[R] prediction interval for multiple future observations
Hi! '?predict.lm' says that the prediction intervals returned by predict() are for single observation only. Is there a way to specify the desired number of observations to construct the interval for? R version 2.4.1 (2006-12-18) -- Vlad Skvortsov, [EMAIL PROTECTED], http://vss.73rus.com __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] prediction interval for multiple future observations
Vlad Skvortsov wrote: '?predict.lm' says that the prediction intervals returned by predict() are for single observation only. Is there a way to specify the desired number of observations to construct the interval for? You can generate the desired sequence of new values using seq or sequence and submit them to predict.lm in the newdata argument. -- View this message in context: http://www.nabble.com/prediction-interval-for-multiple-future-observations-tf4303243.html#a12249044 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] rv package, rvnorm function
John P. Burkett burkett at uri.edu writes: In an attempt to learn to use the rv package, I have been working through the examples in Jouni Kerman and Andrew Gelman's Using Random Variables to Manipulate and Summarize Simulations in R (July 4, 2007). Perhaps Jouni can help you? __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] prediction interval for multiple future observations
On Mon, 20 Aug 2007, Vlad Skvortsov wrote: Hi! '?predict.lm' says that the prediction intervals returned by predict() are for single observation only. Is there a way to specify the desired number of observations to construct the interval for? What it says in full is The prediction intervals are for a single observation at each case in 'newdata' (or by default, the data used for the fit) with error variance(s) 'pred.var'. I think you misunderstand: predict.lm returns a prediction interval for each row of 'newdata'. The comment in part means that those intervals are to be considered individually, and not as a joint prediction region for all the future observations. If you want, say, a prediction interval for the average of 10 indepedent observations at a case, use 'pred.var' to specify the error variance. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 mailinglist ETH website now with trusted certificate
Some of you, notably I think users of MS Internet Explorer, may be happy to learn that since yesterday, the web server of the R mailing lists https://stat.ethz.ch/ now runs an ``official'' / trusted certificate as opposed to the inofficial one (Math deparment ETH) that we have had for years instead. In particular, this should make access to the (first but by far not only) mailing list archives more convenient to you, e.g. for this month, for R-help, https://stat.ethz.ch/pipermail/r-help/2007-August/thread.html In parallel, the R Foundation is getting (buying) certificates for several R-project.org servers, notably also the subversion (R source) server, and these will hopefully be put in place as well with the next few weeks. Martin Maechler, ETH Zurich R core __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Ansari vs wilcoxon vs ks test
Hi all, This is a statistics question, I hope someone out there will be able to help me. I have one population (oligonucleotide probes spotted on a Nimblegen array). I measured parameter one (intensity after hybridisation) and I have selected a subpopulation of the initial (one tenth of the initial) according to a threshold value. I now measure parameter two of the original population (GC content). The distribution of the population for this value is roughly bell-shaped. I want to see if the subpopulation I selected in the previous step shares the same characteristics with regards to the GC content with the entire population or if selecting for parameter one has messed with parameter two. What I thought was to compare the distributions of this second attribute of the two populations. I believe that the ansari-bradley, wilcoxon and Kolmogorov-Smirnov tests perform such tests but -after searching- I am not sure which is more appropriate (if any). I realize that ansari-bradley and ks are more sensitive to the actual shape of the curve while wilcoxon focuses on testing for a shift of the median. I can not figure out though what is the difference between ansari-bradley and ks . Is there any important difference in the assumptions of these three tests that I should consider before choosing? Finally, and I apologize for the naivity of the question, all the ks.test(), wilcox.test () and ansari.test() expect the raw measurements for the populations and I do not need to pre-process in any way, right? ANY suggestion please? Niki [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Partial comparison in string vector
Hi! seq(along=x) %in% grep(e,x) Steve Powell-4 wrote: I have a vector of strings x=c(w,ex,ee) And I want to get a logical vector showing the positions where my search string e matches the elements partially, i.e. is at least the left-hand part of the target strings, i.e. I want to get a vector FALSE TRUE TRUE. -- View this message in context: http://www.nabble.com/Partial-comparison-in-string-vector-tf4304145.html#a12251593 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Regulatory Computing with R
Dear all - Cody and Bert have some amusing points. The problems with R that Cody states are no different than those that any organization has with any programming work. Period. We've mostly solved them through appropriate approaches, addressing through quality management, some of the issues raised by Cody with respect to 3-rd party packages, etc. Quality Management != Absolute Quality. It's about risk management, as David's presentation of our work at UseR hopefully explained. It's about common sense. Combined, this can result in reasonable statements like (real names used in artificial examples to make a point): Martin M does X with R, I trust Martin, therefore I trust X done with R because the risk of wrong results in Y will have a low impact. If Y happened to have an impact of $500m, then a reasonable approach might be to reconsider and find an additional expert, say Doug B, to confirm. Alternatively, if you don't believe in expert opinions (or subjective probability, or mechanistic modeling), and feel like an empirical frequentist, you might just get a team of monkeys to verify that X in R seems correct, based on a project management strategy that incorporates someone's favorite IT risk mitigation approach. With respect to Bert's points about 21CFR Part 11, please read the documents on the R WWW with respect to such things for a pretty informed opinion as to what is really happening. I may not speak for Novartis, but it is possible that we'll be using a non-commercial version of R at some point in the future and we've been looking into the risk management strategies. Some people are annoyed at the packages we will not let people use, but code review suggests that we really don't want people to use them (risk management, again). The supporting infrastructure will be nice, but it'll also be a PITA to build. But it really is just a matter of codification of common sense -- you should always put anything that you want to reproduce under version control, you should always have test cases to confirm that the implementations that you are using work in a few average cases (no one can cover every corner case) and you should make sure that you align your data and computer code with your reporting workflow. It's the implementation of common sense that seems to be hard, as most R-help readers should be aware of by now. I can't claim to implement it all the time, as readers of this list are probably additionally aware. Best regards / Mit freundlichen Grüssen, Anthony (Tony) Rossini Novartis Pharma AG MODELING SIMULATION Group Head a.i., -- EU Statistical Modeling CHBS, WSJ-027.1.012 Novartis Pharma AG Lichtstrasse 35 CH-4056 Basel Switzerland Phone: +41 61 324 4186 Fax: +41 61 324 3039 Cell: +41 79 367 4557 (to send an SMS from Lotus Notes put the following: [EMAIL PROTECTED] in the To box : - only the content of the subject is sent) Email : [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Vectors in R (WAS Re: Does anyone.... worth a warning?!? No warning at all)
Dear Ted (and community), You raise a very interesting point - namely, what should and should not be called a vector in R (it's neither a class or mode, formally). I don't know which version of the R Language Definition you were quoting from, but mine (Version 2.5.1 DRAFT), says: Vectors can be thought of as contiguous cells containing data. (doesn't say homogeneous in the version that I have). In that sense it's more analagous to 'lists' in Python, Scheme, etc. (with the additional benefit that the names attribute for R vectors allows you to use them also as 'dictionaries' or 'hash tables'), and less like the 1-D array used in mathematics. (Incidentally, the array class in Python is like the matrix and array classes in R, which do require specification of row or column). In any case, the quote above is more consistent with my understanding of the basic data objects in R, as atomic vectors and lists are both contiguous cells containing data, only that they differ in the value of their mode attributes. I think it can be a bit confusing when they are introduced separately (e.g., in the R Language Definition document with headings, Vectors and Lists in section 2.1) - though I think its origin lies in the pedagogy of the language. For instance, introductory documents often show off R as a calculator and draw the analogy between the vector notation used in mathematics and the application of +() [as an operator rather than a function] on a pair of numeric vectors in R. This is probably due to the background of the audience these documents are intended to address (Python/Scheme, perhaps more computer science; R/S, more statistics or mathematics perhaps). I think this is a bit unfortunate as students can get stuck with the idea that there are (atomic) vectors, and then another thing called a list - and then later he/she is told that a list is a vector as well, and has to reconcile this new bit of information - while conceptually they are similar except that a certain set of functions (e.g., the arithemetic operators and string functions) cannot be applied to vectors of mode list, but many other functions (e.g., extraction, subsetting, replacement) can be applied in the same way. This article was very elucidating: Statistical programming with R, Part 3: Reusable and object-oriented programming http://www.ibm.com/developerworks/linux/library/l-r3.html In it, David Mertz says: 'The main thing to keep in mind about R data is that everything is a vector. Even objects that look superficially distinct from vectors -- matrices, arrays, data.frames, etc. -- are really just vectors with extra (mutable) attributes that tell [generic functions in] R to treat them in special ways.' So matrices, arrays, lists, data frames, (and even factors) are all vectors (used henceforth in the sense of contiguous cells as are lists in Python/Scheme), with additional attributes attached. When these attributes are removed, print() will allow us to view them to us as 1-D objects (a sequence of values; not necessarily a 1-D row or column matrix). One defining attribute besides mode and length is the class attribute, which determines the dispatch method for a generic function. For instance, the [() and [-() functions allow N-D subscripting notation for matrix, array, and data.frame classes, but as they are also still vectors (contiguous cells), and therefore can be subscripted as stated, cells are accessed through indexing operations such as x[5]. This is important in it that it allows one to use many functions not immediately thought of as applicable to data frames (which is a list, which is a vector, etc. http://tolstoy.newcastle.edu.au/R/help/00b/2390.html); for me that would be functions like append(), replace(), etc. For example: df - data.frame(a=1:5,c=11:15,d=16:20) append(df,list(b=6:10),1) $a [1] 1 2 3 4 5 $b [1] 6 7 8 9 10 $c [1] 11 12 13 14 15 $d [1] 16 17 18 19 20 replace(df,c(FALSE,TRUE,FALSE),list(b=21:25)) a c d 1 1 21 16 2 2 22 17 3 3 23 18 4 4 24 19 5 5 25 20 append() returns a list because c() is invoked internally, and this removes all extra attributes except names (including class, row.names, etc.). So, retaining the intrinsic mode list, the append function returns a class list object by default ['If the object does not have a class attribute, it has an implicit class, matrix, array or the result of mode(x)', says ?class] when applied to a data frame. On the other hand, replace() still returns a data frame because only [.-data.frame() is invoked so the returned object retains the class of data.frame. Even factors, which fails the is.vector() test, are actually vectors (IMHO). The R Language definition says, Factors are currently implemented using an integer array [which is a vector] to specify the actual levels and a second array of names [in the levels attribute] that are mapped to the integers. As an example, the following behavior is also predictable in that if we know how each function
[R] Stacked Bar
Hi R Users! Thanks in advance. I am using R-2.5.1 on Windows XP. I am trying to do a stacked bar plot, but could not get through the following problem. The code is given below. 1. How can I provide 15 different colors for each method with 15 Rows? 2. How can I put the legend in a particular position (eg., in the top or bottom or right or left)? How can I put legend using a number of rows (eg., using two or three rows)? Once again thank you very much for your time. Regards, Debabrata (Deb) Statistician NSW Department of CommerceSydney, Australia. The Code: library(lattice) library(graphics) x - matrix(1:75, ncol= 5) dimnames(x)[[2]] - paste(Method, 1:5, sep=) dimnames(x)[[1]] - paste(Row, 1:15, sep=) # library: graphics barplot(x, beside=FALSE, col= 1:nrow(x), legend= rownames(x) ) # library: lattice barchart(Freq ~ Var2, data = as.data.frame.table(x), groups = Var1, stack = TRUE, auto.key = TRUE) - Park yourself in front of a world of choices in alternative vehicles. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Ask for functions to obtain partial R-square (squared partial correlation coefficients)
Many thanks to Frank. The anova.Design function is advanced and very powerful. Sometimes, I want to get the PCORR2 and SCORR2, any functions available now in R packages? Frank E Harrell Jr wrote: Ye Xingwang wrote: The partial R-square (or coefficient of partial determination, or squared partial correlation coefficients) measures the marginal contribution of one explanatory variable when all others are already included in multiple linear regression model. The following link has very clear explanations on partial and semi-partial correlation: http://www.psy.jhu.edu/~ashelton/courses/stats315/week2.pdf In SAS, the options is PCORR2 and SCORR2. For example(from http://www.ats.ucla.edu/stat/sas/examples/alsm/alsmsasch7.htm) data ch7tab01; input X1 X2 X3 Y; label x1 = 'Triceps' x2 = 'Thigh cir.' x3 = 'Midarm cir.' y = 'body fat'; cards; 19.5 43.1 29.1 11.9 24.7 49.8 28.2 22.8 30.7 51.9 37.0 18.7 29.8 54.3 31.1 20.1 19.1 42.2 30.9 12.9 25.6 53.9 23.7 21.7 31.4 58.5 27.6 27.1 27.9 52.1 30.6 25.4 22.1 49.9 23.2 21.3 25.5 53.5 24.8 19.3 31.1 56.6 30.0 25.4 30.4 56.7 28.3 27.2 18.7 46.5 23.0 11.7 19.7 44.2 28.6 17.8 14.6 42.7 21.3 12.8 29.5 54.4 30.1 23.9 27.7 55.3 25.7 22.6 30.2 58.6 24.6 25.4 22.7 48.2 27.1 14.8 25.2 51.0 27.5 21.1 ; run; proc reg data = ch7tab01; model y = x1 x2 / pcorr2 SCORR2; model y = x1-x3 / pcorr2 SCORR2; run; quit; There has been a post in http://tolstoy.newcastle.edu.au/R/help/05/03/0437.html It will be great appreciated if someone could write a general function to work with class lm or glm to obtain the pcorr2 (squared partial correlation coefficients using Type II sums of squares) and scorr2 (squared semi-partial correlation coefficients using Type II sums of squares) for all independent variables (3 variables) simultaneously? Thank you. Xingwang Ye library(Design) # requires Hmisc f - ols(y ~ x1 + x2) p - plot(anova(f), what='partial R2') p The anova.Design function called above handles pooling related degrees of freedom and pools main effects with related interaction effects to get the total partial effect. Frank __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Stacked Bar
Deb Midya wrote: Hi R Users! Thanks in advance. I am using R-2.5.1 on Windows XP. I am trying to do a stacked bar plot, but could not get through the following problem. The code is given below. 1. How can I provide 15 different colors for each method with 15 Rows? 2. How can I put the legend in a particular position (eg., in the top or bottom or right or left)? How can I put legend using a number of rows (eg., using two or three rows)? Hi Deb, As you have probably noticed, the integer coded colors repeat too quickly for the number of colors you want. You can use the rainbow() function to generate colors like this: barplot(x,beside=FALSE,col=rainbow(nrow(x))) or there are lots of other color generating functions in the grDevices or plotrix packages. Here's how to get your legend in an empty space for your plot. There is also an emptyspace() function in the plotrix package that tries to find the biggest empty space in a plot, although it probably wouldn't work in this case. legend(0,1000,rownames(x),fill=rainbow(nrow(x))) Jim __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] runing .r file from C#
Hi, I know that the general subject calling R from C has been discused but I have been reading the manuals and also scouting the lists and I can not seam to find a working solution for my problem. I want to call a R script ( let's call it test.r ) from within C# code. After reading about this topic I am trying to do this : System.Diagnostics.Process proc = new System.Diagnostics.Process(); proc.StartInfo.FileName = E:/R/R-2.5.1 /bin/Rterm.exe; proc.StartInfo.Arguments = 'test.r' --no-save; proc.StartInfo.UseShellExecute = false; proc.StartInfo.RedirectStandardOutput = false; proc.Start(); bun when Rterm starts it shows parameter test.r ignored When I try to do the same from a command line shell it DOES work just fine : Rterm.exe test.r --no-save runs the file without any problems. Do you have any idea how to make it not to ignore the input file? Or is there other way to just execute a .r file from C# code? Thank you very much, Alex -- Alexandru Maruseac BEST Bucharest E-mail: [EMAIL PROTECTED] Tel:+40 0722 329 083 www.BEST.eu.org www.BESTBc.pub.ro [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Stacked Bar
Jim, Thanks for such a quick response. It works well. Is it possible to fill the bars with patterns and colours? Regards, Deb Jim Lemon [EMAIL PROTECTED] wrote: Deb Midya wrote: Hi R Users! Thanks in advance. I am using R-2.5.1 on Windows XP. I am trying to do a stacked bar plot, but could not get through the following problem. The code is given below. 1. How can I provide 15 different colors for each method with 15 Rows? 2. How can I put the legend in a particular position (eg., in the top or bottom or right or left)? How can I put legend using a number of rows (eg., using two or three rows)? Hi Deb, As you have probably noticed, the integer coded colors repeat too quickly for the number of colors you want. You can use the rainbow() function to generate colors like this: barplot(x,beside=FALSE,col=rainbow(nrow(x))) or there are lots of other color generating functions in the grDevices or plotrix packages. Here's how to get your legend in an empty space for your plot. There is also an emptyspace() function in the plotrix package that tries to find the biggest empty space in a plot, although it probably wouldn't work in this case. legend(0,1000,rownames(x),fill=rainbow(nrow(x))) Jim - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] library(fCalendar) timeDate(12.03.2005, format=%d.%m.%Y)
Thanks! Seems to work fine now! Regards, Ola Martin Becker wrote: Dear Ola, I think you spotted a small bug in *package* fCalendar. Explicit specification should prevent autodetection of the date format, which is not the case for fCalendar v251.70, instead autodetection is done at least once (twice, if actually appropriate). With the following patch, things should work ok: diff --recursive fCalendar.orig/R/3A-TimeDateClass.R fCalendar/R/3A-TimeDateClass.R 433c433 charvec = format(strptime(charvec, .whichFormat(charvec)), isoFormat) --- charvec = format(strptime(charvec, format), isoFormat) You did not provide the output of sessionInfo() (which you are asked for in the posting guide). If you are using Windows and don't know how to apply the patch, you can download a patched binary version here: http://www.saar-gate.net/download/fCalendar_251.70.zip Regards, Martin PS: Maybe r-sig-finance is more appropriate for questions concerning Rmetrics. Ola Lindqvist wrote: Dear R users, I have problem with the library fCalendar. I am not using the US standard format notations. It seems like it is not possible to have different format than the US standards. Anyone how knows a way to go around this problem? Here is the code I enter: myDate = 12.03.2005 timeDate(myDate, format = %d.%m.%Y) And I get following error message: Error in if (sum(lt$sec + lt$min + lt$hour) == 0) isoFormat = %Y-%m-%d : missing value where TRUE/FALSE needed Thanks, Ola __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Stacked Bar
I think you want to use the 'density' argument. For example: barplot(1:5,col=1) legend(topleft,fill=1,legend=text,cex=1.2) par(new=TRUE) barplot(1:5,density=5,col=2) legend(topleft,fill=2,density=20,legend=text,bty=n,cex=1.2) (if you wanted to overlay solid colors with hatching) Here's the lattice alternative of the bar graph, though the help page says 'density' is currently unimplemented (Package lattice version 0.16-2). To get the legend into columns, I followed the suggestion described here: http://tolstoy.newcastle.edu.au/R/help/05/04/2529.html Essentially I use mapply() and the line following to create a list with alternating 'text' and 'rect' arguments (3 times to get 3 columns). === x - matrix(1:75, ncol= 5) dimnames(x)[[2]] - paste(Method, 1:5, sep=) dimnames(x)[[1]] - paste(Row, 1:15, sep=) u - mapply(function(x,y) list(text=list(lab=x),rect=list(col=y)), x = as.data.frame(matrix(levels(as.data.frame.table(x)$Var1), ncol=3)), y = as.data.frame(matrix(rainbow(nrow(x)), ncol=3)), SIMPLIFY=FALSE) key - c(rep=FALSE,space=bottom,unlist(names-(u,NULL),rec=FALSE)) barchart(Freq ~ Var2, data = as.data.frame.table(x), groups = Var1, stack = TRUE, col=rainbow(nrow(x)),density=5, key = key ) === (I often use tim.colors() in the 'fields' package, if you wanted other ideas for color schemes). --- Deb Midya [EMAIL PROTECTED] wrote: Jim, Thanks for such a quick response. It works well. Is it possible to fill the bars with patterns and colours? Regards, Deb Jim Lemon [EMAIL PROTECTED] wrote: Deb Midya wrote: Hi R Users! Thanks in advance. I am using R-2.5.1 on Windows XP. I am trying to do a stacked bar plot, but could not get through the following problem. The code is given below. 1. How can I provide 15 different colors for each method with 15 Rows? 2. How can I put the legend in a particular position (eg., in the top or bottom or right or left)? How can I put legend using a number of rows (eg., using two or three rows)? Hi Deb, As you have probably noticed, the integer coded colors repeat too quickly for the number of colors you want. You can use the rainbow() function to generate colors like this: barplot(x,beside=FALSE,col=rainbow(nrow(x))) or there are lots of other color generating functions in the grDevices or plotrix packages. Here's how to get your legend in an empty space for your plot. There is also an emptyspace() function in the plotrix package that tries to find the biggest empty space in a plot, although it probably wouldn't work in this case. legend(0,1000,rownames(x),fill=rainbow(nrow(x))) Jim - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Comedy with an Edge to see what's on, when. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] runing .r file from C#
On Tue, 21 Aug 2007, Alex MD wrote: Hi, I know that the general subject calling R from C has been discused but I have been reading the manuals and also scouting the lists and I can not seam to find a working solution for my problem. It's a C# issue. I want to call a R script ( let's call it test.r ) from within C# code. After reading about this topic I am trying to do this : System.Diagnostics.Process proc = new System.Diagnostics.Process(); proc.StartInfo.FileName = E:/R/R-2.5.1 /bin/Rterm.exe; proc.StartInfo.Arguments = 'test.r' --no-save; proc.StartInfo.UseShellExecute = false; proc.StartInfo.RedirectStandardOutput = false; proc.Start(); bun when Rterm starts it shows parameter test.r ignored When I try to do the same from a command line shell it DOES work just fine : Rterm.exe test.r --no-save runs the file without any problems. Do you have any idea how to make it not to ignore the input file? Or is there other way to just execute a .r file from C# code? You need a shell for redirection ( |) to work, and 'system' commands in Windows do not usually use one (as in C, C++, R, Perl): you seem to have turned off using a shell in C#. However, I think you should be using RScript.exe, where this is not an issue. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] runing .r file from C#
Thank you very much for your help. It did work. So here is the code that I've used case anybody else is seaching for an answer to this problem : System.Diagnostics.Process proc = new System.Diagnostics.Process(); proc.StartInfo.FileName = E:/R/bin/Rscript.exe; proc.StartInfo.WorkingDirectory = E:/R/bin/; proc.StartInfo.Arguments = E:/R/bin/test.r; proc.StartInfo.UseShellExecute = true; proc.StartInfo.RedirectStandardOutput = false; proc.Start(); Thanks again for your help, good luck. Alex On 8/21/07, Prof Brian Ripley [EMAIL PROTECTED] wrote: On Tue, 21 Aug 2007, Alex MD wrote: Hi, I know that the general subject calling R from C has been discused but I have been reading the manuals and also scouting the lists and I can not seam to find a working solution for my problem. It's a C# issue. I want to call a R script ( let's call it test.r ) from within C# code. After reading about this topic I am trying to do this : System.Diagnostics.Process proc = new System.Diagnostics.Process(); proc.StartInfo.FileName = E:/R/R-2.5.1 /bin/Rterm.exe; proc.StartInfo.Arguments = 'test.r' --no-save; proc.StartInfo.UseShellExecute = false; proc.StartInfo.RedirectStandardOutput = false; proc.Start(); bun when Rterm starts it shows parameter test.r ignored When I try to do the same from a command line shell it DOES work just fine : Rterm.exe test.r --no-save runs the file without any problems. Do you have any idea how to make it not to ignore the input file? Or is there other way to just execute a .r file from C# code? You need a shell for redirection ( |) to work, and 'system' commands in Windows do not usually use one (as in C, C++, R, Perl): you seem to have turned off using a shell in C#. However, I think you should be using RScript.exe, where this is not an issue. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 -- Alexandru Maruseac BEST Bucharest E-mail: [EMAIL PROTECTED] Tel:+40 0722 329 083 www.BEST.eu.org www.BESTBc.pub.ro [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] standardized cronbach's alpha?
Hi list members Any easy way to get standardized cronbach's alpha for a scale, as in SPSS? Thanks Steve Powell proMENTE social research research | evaluation | training consulting Kranjčevićeva 35, 71000 Sarajevo mobile: +387 61 215 997 | office: +387 33 556 865 | fax: +387 33 556 866 Checked by AVG Free Edition. 17:44 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] library(fCalendar) timeDate(12.03.2005, format=%d.%m.%Y)
OL == Ola Lindqvist [EMAIL PROTECTED] on Tue, 21 Aug 2007 14:32:19 +0200 writes: OL Thanks! OL Seems to work fine now! Well, for your example. But sorry to say, the patch breaks other cases. I'm investigating further (and will hopefully contribute to a new CRAN release of fCalendar once Diethelm Wuertz is back from wherever; I've already made more changes) Martin Maechler, ETH Zurich [but different department than D.Wuertz] OL Martin Becker wrote: Dear Ola, I think you spotted a small bug in *package* fCalendar. Explicit specification should prevent autodetection of the date format, which is not the case for fCalendar v251.70, instead autodetection is done at least once (twice, if actually appropriate). With the following patch, things should work ok: diff --recursive fCalendar.orig/R/3A-TimeDateClass.R fCalendar/R/3A-TimeDateClass.R 433c433 charvec = format(strptime(charvec, .whichFormat(charvec)), isoFormat) --- charvec = format(strptime(charvec, format), isoFormat) You did not provide the output of sessionInfo() (which you are asked for in the posting guide). If you are using Windows and don't know how to apply the patch, you can download a patched binary version here: http://www.saar-gate.net/download/fCalendar_251.70.zip Regards, Martin PS: Maybe r-sig-finance is more appropriate for questions concerning Rmetrics. Ola Lindqvist wrote: Dear R users, I have problem with the library fCalendar. I am not using the US standard format notations. It seems like it is not possible to have different format than the US standards. Anyone how knows a way to go around this problem? Here is the code I enter: myDate = 12.03.2005 timeDate(myDate, format = %d.%m.%Y) And I get following error message: Error in if (sum(lt$sec + lt$min + lt$hour) == 0) isoFormat = %Y-%m-%d : missing value where TRUE/FALSE needed Thanks, Ola __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. OL __ OL R-help@stat.math.ethz.ch mailing list OL https://stat.ethz.ch/mailman/listinfo/r-help OL PLEASE do read the posting guide http://www.R-project.org/posting-guide.html OL and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Variable c and function c
I have found the error in my script which was semi-automatically translated from the other person's MATLAB code. The error is that c was assigned a value inside a function. That is the function body contained the following instructions c-nw*czr d-nw*cz rFren-0.5*(abs((cz-c)/(cz+c))^2+abs((d-czr)/(d+czr))^2) firstguess-c( 0,0,0,3,0.5, 0 , 0 , 0.01) I have already run this function and obtained the results, it was rather long process, therefore I don't want to rerun it. I was not given any warnings. How did the interpreter treat this? Will the result change, if I change the variable from c to, say, c. ? This variable is not used anywhere else in the function. -- View this message in context: http://www.nabble.com/Variable-c-and-function-c-tf4305781.html#a12256590 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] standardized cronbach's alpha?
look at: help(cronbach.alpha, package = ltm) Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm - Original Message - From: Steve Powell [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent: Tuesday, August 21, 2007 4:02 PM Subject: [R] standardized cronbach's alpha? Hi list members Any easy way to get standardized cronbach's alpha for a scale, as in SPSS? Thanks Steve Powell proMENTE social research research | evaluation | training consulting Kranjčevićeva 35, 71000 Sarajevo mobile: +387 61 215 997 | office: +387 33 556 865 | fax: +387 33 556 866 Checked by AVG Free Edition. 17:44 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Subsetting zoo object with a vector of time values.
I have a zoo object for which I would like to subset using a vector of time values. For example, I have the following time values represented in my zoo object. -50.000 -49.996 -49.995 -49.960 -49.956 -49.955 -49.920 -49.916 -49.915 -49.880 and would like to get observations corresponding to times -50 -49.96 -49.92 -49.88. What can I do without using the lapply or which functions? Thank you. Todd Remund __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Odp: Variable c and function c
Hi
[EMAIL PROTECTED] napsal dne 21.08.2007 17:07:05:
I have found the error in my script which was semi-automatically
translated
from the other person's MATLAB code.
The error is that c was assigned a value inside a function.
That is the function body contained the following instructions
c-nw*czr
d-nw*cz
rFren-0.5*(abs((cz-c)/(cz+c))^2+abs((d-czr)/(d+czr))^2)
My guess is that rFren is computed with c defined above and
firstguess-c( 0,0,0,3,0.5, 0 , 0 , 0.01)
firstguess is assigned above numbers
See
fff-function(nw=10, czr=5, cz=3) {
c-nw*czr
d-nw*cz
rFren-0.5*(abs((cz-c)/(cz+c))^2+abs((d-czr)/(d+czr))^2)
firstguess-c( 0,0,0,3,0.5, 0 , 0 , 0.01)
list(rFren, firstguess)
}
fff()
[[1]]
[1] 0.6483025
[[2]]
[1] 0e+00 0e+00 0e+00 3e+00 5e-01 0e+00 0e+00 1e-06
fff(1)
[[1]]
[1] 0.0625
[[2]]
[1] 0e+00 0e+00 0e+00 3e+00 5e-01 0e+00 0e+00 1e-06
Regards
Petr
I have already run this function and obtained the results, it was rather
long process, therefore I don't want to rerun it.
I was not given any warnings.
How did the interpreter treat this?
Will the result change, if I change the variable from c to, say, c.
?
This variable is not used anywhere else in the function.
--
View this message in context:
http://www.nabble.com/Variable-c-and-function-c-
tf4305781.html#a12256590
Sent from the R help mailing list archive at Nabble.com.
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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Re: [R] Subsetting zoo object with a vector of time values.
On Tue, 21 Aug 2007, Todd Remund wrote: I have a zoo object for which I would like to subset using a vector of time values. For example, I have the following time values represented in my zoo object. -50.000 -49.996 -49.995 -49.960 -49.956 -49.955 -49.920 -49.916 -49.915 -49.880 and would like to get observations corresponding to times -50 -49.96 -49.92 -49.88. What can I do without using the lapply or which functions? Use the window() function: z - zoo(1:10, c(-50.000, -49.996, -49.995, -49.960, -49.956, -49.955, -49.920, -49.916, -49.915, -49.880)) window(z, c(-50, -49.96, -49.92, -49.88)) hth, Z Thank you. Todd Remund __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Subsetting zoo object with a vector of time values.
See ?window.zoo e.g. library(zoo) # create test data tt - c(-50, -49.996, -49.995, -49.96, -49.956, -49.955, -49.92, -49.916, -49.915, -49.88) z - zoo(seq_along(tt), tt) window(z, c(-50, -49.96, -49.92, -49.88)) On 8/21/07, Todd Remund [EMAIL PROTECTED] wrote: I have a zoo object for which I would like to subset using a vector of time values. For example, I have the following time values represented in my zoo object. -50.000 -49.996 -49.995 -49.960 -49.956 -49.955 -49.920 -49.916 -49.915 -49.880 and would like to get observations corresponding to times -50 -49.96 -49.92 -49.88. What can I do without using the lapply or which functions? Thank you. Todd Remund __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Selective transformation
I am looking for a way to transform select observations based on a value based criteria. Why - Am learning r and would like to perform regression analysis of given variables of the babies dataset (part of UsingR) for example babies$wt1, the data in the variables does contain values which should be interpreted as unknown, some variables have 999 for unknown and some have 99 for the same, since lm() expects not available data to be marked using NA. I would like to use a solution that does not employ loops (I think it may not be the ideal way) I am looking at using apply() and supply the name of my function responsible for transformation, but am unable to know now to reference the element of the vector/list being currently processed by apply() so I may do in place substitution (if value is 99 or 999) of the value with NA. Pinpoint customers who are looking for what you sell. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] extracting month from date in numeric form
Hi, Anyone knows what would be a short way of extracting a month from a date in numeric or integer format? months(1979-12-20) returns December in character format. How could I get 12 in numeric or integer format? Thanks! G. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Variable c and function c
On 8/21/2007 11:07 AM, Vladimir Eremeev wrote: I have found the error in my script which was semi-automatically translated from the other person's MATLAB code. The error is that c was assigned a value inside a function. That is the function body contained the following instructions c-nw*czr d-nw*cz rFren-0.5*(abs((cz-c)/(cz+c))^2+abs((d-czr)/(d+czr))^2) firstguess-c( 0,0,0,3,0.5, 0 , 0 , 0.01) I have already run this function and obtained the results, it was rather long process, therefore I don't want to rerun it. I was not given any warnings. How did the interpreter treat this? Will the result change, if I change the variable from c to, say, c. ? This variable is not used anywhere else in the function. When looking for a function, R will ignore variables that are not functions. So what you show above is not an error, though it can be confusing, so it's not a good idea. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] extracting month from date in numeric form
Hi, perhaps: format.Date(as.Date(1979-12-20), %m) -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O On 21/08/07, Gonçalo Ferraz [EMAIL PROTECTED] wrote: Hi, Anyone knows what would be a short way of extracting a month from a date in numeric or integer format? months(1979-12-20) returns December in character format. How could I get 12 in numeric or integer format? Thanks! G. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] extracting month from date in numeric form
On 8/21/07, Gonçalo Ferraz [EMAIL PROTECTED] wrote: Hi, Anyone knows what would be a short way of extracting a month from a date in numeric or integer format? months(1979-12-20) returns December in character format. How could I get 12 in numeric or integer format? Here are a few solutions: format(as.Date(1979-12-20), %m) as.POSIXlt(as.Date(1979-12-20))$mo + 1 as.numeric(substring(1979-12-20, 6, 7)) as.numeric(factor(months(as.Date(1979-12-20), abbrev = TRUE), levels = month.abb)) See R News 4/1 Help Desk article for more on dates. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Optimization problem
Hello Folks, Very new to R so bear with me, running 5.2 on XP. Trying to do a zero-inflated negative binomial regression on placental scar data as dependent. Lactation, location, number of tick larvae present and mass of mouse are independents. Dataframe and attributes below: Location Lac Scars Lar Mass Lacfac 1 Tullychurry 0 0 15 13.87 0 2 Somerset 0 0 0 15.60 0 3 Tollymore 0 0 3 16.43 0 4 Tollymore 0 0 0 16.55 0 5 Caledon 0 0 0 17.47 0 6 Hillsborough 1 5 0 18.18 1 7 Caledon 0 0 1 19.06 0 8 Portglenone 0 4 0 19.10 0 9 Portglenone 0 5 0 19.13 0 10Tollymore 0 5 3 19.50 0 11 Hillsborough 1 5 0 19.58 1 12 Portglenone 0 4 0 19.76 0 13 Caledon 0 8 0 19.97 0 14 Hillsborough 1 4 0 20.02 1 15 Tullychurry 0 3 3 20.13 0 16 Hillsborough 1 5 0 20.18 1 17 LoughNavar 1 5 0 20.20 1 18Tollymore 0 0 1 20.24 0 19 Hillsborough 1 5 0 20.48 1 20 Caledon 0 4 1 20.56 0 21 Caledon 0 3 2 20.58 0 22Tollymore 0 4 3 20.58 0 23Tollymore 0 0 2 20.88 0 24 Hillsborough 1 0 0 21.01 1 25 Portglenone 0 5 0 21.08 0 26 Tullychurry 0 2 5 21.28 0 27 Ballysallagh 1 4 0 21.59 1 28 Caledon 0 0 1 21.68 0 29 Hillsborough 1 5 0 22.09 1 30 Tullychurry 0 5 5 22.28 0 31 Tullychurry 1 6 75 22.43 1 32 Ballysallagh 1 5 0 22.57 1 33 Ballysallagh 1 4 0 22.67 1 34 LoughNavar 1 5 3 22.71 1 35 Hillsborough 1 4 0 23.01 1 36 Caledon 0 0 3 23.08 0 37 LoughNavar 1 5 0 23.53 1 38 Ballysallagh 1 4 0 23.55 1 39 Portglenone 1 6 0 23.61 1 40 Mt.Stewart 0 3 0 23.70 0 41 Somerset 0 5 0 23.83 0 42 Ballysallagh 1 5 0 23.93 1 43 Ballysallagh 1 5 0 24.01 1 44 Caledon 0 0 3 24.14 0 45 LoughNavar 0 6 0 24.30 0 46 LoughNavar 1 5 0 24.34 1 47 Hillsborough 1 4 0 24.45 1 48 Caledon 0 3 2 24.55 0 49 Tullychurry 0 5 44 24.83 0 50 Hillsborough 1 5 0 24.86 1 51 Ballysallagh 1 5 0 25.02 1 52 Tullychurry 0 0 9 25.27 0 53 Mt.Stewart 0 5 0 25.31 0 54 LoughNavar 1 4 8 25.43 1 55 Somerset 1 0 0 25.58 1 56 Hillsborough 1 5 0 25.82 1 57 Portglenone 1 2 0 26.02 1 58 Ballysallagh 1 5 0 26.19 1 59 Mt.Stewart 1 0 0 26.66 1 60 Randalstown 1 0 1 26.70 1 61 Somerset 0 4 0 27.01 0 62 Mt.Stewart 0 4 0 27.05 0 63 Somerset 0 3 0 27.10 0 64 Somerset 0 6 0 27.34 0 65 Somerset 0 0 0 27.87 0 66 LoughNavar 1 5 1 28.01 1 67 Tullychurry 1 6 42 28.55 1 68 Hillsborough 1 5 0 28.84 1 69 Portglenone 1 4 0 29.00 1 70 Somerset 1 4 0 31.87 1 71 Ballysallagh 1 5 0 33.06 1 72 LoughNavar 1 4 0 33.24 1 73 Somerset 1 4 0 33.36 1 alan : 'data.frame':73 obs. of 6 variables: $ Location: Factor w/ 10 levels Ballysallagh,..: 10 8 9 9 2 3 2 6 6 9 ... $ Lac : int 0 0 0 0 0 1 0 0 0 0 ... $ Scars : int 0 0 0 0 0 5 0 4 5 5 ... $ Lar : int 15 0 3 0 0 0 1 0 0 3 ... $ Mass: num 13.9 15.6 16.4 16.6 17.5 ... $ Lacfac : Factor w/ 2 levels 0,1: 1 1 1 1 1 2 1 1 1 1 ... The syntax I used to create the model is: zinb.zc - zicounts(resp=Scars~.,x =~Location + Lar + Mass + Lar:Mass + Location:Mass,z =~Location + Lar + Mass + Lar:Mass + Location:Mass, data=alan) The error given is: Error in optim(par = parm, fn = neg.like, gr = neg.grad, hessian = TRUE, : non-finite value supplied by optim In addition: Warning message: fitted probabilities numerically 0 or 1 occurred in: glm.fit(zz, 1 - pmin(y, 1), family = binomial()) I understand this is a problem with the model I specified, could anyone help out?? Many thanks Alan Harrison Quercus Queen's University Belfast MBC, 97 Lisburn Road Belfast BT9 7BL T: 02890 972219 M: 07798615682 [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Formatting Sweave in R-News
Hi,
I am editing a document for submission to the R-news newsletter, and
in my article my Sweave code inserts a dynamically generated PDF
report that my R program generates.
However, when I insert the PDF using the following Sweave code:
\newpage
\includegraphics[scale=1.0]{\Sexpr{print(location)}}
\newpage
(in tex this looks like):
\newpage
\includegraphics[scale=1.0]{/home/arjun/sample.pdf}
\newpage
However, the r-news style package over-rides everything that I can set
(including using the minipage option) to make my included PDF small
sized. Part of the problem is that the R-news style specifies a
two-column formatting, and so the PDF is shrunk to fit in one column.
How can I, for just one page, over-ride the styles to include the PDF?
Even if I hard-hack the graphics to be scaled up in size, that does
not get rid of the vertical line that in between the two columns, and
thus breaking my image.
I realise that this is not an R problem, but more a latex problem, but
I am hoping that somebody has faced similar problems with the Rnews
styles and has an idea on how to do this.
Thank you,
Yours sincerely,
--
Arjun Ravi Narayan
[EMAIL PROTECTED]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Selective transformation
Allan Kamau wrote: I am looking for a way to transform select observations based on a value based criteria. Why - Am learning r and would like to perform regression analysis of given variables of the babies dataset (part of UsingR) for example babies$wt1, the data in the variables does contain values which should be interpreted as unknown, some variables have 999 for unknown and some have 99 for the same, since lm() expects not available data to be marked using NA. I would like to use a solution that does not employ loops (I think it may not be the ideal way) I am looking at using apply() and supply the name of my function responsible for transformation, but am unable to know now to reference the element of the vector/list being currently processed by apply() so I may do in place substitution (if value is 99 or 999) of the value with NA. Does this do what you want? babies$wt1 - with(babies, replace(wt1, wt1 == 999, NA)) ?replace Pinpoint customers who are looking for what you sell. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Chuck Cleland, Ph.D. NDRI, Inc. 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] divided scatter plots
I have a data set which contains two columns. The first column is a list of countries, and the second column contains their political risk ratings. I would like to create one large plot that contains 5 different sections, each with a scatter plot. To clarify, I have divided the countries into 5 groups. For each group (continent), I would like to have the name of the continent on the x-axis, and points representing countries and their risk rating on the y-axis. However, I want all 5 scatter plots to be in one large plot. What function should I use to do this? Also, is it possible to label each point? thanks for any help! __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Selective transformation
- Original Message From: Chuck Cleland [EMAIL PROTECTED] To: Allan Kamau [EMAIL PROTECTED] Cc: r-help@stat.math.ethz.ch Sent: Tuesday, August 21, 2007 6:48:53 PM Subject: Re: [R] Selective transformation Allan Kamau wrote: I am looking for a way to transform select observations based on a value based criteria. Why - Am learning r and would like to perform regression analysis of given variables of the babies dataset (part of UsingR) for example babies$wt1, the data in the variables does contain values which should be interpreted as unknown, some variables have 999 for unknown and some have 99 for the same, since lm() expects not available data to be marked using NA. I would like to use a solution that does not employ loops (I think it may not be the ideal way) I am looking at using apply() and supply the name of my function responsible for transformation, but am unable to know now to reference the element of the vector/list being currently processed by apply() so I may do in place substitution (if value is 99 or 999) of the value with NA. Does this do what you want? babies$wt1 - with(babies, replace(wt1, wt1 == 999, NA)) ?replace Thanks Chuck, the replace command is just what I was looking for. wt1-babies$wt1 wt1-replace(wt1,wt1==999,NA) I get NA in wt1 vector in place of 999 Pinpoint customers who are looking for what you sell. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Chuck Cleland, Ph.D. NDRI, Inc. 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 Luggage? GPS? Comic books? __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] pvals.fnc unhappy about lmer objects
Dear folks (or Dear Professor Bates), I'm quite confused as to the current status of some of the available functions applicable to lmer objects. Following the examples in Baayen, Davidson, Bates (2006), my plan is to run mcmcsamp on a random effect model created by lmer in package lme4, then use the (perhaps outdated) pvals to estimate p-value. But then I couldn't find pvals anywhere. So question number one is : Has pvals been replaced? by pvals.fnc in the package languageR perhaps? From the help page, it's stated that pvals.fnc takes a model fitted with lmer, in contrast with the pvals I've read about in Baayen et al, which takes an mcmc object. So I then tried, hr.lmer = lmer(hr ~ tg + spl + (1+tg|M), data=dat) pvals.fnc(hr.lmer) An error is reported, Error in get(x, envir, mode, inherits) : variable Gen.2001 of mode function was not found First I suspected it must be my model, so I run through the example in pvals.fnc help page, as follow, data(primingHeid) primingHeid = primingHeid[primingHeid$RT 7.1,] primingHeid = primingHeid[primingHeid$RT 7.1,] primingHeid.lmer = lmer(RT ~ RTtoPrime * ResponseToPrime + + Condition + (1|Subject) + (1|Word), data = primingHeid) primingHeid.pvals = pvals.fnc(primingHeid.lmer) The same error is encountered, Error in get(x, envir, mode, inherits) : variable Gen.2001 of mode function was not found Thanks in advance. Horace W. Tso *- PS: I'm on Windows XP, with R2.5.1 (2007-06-27) sessionInfo() R version 2.5.1 (2007-06-27) i386-pc-mingw32 locale: LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252 attached base packages: [1] splines graphics grDevices datasets utils stats methods base other attached packages: languageRrpart MASS Design survival Hmisce1071class cluster 0.2 3.1-37 7.2-34 2.1-1 2.32 3.4-2 1.5-16 7.2-34 1.11.7 zipfR coda xlsReadWrite lme4 Matrix lattice R.oo zoo 0.6-0 0.12-1 1.3.2 0.99875-7 0.999375-1 0.15-11 1.2.7 1.2-2 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] small issue with densityplot
Hi folks, This is really minor but to someone not familiar with the various tentacles of the lmer package it could be really annoying. I was trying to plot the posterior density of the fixed effect parameters of a lmer model, hr.mcmc = mcmcsamp(hr.lmer, n=5) densityplot(hr.mcmc, plot.points=F) There is this error, Error in densityplot(hr.mcmc, plot.points = F) : no applicable method for densityplot It kind of smells like something I've come across before. So I checked the mcmcsamp help page, and alas, the example suggests that the package coda is needed. From the help page of densityplot alone, there is no way one could figure out this dependency. It says, together with histogram, it is part of lattice. Could the function author *please* make clarification in future editions of lattice. Thanks. Horace W. Tso __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] small issue with densityplot
On 8/21/07, Horace Tso [EMAIL PROTECTED] wrote: Hi folks, This is really minor but to someone not familiar with the various tentacles of the lmer package it could be really annoying. I was trying to plot the posterior density of the fixed effect parameters of a lmer model, hr.mcmc = mcmcsamp(hr.lmer, n=5) densityplot(hr.mcmc, plot.points=F) There is this error, Error in densityplot(hr.mcmc, plot.points = F) : no applicable method for densityplot It kind of smells like something I've come across before. So I checked the mcmcsamp help page, and alas, the example suggests that the package coda is needed. From the help page of densityplot alone, there is no way one could figure out this dependency. It says, together with histogram, it is part of lattice. Could the function author *please* make clarification in future editions of lattice. There is nothing to clarify. densityplot() is a generic function, and it is not possible for the author of the generic function to anticipate and document all possible methods, especially those in other packages. I would say that since you are using mcmcsamp(), it's perfectly reasonable to expect you to look at its help page to figure out what you can do with the results. What gave you the idea that densityplot would work on the result of mcmcsamp in the first place? -Deepayan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Stacked Bar
On 8/21/07, Stephen Tucker [EMAIL PROTECTED] wrote: I think you want to use the 'density' argument. For example: barplot(1:5,col=1) legend(topleft,fill=1,legend=text,cex=1.2) par(new=TRUE) barplot(1:5,density=5,col=2) legend(topleft,fill=2,density=20,legend=text,bty=n,cex=1.2) (if you wanted to overlay solid colors with hatching) Here's the lattice alternative of the bar graph, though the help page says 'density' is currently unimplemented (Package lattice version 0.16-2). Yes, and that's unlikely to change unless grid begins to support it. To get the legend into columns, I followed the suggestion described here: http://tolstoy.newcastle.edu.au/R/help/05/04/2529.html Essentially I use mapply() and the line following to create a list with alternating 'text' and 'rect' arguments (3 times to get 3 columns). === x - matrix(1:75, ncol= 5) dimnames(x)[[2]] - paste(Method, 1:5, sep=) dimnames(x)[[1]] - paste(Row, 1:15, sep=) u - mapply(function(x,y) list(text=list(lab=x),rect=list(col=y)), x = as.data.frame(matrix(levels(as.data.frame.table(x)$Var1), ncol=3)), y = as.data.frame(matrix(rainbow(nrow(x)), ncol=3)), SIMPLIFY=FALSE) key - c(rep=FALSE,space=bottom,unlist(names-(u,NULL),rec=FALSE)) barchart(Freq ~ Var2, data = as.data.frame.table(x), groups = Var1, stack = TRUE, col=rainbow(nrow(x)),density=5, key = key ) === A more transparent solution (IMO) is something like barchart(Freq ~ Var2, data = as.data.frame.table(x), groups = Var1, stack = TRUE, par.settings = list(superpose.polygon = list(col=rainbow(nrow(x, auto.key = list(space = right, columns = 2) ) -Deepayan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] small issue with densityplot
Deepayan, you're right. Now I realize anyone could write a densityplot function to apply on a different class of objects. I guess I should write to the author of lme4 which from what I could see does not describe a densityplot. By the way, it is used in Baayen, Davidson, and Bates (2006). H. Deepayan Sarkar [EMAIL PROTECTED] 8/21/2007 11:17:39 AM On 8/21/07, Horace Tso [EMAIL PROTECTED] wrote: Hi folks, This is really minor but to someone not familiar with the various tentacles of the lmer package it could be really annoying. I was trying to plot the posterior density of the fixed effect parameters of a lmer model, hr.mcmc = mcmcsamp(hr.lmer, n=5) densityplot(hr.mcmc, plot.points=F) There is this error, Error in densityplot(hr.mcmc, plot.points = F) : no applicable method for densityplot It kind of smells like something I've come across before. So I checked the mcmcsamp help page, and alas, the example suggests that the package coda is needed. From the help page of densityplot alone, there is no way one could figure out this dependency. It says, together with histogram, it is part of lattice. Could the function author *please* make clarification in future editions of lattice. There is nothing to clarify. densityplot() is a generic function, and it is not possible for the author of the generic function to anticipate and document all possible methods, especially those in other packages. I would say that since you are using mcmcsamp(), it's perfectly reasonable to expect you to look at its help page to figure out what you can do with the results. What gave you the idea that densityplot would work on the result of mcmcsamp in the first place? -Deepayan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] clusterCall with replicate function
I am trying to run a monte carlo process using snow with a MPI cluster. I have ~thirty processors to run the algorithm on and I want to run it 5000 times and take the average of the output. A very simple way to do this is to divide 5000 by the number of processors to get a number n and tell each processor to run the algorithm n times. I realize there are more efficient ways to manage the parallelization. To implement this I used the clusterCall command with the replicate function along the lines of clusterCall(cl, replicate, n, function(args)). Because my function is a monte carlo process it relies on drawing from random distributions to generate output. When I do this, all of my processors generate the same random numbers. I copied the following from the command space for a simple example: cl-makeCluster(cl, replicate,1,runif(2)) clusterCall(cl, replicate, 2, runif(2)) [[1]] 0.65339590.6533959 0.10710510.1071051 [[2]] 0.65339590.6533959 0.10710510.1071051 This is not alleviated by using clusterApply to set a random seed for each processor and seems to be related to the use of the replicate function within clusterCall. I have rearranged the function so that replicate is used to call the clusterCall function (ie. replicate(2, clusterCall(cl, runif,2),simplify=F) ) and resolved the random number issue. However, this also involves much more communication between master and slaves and results in slower computation time. Will rsprng fix this problem? Is there a better way to do this without using replicate? I hope this is somewhat clear. Thanks, Mike __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how do i use the get function to obtain an element from a list...
my problem can be explained with the following example: x - 1:12 y - 13:24 a - data.frame(x = x, y = y) ## if i write a$x ## it returns [1] 1 2 3 4 5 6 7 8 9 10 11 12 ## but the function get doesn't recognize a$x. Instead it produces the following error: get(a$x) Error in get(x, envir, mode, inherits) : variable a$x was not found i intend to do it inside a loop, using a new object (and hence, a new name) for each iteration (i.e., instead of a$x, it would be a$1, a$2, a$3, and so on, for a million times). i would greatly appreciate it if someone could help me on this issue, thanks in advance, Juan Manuel Barreneche, Zoología de Vertebrados, Facultad de Ciencias, UDELAR, Uruguay. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how do i use the get function to obtain an element from alist...
eval(parse(text=(a$x). You can only use get when it's an object. a$x isn't.
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Juan Manuel
Barreneche
Sent: Tuesday, August 21, 2007 3:35 PM
To: r-help@stat.math.ethz.ch
Subject: [R] how do i use the get function to obtain an element from alist...
my problem can be explained with the following example:
x - 1:12
y - 13:24
a - data.frame(x = x, y = y)
## if i write
a$x
## it returns
[1] 1 2 3 4 5 6 7 8 9 10 11 12
## but the function get doesn't recognize a$x. Instead it produces the
following error:
get(a$x)
Error in get(x, envir, mode, inherits) : variable a$x was not found
i intend to do it inside a loop, using a new object (and hence, a new
name) for each iteration (i.e., instead of a$x, it would be a$1, a$2, a$3, and
so on, for a million times).
i would greatly appreciate it if someone could help me on this issue,
thanks in advance,
Juan Manuel Barreneche,
Zoología de Vertebrados,
Facultad de Ciencias,
UDELAR, Uruguay.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
This is not an offer (or solicitation of an offer) to buy/se...{{dropped}}
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] how do i use the get function to obtain an element from a list...
On 8/21/2007 3:35 PM, Juan Manuel Barreneche wrote: my problem can be explained with the following example: x - 1:12 y - 13:24 a - data.frame(x = x, y = y) ## if i write a$x ## it returns [1] 1 2 3 4 5 6 7 8 9 10 11 12 ## but the function get doesn't recognize a$x. Instead it produces the following error: get(a$x) Error in get(x, envir, mode, inherits) : variable a$x was not found a$x is an expression, which you could evaluate, not a variable, which you could get. i intend to do it inside a loop, using a new object (and hence, a new name) for each iteration (i.e., instead of a$x, it would be a$1, a$2, a$3, and so on, for a million times). i would greatly appreciate it if someone could help me on this issue, Why name things? I'd use something like for (i in 1:100) print(a[[i]]) Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Random Sampling from a Matrix
Dear Friends, I have a matrix of size 5000 X 20. The first two columns are indicator variables taking the value of either 0 or 1. Let us call the first two columns Y1 and Y2. I need to randomly sample 1000 rows with all the associated columns, in other words my new matrix should be of size 1000 X 20. I realize that using this command newmat - mainmat[sample(1000,replace=F),] achieves this. However, I would like to make sure that both Y1 and Y2 have more or less an equal amount of 0's and 1's. At present when I sample, I get cases where sometimes all my Y2's are 0. Is there any way to accomodate this problem. Thanks in advance. Regards Anup - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how do i use the get function to obtain an element from a list...
Hi, you can try this: eval(parse(text=a$x)) -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O On 21/08/07, Juan Manuel Barreneche [EMAIL PROTECTED] wrote: my problem can be explained with the following example: x - 1:12 y - 13:24 a - data.frame(x = x, y = y) ## if i write a$x ## it returns [1] 1 2 3 4 5 6 7 8 9 10 11 12 ## but the function get doesn't recognize a$x. Instead it produces the following error: get(a$x) Error in get(x, envir, mode, inherits) : variable a$x was not found i intend to do it inside a loop, using a new object (and hence, a new name) for each iteration (i.e., instead of a$x, it would be a$1, a$2, a$3, and so on, for a million times). i would greatly appreciate it if someone could help me on this issue, thanks in advance, Juan Manuel Barreneche, Zoología de Vertebrados, Facultad de Ciencias, UDELAR, Uruguay. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Optimization problem
Lac and Lacfac are the same. On 8/21/07, Alan Harrison [EMAIL PROTECTED] wrote: Hello Folks, Very new to R so bear with me, running 5.2 on XP. Trying to do a zero-inflated negative binomial regression on placental scar data as dependent. Lactation, location, number of tick larvae present and mass of mouse are independents. Dataframe and attributes below: Location Lac Scars Lar Mass Lacfac 1 Tullychurry 0 0 15 13.87 0 2 Somerset 0 0 0 15.60 0 3 Tollymore 0 0 3 16.43 0 4 Tollymore 0 0 0 16.55 0 5 Caledon 0 0 0 17.47 0 6 Hillsborough 1 5 0 18.18 1 7 Caledon 0 0 1 19.06 0 8 Portglenone 0 4 0 19.10 0 9 Portglenone 0 5 0 19.13 0 10Tollymore 0 5 3 19.50 0 11 Hillsborough 1 5 0 19.58 1 12 Portglenone 0 4 0 19.76 0 13 Caledon 0 8 0 19.97 0 14 Hillsborough 1 4 0 20.02 1 15 Tullychurry 0 3 3 20.13 0 16 Hillsborough 1 5 0 20.18 1 17 LoughNavar 1 5 0 20.20 1 18Tollymore 0 0 1 20.24 0 19 Hillsborough 1 5 0 20.48 1 20 Caledon 0 4 1 20.56 0 21 Caledon 0 3 2 20.58 0 22Tollymore 0 4 3 20.58 0 23Tollymore 0 0 2 20.88 0 24 Hillsborough 1 0 0 21.01 1 25 Portglenone 0 5 0 21.08 0 26 Tullychurry 0 2 5 21.28 0 27 Ballysallagh 1 4 0 21.59 1 28 Caledon 0 0 1 21.68 0 29 Hillsborough 1 5 0 22.09 1 30 Tullychurry 0 5 5 22.28 0 31 Tullychurry 1 6 75 22.43 1 32 Ballysallagh 1 5 0 22.57 1 33 Ballysallagh 1 4 0 22.67 1 34 LoughNavar 1 5 3 22.71 1 35 Hillsborough 1 4 0 23.01 1 36 Caledon 0 0 3 23.08 0 37 LoughNavar 1 5 0 23.53 1 38 Ballysallagh 1 4 0 23.55 1 39 Portglenone 1 6 0 23.61 1 40 Mt.Stewart 0 3 0 23.70 0 41 Somerset 0 5 0 23.83 0 42 Ballysallagh 1 5 0 23.93 1 43 Ballysallagh 1 5 0 24.01 1 44 Caledon 0 0 3 24.14 0 45 LoughNavar 0 6 0 24.30 0 46 LoughNavar 1 5 0 24.34 1 47 Hillsborough 1 4 0 24.45 1 48 Caledon 0 3 2 24.55 0 49 Tullychurry 0 5 44 24.83 0 50 Hillsborough 1 5 0 24.86 1 51 Ballysallagh 1 5 0 25.02 1 52 Tullychurry 0 0 9 25.27 0 53 Mt.Stewart 0 5 0 25.31 0 54 LoughNavar 1 4 8 25.43 1 55 Somerset 1 0 0 25.58 1 56 Hillsborough 1 5 0 25.82 1 57 Portglenone 1 2 0 26.02 1 58 Ballysallagh 1 5 0 26.19 1 59 Mt.Stewart 1 0 0 26.66 1 60 Randalstown 1 0 1 26.70 1 61 Somerset 0 4 0 27.01 0 62 Mt.Stewart 0 4 0 27.05 0 63 Somerset 0 3 0 27.10 0 64 Somerset 0 6 0 27.34 0 65 Somerset 0 0 0 27.87 0 66 LoughNavar 1 5 1 28.01 1 67 Tullychurry 1 6 42 28.55 1 68 Hillsborough 1 5 0 28.84 1 69 Portglenone 1 4 0 29.00 1 70 Somerset 1 4 0 31.87 1 71 Ballysallagh 1 5 0 33.06 1 72 LoughNavar 1 4 0 33.24 1 73 Somerset 1 4 0 33.36 1 alan : 'data.frame':73 obs. of 6 variables: $ Location: Factor w/ 10 levels Ballysallagh,..: 10 8 9 9 2 3 2 6 6 9 ... $ Lac : int 0 0 0 0 0 1 0 0 0 0 ... $ Scars : int 0 0 0 0 0 5 0 4 5 5 ... $ Lar : int 15 0 3 0 0 0 1 0 0 3 ... $ Mass: num 13.9 15.6 16.4 16.6 17.5 ... $ Lacfac : Factor w/ 2 levels 0,1: 1 1 1 1 1 2 1 1 1 1 ... The syntax I used to create the model is: zinb.zc - zicounts(resp=Scars~.,x =~Location + Lar + Mass + Lar:Mass + Location:Mass,z =~Location + Lar + Mass + Lar:Mass + Location:Mass, data=alan) The error given is: Error in optim(par = parm, fn = neg.like, gr = neg.grad, hessian = TRUE, : non-finite value supplied by optim In addition: Warning message: fitted probabilities numerically 0 or 1 occurred in: glm.fit(zz, 1 - pmin(y, 1), family = binomial()) I understand this is a problem with the model I specified, could anyone help out?? Many thanks Alan Harrison Quercus Queen's University Belfast MBC, 97 Lisburn Road Belfast BT9 7BL T: 02890 972219 M: 07798615682 [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal,
Re: [R] standardized cronbach's alpha?
You can get this using alpha() or alpha.Summary() in the MiscPsycho package. Stratified alpha coefficients are coming for the next release, BTW. -Original Message- From: [EMAIL PROTECTED] on behalf of Steve Powell Sent: Tue 8/21/2007 10:02 AM To: r-help@stat.math.ethz.ch Subject: [R] standardized cronbach's alpha? Hi list members Any easy way to get standardized cronbach's alpha for a scale, as in SPSS? Thanks Steve Powell proMENTE social research research | evaluation | training consulting Kranjceviceva 35, 71000 Sarajevo mobile: +387 61 215 997 | office: +387 33 556 865 | fax: +387 33 556 866 Checked by AVG Free Edition. 17:44 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] summing columns of data frame by group
I have a data frame and one separate vector that is a grouping variable for the data frame. I would like to take all rows of the data frame belonging to each group and then sum the columns with out using a for statement. Something like: take all rows of group 1 then apply(group1.data,1,sum), but do this without having to do it separately for each group. Any ideas? thank you. Dan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] clusterCall with replicate function
replicate uses non-standard evaluation of its expr argument so that doesn't work well with clusterCall or clusterApply. The way you are using it you will get the same answer on all nodes no patter what you do about the generators because the computations are all being done on the master. You can use clusterEvalQ, as in clusterEvalQ(cl, replicate(2, runif(2))) Or you can use something like clusterCall(cl, function(n) replicate(n, runif(2)), 2) This will still probably give identical results most of the time if you don't do something about the RNG. clusterSetupRNG will help you with that -- just install the rlecuyer package and call clusterSetupRNG. Best, luke On Tue, 21 Aug 2007, Michael Gormley wrote: I am trying to run a monte carlo process using snow with a MPI cluster. I have ~thirty processors to run the algorithm on and I want to run it 5000 times and take the average of the output. A very simple way to do this is to divide 5000 by the number of processors to get a number n and tell each processor to run the algorithm n times. I realize there are more efficient ways to manage the parallelization. To implement this I used the clusterCall command with the replicate function along the lines of clusterCall(cl, replicate, n, function(args)). Because my function is a monte carlo process it relies on drawing from random distributions to generate output. When I do this, all of my processors generate the same random numbers. I copied the following from the command space for a simple example: cl-makeCluster(cl, replicate,1,runif(2)) clusterCall(cl, replicate, 2, runif(2)) [[1]] 0.65339590.6533959 0.10710510.1071051 [[2]] 0.65339590.6533959 0.10710510.1071051 This is not alleviated by using clusterApply to set a random seed for each processor and seems to be related to the use of the replicate function within clusterCall. I have rearranged the function so that replicate is used to call the clusterCall function (ie. replicate(2, clusterCall(cl, runif,2),simplify=F) ) and resolved the random number issue. However, this also involves much more communication between master and slaves and results in slower computation time. Will rsprng fix this problem? Is there a better way to do this without using replicate? I hope this is somewhat clear. Thanks, Mike __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Luke Tierney Chair, Statistics and Actuarial Science Ralph E. Wareham Professor of Mathematical Sciences University of Iowa Phone: 319-335-3386 Department of Statistics andFax: 319-335-3017 Actuarial Science 241 Schaeffer Hall email: [EMAIL PROTECTED] Iowa City, IA 52242 WWW: http://www.stat.uiowa.edu __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] problems installing updated version of vars package
All, I was looking onlin and noticed that the vars package (by Bernhard Pfaff) was recently updated (update date listed Aug 6, 2007) The updated packages has some features that I would find very useful. I have used the update packages function and vars was one of the packages identified as needing an update. I was able to updated and it appeared to work, however when I load the package it does not seem to be the most recent version? Has anyone else had similar problems? Or does anyone have any suggestions? System Info: R 2.4.1 Windows XP install mirror: USA 3 (UCLA I think) thanks, Spencer [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] small issue with densityplot
On 8/21/07, Horace Tso [EMAIL PROTECTED] wrote: Deepayan, you're right. Now I realize anyone could write a densityplot function to apply on a different class of objects. I guess I should write to the author of lme4 which from what I could see does not describe a densityplot. By the way, it is used in Baayen, Davidson, and Bates (2006). You're right. It's my bad for not documenting that the coda package is required for plots of the results of mcmcsamp. I was trying to piggy-back on the work that the authors of the coda package had done on diagnostics, etc. It was actually Deepayan and I who added the lattice-based plots to the coda package for exactly this purpose (did you remember that, Deepayan?). In some ways it might be better to remove the dependence on coda and write the classes and methods in the lme4 package. That way I can use S4 classes and irritate all those people who rail against S4 classes and methods (and you know who you are). Deepayan Sarkar [EMAIL PROTECTED] 8/21/2007 11:17:39 AM On 8/21/07, Horace Tso [EMAIL PROTECTED] wrote: Hi folks, This is really minor but to someone not familiar with the various tentacles of the lmer package it could be really annoying. I was trying to plot the posterior density of the fixed effect parameters of a lmer model, hr.mcmc = mcmcsamp(hr.lmer, n=5) densityplot(hr.mcmc, plot.points=F) There is this error, Error in densityplot(hr.mcmc, plot.points = F) : no applicable method for densityplot It kind of smells like something I've come across before. So I checked the mcmcsamp help page, and alas, the example suggests that the package coda is needed. From the help page of densityplot alone, there is no way one could figure out this dependency. It says, together with histogram, it is part of lattice. Could the function author *please* make clarification in future editions of lattice. There is nothing to clarify. densityplot() is a generic function, and it is not possible for the author of the generic function to anticipate and document all possible methods, especially those in other packages. I would say that since you are using mcmcsamp(), it's perfectly reasonable to expect you to look at its help page to figure out what you can do with the results. What gave you the idea that densityplot would work on the result of mcmcsamp in the first place? -Deepayan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help 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 use the get function to obtain an element from a list...
One simple way that I haven't seen mentioned yet is to do: get(a)$x (which of course allows further variants such as get(a)$x[3:6] ...) -- Tony Plate Juan Manuel Barreneche wrote: my problem can be explained with the following example: x - 1:12 y - 13:24 a - data.frame(x = x, y = y) ## if i write a$x ## it returns [1] 1 2 3 4 5 6 7 8 9 10 11 12 ## but the function get doesn't recognize a$x. Instead it produces the following error: get(a$x) Error in get(x, envir, mode, inherits) : variable a$x was not found i intend to do it inside a loop, using a new object (and hence, a new name) for each iteration (i.e., instead of a$x, it would be a$1, a$2, a$3, and so on, for a million times). i would greatly appreciate it if someone could help me on this issue, thanks in advance, Juan Manuel Barreneche, Zoología de Vertebrados, Facultad de Ciencias, UDELAR, Uruguay. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] (Most efficient) way to make random sequences of random sequences
Hi, I was wondering the what would be the (most efficient) way to generate a sequence of sequences, i mean: if I have 1,2 and 3. I'd like to generate a sequence of length N*3 (N ~ 1,000,000 or more) Where random permutations of the sequence 1,2,3 follow each other. i.e 1,2,3,1,3,2,3,2,1 /!\ The thing is that there should never be twice the same number of in the same sub-sequence, meaning that this is different from generating a vector with the numbers 1,2 and 3 randomly distributed. Any suggestion very welcome! Thanks, Emmanuel __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] summing columns of data frame by group
?by perhaps --- Daniel O'Shea [EMAIL PROTECTED] wrote: I have a data frame and one separate vector that is a grouping variable for the data frame. I would like to take all rows of the data frame belonging to each group and then sum the columns with out using a for statement. Something like: take all rows of group 1 then apply(group1.data,1,sum), but do this without having to do it separately for each group. Any ideas? thank you. Dan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] summing columns of data frame by group
Here's one way,
lapply(split(DF, your.vector), function(x) {apply(x, 2, sum)})
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Daniel O'Shea
Sent: Tuesday, August 21, 2007 3:53 PM
To: r-help@stat.math.ethz.ch
Subject: [R] summing columns of data frame by group
I have a data frame and one separate vector that is a
grouping variable for the data frame. I would like to take
all rows of the data frame belonging to each group and then
sum the columns with out using a for statement.
Something like: take all rows of group 1 then
apply(group1.data,1,sum), but do this without having to do it
separately for each group. Any ideas? thank you.
Dan
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Re: [R] small issue with densityplot
On 22/08/2007, at 8:44 AM, Douglas Bates wrote:
In some ways it might be better to remove the dependence on coda and
write the classes and methods in the lme4 package. That way I can use
S4 classes and irritate all those people who rail against S4 classes
and methods (and you know who you are).
That --- i.e. causing such irritation --- would indeed seem to be
the only
purpose/justification for using the (utterly incomprehensible to the
human
mind, which is what I'm equipped with) S4 classes and methods!
cheers,
Rolf
##
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[R] quantile() returns a value outside the data range
Hello,
I am getting an unexpected result from quantile(). Specifically, the
return value falls outside the range of the data, which I wouldn't
have thought possible for a weighted average of 2 order statistics.
Is this an unintended accuracy issue or am I being too casual in my
comparison (is there some analogue of 'all.equal' for =)?
Small example:
foo - h2[,17][h2$plate==222] # some data I have, details not
interesting
foo - sort(foo)[1:3]# can demo with the 3 smallest
values
yo - data.frame(matrix(foo,nrow=length(foo),ncol=10))
fooQtile - rep(0,9)
for(i in 1:9) { # compute 0.01 quantile, for all
'types'
+ fooQtile[i] - quantile(foo,probs=0.01,type=i)
+ yo[,i+1] - foo = fooQtile[i]
+ }
names(yo) - c(myData,paste(qType,1:9,sep=))
yo
myData qType1 qType2 qType3 qType4 qType5 qType6 qType7 qType8
qType9
1 6.402611 TRUE TRUE TRUE TRUE TRUE TRUE FALSE FALSE
TRUE
2 6.402611 TRUE TRUE TRUE TRUE TRUE TRUE FALSE FALSE
TRUE
3 6.420587 FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
FALSE
fooQtile
[1] 6.40261 6.40261 6.40261 6.40261 6.40261 6.40261 6.40261 6.40261
6.40261
table(fooQtile)
fooQtile
6.40261053520674 6.40261053520674
27
I expected the returned quantile to be either equal to or greater than
the minimum observed value and that is the case for types 1-6 and
9. But for types 7 (the default) and 8, the returned quantile is less
than the minimum observed value. The difference between the type
(1-6,9) and type (7,8) return values and between the returned quantile
and the minimum is obviously very very small.
Is there any choice for 'type' that is guaranteed to return values
inside the observed range?
Thanks, Jenny
Dr. Jennifer Bryan
Assistant Professor
Department of Statistics and
the Michael Smith Laboratories
University of British Columbia
333-6356 Agricultural Road
Vancouver, BC Canada
V6T 1Z2
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Re: [R] (Most efficient) way to make random sequences of random sequences
One way: N - 10 s - c(apply(matrix(rep(1:3,N),3,N),2,sample)) url:www.econ.uiuc.edu/~rogerRoger Koenker email[EMAIL PROTECTED]Department of Economics vox: 217-333-4558University of Illinois fax: 217-244-6678Champaign, IL 61820 On Aug 21, 2007, at 3:49 PM, Emmanuel Levy wrote: Hi, I was wondering the what would be the (most efficient) way to generate a sequence of sequences, i mean: if I have 1,2 and 3. I'd like to generate a sequence of length N*3 (N ~ 1,000,000 or more) Where random permutations of the sequence 1,2,3 follow each other. i.e 1,2,3,1,3,2,3,2,1 /!\ The thing is that there should never be twice the same number of in the same sub-sequence, meaning that this is different from generating a vector with the numbers 1,2 and 3 randomly distributed. Any suggestion very welcome! Thanks, Emmanuel __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] (Most efficient) way to make random sequences of random sequences
On 22/08/2007, at 8:49 AM, Emmanuel Levy wrote:
Hi,
I was wondering the what would be the (most efficient) way to generate
a sequence
of sequences, i mean:
if I have 1,2 and 3.
I'd like to generate a sequence of length N*3 (N ~ 1,000,000 or more)
Where random permutations of the sequence 1,2,3 follow each other.
i.e 1,2,3,1,3,2,3,2,1
/!\ The thing is that there should never be twice the same number of
in the same sub-sequence, meaning that this is different from
generating a vector with the numbers 1,2 and 3 randomly distributed.
require(gtools)
m - permutations(3,3)
M - m[sample(1:6,100,TRUE),]
x - as.vector(t(M))
##
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] small issue with densityplot
Thank you Prof. Bates and Deepayan. As the title of this thread suggests, it is a minor issue once the user is familar with where to look for hints. But better documentation certainly won't hurt. H. Douglas Bates [EMAIL PROTECTED] 8/21/2007 1:44:11 PM On 8/21/07, Horace Tso [EMAIL PROTECTED] wrote: Deepayan, you're right. Now I realize anyone could write a densityplot function to apply on a different class of objects. I guess I should write to the author of lme4 which from what I could see does not describe a densityplot. By the way, it is used in Baayen, Davidson, and Bates (2006). You're right. It's my bad for not documenting that the coda package is required for plots of the results of mcmcsamp. I was trying to piggy-back on the work that the authors of the coda package had done on diagnostics, etc. It was actually Deepayan and I who added the lattice-based plots to the coda package for exactly this purpose (did you remember that, Deepayan?). In some ways it might be better to remove the dependence on coda and write the classes and methods in the lme4 package. That way I can use S4 classes and irritate all those people who rail against S4 classes and methods (and you know who you are). Deepayan Sarkar [EMAIL PROTECTED] 8/21/2007 11:17:39 AM On 8/21/07, Horace Tso [EMAIL PROTECTED] wrote: Hi folks, This is really minor but to someone not familiar with the various tentacles of the lmer package it could be really annoying. I was trying to plot the posterior density of the fixed effect parameters of a lmer model, hr.mcmc = mcmcsamp(hr.lmer, n=5) densityplot(hr.mcmc, plot.points=F) There is this error, Error in densityplot(hr.mcmc, plot.points = F) : no applicable method for densityplot It kind of smells like something I've come across before. So I checked the mcmcsamp help page, and alas, the example suggests that the package coda is needed. From the help page of densityplot alone, there is no way one could figure out this dependency. It says, together with histogram, it is part of lattice. Could the function author *please* make clarification in future editions of lattice. There is nothing to clarify. densityplot() is a generic function, and it is not possible for the author of the generic function to anticipate and document all possible methods, especially those in other packages. I would say that since you are using mcmcsamp(), it's perfectly reasonable to expect you to look at its help page to figure out what you can do with the results. What gave you the idea that densityplot would work on the result of mcmcsamp in the first place? -Deepayan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] (Most efficient) way to make random sequences of random sequences
Similarly:
s - c(replicate(N, sample(3)))
Andy
From: roger koenker
One way:
N - 10
s - c(apply(matrix(rep(1:3,N),3,N),2,sample))
url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
vox: 217-333-4558University of Illinois
fax: 217-244-6678Champaign, IL 61820
On Aug 21, 2007, at 3:49 PM, Emmanuel Levy wrote:
Hi,
I was wondering the what would be the (most efficient) way
to generate
a sequence
of sequences, i mean:
if I have 1,2 and 3.
I'd like to generate a sequence of length N*3 (N ~
1,000,000 or more)
Where random permutations of the sequence 1,2,3 follow each other.
i.e 1,2,3,1,3,2,3,2,1
/!\ The thing is that there should never be twice the same number of
in the same sub-sequence, meaning that this is different from
generating a vector with the numbers 1,2 and 3 randomly distributed.
Any suggestion very welcome! Thanks,
Emmanuel
__
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Re: [R] (Most efficient) way to make random sequences of random sequences
Thanks for all your suggestions!
I guess I'll have to make tests to see which one is faster :)
Will come back to you when it's done.
On 8/21/07, Liaw, Andy [EMAIL PROTECTED] wrote:
Similarly:
s - c(replicate(N, sample(3)))
Andy
From: roger koenker
One way:
N - 10
s - c(apply(matrix(rep(1:3,N),3,N),2,sample))
url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
vox: 217-333-4558University of Illinois
fax: 217-244-6678Champaign, IL 61820
On Aug 21, 2007, at 3:49 PM, Emmanuel Levy wrote:
Hi,
I was wondering the what would be the (most efficient) way
to generate
a sequence
of sequences, i mean:
if I have 1,2 and 3.
I'd like to generate a sequence of length N*3 (N ~
1,000,000 or more)
Where random permutations of the sequence 1,2,3 follow each other.
i.e 1,2,3,1,3,2,3,2,1
/!\ The thing is that there should never be twice the same number of
in the same sub-sequence, meaning that this is different from
generating a vector with the numbers 1,2 and 3 randomly distributed.
Any suggestion very welcome! Thanks,
Emmanuel
__
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Re: [R] small issue with densityplot
On 8/21/07, Douglas Bates [EMAIL PROTECTED] wrote: On 8/21/07, Horace Tso [EMAIL PROTECTED] wrote: Deepayan, you're right. Now I realize anyone could write a densityplot function to apply on a different class of objects. I guess I should write to the author of lme4 which from what I could see does not describe a densityplot. By the way, it is used in Baayen, Davidson, and Bates (2006). You're right. It's my bad for not documenting that the coda package is required for plots of the results of mcmcsamp. I was trying to piggy-back on the work that the authors of the coda package had done on diagnostics, etc. It was actually Deepayan and I who added the lattice-based plots to the coda package for exactly this purpose (did you remember that, Deepayan?). I do remember that, but my point remains: (1) densityplot is a generic in lattice and documented as such. (2) mcmcsamp produces an object of class mcmc defined in coda, and this is documented in help(mcmcsamp), which even uses densityplot for its examples after attaching coda. (3) densityplot.mcmc is documented in coda. which all seems to be as it should be (except maybe ?densityplot.mcmc should additionally have an alias for densityplot). The only confusing aspect I see is that lme4 does not require or import coda, even though mcmcsamp produces objects of a class that is supposedly from coda. I realize that ownership of S3 classes is not well defined, but wouldn't that be the logical thing to do? The original problem wouldn't have come up if this were true. -Deepayan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Simulation of logistic equation
Dear all, I would like to simulate a study in which the prevalence of an event is 1% (p = 0.01), and the risk of the event is associated with two risk factors: x1 and x2. The odds ratio per standard deviation increase in x1 is 1.8 (95% CI: 1.1 to 2.8), and the odds ratio for x2 being present is 2.1 (1.8 - 3.1). x1 is a continuous variable with normal distribution (mean = 1.0, SD=0.10). x2 is a categorical variable (with two values 0 and 1), and the prevalence of x2=1 is around 10%. I have looked at the codes given in Design package, but I am not sure how to simulate the prevalence. Any help is highly appreciated. Regard, Nguyen Nguyen Dinh Nguyen Bone and Mineral Research Program Garvan Institute of Medical Research St Vincent's Hospital 384 Victoria Street, Darlinghurst Sydney, NSW 2010 Australia Tel; 61-2-9295 8274 Fax: 61-2-9295 8241 E-mail: [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Output from while and for loop
Hello,
I am new and am having a hard time getting the proper syntax for output
from loops. I am working on a simulation to generate a null expectation of
bee behavior. Pieces of it work. The part that I am having specific
difficulty is in output of a vector from within the while loop that I am
using. Basically the simulation works as such: I have a starting point
and a neighbor matrix and a certain threshold distance for travel. In the
while loop the bee moves to a randomly chosen neighbor location. I want
to be able to record the elevations of these points (including the starting
point) so that I can look at variance in elevation and mean elevation. The
loop itself works as does the calculation of the final elevation list,
change in elevation list, and true total distance traveled. I have looked
in all of the email archives but have not come across a correct way of
doing it. Code below:
start.elev.list-list()
final.mean.elev.list-list()
final.elev.list-list()
final.distance.list-list()
final.delta.elev.list-list()
final.var.elev-list()
b-length(Bees.Day.1$bee)
for (bee in 1:b){
#this is for number of bees that are trackable in the day with starting
points and threshold distances
elev.current.vector-vector(mode=numeric, length=0)
count-1
ElevSS-0
d.traveled-0
thresh-Bees.Day.1$cum.dist[bee]
n-Bees.Day.1$grid.pt[bee]
#I'm making this up for the threshold, want to be bee specific
#current.point-round(runif(1,1,n)) #random starting point
current.point-Day.1.neighbor.matrix[1,n]
#I want to specify the first point in the matrix
Elev.Sum-Day.1.elev.vector[current.point]
while(d.traveledthresh){
#which of the four options will be selection
transition-round(runif(1,1,4))
#so, what's the new point?
new.point- Day.1.neighbor.matrix[transition,n]
#what is the variance in elevation changed
Elev.current-Day.1.elev.vector[current.point]
elev.current.vector[i]-Elev.current
Elev.new-Day.1.elev.vector[new.point]
Elev.Sum-(Elev.Sum+Elev.new)
#how far will bee travelled
current.travel- Day.1.distance.matrix[current.point, new.point]
d.traveled- current.travel + d.traveled
current.point- new.point
#Number of iterations until we reach the threshold
count-count+1
}
print(count)
print(elev.current.vector)
mean.elev-Elev.Sum/count
print(paste(Final mean elev for bee, bee, is, mean.elev, sep= ))
final.mean.elev.list[bee]-list(mean.elev)
#What was the start elevation?
start.elev-Day.1.elev.vector[n]
print(paste(Start elev for bee,bee,is,start.elev, sep= ))
start.elev.list[bee]-list(start.elev)
#what is the final elevation?
final.elev-Day.1.elev.vector[current.point]
print(paste(Final elev for bee,bee,is, final.elev,sep= ))
final.elev.list[bee]-list(final.elev)
print(paste(Final travel distance for bee, bee,is, d.traveled, sep= ))
final.distance.list[bee]-list(d.traveled)
net.delta.elev-(final.elev-Day.1.elev.vector[n])
print(paste(Final net change in elevation for bee,bee,is,
net.delta.elev,sep= ))
final.delta.elev.list[bee]-list(net.delta.elev)
}
~~
Ryan D. Briscoe Runquist
Population Biology Graduate Group
University of California, Davis
[EMAIL PROTECTED]
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Re: [R] Partial comparison in string vector
Use grep(^e,x) if you are looking for the entries where e is the first character. --- Vladimir Eremeev [EMAIL PROTECTED] wrote: Hi! seq(along=x) %in% grep(e,x) Steve Powell-4 wrote: I have a vector of strings x=c(w,ex,ee) And I want to get a logical vector showing the positions where my search string e matches the elements partially, i.e. is at least the left-hand part of the target strings, i.e. I want to get a vector FALSE TRUE TRUE. -- View this message in context: http://www.nabble.com/Partial-comparison-in-string-vector-tf4304145.html#a12251593 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] rectify a program of seasonal dummies matrix
Hi friends,
I would like to construct a matrix of seasonal dummies with number of rows
(observations)=100. such matrix is written as follows:[1 0 0 0;0 1 0 0;0 0 1
0;0 0 0 1;1 0 0 0;0 1 0 0;0 0 1 0;0 0 0 1;etc...] . I wrote the following
program:
T=100
br=matrix(0,T,4)
{
for (i in 1:T)
for (j in 1:4)
if i==j
br[i,j]=1
if else (abs(i-j)%%4==0
br[i,j]=1
else
br[i,j]=0
}
z-br
z
but unfortunately I obtained from the console the following message:
{
+ for (i in 1:T)
+ for (j in 1:4)
+ (if i==j)
Erreur : syntax error, unexpected SYMBOL, expecting '(' dans :
br[i,j]=1
Erreur dans br[i, j] = 1 : objet i non trouvé
(if else (abs(i-j)%%4==0)
Erreur : syntax error, unexpected ELSE, expecting '(' dans (if else
br[i,j]=1
Erreur dans br[i, j] = 1 : objet i non trouvé
else
Erreur : syntax error, unexpected ELSE dans else
br[i,j]=0
Erreur dans br[i, j] = 0 : objet i non trouvé
}
Erreur : syntax error, unexpected '}' dans }
Can you please rectify my smal program, I tried to rectify it but I can't. Many
thanks in advance.
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[R] tackle memory insufficiency for large dataset using save() load()??
Hello List, i have been agonizing over this for days, any reply would be greatly appreciated! Situation:___ My original dataset is a .csv dataset (w/ 2M records) with 4 variables: job_id (Primary key, won't be used for analysis, just used for join tables), sector_id (categorical variable, for 19 industry sectors), sqft (con't variable for square footage), building_type (categorical, for 2 building types) some values of sqft were inputed wrong, so i'd like to set sqft1 to NA and then use aregImpute() to impute those NAs. Problem: the origianl dataset(.csv format) is too large. though i could read that dataset into R, i could not get aregImpute() run even i set the memory limit to 3G ! (yes, i did the switch in windows to reach 3G rather than 2G) Goal: try to find a way to slim down my dataset so as to get aregImpute() running. What i did: i searched in the archive, and found someone said, as R tends to inflate memory, it is a good idea to first read the original dataset into R-- then save it as a more compact binary file using save() -- and then reload the compact binary file back into R using load(). this way would reduce the memory allocation. HOWEVER, after i saved my original dataset into a compact binary file using save(), and used load(filename.Rdata) to reload the new compact data format into R, I could not figure out how to retrive all my variables!!! R shows the new dataset is not a list, nor a matrix, or a dataframe, but just a character with length 1 !!! and there is no way i could do attach(). i generated a 1K-row subset out of my original dataset to illustrate my problem (does anyone know how to get my four variables back from this compact binary new dataset? what did i do wrong?): data - read.table (file.choose(),header=T,sep=,) summary(data) job_id sector_id sqftbuilding_type Min. : 1.0 Min. : 6.000 Min. : 0.00 Min. :1.000 1st Qu.: 250.8 1st Qu.: 6.000 1st Qu.: 3.00 1st Qu.:2.000 Median : 500.5 Median :11.000 Median : 4.00 Median :2.000 Mean : 500.5 Mean : 9.455 Mean : 12.49 Mean :1.996 3rd Qu.: 750.3 3rd Qu.:11.000 3rd Qu.: 4.00 3rd Qu.:2.000 Max. :1000.0 Max. :12.000 Max. :192.00 Max. :2.000 attach(data) sqft[sqft1] - NA sector.f - as.factor(sector_id) building_type.f - as.factor (building_type) d - data.frame(job_id,sector.f,sqft, building_type.f) summary (d) job_id sector.f sqftbuilding_type.f Min. : 1.0 6 :340 Min. : 3.00 1: 4 1st Qu.: 250.8 11:505 1st Qu.: 4.00 2:996 Median : 500.5 12:155 Median : 4.00 Mean : 500.5Mean : 14.16 3rd Qu.: 750.33rd Qu.: 17.00 Max. :1000.0Max. :192.00 NA's :118.00 save (d, file=compact_d.Rdata, ascii=FALSE) newdata - load (compact_d.Rdata) summary(newdata) Length Class Mode 1 character character attach(newdata) Error in attach(newdata) : file 'd' not found is.data.frame (newdata) [1] FALSE is.list (newdata) [1] FALSE is.matrix (newdata) [1] FALSE _ btw, i also tried to just save (into compact binary) and reload (the new compact binary data format) (as i could do the NA stuff in sql anyhow). however, i still got stucked at the same spot: data - read.table (file.choose(),header=T,sep=,) summary(data) job_id sector_id sqftbuilding_type Min. : 1.0 Min. : 6.000 Min. : 0.00 Min. :1.000 1st Qu.: 250.8 1st Qu.: 6.000 1st Qu.: 3.00 1st Qu.:2.000 Median : 500.5 Median :11.000 Median : 4.00 Median :2.000 Mean : 500.5 Mean : 9.455 Mean : 12.49 Mean :1.996 3rd Qu.: 750.3 3rd Qu.:11.000 3rd Qu.: 4.00 3rd Qu.:2.000 Max. :1000.0 Max. :12.000 Max. :192.00 Max. :2.000 save (data, file=compact_data.Rdata, ascii=FALSE) newdata - load (compact_data.Rdata) summary(newdata) Length Class Mode 1 character character attach(newdata) Error: restore file may be empty -- no data loaded In addition: Warning message: file 'data' has magic number '' Use of save versions prior to 2 is deprecated is.data.frame (newdata) [1] FALSE is.list (newdata) [1] FALSE is.matrix (newdata) [1] FALSE - Building a website is a piece of cake. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained,
Re: [R] tackle memory insufficiency for large dataset using save() load()?
See ?save . The ... arguments are the ***names*** of the objects, not the objects so you want save(d, ...whatever...) not save(d, ...whatever...) . Also don't use attach and detach and read this about factors which applies if your factor has many levels but can be ignored if not: http://www.mail-archive.com/r-help@stat.math.ethz.ch/msg92970.html On 8/21/07, Jessica Z [EMAIL PROTECTED] wrote: Hello List, i have been agonizing over this for days, any reply would be greatly appreciated! Situation:___ My original dataset is a .csv dataset (w/ 2M records) with 4 variables: job_id (Primary key, won't be used for analysis, just used for join tables), sector_id (categorical variable, for 19 industry sectors), sqft (con't variable for square footage), building_type (categorical, for 2 building types) some values of sqft were inputed wrong, so i'd like to set sqft1 to NA and then use aregImpute() to impute those NAs. Problem: the origianl dataset(.csv format) is too large. though i could read that dataset into R, i could not get aregImpute() run even i set the memory limit to 3G ! (yes, i did the switch in windows to reach 3G rather than 2G) Goal: try to find a way to slim down my dataset so as to get aregImpute() running. What i did: i searched in the archive, and found someone said, as R tends to inflate memory, it is a good idea to first read the original dataset into R-- then save it as a more compact binary file using save() -- and then reload the compact binary file back into R using load(). this way would reduce the memory allocation. HOWEVER, after i saved my original dataset into a compact binary file using save(), and used load(filename.Rdata) to reload the new compact data format into R, I could not figure out how to retrive all my variables!!! R shows the new dataset is not a list, nor a matrix, or a dataframe, but just a character with length 1 !!! and there is no way i could do attach(). i generated a 1K-row subset out of my original dataset to illustrate my problem (does anyone know how to get my four variables back from this compact binary new dataset? what did i do wrong?): data - read.table (file.choose(),header=T,sep=,) summary(data) job_id sector_id sqftbuilding_type Min. : 1.0 Min. : 6.000 Min. : 0.00 Min. :1.000 1st Qu.: 250.8 1st Qu.: 6.000 1st Qu.: 3.00 1st Qu.:2.000 Median : 500.5 Median :11.000 Median : 4.00 Median :2.000 Mean : 500.5 Mean : 9.455 Mean : 12.49 Mean :1.996 3rd Qu.: 750.3 3rd Qu.:11.000 3rd Qu.: 4.00 3rd Qu.:2.000 Max. :1000.0 Max. :12.000 Max. :192.00 Max. :2.000 attach(data) sqft[sqft1] - NA sector.f - as.factor(sector_id) building_type.f - as.factor (building_type) d - data.frame(job_id,sector.f,sqft, building_type.f) summary (d) job_id sector.f sqftbuilding_type.f Min. : 1.0 6 :340 Min. : 3.00 1: 4 1st Qu.: 250.8 11:505 1st Qu.: 4.00 2:996 Median : 500.5 12:155 Median : 4.00 Mean : 500.5Mean : 14.16 3rd Qu.: 750.33rd Qu.: 17.00 Max. :1000.0Max. :192.00 NA's :118.00 save (d, file=compact_d.Rdata, ascii=FALSE) newdata - load (compact_d.Rdata) summary(newdata) Length Class Mode 1 character character attach(newdata) Error in attach(newdata) : file 'd' not found is.data.frame (newdata) [1] FALSE is.list (newdata) [1] FALSE is.matrix (newdata) [1] FALSE _ btw, i also tried to just save (into compact binary) and reload (the new compact binary data format) (as i could do the NA stuff in sql anyhow). however, i still got stucked at the same spot: data - read.table (file.choose(),header=T,sep=,) summary(data) job_id sector_id sqftbuilding_type Min. : 1.0 Min. : 6.000 Min. : 0.00 Min. :1.000 1st Qu.: 250.8 1st Qu.: 6.000 1st Qu.: 3.00 1st Qu.:2.000 Median : 500.5 Median :11.000 Median : 4.00 Median :2.000 Mean : 500.5 Mean : 9.455 Mean : 12.49 Mean :1.996 3rd Qu.: 750.3 3rd Qu.:11.000 3rd Qu.: 4.00 3rd Qu.:2.000 Max. :1000.0 Max. :12.000 Max. :192.00 Max. :2.000 save (data, file=compact_data.Rdata, ascii=FALSE) newdata - load (compact_data.Rdata) summary(newdata) Length Class Mode 1 character character attach(newdata) Error: restore file may be empty -- no data loaded In addition: Warning message: file 'data' has magic number '' Use of save versions prior to 2 is deprecated is.data.frame (newdata) [1] FALSE is.list (newdata) [1] FALSE is.matrix (newdata) [1] FALSE - Building a website is a piece of cake. [[alternative
Re: [R] R on a flash drive
I often run R via a Ceedo virtualisation on a USB drive (http://www.ceedo.com/) with XP. It costs a few dollars to it this way, but is a very low stress installation and has worked flawlessly, albeit a little slower (barely noticeable). Very handy if you are often working on various machines without administrator rights (as I do in clinic) - just plug in your USB and go directly back to your project. It then removes any trace of you (so they say) when you log out. And you can use it for other software (within limits though) you might want to carry around. Hope that helps. Scott Scott Williams MD Peter MacCallum Cancer Centre Melbourne Australia -Original Message- From: John Kane [mailto:[EMAIL PROTECTED] Sent: Tuesday, 21 August 2007 12:28 AM To: John Kane; Erin Hodgess; r-help@stat.math.ethz.ch Subject: Re: [R] R on a flash drive Oops meant to send this to the list. --- John Kane [EMAIL PROTECTED] wrote: --- Erin Hodgess [EMAIL PROTECTED] wrote: Dear R People: Has anyone run R from a flash drive, please? If so, how did it work, please? Yes I run R, occasionally, on a USB with no problem on WindowsXP. It works well, albeit a bit more slowly than from the hard drive which is as you would expect. The last time I upgraded the USB (to 2.5.0 ?) I simply downloaded R and installed it on the USB drive rather than the C: drive and then installed all my usual optional packages using the normal Rgui interface. I usually have R, Tinn-R and portable versions of OpenOoffice.org, and Firefox installed on the USB. Get news delivered with the All new Yahoo! Mail. Enjoy RSS feeds right on your Mail page. Start today at http://mrd.mail.yahoo.com/try_beta?.intl=ca __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Optimization problem
(Hope this gets threaded properly. Sorry if it doesn't.) Gabor: Lac and Lacfac being the same is irrelevant, wouldn't produce NAs (but would produce something like a singular Hessian and maybe other problems) -- but they're not even specified in this model. The bottom line is that you have a location with a single observation, so the GLM that zicounts runs to get the initial parameter values has an unestimable location:mass interaction for one location, so it gives an NA, so optim complains. In gruesome detail: ## set up data scardat = read.table(scars.dat,header=TRUE) library(zicounts) ## try to run model zinb.zc - zicounts(resp=Scars~., x =~Location + Lar + Mass + Lar:Mass + Location:Mass, z =~Location + Lar + Mass + Lar:Mass + Location:Mass, data=scardat) ## tried to debug this by dumping zicounts.R to a file, modifying ## it to put a trace argument in that would print out the parameters ## and log-likelihood for every call to the log-likelihood function. dump(zicounts,file=zicounts.R) source(zicounts.R) zinb.zc - zicounts(resp=Scars~., x =~Location + Lar + Mass + Lar:Mass + Location:Mass, z =~Location + Lar + Mass + Lar:Mass + Location:Mass, data=scardat,trace=TRUE) ## this actually didn't do any good because the negative log-likelihood ## function never gets called -- as it turns out optim() barfs when it ## gets its initial values, before it ever gets to evaluating the log-likelihood ## check the glm -- this is the equivalent of what zicounts does to ## get the initial values of the x parameters p1 - glm(Scars~Location + Lar + Mass + Lar:Mass + Location:Mass, data=scardat,family=poisson) which(is.na(coef(p1))) ## find out what the deal is table(scardat$Location) scar2 = subset(scardat,Location!=Randalstown) ## first step to removing the bad point from the data set -- but ... table(scar2$Location) ## it leaves the Location factor with the same levels, so ## now we have ZERO counts for one location: ## redefine the factor to drop unused levels scar2$Location - factor(scar2$Location) ## OK, looks fine now table(scar2$Location) zinb.zc - zicounts(resp=Scars~., x =~Location + Lar + Mass + Lar:Mass + Location:Mass, z =~Location + Lar + Mass + Lar:Mass + Location:Mass, data=scar2) ## now we get another error (system is computationally singular when ## trying to compute Hessian -- overparameterized?) Not in any ## trivial way that I can see. It would be nice to get into the guts ## of zicounts and stop it from trying to invert the Hessian, which is ## I think where this happens. In the meanwhile, I have some other ideas about this analysis (sorry, but you started it ...) Looking at the data in a few different ways: library(lattice) xyplot(Scars~Mass,groups=Location,data=scar2,jitter=TRUE, auto.key=list(columns=3)) xyplot(Scars~Mass|Location,data=scar2,jitter=TRUE) xyplot(Scars~Lar,groups=Location,data=scar2, auto.key=list(columns=3)) xyplot(Scars~Mass|Lar,data=scar2) xyplot(Scars~Lar|Location,data=scar2) Some thoughts: (1) I'm not at all sure that zero-inflation is necessary (see Warton 2005, Environmentrics). This is a fairly small, noisy data set without huge numbers of zeros -- a plain old negative binomial might be fine. I don't actually see a lot of signal here, period (although there may be some) ... there's not a huge range in Lar (whatever it is -- the rest of the covariates I think I can interpret). It would be tempting to try to fit location as a random effect, because fitting all those extra degrees of freedom is going to kill you. On the other hand, GLMMs are a bit hairy. cheers Ben __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] C code generators
Dear R-helpers Are there any established R packages that include a C code generator -- that generates new C language files and compiles them? To be precise what I'm looking for is a process that takes text input in some format (it might be pseudocode, fragments of C code, etc) and creates a valid C language source file that can be compiled by R CMD COMPILE. Ideally the procedure should also cause the C code to be compiled and dynamically loaded. To give a trivial example, suppose I want to be able to perform image filtering operations on matrices. All filters have the same structure: each position [i,j] in the matrix is visited in a double loop; a calculation (depending on the filter) is performed using the value at [i,j] and also the values at neighbouring positions [i+1,j] , [i+1,j+1] etc; the result is written to the output matrix at the position [i,j]. The C code for the loop is always the same; only a few lines of code that perform the filter calculation will change. I would like a procedure that accepts a text file containing just a few lines of C code or pseudocode, and inserts these lines into the appropriate place in the loop, producing a valid C language routine which can then be compiled by R CMD COMPILE and dynamically loaded. (Of course, it can get more complicated than just inserting a single text fragment!) I once implemented such a feature in an image processing package, so I know it's not hard. Before dusting off this ancient code I would like to learn whether there's an R package or other open source program that already does it. Any pointers would be welcome thanks Adrian Baddeley __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] polar.plot orientation and scale in plotrix
Thanks for that Stephen. It works nicely. It seems that my expectations for the plot defaults were non-standard. From your remarks and those of a colleague with an engineering background, I guess the defaults are standard. As a biologist, it seemed sensible to me to expect it to work like a compass though. Future updates to the plotrix package apparently might include options to easily customize these features (thanks to the package maintainer). Best regards, Quoting Stephen Tucker [EMAIL PROTECTED]: I think that's the standard presentation for polar plots (theta measured from positive x-axis) - that I've seen, anyway. But for customization you can shift your origin for theta and define your own labels. For example, here is a modification to the example in the help page for polar.plot(): testlen-c(rnorm(36)*2+5) testpos-seq(0,350,by=10) polar.plot(testlen,360-(testpos+90), main=Test Polar Plot,lwd=3,line.col=4, labels=seq(0,359,by=45)[c(3:1,8:4)], label.pos=seq(0,359,by=45)) --- Tim Sippel [EMAIL PROTECTED] wrote: Hello all- I would like to orient my polar.plot (from package plotrix) so that the circular scale runs clockwise and the origin (ie. 0 degrees) starts at the top of the plot. The defaults of running the scale counter-clockwise and beginning with 90 degrees at the top of the graph seems counter-intuitive to me. I'm using R 2.5.0, and plotrix version 2.2-4. Many thanks, Tim [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Take the Internet to Go: Yahoo!Go puts the Internet in your pocket: mail, news, photos more. http://mobile.yahoo.com/go?refer=1GNXIC __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] rectify a program of seasonal dummies matrix
Your syntax is wrong; e.g.,
if i==j
should be
if (i == j)
same with your use of 'if else'. You need to use the correct syntax.
Your example is hard to follow without the correct indentation since
you are using the incorrect syntax.
On 8/21/07, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
Hi friends,
I would like to construct a matrix of seasonal dummies with number of rows
(observations)=100. such matrix is written as follows:[1 0 0 0;0 1 0 0;0 0 1
0;0 0 0 1;1 0 0 0;0 1 0 0;0 0 1 0;0 0 0 1;etc...] . I wrote the following
program:
T=100
br=matrix(0,T,4)
{
for (i in 1:T)
for (j in 1:4)
if i==j
br[i,j]=1
if else (abs(i-j)%%4==0
br[i,j]=1
else
br[i,j]=0
}
z-br
z
but unfortunately I obtained from the console the following message:
{
+ for (i in 1:T)
+ for (j in 1:4)
+ (if i==j)
Erreur : syntax error, unexpected SYMBOL, expecting '(' dans :
br[i,j]=1
Erreur dans br[i, j] = 1 : objet i non trouvé
(if else (abs(i-j)%%4==0)
Erreur : syntax error, unexpected ELSE, expecting '(' dans (if else
br[i,j]=1
Erreur dans br[i, j] = 1 : objet i non trouvé
else
Erreur : syntax error, unexpected ELSE dans else
br[i,j]=0
Erreur dans br[i, j] = 0 : objet i non trouvé
}
Erreur : syntax error, unexpected '}' dans }
Can you please rectify my smal program, I tried to rectify it but I can't.
Many thanks in advance.
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--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
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Re: [R] tackle memory insufficiency for large dataset using save() load()?
On 22/08/2007, at 1:48 PM, Gabor Grothendieck wrote:
See ?save . The ... arguments are the ***names*** of the objects, not
the objects
so you want save(d, ...whatever...) not save(d, ...whatever...) .
I think this is wrong. You want the objects not their names.
If you want to make use of object names, use the list argument.
I.e.
save(melvin,clyde,file=irving)
and
save(list=c(melvin,clyde),file=irving)
accomplish the same thing.
cheers,
Rolf Turner
##
Attention:\ This e-mail message is privileged and confidenti...{{dropped}}
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] tackle memory insufficiency for large dataset using save() load()?
?save says its the names (not the objects) although I just
tried it and both save(iris, file = /iris.Rdata) and
save(iris, file = /iris.Rdata) seemed to work so you are
right that it seems to work with the objects, not just the names,\
although its not documented to do so.
Usage
save(..., list = character(0),
file = stop('file' must be specified),
ascii = FALSE, version = NULL, envir = parent.frame(),
compress = !ascii, eval.promises = TRUE)
save.image(file = .RData, version = NULL, ascii = FALSE,
compress = !ascii, safe = TRUE)
Arguments
... the names of the objects to be saved.
list A character vector containing the names of objects to be saved.
On 8/21/07, Rolf Turner [EMAIL PROTECTED] wrote:
On 22/08/2007, at 1:48 PM, Gabor Grothendieck wrote:
See ?save . The ... arguments are the ***names*** of the objects, not
the objects
so you want save(d, ...whatever...) not save(d, ...whatever...) .
I think this is wrong. You want the objects not their names.
If you want to make use of object names, use the list argument.
I.e.
save(melvin,clyde,file=irving)
and
save(list=c(melvin,clyde),file=irving)
accomplish the same thing.
cheers,
Rolf Turner
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R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] tackle memory insufficiency for large dataset using save() load()?
On 22/08/2007, at 2:52 PM, Gabor Grothendieck wrote:
?save says its the names (not the objects) although I just
tried it and both save(iris, file = /iris.Rdata) and
save(iris, file = /iris.Rdata) seemed to work so you are
right that it seems to work with the objects, not just the names,\
although its not documented to do so.
Yeah, you're right --- it works both ways. I just looked at the
code and it converts the ... argument to a list of names using
as.character(substitute(list(...)))[-1].
These R Core people are sneaky-clever!
cheers,
Rolf Turner
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R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] rectify a program of seasonal dummies matrix
Hello,
the main problem seems to be the if else, should be else if.
Your code is hard to read, maybe you should consider using more () {}:
T - 100;
br - matrix(0,T,4);
for (i in 1:T) {
for (j in 1:4) {
if (i==j) {
br[i,j] - 1;
}
else if ((abs(i-j)%%4)==0) {
br[i,j] - 1;
}
else {
br[i,j] - 0;
}
}
}
A simpler approach is creating a diagonal matrix and multply it :
# create small diagonal matrix
mat = diag(x=1, nrow=4, ncol=4);
mat
# multiply diagonal matrix and re-dimension it to 4 cols
br - rep(mat, 25);
dim(br) - c(100, 4);
br;
Hope this helps,
FS
--
Friedrich Schuster
mail at friedrich-schuster.de
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] problems installing updated version of vars package
On Tue, 21 Aug 2007, sj wrote: All, I was looking onlin and noticed that the vars package (by Bernhard Pfaff) was recently updated (update date listed Aug 6, 2007) The updated packages has some features that I would find very useful. I have used the update packages function and vars was one of the packages identified as needing an update. I was able to updated and it appeared to work, however when I load the package it does not seem to be the most recent version? Has anyone else had similar problems? Or does anyone have any suggestions? System Info: R 2.4.1 You need to update your R (as the posting guide asked you to before posting) or install the package from sources. The binary builds are not updated for obsolete versions of R: the builds for 2.4.x stopped on 28 June. Windows XP install mirror: USA 3 (UCLA I think) thanks, Spencer [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. PLEASE do! -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] open/execute/call/run an external file
I'm trying to figure out how to trigger a process from within R. I have an exectuable file that runs a Fortran model, but ideally, would like to run it from R. Note that I'm not talking about importing the function at all, passing variables, or anything complicated like that. I basically just want a script that double-clicks on a particular file and opens/runs it for me. The idea here is that the executable Fortran file, when double clicked, simply draws all necessary inputs from text files within the same directory and I have no need to change this. So I've used R to summarize some raw data and format these required text input files in the way the Fortran executable requires, and also have scripts to interpret the Fortran text file outputs and summarize/plot them in R. The problem is I must run the first part of the R script to send data from R to the model, then double click the Fortran executable, then run the second part of the R script to get the model outputs into R, in three separate steps. Given that I may be doing this hundereds of times, I'd prefer to do it all in one step. Any thoughts?---steve __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help 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 on a flash drive
On Wed, 22 Aug 2007, Williams Scott wrote: I often run R via a Ceedo virtualisation on a USB drive (http://www.ceedo.com/) with XP. It costs a few dollars to it this way, but is a very low stress installation and has worked flawlessly, albeit It is not necessary though, as R does not need 'virtualisation'. For Windows this is covered in the rw-FAQ Q2.6. a little slower (barely noticeable). Perhaps the overhead of Credo? Once R starts up (which does take longer on a slow drive) I found no time-able difference in 2.1.x (all the files frequently used from disc are cached on startup). It would be nice to give the R developers the credit for writing R in such a way that it works well from slow media, instead of it being credited to an unnecessary commercial product. Very handy if you are often working on various machines without administrator rights (as I do in clinic) - just plug in your USB and go directly back to your project. It then removes any trace of you (so they say) when you log out. And you can use it for other software (within limits though) you might want to carry around. Many sites would not allow programs to be run from a USB drive or make it a breach of usage conditions to do so. Hope that helps. Scott Scott Williams MD Peter MacCallum Cancer Centre Melbourne Australia -Original Message- From: John Kane [mailto:[EMAIL PROTECTED] Sent: Tuesday, 21 August 2007 12:28 AM To: John Kane; Erin Hodgess; r-help@stat.math.ethz.ch Subject: Re: [R] R on a flash drive Oops meant to send this to the list. --- John Kane [EMAIL PROTECTED] wrote: --- Erin Hodgess [EMAIL PROTECTED] wrote: Dear R People: Has anyone run R from a flash drive, please? If so, how did it work, please? Yes I run R, occasionally, on a USB with no problem on WindowsXP. It works well, albeit a bit more slowly than from the hard drive which is as you would expect. The last time I upgraded the USB (to 2.5.0 ?) I simply downloaded R and installed it on the USB drive rather than the C: drive and then installed all my usual optional packages using the normal Rgui interface. I usually have R, Tinn-R and portable versions of OpenOoffice.org, and Firefox installed on the USB. Get news delivered with the All new Yahoo! Mail. Enjoy RSS feeds right on your Mail page. Start today at http://mrd.mail.yahoo.com/try_beta?.intl=ca __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] open/execute/call/run an external file
On Tue, 21 Aug 2007, STEPHEN M POWERS wrote: I'm trying to figure out how to trigger a process from within R. I have an exectuable file that runs a Fortran model, but ideally, would like to run it from R. Note that I'm not talking about importing the function at all, passing variables, or anything complicated like that. I basically just want a script that double-clicks on a particular file and opens/runs it for me. The idea here is that the executable Fortran file, when double clicked, simply draws all necessary inputs from text files within the same directory and I have no need to change this. So I've used R to summarize some raw data and format these required text input files in the way the Fortran executable requires, and also have scripts to interpret the Fortran text file outputs and summarize/plot them in R. The problem is I must run the first part of the R script to send data from R to the model, then double click the Fortran executable, then run the second part of the R script to get the model outputs into R, in three separate steps. Given that I may be doing this hundereds of times, I'd prefer to do it all in one step. Any thoughts?---steve It is described in the relevant manual: Writing R Extensions. ?system, and if this is Windows also ?shell and ?shell.exec. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. When you do, you will see that we asked for your OS which is relevant here. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
