I know too Statistica y SPSS in any version. You are free to give advice to me. Im applying in so many web pages, but does not work. I dont have any problem with visa to be in USA, just I need some help. I will send you my resume if you ask me Best Regards, O__ José Bustos M. c/ /'_ --- Master in Applied Statistics (University of Concepcion) (*) \(*) -- B.S. in Marine Biology (Catholic University of Concepcion) -Science Faculty -Catholic University of Concepcion -Alonso de Ribera 2850 Concepción, Casilla 297 -Cell phone: +56 9 9 5939144 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Plot error
Hello, I am an R amateur. I want to plot data such that the 3 time points(a,b,c) lie on the X-axis and the values of these times points are on Y-axis for n samples (e.g.100). So, I have an object x, dim 100 4, it is a dataframe (when checked the class) x = name a b c 10.11 1.11 0.86 2 . . . 3 . . . . . . 100 so when i say: plot(1:3, x[,2:4], type=l) - I get the error below Error in xy.coords(x, y, xlabel, ylabel, log) : (list) object cannot be coerced to type 'double' However if I do: plot(1:3, x[1,2:4], type=l) -- It works for the 1st row, and each individual row BUT NOT ALL ROWS Please could someone explain what is happening here? I wonder if I need to use 'lines' for the remaining, BUT I have another dataset y with same dimensions as x, which I want to plot on the same graph/plot to see the difference between x and y. Thanks, NS -- View this message in context: http://www.nabble.com/Plot-error-tp23761408p23761408.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] Plot error
I want to plot data such that the 3 time points(a,b,c) lie on the X-axis
and
the values of these times points are on Y-axis for n samples (e.g.100).
So, I have an object x, dim 100 4, it is a dataframe (when checked the
class)
x =
name a b c
10.11 1.11 0.86
2 . . .
3 . . .
.
.
.
100
so when i say:
plot(1:3, x[,2:4], type=l) - I get the error below
Error in xy.coords(x, y, xlabel, ylabel, log) :
(list) object cannot be coerced to type 'double'
However if I do:
plot(1:3, x[1,2:4], type=l) -- It works for the 1st row, and
each
individual row BUT NOT ALL ROWS
Please could someone explain what is happening here?
I wonder if I need to use 'lines' for the remaining, BUT I have another
dataset y with same dimensions as x, which I want to plot on the same
graph/plot to see the difference between x and y.
Your data looks like this:
x - data.frame(name=sample(letters, 10), a=runif(10), b=rnorm(10),
c=rlnorm(10))
The problem is that the subset x[,2:4] is also a data frame, not a matrix.
class(x[,2:4]) #[1] data.frame
The simplest thing is probably to use lines, as you say.
row - seq_len(nrow(x))
xx - x[,2:4]
plot(row, xx$a, ylim=range(xx), type=l)
lines(row, xx$b, col=blue)
lines(row, xx$c, col=green)
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
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Re: [R] question about using a remote system
Will R be able to send one batch of commands to the remote machine, then wait for the output? Or will there be back and forth with new commands based on the output from the first commands? You may want to look at the nws package for one way to have jobs run on a remote machine. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of Erin Hodgess Sent: Thursday, May 28, 2009 6:56 AM To: Mark Wardle Cc: R help Subject: Re: [R] question about using a remote system My goal is for a user to sit down at a Linux laptop, get to an Rcmdr type screen, submit jobs on a remote system and then get the results back in R. We will assume that the user is naive, and the only thing he/she can do is get to the Rcmdr screen. The Rcmdr plugin will have a submit jobs menu. The user presses that option, and gets results, without having to know the underside. thanks, Erin On Thu, May 28, 2009 at 1:54 AM, Mark Wardle m...@wardle.org wrote: Hi. Do you need an interactive session at the remote machine, or are you simply wanting to run a pre-written script? If the latter, then you can ask ssh to execute a remote command, which conceivably could be R CMD x If you explain exactly why and what you are trying to do, then perhaps there's a better solution bw Mark 2009/5/28 Erin Hodgess erinm.hodg...@gmail.com: Dear R People: I would like to set up a plug-in for Rcmdr to do the following: I would start on a Linux laptop. Then I would log into another outside system and run a some commands. Now, when I tried to do system(ssh e...@xxx.edu) password xx It goes to the remote system. how do I continue to issue commands from the Linux laptop please? (hope this makes sense) thanks, Erin -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: erinm.hodg...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Dr. Mark Wardle Specialist registrar, Neurology Cardiff, UK -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: erinm.hodg...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] maximum over one dimension of a 3-dimensional array
Hi, I'm running R 2.7.2 on windows XP. I'd like to find the maximum of a 3-d array over it's third index to create a 2-d array. For example: x - array(c(1,2,3,10,11,12,3:8),c(2,3,2)) x , , 1 [,1] [,2] [,3] [1,]13 11 [2,]2 10 12 , , 2 [,1] [,2] [,3] [1,]357 [2,]468 x1 - x[,,1] x2 - x[,,2] pmax(x1,x2) [,1] [,2] [,3] [1,]35 11 [2,]4 10 12 Is there a pre-defined function that I can use to do this without using a for-loop? Also, the third index can be long and of variable length, so I don't want to explicitly write out x1, x2,... Thanks in advance for your help. eric __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] sample unique pairs from a matrix
Your last step will either be a single number (not really a sampling
operation) or a
non-positive number. So at best you really only have an number that
depends
entirely on the prior sequence of draws.
--
David.
On May 28, 2009, at 8:21 AM, jos matejus wrote:
Dear Ritchie and David,
Thanks very much for your advice. I had thought of this potential
solution, however it doesn't really fullfill my second criteria which
is that once a particular cell has been sampled, the row and column of
that cell can't be sampled from subsequently. In other words, the next
sample would be taken from a 5x5 matrix, and then a 4x4 matrix and so
on until I have my 6 values.
I will keep thinking!
Cheers
Jos
2009/5/28 David Winsemius dwinsem...@comcast.net:
On May 28, 2009, at 6:33 AM, jos matejus wrote:
Dear R users,
I have a matrix of both negative and positive values that I would
like
to randomly sample with the following 2 conditions:
1. only sample positive values
2. once a cell in the matrix has been sampled the row and column of
that cell cannot be sampled from again.
#some dummy data
set.seed(101)
dataf - matrix(rnorm(36,1,2), nrow=6)
I can do this quite simply if all the values are positive by using
the
sample function without replacement on the column and row indices.
samrow - sample(6,replace=F)
samcol - sample(6,replace=F)
values - numeric(6)
for(i in 1:6){
values[i] - dataf[samrow[i], samcol[i]]
}
However, I am not sure how to include the logical condition to only
include postitive values
Any help would be gratefully received.
Jos
M - matrix(rnorm(36),nrow=6)
M[M[,]0]
[1] 1.65619781 0.56182830 0.23812890 0.81493915 1.01279243
1.29188874
0.64252343 0.53748655 0.31503112
[10] 0.37245358 0.07942883 0.56834586 0.62200056 0.39478167
0.02374574
0.04974857 0.56219171 0.52901658
sample(M[M[,]0],6,replace=F)
[1] 0.56834586 0.07942883 0.31503112 0.62200056 0.02374574 0.64252343
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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Re: [R] maximum over one dimension of a 3-dimensional array
On May 28, 2009, at 11:25 AM, eric lee wrote: Hi, I'm running R 2.7.2 on windows XP. I'd like to find the maximum of a 3-d array over it's third index to create a 2-d array. For example: x - array(c(1,2,3,10,11,12,3:8),c(2,3,2)) x , , 1 [,1] [,2] [,3] [1,]13 11 [2,]2 10 12 , , 2 [,1] [,2] [,3] [1,]357 [2,]468 x1 - x[,,1] x2 - x[,,2] pmax(x1,x2) [,1] [,2] [,3] [1,]35 11 [2,]4 10 12 Consider: apply(x, 1:2, max) [,1] [,2] [,3] [1,]35 11 [2,]4 10 12 or apply(x, c(1,2), max) # not necessarily adjacent dimensions [,1] [,2] [,3] [1,]35 11 [2,]4 10 12 Is there a pre-defined function that I can use to do this without using a for-loop? Also, the third index can be long and of variable length, so I don't want to explicitly write out x1, x2,... Thanks in advance for your help. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] String replacement in an expression
Dear R-experts, I need to replace in an expression the character Cl by Cl+beta But in the following case: form-expression((Cl-(V *ka) ) +(V *Vm *exp(-(Clm/Vm) *t))) gsub(Cl,(Cl+beta),as.character(form)) We obtain: [1] ((Cl+beta) - (V * ka)) + (V * Vm * exp(-((Cl+beta)m/Vm) * t)) the character Clm has been also replaced. How could I avoid this unwanted replacement ? Thank you in advance for any help. -- - Caroline BAZZOLI INSERM U738 - Université PARIS 7 UFR de Medecine - Site Bichat 16 rue Henri Huchard 75018 PARIS, FRANCE email: caroline.bazz...@inserm.fr www.biostat.fr PFIM: www.pfim.biostat.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.
Re: [R] String replacement in an expression
Try matching on word boundaries as well: gsub(\\bCl\\b,(Cl+beta),as.character(form)) [1] ((Cl+beta) - (V * ka)) + (V * Vm * exp(-(Clm/Vm) * t)) See ?regexp On Thu, May 28, 2009 at 11:41 AM, Caroline Bazzoli caroline.bazz...@inserm.fr wrote: Dear R-experts, I need to replace in an expression the character Cl by Cl+beta But in the following case: form-expression((Cl-(V *ka) ) +(V *Vm *exp(-(Clm/Vm) *t))) gsub(Cl,(Cl+beta),as.character(form)) We obtain: [1] ((Cl+beta) - (V * ka)) + (V * Vm * exp(-((Cl+beta)m/Vm) * t)) the character Clm has been also replaced. How could I avoid this unwanted replacement ? Thank you in advance for any help. -- - Caroline BAZZOLI INSERM U738 - Université PARIS 7 UFR de Medecine - Site Bichat 16 rue Henri Huchard 75018 PARIS, FRANCE email: caroline.bazz...@inserm.fr www.biostat.fr PFIM: www.pfim.biostat.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-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] getTree visualization (randomForest)
Dear All,
I would like to visualize a tree extracted from a random forest using
getTree {randomForest}.
I'm wondering if there is any way to do it directly or to convert my tree
to any other class of tree repesentation, which then can be plotted?
Thank you in advance,
Karoly
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Re: [R] can you help me please :)
On Thu, 28 May 2009 15:47:18 +0300 NOUF AL NUMAIR
noufalnum...@hotmail.com wrote:
NAN barplot(zz,width = 4,names.arg= xx,axes = TRUE, axisnames = TRUE,
NAN main=title,xlab=xlab,ylab=ylab,ylim=c(0,1175),xlim=c(0,226),col =
NAN c(for (i in zz){if i70 col= lightblue else col= mistyrose))
NAN
NAN
NAN i tried to get red of zero then plot it
? I have no idea what this means...
NAN i want to color different result with different colors ?
NAN
NAN any body can help plaaaeee
If you want help try to be more clear about what you want. It is hard
with your mail because there is so much what does not belong to your
current problem, one has to scroll over plenty of numbers and still I am
not sure whether I understood what you want.
Hence you want your bar color conditional on the value or the height,
it is helpful to create a vector beforehand.(for control purposes)
y-c(1175, 483, 240, 170, 99, 79, 76, 45, 38, 35,
21, 16, 14, 19, 16)
x-c( 1, 2, 3, 4, 5, 6, 7,
8, 9, 10, 11, 12, 13, 14, 15)
d1-data.frame(x=x,y=y)
Color-ifelse(d1$y70,blue,red)
barplot(as.matrix(d1$y),names.arg=d1$x,col=Color,beside=T)
as an example.
hth
Stefan
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Re: [R] Neural Network resource
The package AMORE appears to be more flexible, but I got very poor
results using it when I tried to improve the predictive accuracy of a
regression model. I don't understand all the options well enough to be
able to fine tune it to get better predictions. However, using the
nnet() function in package VR gave me decent results and is pretty easy
to use (see the Venables and Ripley book, Modern Applied Statistics with
S, pages 243 to 249, for more details). I tried using package neuralnet
as well but the neural net failed to converge. I could not figure out
how to set the threshold option (or other options) to get the neural net
to converge. I explored package neural as well. Of all these 4 packages,
the nnet() function in package VR worked the best for me.
As another R user commented as well, you have too many hidden layers and
too many neurons. In general you do not need more than 1 hidden layer.
One hidden layer is sufficient for the universal approximator property
of neural networks to hold true. As you keep adding neurons to the one
hidden layer, the problem becomes more and more non-linear. If you add
too many neurons you will overfit. In general, you do not need to add
more than 10 neurons. The activation function in the hidden layer of
Venables and Ripley's nnet() function is logistic, and you can specify
the activation function in the output layer to be linear using linout =
T in nnet(). Using one hidden layer, and starting with one hidden neuron
and working up to 10 hidden neurons, I built several neural nets (4,000
records) and computed the training MSE. I also computed the validation
MSE on a holdout sample of over 1,000 records. I also started with 2
variables and worked up to 15 variables in a for loop, so in all, I
built 140 neural nets using 2 for loops, and stored the results in
lists. I arranged my variables in the data frame based on correlations
and partial correlations so that I could easily add variables in a for
loop. This was my crude attempt to simulate variable selection since,
from what I have seen, neural networks do not have variable selection
methods. In my particular case, neural networks gave me marginally
better results than regression. It all depends on the problem. If the
data has non-linear patterns, neural networks will be better than linear
regression.
My code is below. You can modify it to suit your needs if you find it
useful. There are probably lines in the code that are redundant which
can be deleted.
HTH.
Jude Ryan
My code:
# set order in data frame train2 based on correlations and partial
correlations
train2 - train[, c(5,27,19,20,25,26,4,9,3,10,16,6,2,14,21,28)]
dim(train2)
names(train2)
library(nnet)
# skip = T
# train 10 neural networks in a loop and find the one with the minimum
test and validation error
# create various lists to store the results of the neural network
running in two for loops
# The Column List is for the outer for loop, which loops over variables
# The Row List is for the inner for loop, which loops over number of
neurons in the hidden layer
col_nn - list() # stores the results of nnet() over variables - outer
loop
row_nn - list() # stores the results of nnet() over neurons - inner
loop
col_mse - list()
# row_mse - list() # not needed because nn.mse is a data frame with
rows
col_sum - list()
row_sum - list()
col_vars - list()
row_vars - list()
col_wts - list()
row_wts - list()
df_dim - dim(train2)
df_dim[2] # number of variables
df_dim[2] - 1
num_of_neurons - 10
# build data frame to store results of neural net for each run
nn.mse - data.frame(Train_MSE=seq(1:num_of_neurons),
Valid_MSE=seq(1:num_of_neurons))
# open log file and redirect output to log file
sink(D:\\XXX\\YYY\\ Programs\\Neural_Network_v8_VR_log.txt)
# outer loop - loop over variables
for (i in 3:df_dim[2]) { # df_dim[2]
# inner loop - loop over number of hidden neurons
for (j in 1:num_of_neurons) { # upto 10 neurons in the hidden layer
# need to create a new data frame with just the predictor/input
variables needed
train3 - train2[,c(1:i)]
coreaff.nn - nnet(dep_var ~ ., train3, size = j, decay = 1e-3,
linout = T, skip = T, maxit = 1000, Hess = T)
# row_vars[[j]] - coreaff.nn$call # not what we want
# row_vars[[j]] - names(train3)[c(2:i)] # not needed in inner loop
- same number of variables for all neurons
row_sum[[j]] - summary(coreaff.nn)
row_wts[[j]] - coreaff.nn$wts
rownames(nn.mse)[j] - paste(H, j, sep=)
nn.mse[j, Train_MSE] - mean((train3$dep_var -
predict(coreaff.nn))^2)
nn.mse[j, Valid_MSE] - mean((valid$dep_var - predict(coreaff.nn,
valid))^2)
}
col_vars[[i-2]] - names(train3)[c(2:i)]
col_sum[[i-2]] - row_sum
col_wts[[i-2]] - row_wts
col_mse[[i-2]] - nn.mse
}
# cbind(col_vars[1],col_vars[2])
col_vars
col_sum
col_wts
sink()
cbind(col_mse[[1]],col_mse[[2]],col_mse[[3]],col_mse[[4]],col_mse[[5]],c
ol_mse[[6]],col_mse[[7]],
Re: [R] String replacement in an expression
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Caroline Bazzoli Sent: Thursday, May 28, 2009 8:41 AM To: r-help@r-project.org Subject: [R] String replacement in an expression Dear R-experts, I need to replace in an expression the character Cl by Cl+beta But in the following case: form-expression((Cl-(V *ka) ) +(V *Vm *exp(-(Clm/Vm) *t))) gsub(Cl,(Cl+beta),as.character(form)) We obtain: [1] ((Cl+beta) - (V * ka)) + (V * Vm * exp(-((Cl+beta)m/Vm) * t)) the character Clm has been also replaced. How could I avoid this unwanted replacement ? I like to use substitute(), which works with the expression as an expression, instead of converting it to a string, changing it, and then parsing the result to make the new expression. E.g., form-expression((Cl-(V *ka) ) +(V *Vm *exp(-(Clm/Vm) *t))) do.call(substitute, list(form[[1]], list(Cl=Quote(Cl+beta (Cl + beta - (V * ka)) + (V * Vm * exp(-(Clm/Vm) * t)) Note how it adds parentheses where appropriate: do.call(substitute, list(form[[1]], list(Cl=Quote(Cl+beta), Clm=Quote(Clm+gamma (Cl + beta - (V * ka)) + (V * Vm * exp(-((Clm + gamma)/Vm) * t)) The do.call() is needed in this case because substitute() does not evaluate its first argument and the do.call takes care of evaluating the variable 'form' so substitute sees its value. (S+'s substitute() has an evaluator=FALSE/TRUE argument to let you avoid the do.call().) R's substitute() doesn't seem to go into expression objects, hence I passed it the first element of the expression and it returned the converted first element. You could wrap the output with as.expression if you really wanted the expression object output. Bill Dunlap TIBCO Software Inc - Spotfire Division wdunlap tibco.com Thank you in advance for any help. -- - Caroline BAZZOLI INSERM U738 - Université PARIS 7 UFR de Medecine - Site Bichat 16 rue Henri Huchard 75018 PARIS, FRANCE email: caroline.bazz...@inserm.fr www.biostat.fr PFIM: www.pfim.biostat.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-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] boxplot
amor Gandhi wrote: Hi gues, This should read: Hi, guess what I want amor Gandhi wrote: Is there any function in R for boxplot with different time points? t1 - c(rep(1,20),rep(2,20)) t2 - c(rep(1,10),rep(2,10),rep(1,10),rep(2,10)) x - rnorm(40,5,1) dat - data.frame(t1,t2,x) boxplot(x~t1,t2) This might come close library(lattice) t1 - c(rep(1,20),rep(2,20)) t2 - c(rep(1,10),rep(2,10),rep(1,10),rep(2,10)) x - rnorm(40,5,1) dat - data.frame(t1=as.factor(t1),t2=as.factor(t2),x) bwplot(x~t1|t2,data=dat) bwplot(t1+t2~x,data=dat,outer=TRUE) -- View this message in context: http://www.nabble.com/boxplot-tp23752278p23767111.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] Hmisc package: deff() command's formula for the design effect
One additional question: It seems that in practice the design effect is often calculated on the response variable or on individual predictor variables. I have some OLS models using observational data that are clustered. Does it make sense to calculate the design effect on the residuals of one of these models to see the extent to which there is still clustering left even after including covariates? Thomas Lumley wrote: The formula in Hmisc is correct (if the correlation doesn't vary with the cluster size). If you think of the formula for the variance of a sum, it involves adding up all the variances and covariances. A cluster of size k has k^2-k covariances between members, so the total number of covariances is sum(k^2-k) over all the clusters, plus the sum(k) variances. Another way to think of it is that the larger clusters get too much weight, so in addition to the rho*(B-1) factor that you would have for equal-sized clusters there is an additional loss of efficiency due to giving too much weight to the larger clusters. -thomas __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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/Hmisc-package%3A-deff%28%29-command%27s-formula-for-the-design-effect-tp23415477p23767287.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] ggplot2 legend
Hi: I need some help with the legend. I got 14 samples(Muestreo) and I am trying to plot a smooth line for each sample. I am able to accomplish that but the problem is that the legend only displays every other sample. How can I force the legend to show all of my Muestreos? Thanks in advance. fish_ByMuestreo - structure(list(data = structure(list(SampleDate = structure(c(3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 16L, 16L, 16L, 7L, 7L, 7L, 7L, 10L, 10L, 10L, 13L, 13L, 13L, 13L, 13L, 27L, 27L, 27L, 27L, 27L, 17L, 17L, 17L, 17L, 20L, 20L, 20L, 20L, 24L, 24L, 24L, 24L, 24L, 30L, 30L, 30L, 30L, 42L, 42L, 42L, 42L, 42L, 33L, 33L, 33L, 33L, 36L, 36L, 36L, 36L, 36L, 39L, 39L, 39L, 39L, 39L, 39L, 14L, 14L, 14L, 14L, 14L, 14L, 5L, 5L, 5L, 8L, 8L, 8L, 11L, 11L, 11L, 11L, 15L, 15L, 15L, 28L, 28L, 28L, 28L, 28L, 18L, 18L, 18L, 18L, 18L, 22L, 22L, 22L, 22L, 25L, 25L, 25L, 25L, 40L, 40L, 40L, 40L, 40L, 31L, 31L, 31L, 31L, 37L, 37L, 37L, 37L, 1L, 1L, 1L, 1L, 1L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 16L, 16L, 16L, 7L, 7L, 7L, 7L, 10L, 10L, 10L, 13L, 13L, 13L, 13L, 13L, 27L, 27L, 27L, 27L, 27L, 17L, 17L, 17L, 17L, 20L, 20L, 20L, 20L, 24L, 24L, 24L, 24L, 24L, 30L, 30L, 30L, 30L, 42L, 42L, 42L, 42L, 42L, 33L, 33L, 33L, 33L, 36L, 36L, 36L, 36L, 36L, 14L, 14L, 14L, 14L, 14L, 14L, 5L, 5L, 5L, 8L, 8L, 8L, 11L, 11L, 11L, 11L, 15L, 15L, 15L, 28L, 28L, 28L, 28L, 28L, 18L, 18L, 18L, 18L, 18L, 22L, 22L, 22L, 22L, 25L, 25L, 25L, 25L, 40L, 40L, 40L, 40L, 40L, 31L, 31L, 31L, 31L, 34L, 34L, 34L, 34L, 34L, 37L, 37L, 37L, 37L, 1L, 1L, 1L, 1L, 1L, 6L, 6L, 6L, 6L, 6L, 6L, 9L, 9L, 9L, 12L, 12L, 12L, 21L, 21L, 21L, 21L, 29L, 29L, 29L, 19L, 19L, 19L, 19L, 19L, 23L, 23L, 23L, 23L, 23L, 26L, 26L, 26L, 26L, 41L, 41L, 41L, 41L, 32L, 32L, 32L, 32L, 32L, 35L, 35L, 35L, 35L, 38L, 38L, 38L, 38L, 38L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 16L, 16L, 16L, 7L, 7L, 7L, 7L, 10L, 10L, 10L, 13L, 13L, 13L, 13L, 13L, 27L, 27L, 27L, 27L, 27L, 17L, 17L, 17L, 17L, 20L, 20L, 20L, 20L, 24L, 24L, 24L, 24L, 24L, 30L, 30L, 30L, 30L, 42L, 42L, 42L, 42L, 42L, 33L, 33L, 33L, 33L, 36L, 36L, 36L, 36L, 36L, 39L, 39L, 39L, 39L, 39L, 39L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 16L, 16L, 16L, 7L, 7L, 7L, 7L, 10L, 10L, 10L, 13L, 13L, 13L, 13L, 13L, 27L, 27L, 27L, 27L, 27L, 17L, 17L, 17L, 17L, 20L, 20L, 20L, 20L, 24L, 24L, 24L, 24L, 24L, 30L, 30L, 30L, 30L, 42L, 42L, 42L, 42L, 42L, 33L, 33L, 33L, 33L, 36L, 36L, 36L, 36L, 36L, 39L, 39L, 39L, 39L, 39L, 39L), .Label = c(10/2/2002, 10/4/2002, 6/23/2002, 6/30/2002, 7/10/2002, 7/12/2002, 7/14/2002, 7/17/2002, 7/19/2002, 7/21/2002, 7/24/2002, 7/26/2002, 7/28/2002, 7/3/2002, 7/31/2002, 7/7/2002, 8/11/2002, 8/14/2002, 8/16/2002, 8/18/2002, 8/2/2002, 8/21/2002, 8/23/2002, 8/25/2002, 8/28/2002, 8/30/2002, 8/4/2002, 8/7/2002, 8/9/2002, 9/1/2002, 9/11/2002, 9/13/2002, 9/15/2002, 9/18/2002, 9/20/2002, 9/22/2002, 9/25/2002, 9/27/2002, 9/29/2002, 9/4/2002, 9/6/2002, 9/8/2002), class = factor), PondName = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L), .Label = c(Pond01, Pond02, Pond03, Pond04, Pond05, Pond06, Pond07), class = factor), Muestreo = c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 5L, 5L, 5L, 6L, 6L, 6L, 6L, 6L, 7L, 7L, 7L, 7L, 7L, 8L, 8L, 8L, 8L, 9L, 9L, 9L, 9L, 10L, 10L, 10L, 10L, 10L, 11L, 11L, 11L, 11L, 12L, 12L, 12L, 12L, 12L, 13L,
Re: [R] Changing point color/character in qqmath
Apologies once more, there was a slight error in our example. Here is a
correct version of how to color the points in qqmath according to another
variable.
qqmath(~ yield | variety, data = barley, groups=year,
auto.key=TRUE,
prepanel = function(x, ...) {
list(xlim = range(qnorm(ppoints(length(x)
},
panel = function(x, ...) {
qx - qnorm(ppoints(length(x)))[rank(x)]
panel.xyplot(qx, x, ...)
})
Kevin
On Wed, May 27, 2009 at 4:39 PM, Kevin W kw.st...@gmail.com wrote:
Thanks to Deepayan, I have a corrected version of how to color points in a
qqmath plot according to another variable. Using the barley data for a more
concise example::
qqmath(~ yield | variety, data = barley, groups=year,
auto.key=TRUE,
prepanel = function(x, ...) {
list(xlim = range(qnorm(ppoints(length(x)
},
panel = function(x, ...) {
xx - qnorm(ppoints(length(x)))[order(x)]
panel.xyplot(x = xx, y = x, ...)
})
The example I posted previously (and shown below) is not correct. My
apologies.
Kevin
On Wed, May 27, 2009 at 11:05 AM, Kevin W kw.st...@gmail.com wrote:
Having solved this problem, I am posting this so that the next time I
search for how to do this I will find an answer...
Using qqmath(..., groups=num) creates a separate qq distribution for each
group (within a panel). Using the 'col' or 'pch' argument does not
(usually) work because panel.qqmath sorts the data (but not 'col' or 'pch')
before plotting. Sorting the data before calling qqmath will ensure that
the sorting does not change the order of the data.
For example, to obtain one distribution per voice part and color the point
by part 1 or part 2:
library(lattice)
singer - singer
singer - singer[order(singer$height),]
singer$part - factor(sapply(strsplit(as.character(singer$voice.part),
split = ), [, 1),
levels = c(Bass, Tenor, Alto, Soprano))
singer$num - factor(sapply(strsplit(as.character(singer$voice.part),
split = ), [, 2))
qqmath(~ height | part, data = singer,
col=singer$num,
layout=c(4,1))
Kevin
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[R] plotting time series with data gap using type line- but do not want to connect gap with line
I have a time series of about 1500 measurements. There are sporadic data gaps. I would like to plot the time series with type line. I only want a line to connect the periods with data; I don't want a line to connect the points across a data gap. Is there a function or recommended method to deal with this? For example there are 2000 days days 1:1000 each have one measurement of temperature days 1001: 1199 have no data days 1200:2000 each have one measurement of temperature I want the plot to connect points 1:1000 with a line, and to connect points 1200: 2000 with a different line. I don't want a line to connect points 1001:1200. I searched the help archives and google but couldn't find anything. Any advice? JK [[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] plotting time series with data gap using type line- but do not want to connect gap with line
Just place a point with an NA value between the two segments. Here is one way to do it in zoo: set.seed(123) library(zoo) # create sample data tt - c(1:1000, 1200:2000) z - zoo(rnorm(length(tt)), tt) # this will fill in omitted values with NAs z - as.zoo(as.ts(z)) plot(z) On Thu, May 28, 2009 at 3:05 PM, josef.kar...@phila.gov wrote: I have a time series of about 1500 measurements. There are sporadic data gaps. I would like to plot the time series with type line. I only want a line to connect the periods with data; I don't want a line to connect the points across a data gap. Is there a function or recommended method to deal with this? For example there are 2000 days days 1:1000 each have one measurement of temperature days 1001: 1199 have no data days 1200:2000 each have one measurement of temperature I want the plot to connect points 1:1000 with a line, and to connect points 1200: 2000 with a different line. I don't want a line to connect points 1001:1200. I searched the help archives and google but couldn't find anything. Any advice? JK [[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] installing gRain package on ubuntu 8.10
Dear all I am trying to install the gRain package on ubuntu 8.10, running R Vers. 2.9. Unfortunately, I get a bad Exit-Status. What could cause that problem? Any help is appreciated. Many thanks in advance. A. Biedermann install.packages(gRain) Warnung in install.packages(gRain) : Argument 'lib' fehlt: nutze '/home/user1/R/i486-pc-linux-gnu-library/2.9' versuche URL 'http://cran.ch.r-project.org/src/contrib/gRain_0.8.0.tar.gz' Content type 'application/x-gzip' length 133482 bytes (130 Kb) URL geöffnet == downloaded 130 Kb * Installing *source* package ‘gRain’ ... ** R ** demo ** inst ** preparing package for lazy loading Error in library(pkg, character.only = TRUE, logical.return = TRUE, lib.loc = lib.loc) : 'gRbase' is not a valid installed package Fehler: lazy loading failed für Paket ‘gRain’ * Removing ‘/home/user1/R/i486-pc-linux-gnu-library/2.9/gRain’ Die heruntergeladenen Pakete sind in ‘/tmp/RtmpgVRO21/downloaded_packages’ Warning message: In install.packages(gRain) : Installation des Pakets 'gRain' hatte Exit-Status ungleich 0 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 add 'X' before numeric colnames when read.table
Hi all, Is there any way to keep numeric colnames as is? Any hint will be appreicated. Xin Zheng [[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] String replacement in an expression
Caroline Bazzoli wrote: Dear R-experts, I need to replace in an expression the character Cl by Cl+beta But in the following case: form-expression((Cl-(V *ka) ) +(V *Vm *exp(-(Clm/Vm) *t))) gsub(Cl,(Cl+beta),as.character(form)) We obtain: [1] ((Cl+beta) - (V * ka)) + (V * Vm * exp(-((Cl+beta)m/Vm) * t)) the character Clm has been also replaced. How could I avoid this unwanted replacement ? try '\\bCl\\b' as the pattern, which says 'match Cl as a separate word'. vQ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 add 'X' before numeric colnames when read.table
?read.table check.names=FALSE On Thu, May 28, 2009 at 3:25 PM, Zheng, Xin (NIH) [C] zheng...@mail.nih.gov wrote: Hi all, Is there any way to keep numeric colnames as is? Any hint will be appreicated. Xin Zheng [[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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to avoid add 'X' before numeric colnames when read.table
This really depends on where the 'X' is coming from, which you did not tell us (see the posting guide). For example, if the 'data.frame' function is the one adding the 'X', then you can use the 'check.names' argument to prevent the addition (or you can use data.frame with the argument to explicitly create your own data frame with the names you want, then pass that to the function that is doing the conversion currently). If a different function is adding the 'X', then look at the help page for that function, or give us more to go on. Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of Zheng, Xin (NIH) [C] Sent: Thursday, May 28, 2009 1:26 PM To: r-help@r-project.org Subject: [R] how to avoid add 'X' before numeric colnames when read.table Hi all, Is there any way to keep numeric colnames as is? Any hint will be appreicated. Xin Zheng [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to avoid add 'X' before numeric colnames when read.table
Thank you guys. But I also want 'check.names' to help keeping colnames unique. It seems there's no proper option for that in 'read.table'. I have to alter colnames with strsplit or other similar functions. From: jim holtman [mailto:jholt...@gmail.com] Sent: Thursday, May 28, 2009 3:40 PM To: Zheng, Xin (NIH) [C] Cc: r-help@r-project.org Subject: Re: [R] how to avoid add 'X' before numeric colnames when read.table ?read.table check.names=FALSE On Thu, May 28, 2009 at 3:25 PM, Zheng, Xin (NIH) [C] zheng...@mail.nih.govmailto:zheng...@mail.nih.gov wrote: Hi all, Is there any way to keep numeric colnames as is? Any hint will be appreicated. Xin Zheng [[alternative HTML version deleted]] __ R-help@r-project.orgmailto:R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How this addition works?
I have following addition : 1:2 + 1:10 [1] 2 4 4 6 6 8 8 10 10 12 I could not understand how R adding those two unequal vector? Any help? [[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 add 'X' before numeric colnames when read.table
substring(header, 2,) works for the purpose perfectly. Xin Zheng [[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] filter function speed
I am trying to use a linear filter to reduce loops and thereby increase the speed of an existing program. However, while the filter function (stats package) should have reduced the looping by about 30-fold, the time to complete the program remained about the same. This surprised me, because I had made an analogous change to a Matlab version of the same program using Matlab's filter function, and that change made the program run about 9 times as fast. In R, is there another function that would be more efficient than filter in the stats package? Any advice would be appreciated, as I would hate to see Matlab win this speed battle. Relevant line from R code: evTemp=filter(lambda*tempterm,1-lambda,method=recursive) #lambda=scalar between 0 and 1 (representing learning rate in a reinforcement learning model) #tempterm=vector with from 1 to 150 elements (representing reinforcement value data) Thank you. Anthony Bishara Department of Psychology College of Charleston http://bisharaa.people.cofc.edu/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] String replacement in an expression
Bill Dunlap TIBCO Software Inc - Spotfire Division wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Wacek Kusnierczyk Sent: Thursday, May 28, 2009 12:31 PM To: Caroline Bazzoli Cc: r-help@r-project.org Subject: Re: [R] String replacement in an expression Caroline Bazzoli wrote: Dear R-experts, I need to replace in an expression the character Cl by Cl+beta But in the following case: form-expression((Cl-(V *ka) ) +(V *Vm *exp(-(Clm/Vm) *t))) gsub(Cl,(Cl+beta),as.character(form)) We obtain: [1] ((Cl+beta) - (V * ka)) + (V * Vm * exp(-((Cl+beta)m/Vm) * t)) the character Clm has been also replaced. How could I avoid this unwanted replacement ? try '\\bCl\\b' as the pattern, which says 'match Cl as a separate word'. That works in this case, but \\b idea of what a word is not same as R's idea of what a name is. E..g, \\b thinks that a period is not in a word but R thinks periods in names are fine. gsub(\\bC1\\b, (C1+beta), C1 * exp(C1.5 / C2.5)) [1] (C1+beta) * exp((C1+beta).5 / C2.5) This is one more reason to use substitute(), which directly edits an expression to produce a new one. It avoids the deparse-edit-parse cycle that can corrupt things (even if you don't do any editing). substitute(C1 * exp(C1.5 / C2.5), list(C1=Quote(C1+beta))) (C1 + beta) * exp(C1.5/C2.5) Bill Dunlap TIBCO Software Inc - Spotfire Division wdunlap tibco.com vQ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Labeling barplot bars by multiple factors
Ah, that makes sense. But now another two issues have arisen. Firstly, the error bars look like confidence intervals, and I'm pretty sure that they are but does some document verify this? I suppose I could check the code too. Secondly, I just read about how dynamite plots should be avoided. It's quite easy to turn the dynamite plots into dot plots with Inkscape, but is there an equivalent function that generates _hierarchical_ dot plots? Tom On Thu, May 28, 2009 at 12:32 PM, William Dunlap wdun...@tibco.com wrote: -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Thomas Levine Sent: Thursday, May 28, 2009 5:04 AM To: Jim Lemon Cc: r-help@r-project.org Subject: Re: [R] Labeling barplot bars by multiple factors Both of those worked, but hierobarp looked a bit easier, so I used that. The one annoying thing is that it sorts alphabetically. Tom The sorts of functions almost always order things by the order of the levels of your factors. The default ordering is alphabetical (or increasing numeric, if your factor was made from numerical data). To change the order remake the factor and supply the levels argument. E.g., to reverse the order use rev: data$someFactor - factor(data$someFactor, levels=rev(levels(data$someFactor))) Bill Dunlap TIBCO Software Inc - Spotfire Division wdunlap tibco.com On Thu, May 28, 2009 at 6:46 AM, Jim Lemon j...@bitwrit.com.au wrote: Thomas Levine wrote: I want to plot quantitative data as a function of three two-level factors. How do I group the bars on a barplot by level through labeling and spacing? Here http://www.thomaslevine.org/sample_multiple-factor_barplot.png's what I'm thinking of. Also, I'm pretty sure that I want a barplot, but there may be something better. Hi Tom, You may find that the hierobarp function in the plotrix package will do what you want. Jim [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How this addition works?
On 29/05/2009, at 8:00 AM, bogaso.christofer wrote:
I have following addition :
1:2 + 1:10
[1] 2 4 4 6 6 8 8 10 10 12
I could not understand how R adding those two unequal vector? Any
help?
Look at the help for ``+'' (?+) and look at ``Value''. There you will
see:
These operators return vectors containing the result of the
element by element operations. The elements of shorter vectors
are recycled as necessary (with a 'warning' when they are recycled
only _fractionally_). The operators are '+' for addition, '-' for
subtraction, '*' for multiplication, '/' for division and '^' for
exponentiation.
The key word is ``recycled''.
cheers,
Rolf Turner
##
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Re: [R] How this addition works?
recycling rule: repeat the shorter element as many times as necessary, all.equal(1:2 + 1:10 , rep(1:2, length=10) + 1:10) # TRUE HTH, baptiste On 28 May 2009, at 22:00, bogaso.christofer wrote: I have following addition : 1:2 + 1:10 [1] 2 4 4 6 6 8 8 10 10 12 I could not understand how R adding those two unequal vector? Any help? [[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. _ 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] How this addition works?
R recycles the shorter one to match the longer one: 1 2 3 4 5 6 7 8 9 10 + 1 2 1 2 1 2 1 2 1 2 = 2 4 4 6 6 8 8 10 10 12 R does this recycling in many cases, and it can sometimes trap the unwary. Sarah On Thu, May 28, 2009 at 4:00 PM, bogaso.christofer bogaso.christo...@gmail.com wrote: I have following addition : 1:2 + 1:10 [1] 2 4 4 6 6 8 8 10 10 12 I could not understand how R adding those two unequal vector? Any help? -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ggplot2 legend
First, your example didn't work because fish_ByMuestreo didn't build properly (deleting the first structure(list(data = bit solves this). To solve your plotting problem, note that as constructed, the Muestreo column is numeric, whereas you seem to want to treat it as a factor. Solution: convert to factor: fish_ByMuestreo$Muestreo=factor(fish_ByMuestreo$Muestreo) On Thu, May 28, 2009 at 3:47 PM, Felipe Carrillo mazatlanmex...@yahoo.com wrote: Hi: I need some help with the legend. I got 14 samples(Muestreo) and I am trying to plot a smooth line for each sample. I am able to accomplish that but the problem is that the legend only displays every other sample. How can I force the legend to show all of my Muestreos? Thanks in advance. fish_ByMuestreo - structure(list(data = structure(list(SampleDate = structure(c(3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 16L, 16L, 16L, 7L, 7L, 7L, 7L, 10L, 10L, 10L, 13L, 13L, 13L, 13L, 13L, 27L, 27L, 27L, 27L, 27L, 17L, 17L, 17L, 17L, 20L, 20L, 20L, 20L, 24L, 24L, 24L, 24L, 24L, 30L, 30L, 30L, 30L, 42L, 42L, 42L, 42L, 42L, 33L, 33L, 33L, 33L, 36L, 36L, 36L, 36L, 36L, 39L, 39L, 39L, 39L, 39L, 39L, 14L, 14L, 14L, 14L, 14L, 14L, 5L, 5L, 5L, 8L, 8L, 8L, 11L, 11L, 11L, 11L, 15L, 15L, 15L, 28L, 28L, 28L, 28L, 28L, 18L, 18L, 18L, 18L, 18L, 22L, 22L, 22L, 22L, 25L, 25L, 25L, 25L, 40L, 40L, 40L, 40L, 40L, 31L, 31L, 31L, 31L, 37L, 37L, 37L, 37L, 1L, 1L, 1L, 1L, 1L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 16L, 16L, 16L, 7L, 7L, 7L, 7L, 10L, 10L, 10L, 13L, 13L, 13L, 13L, 13L, 27L, 27L, 27L, 27L, 27L, 17L, 17L, 17L, 17L, 20L, 20L, 20L, 20L, 24L, 24L, 24L, 24L, 24L, 30L, 30L, 30L, 30L, 42L, 42L, 42L, 42L, 42L, 33L, 33L, 33L, 33L, 36L, 36L, 36L, 36L, 36L, 14L, 14L, 14L, 14L, 14L, 14L, 5L, 5L, 5L, 8L, 8L, 8L, 11L, 11L, 11L, 11L, 15L, 15L, 15L, 28L, 28L, 28L, 28L, 28L, 18L, 18L, 18L, 18L, 18L, 22L, 22L, 22L, 22L, 25L, 25L, 25L, 25L, 40L, 40L, 40L, 40L, 40L, 31L, 31L, 31L, 31L, 34L, 34L, 34L, 34L, 34L, 37L, 37L, 37L, 37L, 1L, 1L, 1L, 1L, 1L, 6L, 6L, 6L, 6L, 6L, 6L, 9L, 9L, 9L, 12L, 12L, 12L, 21L, 21L, 21L, 21L, 29L, 29L, 29L, 19L, 19L, 19L, 19L, 19L, 23L, 23L, 23L, 23L, 23L, 26L, 26L, 26L, 26L, 41L, 41L, 41L, 41L, 32L, 32L, 32L, 32L, 32L, 35L, 35L, 35L, 35L, 38L, 38L, 38L, 38L, 38L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 16L, 16L, 16L, 7L, 7L, 7L, 7L, 10L, 10L, 10L, 13L, 13L, 13L, 13L, 13L, 27L, 27L, 27L, 27L, 27L, 17L, 17L, 17L, 17L, 20L, 20L, 20L, 20L, 24L, 24L, 24L, 24L, 24L, 30L, 30L, 30L, 30L, 42L, 42L, 42L, 42L, 42L, 33L, 33L, 33L, 33L, 36L, 36L, 36L, 36L, 36L, 39L, 39L, 39L, 39L, 39L, 39L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 16L, 16L, 16L, 7L, 7L, 7L, 7L, 10L, 10L, 10L, 13L, 13L, 13L, 13L, 13L, 27L, 27L, 27L, 27L, 27L, 17L, 17L, 17L, 17L, 20L, 20L, 20L, 20L, 24L, 24L, 24L, 24L, 24L, 30L, 30L, 30L, 30L, 42L, 42L, 42L, 42L, 42L, 33L, 33L, 33L, 33L, 36L, 36L, 36L, 36L, 36L, 39L, 39L, 39L, 39L, 39L, 39L), .Label = c(10/2/2002, 10/4/2002, 6/23/2002, 6/30/2002, 7/10/2002, 7/12/2002, 7/14/2002, 7/17/2002, 7/19/2002, 7/21/2002, 7/24/2002, 7/26/2002, 7/28/2002, 7/3/2002, 7/31/2002, 7/7/2002, 8/11/2002, 8/14/2002, 8/16/2002, 8/18/2002, 8/2/2002, 8/21/2002, 8/23/2002, 8/25/2002, 8/28/2002, 8/30/2002, 8/4/2002, 8/7/2002, 8/9/2002, 9/1/2002, 9/11/2002, 9/13/2002, 9/15/2002, 9/18/2002, 9/20/2002, 9/22/2002, 9/25/2002, 9/27/2002, 9/29/2002, 9/4/2002, 9/6/2002, 9/8/2002), class = factor), PondName = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L,
Re: [R] How this addition works?
bogaso.christofer wrote: I have following addition : 1:2 + 1:10 [1] 2 4 4 6 6 8 8 10 10 12 I could not understand how R adding those two unequal vector? Any help? [[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. Hi, R recycles the shorter vector to yield: 1+1 = 2 2+2 = 4 3+1 = 4 4+2 = 6 5+1 = 6 6+2 = 8 and so on. Is this what you wanted to know? Cheers, -- *Luc Villandré* /Biostatistician McGill University Health Center - Montreal Children's Hospital Research Institute/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] max.col weirdness
Hi, I think there's some rounding issue with returning the max column. (running 2.9.0 on an Apple, but my buddy found it on his PC) x - matrix(c(1234.568,1234.569,1234.567),1) max.col(x) [1] 2 x - matrix(c(12345.568,12345.569,12345.567),1) max.col(x) [1] 3 x - matrix(c(112345.568,112345.569,112345.567),1) max.col(x) [1] 3 max.col(-x) [1] 1 Thanks, Daryl __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ggplot2 legend
Thanks for your help Mike, it works like a charm now!! --- On Thu, 5/28/09, Mike Lawrence mike.lawre...@dal.ca wrote: From: Mike Lawrence mike.lawre...@dal.ca Subject: Re: [R] ggplot2 legend To: Felipe Carrillo mazatlanmex...@yahoo.com Cc: r-h...@stat.math.ethz.ch Date: Thursday, May 28, 2009, 1:06 PM First, your example didn't work because fish_ByMuestreo didn't build properly (deleting the first structure(list(data = bit solves this). To solve your plotting problem, note that as constructed, the Muestreo column is numeric, whereas you seem to want to treat it as a factor. Solution: convert to factor: fish_ByMuestreo$Muestreo=factor(fish_ByMuestreo$Muestreo) On Thu, May 28, 2009 at 3:47 PM, Felipe Carrillo mazatlanmex...@yahoo.com wrote: Hi: I need some help with the legend. I got 14 samples(Muestreo) and I am trying to plot a smooth line for each sample. I am able to accomplish that but the problem is that the legend only displays every other sample. How can I force the legend to show all of my Muestreos? Thanks in advance. library(ggplot2) fishplot - qplot(PondName,BodyWeight.g.,data=fish_ByMuestreo,colour=Muestreo,position=jitter)+stat_summary(aes(group=Muestreo),fun.data=mean_cl_normal, colour=green,geom=smooth,fill=NA)+ opts(title=Average weight(grs) by Pond) print(fishplot) Felipe D. Carrillo Supervisory Fishery Biologist Department of the Interior US Fish Wildlife Service California, USA Mike Lawrence Graduate Student Department of Psychology Dalhousie 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] sample unique pairs from a matrix
Dear All,
Many thanks for all your useful suggestions. Much appreciated.
Jos
2009/5/28 David Winsemius dwinsem...@comcast.net:
Your last step will either be a single number (not really a sampling
operation) or a
non-positive number. So at best you really only have an number that depends
entirely on the prior sequence of draws.
--
David.
On May 28, 2009, at 8:21 AM, jos matejus wrote:
Dear Ritchie and David,
Thanks very much for your advice. I had thought of this potential
solution, however it doesn't really fullfill my second criteria which
is that once a particular cell has been sampled, the row and column of
that cell can't be sampled from subsequently. In other words, the next
sample would be taken from a 5x5 matrix, and then a 4x4 matrix and so
on until I have my 6 values.
I will keep thinking!
Cheers
Jos
2009/5/28 David Winsemius dwinsem...@comcast.net:
On May 28, 2009, at 6:33 AM, jos matejus wrote:
Dear R users,
I have a matrix of both negative and positive values that I would like
to randomly sample with the following 2 conditions:
1. only sample positive values
2. once a cell in the matrix has been sampled the row and column of
that cell cannot be sampled from again.
#some dummy data
set.seed(101)
dataf - matrix(rnorm(36,1,2), nrow=6)
I can do this quite simply if all the values are positive by using the
sample function without replacement on the column and row indices.
samrow - sample(6,replace=F)
samcol - sample(6,replace=F)
values - numeric(6)
for(i in 1:6){
values[i] - dataf[samrow[i], samcol[i]]
}
However, I am not sure how to include the logical condition to only
include postitive values
Any help would be gratefully received.
Jos
M - matrix(rnorm(36),nrow=6)
M[M[,]0]
[1] 1.65619781 0.56182830 0.23812890 0.81493915 1.01279243 1.29188874
0.64252343 0.53748655 0.31503112
[10] 0.37245358 0.07942883 0.56834586 0.62200056 0.39478167 0.02374574
0.04974857 0.56219171 0.52901658
sample(M[M[,]0],6,replace=F)
[1] 0.56834586 0.07942883 0.31503112 0.62200056 0.02374574 0.64252343
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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Re: [R] max.col weirdness
Hi, Sorry, I was too hasty. I see that the tolerance is in the documentation (and that random is the default tie-breaker). Would it be possible to allow tolerance to be a parameter? thanks, Daryl Daryl Morris wrote: Hi, I think there's some rounding issue with returning the max column. (running 2.9.0 on an Apple, but my buddy found it on his PC) x - matrix(c(1234.568,1234.569,1234.567),1) max.col(x) [1] 2 x - matrix(c(12345.568,12345.569,12345.567),1) max.col(x) [1] 3 x - matrix(c(112345.568,112345.569,112345.567),1) max.col(x) [1] 3 max.col(-x) [1] 1 Thanks, Daryl __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] max.col weirdness
Try reading the man page, which says: Details When ties.method = random, as per default, ties are broken at random. In this case, the determination of a tie assumes that the entries are probabilities: there is a relative tolerance of 1e-5, relative to the largest (in magnitude, omitting infinity) entry in the row. Bert Gunter Genentech Nonclinical Biostatistics -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Daryl Morris Sent: Thursday, May 28, 2009 1:47 PM To: r-help@r-project.org Subject: [R] max.col weirdness Hi, I think there's some rounding issue with returning the max column. (running 2.9.0 on an Apple, but my buddy found it on his PC) x - matrix(c(1234.568,1234.569,1234.567),1) max.col(x) [1] 2 x - matrix(c(12345.568,12345.569,12345.567),1) max.col(x) [1] 3 x - matrix(c(112345.568,112345.569,112345.567),1) max.col(x) [1] 3 max.col(-x) [1] 1 Thanks, Daryl __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] custom sort?
Sounds simple but haven't been able to find it in docs: is it possible to sort a vector using a user-defined comparison function? Seems it must be, but sort doesn't seem to provide that option, nor does order sfaics -- View this message in context: http://www.nabble.com/custom-sort--tp23770565p23770565.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] why warning appears when set both 'scale' and 'breaks' meanwhile in heatmap.2 of gplots
The warning I met with is attached here: Warning message: In heatmap.2(as.matrix(val.sf.ns[-1:-3]), Rowv = F, Colv = T, dendrogram = column, : Using scale=row or scale=column when breaks arespecified can produce unpredictable results.Please consider using only one or the other. What does 'unpredictable results' mean? Could anyone explain that? Thanks a lot. Xin Zheng [[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] avoid a loop
Hello, R users. I have the following code: a=1:10 b=-3:15 n=5 x - rep(0,n) for (i in 1:n) x[i] - sum( outer(a,b, function(s,t) abs(a-b-i)==0) ) Can someone tell me if I could avoid the for command? Thank you in advance. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
