[R] Wholesale and retail data analysis tasks
Hello, We are searching for information resources for wholesale and retail data analysis tasks: the tasks that are relevant and demanded from statists of wholesale and retail companies. The reason for this search for us, a sfotware development company is simple: one wholesale company showed interest in our product that builds associative if-then rules based on purchase transactions, and we are redeveloping the system in order it to be a low cost and effective solution for solving analytical tasks for wholesale and retail. We would be happy if we can find here partners that would like to participate in the project providing their requirements and/or additional information on what functionality they would like to have in such a system and being our alpha testers in the future. We take all costs related to system design and development and would be eager to share final results with the people involved in the project and also consider working out future sales channels with them. Thank you for your time and we hope to get some feedback from the audience. Best regards, Endre Domiczi, CEO Sevana Oy, http://www.sevana.fi Email : [EMAIL PROTECTED] GSM : +372 53485178 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 write a sum of matrixes??
I am a new user and I have been struggling for hours. So finally I decide to ask you: If I have a matrix P, and P.2 = P%*%P, and P.3=P.2%*%P is there a function to calculate the power of a matrix?? if not how can i do: for (i in 1:10) {P.i=P^i} after this I need to sum them up and my problem is to combine P and i to P.i can anyone help me please??? Take a look at FAQ on R 7.21 (How can I turn a string into a variable?). This code will do what you ask (though the first matrix is named P.1, rather than P for convenience): paste0 - function(x) paste(P., x, sep=) P.1 - matrix(rnorm(25), nrow=5) varnames - paste0(1:10) for(i in 2:10) assign(paste0(i), get(paste0(i-1)) %*% P.1) Regards, Richie. Mathematical Sciences Unit HSL ATTENTION: This message contains privileged and confidential inform...{{dropped:20}} __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] ggplo2: x_discrete labels size/direction
Hi everyone, I've got quite a few labels along the x axis and ggplot2 basically just crams them on top of each other. Is it possible to reduce the font size and/or text direction? Stretching the windows device window manually also helps, but I found that setting the parameters for the pdf device (where my scripts should print the data), such as paper=a4r just results in a lot of empty space at both sides. Thanks for your help! Mikhail Mikhail Spivakov, PhD Postdoctoral Fellow EMBL/EBI UK Germany -- View this message in context: http://www.nabble.com/ggplo2%3A-x_discrete-labels-size-direction-tp17077479p17077479.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ggplo2: x_discrete labels size/direction
Hi Mikhail, You can reduce the text size using the grid.gedit approach described at the end of the ggplot book, available on Hadley Wickham's website: http://had.co.nz/ggplot2/book.pdf You could use something like: grid.gedit(gPath(xaxis, labels), gp=gpar(fontsize=6)) I'm not aware of a good way of rotating the text. I tried it but it did not look good, because with grid.gedit, the position of the different elements is not adjusted. So if you rotate labels, maybe they will be drawn on top of the axis title. For the pdf device, you also need to specify the size of the graphic region, with something like: pdf(paper=a4r, width=9, height=6) see ?pdf for details Xavier Mikhail Spivakov a écrit : Hi everyone, I've got quite a few labels along the x axis and ggplot2 basically just crams them on top of each other. Is it possible to reduce the font size and/or text direction? Stretching the windows device window manually also helps, but I found that setting the parameters for the pdf device (where my scripts should print the data), such as paper=a4r just results in a lot of empty space at both sides. Thanks for your help! Mikhail Mikhail Spivakov, PhD Postdoctoral Fellow EMBL/EBI UK Germany -- Xavier Chardon Thésard Institut de l'élevage / INRA Projet ACTA modélisation environnementale des systèmes bovins et porcins [EMAIL PROTECTED] [EMAIL PROTECTED] 02 23 48 50 91 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Model Based Bootstrap
Hello. Has anyone any idea how a function would look like of a model based bootstrap, when the underlying time series follows an ARIMA(1,1,1)-process? A pure AR-process is no problem, but what is, if the time series need to be differentiated of order one or above and the additional MA-part? Sample code for a series, which follows a pure AR-process: #Series y of 192 observations, which follows an AR(1)-process #Fit of an AR(1)-Model to y ar.coef - ar(y)$ar ar.resid- ar(y)$resid #Sampling for mean y_sample- numeric(192) y_sample[1] - y[1] mean_y - numeric(1) for (i in 1:1) { for (j in 1:191) { idx - sample(2:192,1,replace=TRUE) y_sample[j+1] - y_sample[j]*ar.coef+ar.resid[idx] } mean_y[i] - mean(y_sample) } What would the function look like if y follows an ARIMA(1,1,1)-process for example or in general if y is a time series, which need to be differentiated and is best modeled with a mixture of AR and MA? I hope you can help me. Sincerely Andreas. Lesen Sie Ihre E-Mails jetzt einfach von unterwegs. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] adding an argument to ...
Bonjour Christophe, You can pass ... to a list and then extract its names, so something like: if( ! uuu %in% names(list(...)) ) Good luck with your keyboard, Romain [EMAIL PROTECTED] wrote: Hi the list Is it possible to add one argument to the arguments contain in ... ? I would like to do : f - function(ttt,...)[ if(... does not contain the args uuu) [add uuu=3 to ...] else[] g(ttt,...) ] More precisely, my function g is a callNextMethod() Thanks Christophe PS: sorry for using [ instead of accolade, I am not in my contry and I do not manage to get the accolade on this ¤¦²¤£$¤@ keybord... -- Mango Solutions data analysis that delivers Tel: +44(0) 1249 76 77 00 Fax: +44(0) 1249 76 77 07 Mob: +44(0) 7813 52 61 23 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] merge numerous columns of unequal length
As the answers you've received suggest, you can use a list. Or you could have two vectors: one with the data, the other with the group identity. The latter format is likely more convenient for a lot of analyses. Since your data are not inherently rectangular, it is probably best to get the idea of spreadsheet out of your head. (It is probably best anyway.) Patrick Burns [EMAIL PROTECTED] +44 (0)20 8525 0696 http://www.burns-stat.com (home of S Poetry and A Guide for the Unwilling S User) T.D.Rudolph wrote: I have numerous objects, each containing continuous data representing the same variable, movement rate, yet each having a different number of rows. e.g. d1-as.matrix(rnorm(5)) d2-as.matrix(rnorm(3)) d3-as.matrix(rnorm(6)) How can I merge these three columns side-by-side in order to create a table regardless of the difference in length? I wish to analyze the output in a spreadsheet format. Thanks! Tyler __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Regarding nls()
Hi, I have no problem performing the regression using R, and I successfully obtain the parameter estimates using the function nls(). However, how do I obtain the ANOVA output, r, r^2 and adj. r^2? Thanks Regards, Guru [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Lattice problems / cannot load lattice
Hi, My problem is simple: since having updated the lattice package, I cannot load lattice anymore. If I type in the command 'library(lattice)' the loading fails with the following message: --- cut here --- Error in library.dynam(lib, package, package.lib) : shared library 'lattice' not found In addition: Warning messages: 1: In loadNamespace(package, c(which.lib.loc, lib.loc), keep.source = keep.source) : package 'lattice' contains no R code 2: S3 methods ‘[.shingle’, ‘as.data.frame.shingle’, ‘plot.shingle’, ‘print.shingle’, ‘summary.shingle’, ‘as.character.shingleLevel’, ‘print.shingleLevel’, ‘print.trellis’, ‘plot.trellis’, ‘update.trellis’, ‘dim.trellis’, ‘dimnames.trellis’, ‘dimnames-.trellis’, ‘[.trellis’, ‘t.trellis’, ‘summary.trellis’, ‘print.summary.trellis’, ‘barchart.formula’, ‘barchart.array’, ‘barchart.default’, ‘barchart.matrix’, ‘barchart.numeric’, ‘barchart.table’, ‘bwplot.formula’, ‘bwplot.numeric’, ‘densityplot.formula’, ‘densityplot.numeric’, ‘dotplot.formula’, ‘dotplot.array’, ‘dotplot.default’, ‘dotplot.matrix’, ‘dotplot.numeric’, ‘dotplot.table’, ‘histogram.formula’, ‘histogram.factor’, ‘histogram.numeric’, ‘qqmath.formula’, ‘qqmath.numeric’, ‘stripplot.formula’, ‘stripplot.numeric’, ‘qq.formula’, ‘xyplot.formula’, ‘levelplot.formula’, [... truncated] Error: package/namespace load failed for 'lattice' --- cut here --- R-System info: arch x86_64 os linux-gnu system x86_64, linux-gnu status major 2 minor 7.0 year 2008 month 04 day22 svn rev45424 language R version.string R version 2.7.0 (2008-04-22) Any hints? Any ideas? Thanks in advance, Kimmo __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] fCopulae
CA == chockri adnen [EMAIL PROTECTED] on Wed, 30 Apr 2008 13:34:42 +0200 CA My problem in a few words is as folow: CA I used the fCopulae packages because i have 2 series which CA are already CA transformed in the uniform domain (the space of the copulas CA functions) and i CA estimated with type archmList() from 1 to 22, but i don't CA know their CA names:for exemple the type=4 is the Gumbel Copula...and for CA the others i CA can't have any idea about how can i find teir names CA explicitly? CA Please can you help me in this point? CA Thank you. the numbers archmList() corresponds to the copula numbers as defined in An Introduction to Copulas by RB Nelson. I will add the reference in the help page of archmList. hope this helps, Yohan -- PhD student Swiss Federal Institute of Technology Zurich www.ethz.ch www.rmetrics.org NOTE: Rmetrics Workshop: http://www.rmetrics.org/meielisalp.htm June 29th - July 3rd Meielisalp, Lake Thune, Switzerland __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] adaptive optimization of mesh size
my post may have slipped through the bank holiday and be forgotten by now, I'm still hoping for some pointers. Please let me know if I need to clarify the problem. baptiste On 4 May 2008, at 16:39, baptiste Auguié wrote: DeaR list, I'm running an external program that computes some electromagnetic response of a scattering body. The numerical scheme is based on a discretization with a characteristic mesh size y. The smaller y is, the better the result (but obviously the computation will take longer). A convergence study showed the error between the computed values and the exact solution of the problem to be a quadratic in y, with standard error increasing as y^3. I wrote the interface to the program in R, as it is much more user friendly and allows for post- processing analysis. Currently, it only runs with user-defined discretization parameter. I would like to implement an adaptive scheme [1] and provide the following improvements, 1) obtain an estimate of the error by fitting the result against a series of mesh sizes with the quadratic model, and extrapolate at y = 0. (quite straight forward) 2) adapt dynamically the set of mesh sizes to fulfill a final accuracy condition, between a starting value (a rule-of thumb estimate is given by the problem values). The lower limit of y should also be constrained by the resources (again, an empirical rule dictates the computation time and memory usage). I'm looking for advice on this second point (both on the technical aspect, and whether this is sound statistically): - I can foresee that I should always start with a few y values before I can do any extrapolation, but how many of them? 3, 10? How could I know? - once I have enough points (say, 10) to use the fitting procedure and get an estimate of the error, how should I decide the best location of the next y if the error is too important? - in a practical implementation, I would use a while loop and append the successive values to a data.frame(y, value). However, this procedure will be run for different parameters (wavelengths, actually), so the set and number of y values may vary between one run and another. I think I'd be better off using a list with each new run having its own data.frame. Does this make sense? Below are a few lines of code to illustrate the problem, program.result - function(x, p){ # made up function that mimicks the results of the real program y - p[3]*x^2 + p[2]*x + p[1] y * (1 + rnorm(1, mean=0, sd = 0.1 * y^3)) } p0 - c(0.1, 0.1, 2) # set of parameters ## user defined limits of the y parameter (log scale) limits - c(0.1, 0.8) limits.log - (10^limits) y.log - seq(limits.log[1], limits.log[2], l=10) y - log10(y.log) result - sapply(y, function(x) program.result(x, p0)) # results of the program fitting and extrapolation procedure library(gplots) # plot with CI plotCI(y, result, y^3, xlim=c(0, 1), ylim=c(0, 2)) # the data with y^3 errors my.data - data.frame(y = y, value = result) fm - lm(value ~ poly(y, degree=2, raw=TRUE), data = my.data , weights = 1/y^3) lines(y, predict(fm, data.frame(y=y)), col = 2) extrap - summary(fm)$coefficients[1,] # intercept and error on it plotCI(0,extrap[1], 2 * extrap[2], col = 2, add=T) ### my naive take on adaptive runs... ## objective - 1e-3 # stop when the standard error of the extrapolated value is smaller than this err - extrap[2] my.color - 3 while (err objective){ new.value - min(y)/2 # i don't know how to choose this optimally y - c(new.value, y) new.result - program.result(new.value, p0) result - c(new.result, result) points(new.value, new.result, col= my.color) my.data - data.frame(y = y, value = result) fm - lm(value ~ poly(y, degree=2, raw=TRUE), data = my.data , weights = 1/y^3) lines(y, predict(fm, data.frame(y=y)), col = my.color) extrap - summary(fm)$coefficients[1,] # intercept and error on it err - extrap[2] print(err) plotCI(0,extrap[1], 2 * err, col = 2, add=T) my.color - my.color + 1 } err Many thanks in advance for your comments, baptiste [1]: Yurkin et al., Convergence of the discrete dipole approximation. II. An extrapolation technique to increase the accuracy. J. Opt. Soc. Am. A / Vol. 23, No. 10 / October 2006 _ Baptiste Auguié Physics Department University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag http://projects.ex.ac.uk/atto __ _ Baptiste Auguié Physics Department University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag http://projects.ex.ac.uk/atto __ R-help@r-project.org
Re: [R] Lattice problems / cannot load lattice
Try installing again by install.packages(lattice, .Library) (from an account with suitable privileges). If that still fails, we need to see the output produced during installation. On Tue, 6 May 2008, K. Elo wrote: Hi, My problem is simple: since having updated the lattice package, I cannot load lattice anymore. If I type in the command 'library(lattice)' the loading fails with the following message: --- cut here --- Error in library.dynam(lib, package, package.lib) : shared library 'lattice' not found In addition: Warning messages: 1: In loadNamespace(package, c(which.lib.loc, lib.loc), keep.source = keep.source) : package 'lattice' contains no R code 2: S3 methods ‘[.shingle’, ‘as.data.frame.shingle’, ‘plot.shingle’, ‘print.shingle’, ‘summary.shingle’, ‘as.character.shingleLevel’, ‘print.shingleLevel’, ‘print.trellis’, ‘plot.trellis’, ‘update.trellis’, ‘dim.trellis’, ‘dimnames.trellis’, ‘dimnames-.trellis’, ‘[.trellis’, ‘t.trellis’, ‘summary.trellis’, ‘print.summary.trellis’, ‘barchart.formula’, ‘barchart.array’, ‘barchart.default’, ‘barchart.matrix’, ‘barchart.numeric’, ‘barchart.table’, ‘bwplot.formula’, ‘bwplot.numeric’, ‘densityplot.formula’, ‘densityplot.numeric’, ‘dotplot.formula’, ‘dotplot.array’, ‘dotplot.default’, ‘dotplot.matrix’, ‘dotplot.numeric’, ‘dotplot.table’, ‘histogram.formula’, ‘histogram.factor’, ‘histogram.numeric’, ‘qqmath.formula’, ‘qqmath.numeric’, ‘stripplot.formula’, ‘stripplot.numeric’, ‘qq.formula’, ‘xyplot.formula’, ‘levelplot.f! ormula’, [... truncated] Error: package/namespace load failed for 'lattice' --- cut here --- R-System info: arch x86_64 os linux-gnu system x86_64, linux-gnu status major 2 minor 7.0 year 2008 month 04 day22 svn rev45424 language R version.string R version 2.7.0 (2008-04-22) Any hints? Any ideas? Thanks in advance, Kimmo __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lattice problems / cannot load lattice
Hi, thanks for the quick reply :) Prof Brian Ripley kirjoitti viestissään (06.05.2008): Try installing again by install.packages(lattice, .Library) (from an account with suitable privileges). Tried (as root) - not working :( If that still fails, we need to see the output produced during installation. Here we go: -- install.packages(lattice, .Library) --- Please select a CRAN mirror for use in this session --- Loading Tcl/Tk interface ... done trying URL 'http://cran.ch.r-project.org/src/contrib/lattice_0.17-7.tar.gz' Content type 'application/x-gzip' length 275240 bytes (268 Kb) opened URL == downloaded 268 Kb * Installing *source* package 'lattice' ... ** libs gcc -std=gnu99 -I/usr/lib64/R/include -I/usr/local/include-fpic -g -O2 -c init.c -o init.o gcc -std=gnu99 -I/usr/lib64/R/include -I/usr/local/include-fpic -g -O2 -c threeDplot.c -o threeDplot.o gcc -std=gnu99 -shared -L/usr/local/lib64 -o lattice.so init.o threeDplot.o -L/usr/lib64/R/lib -lR ** R ** data ** moving datasets to lazyload DB ** demo ** preparing package for lazy loading ** help Building/Updating help pages for package 'lattice' Formats: text html latex example Lattice texthtmllatex example Rows texthtmllatex axis.default texthtmllatex example banking texthtmllatex example barchart.tabletexthtmllatex example barleytexthtmllatex example cloud texthtmllatex example draw.colorkey texthtmllatex draw.key texthtmllatex environmental texthtmllatex example ethanol texthtmllatex example histogram texthtmllatex example interaction texthtmllatex example lattice.options texthtmllatex example latticeParseFormula texthtmllatex level.colors texthtmllatex example levelplot texthtmllatex example llinestexthtmllatex lset texthtmllatex make.groups texthtmllatex example melanoma texthtmllatex example onewaytexthtmllatex packet.panel.default texthtmllatex example panel.axistexthtmllatex panel.barcharttexthtmllatex panel.bwplot texthtmllatex example panel.cloud texthtmllatex panel.densityplot texthtmllatex panel.dotplot texthtmllatex panel.functions texthtmllatex panel.histogram texthtmllatex panel.levelplot texthtmllatex panel.number texthtmllatex panel.pairs texthtmllatex panel.paralleltexthtmllatex panel.qqmath texthtmllatex panel.qqmathline texthtmllatex panel.stripplot texthtmllatex panel.superpose texthtmllatex panel.violin texthtmllatex example panel.xyplot texthtmllatex example prepanel.default texthtmllatex prepanel.functionstexthtmllatex print.trellis texthtmllatex example qqtexthtmllatex example qqmathtexthtmllatex example rfs texthtmllatex example shingles texthtmllatex example simpleKey texthtmllatex simpleTheme texthtmllatex example singertexthtmllatex example splom texthtmllatex example strip.default texthtmllatex example tmd texthtmllatex example trellis.devicetexthtmllatex trellis.objecttexthtmllatex trellis.par.get texthtmllatex example update.trellistexthtmllatex example utilities.3d texthtmllatex xyplot
Re: [R] Lattice problems / cannot load lattice
Prof Brian Ripley kirjoitti viestissään (06.05.2008): Does starting R --vanilla help? I am wondering if you have another corrupt copy of lattice somewhere. The latter was the problem, many thanks for this! I use Rkward as GUI and obviously some packages have been installed into the user directory instead of being installed into the system directory. This messed up the installation :( After having deleted the local directory and restarted R lattice loaded just fine :) Once again: many thanks to Brian for his kind help! Kind regards, Kimmo __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Regarding nls()
Guru S guru.rcom at rediffmail.com writes: I have no problem performing the regression using R, and I successfully obtain the parameter estimates using the function nls(). However, how do I obtain the ANOVA output, r, r^2 and adj. r^2? This is a feature, not a bug. See Douglas Bates's comments on http://www.ens.gu.edu.au/ROBERTK/R/HELP/00B/0399.HTML Dieter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Spacing between lattice panels
I'm trying to set up a lattice plot with two y-axes for each panel. (Yes, I know that multiple y-axes are generally a bad idea; the graph is for someone else and they want it that way.) I've used a custom yscale.component in xyplot to achieve this: myyscale.component - function(...) { ans - yscale.components.default(...) ans$right - ans$left foo - ans$right$labels$at ans$right$labels$labels - as.character(10*foo) ans } xyplot(Sepal.Length ~ Petal.Length| Species, data = iris, scales = list(y=list(relation=free, alternating=3, rot=0)), yscale.component=myyscale.component) The problem is that the panels are too close together. This appears to be because (according to the help documentation) xyplot ignores the alternating argument when relation is not same. I presume that manually setting some plotting parameters with trellis.par.set() will allow me to adjust the spacing between the panels, though I can't figure out which ones to change. Is there a way to get lattice to properly space the panels, or failing that, how do I manually space them? Regards, Richie. Mathematical Sciences Unit HSL ATTENTION: This message contains privileged and confidential inform...{{dropped:20}} __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Dendrogram label size
Is it possible to resize the labels in a dendrogram without applying circles and triangles to edges? I tried cex.labels: plot(scoreDendogramObj, horiz=TRUE, axes=FALSE, cex.labels=0.8) but that didn't have any effect. Thanks, Alex __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Error in FUN(newX[, i], ...) : no complete element pairs in R 2.7.0 patched
Using R 2.6.0 patched I was able to calculate the variance of each row of a matrix without error, even if some rows had only NAs. I would just get NAs back as the variance for those rows. Now with R 2.7.0 patched I get an error no complete element pairs if any one row has all NAs. Can anyone tell if I can make R 2.7.0 patched behave as R 2.6.0 did so I don't have to change all my code. Or maybe suggest a simple work around. A short example is below. Thanks in advance. dat - data.frame(aa=c(1,2,as.numeric('NA'),4), bb=c(2,3,as.numeric('NA'),5), cc=c(3,4,as.numeric('NA'),6), dd=c(4,5,as.numeric('NA'),7)) apply(dat, 1, var, na.rm=TRUE) dat aa bb cc dd 1 1 2 3 4 2 2 3 4 5 3 NA NA NA NA 4 4 5 6 7 apply(dat, 1, var, na.rm=TRUE) Error in FUN(newX[, i], ...) : no complete element pairs ** * This message is for the named person's use only. It may contain confidential, proprietary or legally privileged information. No right to confidential or privileged treatment of this message is waived or lost by any error in transmission. If you have received this message in error, please immediately notify the sender by e-mail, delete the message and all copies from your system and destroy any hard copies. You must not, directly or indirectly, use, disclose, distribute, print or copy any part of this message if you are not the intended recipient. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 write a sum of matrixes??
pascal vrolijk [EMAIL PROTECTED] wrote in news:[EMAIL PROTECTED]: Hello best helpers, I am a new user and I have been struggling for hours. So finally I decide to ask you: If I have a matrix P, and P.2 = P%*%P, and P.3=P.2%*%P is there a function to calculate the power of a matrix?? if not how can i do: for (i in 1:10) {P.i=P^i} Three methods show up in a search r-help power of a matrix: 1) Use a pre-built function: library(msm) ?MatrixExp 2) https://stat.ethz.ch/pipermail/r-help/2007-May/131330.html 3) Bill Venables offered this about a week ago in this list: -- This is probably as good a way as any way for this kind of problem. First define a binary operator: %^% - function(x, n) with(eigen(x), vectors %*% (values^n * t(vectors))) Your toy example then becomes m - matrix(c(1, 0.4, 0.4, 1), nrow = 2) m [,1] [,2] [1,] 1.0 0.4 [2,] 0.4 1.0 m %^% (-0.5) [,1] [,2] [1,] 1.0680744 -0.2229201 [2,] -0.2229201 1.0680744 - -- David Winsemius after this I need to sum them up and my problem is to combine P and i to P.i can anyone help me please??? Thanks and have a nice day, Pascal. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] rpart for survival fits
When I plot a survival fit using rpart for the classification tree, for each node, there is a decimal based number above the event/total. I tried to see if it's the exponential ratio or logrithmics, neither seem to be the case. I'm wondering if anyone knows what they are. - It is an estimate of the event rate, using a Baysian shrinkage argument (#events +a) / (n +b) *c To understand how the constants a and b are chosen, you need to read the detailed documentation (as pointed out already by Brian R). The constant c is chosen to make the printed rate of the top node equal to 1. This last is done just to make it easier to scan the tree -- one can easily see that the rate is, say, 20% lower in some particular node than for the data set as a whole. Terry Therneau __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Gravity (spatial interaction) models in R
Melanie Murphy mamurphy at turbonet.com writes: I was wondering if anyone has developed (or is developing) an implementation for gravity models (spatial interaction) in R. I conducted several searches on the CRAN website with no luck. Currently I am estimating parameters via linearization. There isn't anything (yet) in the Spatial task view on CRAN, and I'm not aware of the topic coming up on the R-sig-geo list either. The nearest sign of transport planning is perhaps the emme2 package for interfacing EMME/2 software. As you are aware, there are many almost equivalent ways of fitting gravity models, so many R functions are feasible. But no dedicated functions so far, I'm afraid - if I'm wrong, please let me know, and I'll update the task view. Roger Thank you in advance Melanie Murphy __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] single plot statement, multiple plots
Hi R, par(mfrow=c(2,2)) x1=(1:5)^1; x2=(1:5)^2; x3=(1:5)^3; x4=(1:5)^4 I need to write a single plot statement, which creates 4 plots (for x1, x2, x3 and x4) in the graphics window, without using 'for' loop. Is this possible? Does 'do.call' help in this context? Or do I have any option in the 'plot' statement itself to do this? Thanks in advance, Shubha This e-mail may contain confidential and/or privileged i...{{dropped:13}} __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Significance analysis of Microarrays (SAM)
Dear list, I am trying to perform a significance analysis of a microarray experiment with survival data using the {samr} package. I have a matrix containing my data which has 17816 rows corresponding to genes, and 286 columns corresponding to samples. The name of this matrix is data.matrix2. Some of the first values of this matrix are: data.matrix2[1:3,1:5] GSM36777 GSM36778 GSM36779 GSM36780 GSM36781 [1,] 1.009274 1.0740659 1.048540 1.015946 1.022650 [2,] 1.007992 0.8768410 0.962442 1.111742 1.121150 [3,] 0.981853 0.9606492 1.024987 1.053302 1.063408 I also have the time in which each patient-sample is examined for relapse. This information is in vector y, which has length 286, and is declared in months. Indicatively: y[1:5] [1] 101 118 9 106 37 Finally, I have a variable censored, which is 1 if the patient has relapsed when examined at the examined time and 0 if not. Indicatively: censored[1:5] [1] 0 0 1 0 1 I am trying to perform the following sam analysis: d=list(data.matrix2,y,censored) samr.obj=samr(d,resp.type=Survival, nperms=20) When I am running the above commands I get the error: Error in check.format(y, resp.type = resp.type, censoring.status = censoring.status) : Error in input response data: response type Survival specified; error in censoring indicator In addition: Warning message: In is.na(x) : is.na() applied to non-(list or vector) of type 'NULL' I really cannot understand what is wrong with my code. Could anyone please help me with this? Thank you all, Eleni [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Spacing between lattice panels
On 5/6/08, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: I'm trying to set up a lattice plot with two y-axes for each panel. (Yes, I know that multiple y-axes are generally a bad idea; the graph is for someone else and they want it that way.) I've used a custom yscale.component in xyplot to achieve this: myyscale.component - function(...) { ans - yscale.components.default(...) ans$right - ans$left foo - ans$right$labels$at ans$right$labels$labels - as.character(10*foo) ans } xyplot(Sepal.Length ~ Petal.Length| Species, data = iris, scales = list(y=list(relation=free, alternating=3, rot=0)), yscale.component=myyscale.component) The problem is that the panels are too close together. This appears to be because (according to the help documentation) xyplot ignores the alternating argument when relation is not same. I presume that manually setting some plotting parameters with trellis.par.set() will allow me to adjust the spacing between the panels, though I can't figure out which ones to change. Is there a way to get lattice to properly space the panels, No. or failing that, how do I manually space them? The space between panels can be controlled using the 'between' argument. You additionally need some space at the right, for which you need to set a width component. E.g., xyplot(Sepal.Length ~ Petal.Length| Species, data = iris, scales = list(y=list(relation=free, alternating=3, rot=0)), yscale.component=myyscale.component, between = list(x = 2), par.settings = list(layout.widths = list(right.padding = 5))) The full list of components of 'layout.widths' is given by names(trellis.par.get(layout.widths)) -Deepayan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] single plot statement, multiple plots
Try plot.zoo in which case you don't need the par: library(zoo) plot(zoo(cbind(x1, x2, x3, x4)), nc = 2) or plot(zoo(outer(1:5, 1:4, ^)), nc = 2) See ?plot.zoo, ?xyplot.zoo and the three vignettes in the zoo package. On Tue, May 6, 2008 at 9:47 AM, Shubha Vishwanath Karanth [EMAIL PROTECTED] wrote: Hi R, par(mfrow=c(2,2)) x1=(1:5)^1; x2=(1:5)^2; x3=(1:5)^3; x4=(1:5)^4 I need to write a single plot statement, which creates 4 plots (for x1, x2, x3 and x4) in the graphics window, without using 'for' loop. Is this possible? Does 'do.call' help in this context? Or do I have any option in the 'plot' statement itself to do this? Thanks in advance, Shubha This e-mail may contain confidential and/or privileged i...{{dropped:13}} __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] single plot statement, multiple plots
Thank you very much Gabor...Zoo is very powerful... Shubha Karanth | Amba Research Ph +91 80 3980 8031 | Mob +91 94 4886 4510 Bangalore * Colombo * London * New York * San José * Singapore * www.ambaresearch.com -Original Message- From: Gabor Grothendieck [mailto:[EMAIL PROTECTED] Sent: Tuesday, May 06, 2008 7:40 PM To: Shubha Vishwanath Karanth Cc: [EMAIL PROTECTED] Subject: Re: [R] single plot statement, multiple plots Try plot.zoo in which case you don't need the par: library(zoo) plot(zoo(cbind(x1, x2, x3, x4)), nc = 2) or plot(zoo(outer(1:5, 1:4, ^)), nc = 2) See ?plot.zoo, ?xyplot.zoo and the three vignettes in the zoo package. On Tue, May 6, 2008 at 9:47 AM, Shubha Vishwanath Karanth [EMAIL PROTECTED] wrote: Hi R, par(mfrow=c(2,2)) x1=(1:5)^1; x2=(1:5)^2; x3=(1:5)^3; x4=(1:5)^4 I need to write a single plot statement, which creates 4 plots (for x1, x2, x3 and x4) in the graphics window, without using 'for' loop. Is this possible? Does 'do.call' help in this context? Or do I have any option in the 'plot' statement itself to do this? Thanks in advance, Shubha This e-mail may contain confidential and/or privileged i...{{dropped:13}} __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. This e-mail may contain confidential and/or privileged i...{{dropped:10}} __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] single plot statement, multiple plots
Shubha Vishwanath Karanth [EMAIL PROTECTED] wrote in news:[EMAIL PROTECTED]: Hi R, par(mfrow=c(2,2)) x1=(1:5)^1; x2=(1:5)^2; x3=(1:5)^3; x4=(1:5)^4 I need to write a single plot statement, which creates 4 plots (for x1, x2, x3 and x4) in the graphics window, without using 'for' loop. Is this possible? Does 'do.call' help in this context? Or do I have any option in the 'plot' statement itself to do this? Have you considered lapply()? -- David Winsemius __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to put different color in some portion of a surface plot?
Hi all, I have following problem : a = b = seq(1, 5, by=500) v = matrix(0, nrow=length(a), ncol=length(a)) for (i in 1:length(a)) { for (j in 1:length(a)) { d = c(17989*a[i], -18109*b[j]) v[i,j] = t(d) %*% matrix(c(0.0001741, 0.0001280, 0.0001280, 0.0002570), nrow=2) %*% d } } library(rgl) open3d() persp3d(a,b,v,col=green,alpha=0.7,aspect=c(1,1,0.5)) Now I want to shed the portion with different color of the surface which satisfy following condition: 0 (a-b) max(a) Can anyone please tell me how to do that? Regards, - [[elided Yahoo spam]] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] single plot statement, multiple plots
Wonderful...This works... lapply(list(x1,x2,x3,x4),plot,type=l) Thanks a lot! -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of David Winsemius Sent: Tuesday, May 06, 2008 7:55 PM To: [EMAIL PROTECTED] Subject: Re: [R] single plot statement, multiple plots Shubha Vishwanath Karanth [EMAIL PROTECTED] wrote in news:[EMAIL PROTECTED]: Hi R, par(mfrow=c(2,2)) x1=(1:5)^1; x2=(1:5)^2; x3=(1:5)^3; x4=(1:5)^4 I need to write a single plot statement, which creates 4 plots (for x1, x2, x3 and x4) in the graphics window, without using 'for' loop. Is this possible? Does 'do.call' help in this context? Or do I have any option in the 'plot' statement itself to do this? Have you considered lapply()? -- David Winsemius __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. This e-mail may contain confidential and/or privileged i...{{dropped:10}} __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to put different color in some portion of a surface plot?
On 5/6/2008 10:31 AM, Megh Dal wrote: Hi all, I have following problem : a = b = seq(1, 5, by=500) v = matrix(0, nrow=length(a), ncol=length(a)) for (i in 1:length(a)) { for (j in 1:length(a)) { d = c(17989*a[i], -18109*b[j]) v[i,j] = t(d) %*% matrix(c(0.0001741, 0.0001280, 0.0001280, 0.0002570), nrow=2) %*% d } } library(rgl) open3d() persp3d(a,b,v,col=green,alpha=0.7,aspect=c(1,1,0.5)) Now I want to shed the portion with different color of the surface which satisfy following condition: 0 (a-b) max(a) Can anyone please tell me how to do that? The col option can be a matrix of the same shape as v, giving the colour at each vertex. (Note that this is different from persp(), which gives the colour in each cell, i.e. one less in each dimension. The persp3d man page tells how to duplicate the persp() behaviour.) For example, shade - outer(a, b, function(x,y) (0 (x-y)) ((x-y) max(a))) persp3d(a,b,v,col=ifelse(shade, red, green), alpha=0.7,aspect=c(1,1,0.5)) Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] To preserve the class Matrix
Hi, Suppose a=matrix(1:9,3,3) a [,1] [,2] [,3] [1,]147 [2,]258 [3,]369 Now, class(a[1:2,]) [1] matrix class(a[1:3,]) [1] matrix class(a[,1:2]) [1] matrix class(a[,1:3]) [1] matrix But, class(a[1,]) [1] integer class(a[,1]) [1] integer Can in a general way get class(a[1,]) or class(a[,1]) to be matrix only? Shubha Karanth | Amba Research Ph +91 80 3980 8031 | Mob +91 94 4886 4510 Bangalore * Colombo * London * New York * San José * Singapore * www.ambaresearch.com This e-mail may contain confidential and/or privileged i...{{dropped:13}} __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] To preserve the class Matrix
Thank you very much Mark! That worked Just a question, ?[ does give an error to me...how do I find it? Shubha Karanth | Amba Research Ph +91 80 3980 8031 | Mob +91 94 4886 4510 Bangalore * Colombo * London * New York * San José * Singapore * www.ambaresearch.com -Original Message- From: Shubha Vishwanath Karanth Sent: Tuesday, May 06, 2008 8:50 PM To: 'Mark Leeds' Subject: RE: [R] To preserve the class Matrix Thank you very much Mark! That worked Just a question, ?[ does give an error to me...how do I find it? Shubha Karanth | Amba Research Ph +91 80 3980 8031 | Mob +91 94 4886 4510 Bangalore * Colombo * London * New York * San José * Singapore * www.ambaresearch.com -Original Message- From: Mark Leeds [mailto:[EMAIL PROTECTED] Sent: Tuesday, May 06, 2008 8:46 PM To: Shubha Vishwanath Karanth Subject: RE: [R] To preserve the class Matrix Hui Shubha: If I understand what you're asking, you want to use drop=FALSE as in a[1,,drop=FALSE] That retains the dimension ( ie, the matrixness ) of the object. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Shubha Vishwanath Karanth Sent: Tuesday, May 06, 2008 11:07 AM To: [EMAIL PROTECTED] Subject: [R] To preserve the class Matrix Hi, Suppose a=matrix(1:9,3,3) a [,1] [,2] [,3] [1,]147 [2,]258 [3,]369 Now, class(a[1:2,]) [1] matrix class(a[1:3,]) [1] matrix class(a[,1:2]) [1] matrix class(a[,1:3]) [1] matrix But, class(a[1,]) [1] integer class(a[,1]) [1] integer Can in a general way get class(a[1,]) or class(a[,1]) to be matrix only? Shubha Karanth | Amba Research Ph +91 80 3980 8031 | Mob +91 94 4886 4510 Bangalore * Colombo * London * New York * San Josi * Singapore * www.ambaresearch.com This e-mail may contain confidential and/or privileged i...{{dropped:13}} This e-mail may contain confidential and/or privileged i...{{dropped:10}} __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] split and subset
Dear list: I ask for your help in a simple problem in which I'm not figuring out the solution My data looks like: dat- data.frame(date=c(12/12/1980, 03/11/1994, 15/11/1999, 31/10/2000, 20/03/2007, 05/01/2001), var1=c(A, A, B, D, C, A), var2=runif(6)) I was wondering if I could split the column date in 3 new columns in he data frame corresponding to day, month and year. If I decide not to split the date in 3 columns, I'd do: dat$date- as.Date(dat$date, %d/%m/%Y) Now, how can I subset parts of the dataframe, lets say for a given month and year, something like subset(dat, date==1999-11-??) The ?? is obviously wrong. How should I do this? Many thanks -- [EMAIL PROTECTED] -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Significance analysis of Microarrays (SAM)
Hi Eleni -- Although samr is not a Bioconductor package, you might have more luck asking on the Bioconductor mailing list, http://bioconductor.org. The obvious place to start, and probably you have already done this, is to ensure that the class of the objects passed to the function agree with the classes described on the function help page. Martin Eleni Christodoulou [EMAIL PROTECTED] writes: Dear list, I am trying to perform a significance analysis of a microarray experiment with survival data using the {samr} package. I have a matrix containing my data which has 17816 rows corresponding to genes, and 286 columns corresponding to samples. The name of this matrix is data.matrix2. Some of the first values of this matrix are: data.matrix2[1:3,1:5] GSM36777 GSM36778 GSM36779 GSM36780 GSM36781 [1,] 1.009274 1.0740659 1.048540 1.015946 1.022650 [2,] 1.007992 0.8768410 0.962442 1.111742 1.121150 [3,] 0.981853 0.9606492 1.024987 1.053302 1.063408 I also have the time in which each patient-sample is examined for relapse. This information is in vector y, which has length 286, and is declared in months. Indicatively: y[1:5] [1] 101 118 9 106 37 Finally, I have a variable censored, which is 1 if the patient has relapsed when examined at the examined time and 0 if not. Indicatively: censored[1:5] [1] 0 0 1 0 1 I am trying to perform the following sam analysis: d=list(data.matrix2,y,censored) samr.obj=samr(d,resp.type=Survival, nperms=20) When I am running the above commands I get the error: Error in check.format(y, resp.type = resp.type, censoring.status = censoring.status) : Error in input response data: response type Survival specified; error in censoring indicator In addition: Warning message: In is.na(x) : is.na() applied to non-(list or vector) of type 'NULL' I really cannot understand what is wrong with my code. Could anyone please help me with this? Thank you all, Eleni [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Martin Morgan Computational Biology / Fred Hutchinson Cancer Research Center 1100 Fairview Ave. N. PO Box 19024 Seattle, WA 98109 Location: Arnold Building M2 B169 Phone: (206) 667-2793 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Scatter Plot - 3 vectors side by side
Hi, I have 3 vectors, x=rnorm(10); y=rnorm(20); z=rnorm(30). I would like to plot 3 vectors side by side (like a bar plot) with scatter plot something similar to the following: .. *** ;;; .. ** ;;; .. *** ;;; .. *** ;;; ... *** ;; x y z How can I do this with Plot()? Thanks in advance. Best regards, Ezhil [[elided Yahoo spam]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Announcement: support of littler, rkward and rpy on Debian/Ubuntu
Dear useRs, This is to announce that the maintainers of the various distributions have decided to provide experimental up-to-date versions of the following R related packages on Debian stable and Ubuntu (i386 and amd64 architectures): littler rkward python-rpy(not on Ubuntu Dapper) python-rpy-doc (not on Ubuntu Dapper) By experimental we mostly mean that support for these packages may remain somewhat of a moving target for some time. The maintainers are not necessarily users of these packages themselves, so positive and negative feedback through the usual channels (see below) would be appreciated. See the Debian and Ubuntu README files on CRAN for more details: http://cran.r-project.org/bin/linux/debian/ http://cran.r-project.org/bin/linux/ubuntu/ This message is CCd to r-help to reach as wide an audience as possible, but we remind that the best place to report problems with these packages or to ask R questions specific to Debian and Ubuntu is the R-SIG-Debian mailing list. See https://stat.ethz.ch/mailman/listinfo/r-sig-debian for more information. --- Vincent Goulet, Associate Professor École d'actuariat Université Laval, Québec [EMAIL PROTECTED] http://vgoulet.act.ulaval.ca __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] split and subset
Try this: subset(dat, format(date, %Y-%m) == 1999-11) On Tue, May 6, 2008 at 12:25 PM, [EMAIL PROTECTED] wrote: Dear list: I ask for your help in a simple problem in which I'm not figuring out the solution My data looks like: dat- data.frame(date=c(12/12/1980, 03/11/1994, 15/11/1999, 31/10/2000, 20/03/2007, 05/01/2001), var1=c(A, A, B, D, C, A), var2=runif(6)) I was wondering if I could split the column date in 3 new columns in he data frame corresponding to day, month and year. If I decide not to split the date in 3 columns, I'd do: dat$date- as.Date(dat$date, %d/%m/%Y) Now, how can I subset parts of the dataframe, lets say for a given month and year, something like subset(dat, date==1999-11-??) The ?? is obviously wrong. How should I do this? Many thanks -- [EMAIL PROTECTED] -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Scatter Plot - 3 vectors side by side
On 5/6/2008 11:46 AM, A Ezhil wrote: Hi, I have 3 vectors, x=rnorm(10); y=rnorm(20); z=rnorm(30). I would like to plot 3 vectors side by side (like a bar plot) with scatter plot something similar to the following: .. *** ;;; .. ** ;;; .. *** ;;; .. *** ;;; ... *** ;; x y z How can I do this with Plot()? How about something like this: stripchart(c(x,y,z) ~ rep(c(x,y,z), c(10,20,30)), vertical=TRUE, ylab=) Thanks in advance. Best regards, Ezhil [[elided Yahoo spam]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Chuck Cleland, Ph.D. NDRI, Inc. (www.ndri.org) 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Is there in R a function equivalent to the mround, as found in most spreadsheets?
Dear R-users, I have the following problem In a lab experiment I have to mix three solutions to get different concentrations of various molecules in a cuvette I've used R to calculate the necessary µliters for each of the level of the experiment and I must confess that it is more useful and easier to achieve the results than using spreadsheets. But there's a problem. Imagine that for a particular cuvette (I have 112 different cuvettes !!) you have to mix the following volumes of solution A, B, and C respectively. c(1803.02, 193.51, 3.47) Each solution is to be taken with 3 different pipettes (5000, 250 and 10 µL Volume max) and each of those delivers volumes in steps of 50 µL, 5 µL or 1µL, respectively Since the above values would eventually become c(1800, 195, 3) it is then necessary to recalculate all the final concentrations of A, B and C, because the volumes are changed. I know that in most spreadsheets (Calc in Open Office, Gnumeric, Excel and so on) there's a function such as mround(num; num) that give the results I need, but I want to learn more on R functions. I played a little with R functions such as round, signif, ceiling, trunc, and floor but without success. Any hint to solve this problem ? Thanks a lot http://www.openofficetips.com/blog/archives/2005/04/rounding_to_the.html http://www.gnome.org/projects/gnumeric/doc/gnumeric-MROUND.shtml -- Ottorino-Luca Pantani, Università di Firenze Dip. Scienza del Suolo e Nutrizione della Pianta P.zle Cascine 28 50144 Firenze Italia Tel 39 055 3288 202 (348 lab) Fax 39 055 333 273 [EMAIL PROTECTED] http://www4.unifi.it/dssnp/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] merge numerous columns of unequal length
Indeed both options are salable, though I agree the latter may be more convenient. Merci! Patrick Burns wrote: As the answers you've received suggest, you can use a list. Or you could have two vectors: one with the data, the other with the group identity. The latter format is likely more convenient for a lot of analyses. Since your data are not inherently rectangular, it is probably best to get the idea of spreadsheet out of your head. (It is probably best anyway.) Patrick Burns [EMAIL PROTECTED] +44 (0)20 8525 0696 http://www.burns-stat.com (home of S Poetry and A Guide for the Unwilling S User) T.D.Rudolph wrote: I have numerous objects, each containing continuous data representing the same variable, movement rate, yet each having a different number of rows. e.g. d1-as.matrix(rnorm(5)) d2-as.matrix(rnorm(3)) d3-as.matrix(rnorm(6)) How can I merge these three columns side-by-side in order to create a table regardless of the difference in length? I wish to analyze the output in a spreadsheet format. Thanks! Tyler __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/merge-numerous-columns-of-unequal-length-tp17071464p17086657.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] unable to use functions require DLL from package base
Hi all, I have issues using some basic functions in R such as these ones : pp.test(R) (where is a vector of returns) Error in .C(R_approx, as.double(x), as.double(y), as.integer(nx), xout = as.double(xout), : C symbol name R_approx not in DLL for package base boxcox(reg,plotit=T) (where reg is an lm object) Error in .C(spline_coef, method = as.integer(method), n = as.integer(nx), : C symbol name spline_coef not in DLL for package base as I do miss some symbol names. How can I overcome this serious problem ? -- View this message in context: http://www.nabble.com/unable-to-use-functions-require-DLL-from-package-%22base%22-tp17086783p17086783.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regarding nls()
My summary of Bates' comments cited below is as follows: 1. ANOVA is an excellent tool but requires nested models. You can do this fairly easily, but it is not so easily automated. 2. The standard definition of R^2 loses its meaning with nonlinear models. Adjusted R^2 is even worse. Bates' condemnation of R^2 has merit, but I would not go as far as he did in the comment cited below (dated 13 Aug 2000). A standard definition of R^2 is as follows: R^2 = (1 - var(prediction error) / var(obs)). I can name several different ways of getting a negative R^2 in this case. When that happens, it says the model is worse than useless, and you would be better off using the training set mean. If I have an audience who wants an R^2 in an application where it is not clear what it even means, I try to briefly explain some of the difficulties while asking what question they are trying to solve using R^2. Their answers will help me make a recommendation, which may include selecting which of the possible generalizations of R^2 to use. Hope this helps. Spencer Graves Dieter Menne wrote: Guru S guru.rcom at rediffmail.com writes: I have no problem performing the regression using R, and I successfully obtain the parameter estimates using the function nls(). However, how do I obtain the ANOVA output, r, r^2 and adj. r^2? This is a feature, not a bug. See Douglas Bates's comments on http://www.ens.gu.edu.au/ROBERTK/R/HELP/00B/0399.HTML Dieter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] categorical data analysis
hie all i am trying to carry out a categorical data analysis but my problem is that when in i use the chi squared test some of my expected values are less than 5. is there a test that can handle this situation. the data is not a 2*2 table. its more from the social sciences where you have from strongly agree to strongly disagree. i know i can collapse vthe tables but there is a loss of information . is the a test vthat i can for this kind of data. thanks in advancde. ray - [[elided Yahoo spam]] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is there in R a function equivalent to the mround, as found in most spreadsheets?
On 5/6/2008 12:07 PM, Dr. Ottorino-Luca Pantani wrote: Dear R-users, I have the following problem In a lab experiment I have to mix three solutions to get different concentrations of various molecules in a cuvette I've used R to calculate the necessary µliters for each of the level of the experiment and I must confess that it is more useful and easier to achieve the results than using spreadsheets. But there's a problem. Imagine that for a particular cuvette (I have 112 different cuvettes !!) you have to mix the following volumes of solution A, B, and C respectively. c(1803.02, 193.51, 3.47) Each solution is to be taken with 3 different pipettes (5000, 250 and 10 µL Volume max) and each of those delivers volumes in steps of 50 µL, 5 µL or 1µL, respectively Since the above values would eventually become c(1800, 195, 3) it is then necessary to recalculate all the final concentrations of A, B and C, because the volumes are changed. I know that in most spreadsheets (Calc in Open Office, Gnumeric, Excel and so on) there's a function such as mround(num; num) that give the results I need, but I want to learn more on R functions. I played a little with R functions such as round, signif, ceiling, trunc, and floor but without success. Any hint to solve this problem ? I believe this function matches the description in OOO: mround - function(number, multiple) multiple * round(number/multiple) Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] General Plotting Question
f - (structure(list(X = structure(96:97, .Label = c(119DAmm, 119DN, 119DNN, 119DO, 119DOC, 119Flow, 119Nit, 119ON, 119OPhos, 119OrgP, 119Phos, 119TKN, 119TOC, 148DAmm, 148DN, 148DNN, 148DO, 148DOC, 148Flow, 148Nit, 148ON, 148OPhos, 148OrgP, 148Phos, 148TKN, 148TOC, 179DAmm, 179DN, 179DNN, 179DO, 179DOC, 179Flow, 179Nit, 179ON, 179OPhos, 179OrgP, 179Phos, 179TKN, 179TOC, 185DAmm, 185DN, 185DNN, 185DO, 185DOC, 185Flow, 185Nit, 185ON, 185OPhos, 185OrgP, 185Phos, 185TKN, 185TOC, 190DAmm, 190DN, 190DNN, 190DO, 190DOC, 190Flow, 190Nit, 190ON, 190OPhos, 190OrgP, 190Phos, 190TKN, 190TOC, 198DAmm, 198DN, 198DNN, 198DO, 198DOC, 198Flow, 198Nit, 198ON, 198OPhos, 198OrgP, 198Phos, 198TKN, 198TOC, 202DAmm, 202DN, 202DNN, 202DO, 202DOC, 202Flow, 202Nit, 202ON, 202OPhos, 202OrgP, 202Phos, 202TKN, 202TOC, 215DAmm, 215DN, 215DNN, 215DO, 215DOC, 215Flow, 215Nit, 215ON, 215OPhos, 215OrgP, 215Phos, 215TKN, 215TOC, 61DAmm, 61DN, 61DNN, 61DO, 61DOC, 61Flow, 61Nit, 61ON, 61OPhos, 61OrgP, 61Phos, 61TKN, 61TOC, BCOPhos, BCOrgP, BCPhos, BCTKN, BCTOC, HCDAmm, HCDN, HCDNN, HCDO, HCDOC, HCFlow, HCNit, HCON, HCOPhos, HCOrgP, HCPhos, HCTKN, HCTOC, SCDAmm, SCDN, SCDNN, SCDO, SCDOC, SCFlow, SCNit, SCON, SCOPhos, SCOrgP, SCPhos, SCTKN, SCTOC), class = factor), RiverMile = structure(c(8L, 8L), .Label = c(119, 148, 179, 185, 190, 198, 202, 215, 61, BC, HC, SC), class = factor), Constituent = structure(5:6, .Label = c(DAmm, DN, DNN, DO, DOC, Flow, Nit, ON, OPhos, OrgP, Phos, TKN, TOC), class = factor), X2.1.06 = c(2.5, 2494), X4.1.06 = c(2.4, NA), X5.1.06 = c(2.3, NA), X6.1.06 = c(2.1, NA), X7.1.06 = c(2, NA), X8.1.06 = c(1.9, NA), X9.1.06 = c(2, NA), X10.1.06 = c(2.1, NA), X11.1.06 = c(2.9, NA), X12.1.06 = c(2.6, NA), X1.1.07 = c(2.1, 4229.55), X2.1.07 = c(2.1, 1823.5), X3.1.07 = c(2.8, 3617.5), X4.1.07 = c(2.3, NA), X5.1.07 = c(3.5, NA), X6.1.07 = c(2.8, 10974.5), X7.1.07 = c(2.5, 9652), X8.1.07 = c(2.4, 6700.5), X9.1.07 = c(2.2, 11438.5), X10.1.07 = structure(c(34L, 33L), .Label = c(-0.0214, -0.3660, #VALUE!, 0., 0.0010, 0.0040, 0.0060, 0.0080, 0.0086, 0.0090, 0.0100, 0.0110, 0.0120, 0.0140, 0.0220, 0.0230, 0.0240, 0.0280, 0.0300, 0.0720, 0.0890, 0.1000, 0.1200, 0.1280, 0.1400, 0.1500, 0.1880, 0.1910, 0.2000, 0.2080, 0.2200, 0.3900, 1141., 2.2000, 2.3000, 2.5000, 2.6000, 2606., 3., 3.4000, 4.5000, 4.6000, 4710., 4720., 4729., 6.9768, 7.7065, 7.7338, 8.0097), class = factor), X11.1.07 = c(2.2, 10968), X12.1.07 = c(2.2, 7361.5), X1.1.08 = c(2, 6797)), .Names = c(X, RiverMile, Constituent, X2.1.06, X4.1.06, X5.1.06, X6.1.06, X7.1.06, X8.1.06, X9.1.06, X10.1.06, X11.1.06, X12.1.06, X1.1.07, X2.1.07, X3.1.07, X4.1.07, X5.1.07, X6.1.07, X7.1.07, X8.1.07, X9.1.07, X10.1.07, X11.1.07, X12.1.07, X1.1.08), row.names = c(NA, -2L), class = data.frame)) plot(f[2,4:26], f[1,4:26]) #Error in plot.window(...) : need finite 'xlim' values #In addition: Warning messages: #1: In min(x) : no non-missing arguments to min; returning Inf #2: In max(x) : no non-missing arguments to max; returning -Inf #I want a scatterplot like what I thought would happen with the plot command and the rows and columns that contain the data. If there is a NA in either case there is no point . plotted. This is how plot works with the data in columns, I think. This is a small example of a much larger data set are there any suggestions. What am I missing? thanks stephen -- Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] General Plotting Question
Try: x - unlist(f[2, 4:26]) y - unlist(f[1, 4:26]) plot(x, y) On Tue, May 6, 2008 at 12:43 PM, stephen sefick [EMAIL PROTECTED] wrote: f - (structure(list(X = structure(96:97, .Label = c(119DAmm, 119DN, 119DNN, 119DO, 119DOC, 119Flow, 119Nit, 119ON, 119OPhos, 119OrgP, 119Phos, 119TKN, 119TOC, 148DAmm, 148DN, 148DNN, 148DO, 148DOC, 148Flow, 148Nit, 148ON, 148OPhos, 148OrgP, 148Phos, 148TKN, 148TOC, 179DAmm, 179DN, 179DNN, 179DO, 179DOC, 179Flow, 179Nit, 179ON, 179OPhos, 179OrgP, 179Phos, 179TKN, 179TOC, 185DAmm, 185DN, 185DNN, 185DO, 185DOC, 185Flow, 185Nit, 185ON, 185OPhos, 185OrgP, 185Phos, 185TKN, 185TOC, 190DAmm, 190DN, 190DNN, 190DO, 190DOC, 190Flow, 190Nit, 190ON, 190OPhos, 190OrgP, 190Phos, 190TKN, 190TOC, 198DAmm, 198DN, 198DNN, 198DO, 198DOC, 198Flow, 198Nit, 198ON, 198OPhos, 198OrgP, 198Phos, 198TKN, 198TOC, 202DAmm, 202DN, 202DNN, 202DO, 202DOC, 202Flow, 202Nit, 202ON, 202OPhos, 202OrgP, 202Phos, 202TKN, 202TOC, 215DAmm, 215DN, 215DNN, 215DO, 215DOC, 215Flow, 215Nit, 215ON, 215OPhos, 215OrgP, 215Phos, 215TKN, 215TOC, 61DAmm, 61DN, 61DNN, 61DO, 61DOC, 61Flow, 61Nit, 61ON, 61OPhos, 61OrgP, 61Phos, 61TKN, 61TOC, BCOPhos, BCOrgP, BCPhos, BCTKN, BCTOC, HCDAmm, HCDN, HCDNN, HCDO, HCDOC, HCFlow, HCNit, HCON, HCOPhos, HCOrgP, HCPhos, HCTKN, HCTOC, SCDAmm, SCDN, SCDNN, SCDO, SCDOC, SCFlow, SCNit, SCON, SCOPhos, SCOrgP, SCPhos, SCTKN, SCTOC), class = factor), RiverMile = structure(c(8L, 8L), .Label = c(119, 148, 179, 185, 190, 198, 202, 215, 61, BC, HC, SC), class = factor), Constituent = structure(5:6, .Label = c(DAmm, DN, DNN, DO, DOC, Flow, Nit, ON, OPhos, OrgP, Phos, TKN, TOC), class = factor), X2.1.06 = c(2.5, 2494), X4.1.06 = c(2.4, NA), X5.1.06 = c(2.3, NA), X6.1.06 = c(2.1, NA), X7.1.06 = c(2, NA), X8.1.06 = c(1.9, NA), X9.1.06 = c(2, NA), X10.1.06 = c(2.1, NA), X11.1.06 = c(2.9, NA), X12.1.06 = c(2.6, NA), X1.1.07 = c(2.1, 4229.55), X2.1.07 = c(2.1, 1823.5), X3.1.07 = c(2.8, 3617.5), X4.1.07 = c(2.3, NA), X5.1.07 = c(3.5, NA), X6.1.07 = c(2.8, 10974.5), X7.1.07 = c(2.5, 9652), X8.1.07 = c(2.4, 6700.5), X9.1.07 = c(2.2, 11438.5), X10.1.07 = structure(c(34L, 33L), .Label = c(-0.0214, -0.3660, #VALUE!, 0., 0.0010, 0.0040, 0.0060, 0.0080, 0.0086, 0.0090, 0.0100, 0.0110, 0.0120, 0.0140, 0.0220, 0.0230, 0.0240, 0.0280, 0.0300, 0.0720, 0.0890, 0.1000, 0.1200, 0.1280, 0.1400, 0.1500, 0.1880, 0.1910, 0.2000, 0.2080, 0.2200, 0.3900, 1141., 2.2000, 2.3000, 2.5000, 2.6000, 2606., 3., 3.4000, 4.5000, 4.6000, 4710., 4720., 4729., 6.9768, 7.7065, 7.7338, 8.0097), class = factor), X11.1.07 = c(2.2, 10968), X12.1.07 = c(2.2, 7361.5), X1.1.08 = c(2, 6797)), .Names = c(X, RiverMile, Constituent, X2.1.06, X4.1.06, X5.1.06, X6.1.06, X7.1.06, X8.1.06, X9.1.06, X10.1.06, X11.1.06, X12.1.06, X1.1.07, X2.1.07, X3.1.07, X4.1.07, X5.1.07, X6.1.07, X7.1.07, X8.1.07, X9.1.07, X10.1.07, X11.1.07, X12.1.07, X1.1.08), row.names = c(NA, -2L), class = data.frame)) plot(f[2,4:26], f[1,4:26]) #Error in plot.window(...) : need finite 'xlim' values #In addition: Warning messages: #1: In min(x) : no non-missing arguments to min; returning Inf #2: In max(x) : no non-missing arguments to max; returning -Inf #I want a scatterplot like what I thought would happen with the plot command and the rows and columns that contain the data. If there is a NA in either case there is no point . plotted. This is how plot works with the data in columns, I think. This is a small example of a much larger data set are there any suggestions. What am I missing? thanks stephen -- Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is there in R a function equivalent to the mround, as found in most spreadsheets?
Hi Ottorino, You could just use the modulus operator %% as follows: x-c(1803.02, 193.51, 3.47); x-x%%c(50,5,1) #just using the modulus operator [1] 1800 1903 thanks Dr. Ottorino-Luca Pantani wrote: Dear R-users, I have the following problem In a lab experiment I have to mix three solutions to get different concentrations of various molecules in a cuvette I've used R to calculate the necessary µliters for each of the level of the experiment and I must confess that it is more useful and easier to achieve the results than using spreadsheets. But there's a problem. Imagine that for a particular cuvette (I have 112 different cuvettes !!) you have to mix the following volumes of solution A, B, and C respectively. c(1803.02, 193.51, 3.47) Each solution is to be taken with 3 different pipettes (5000, 250 and 10 µL Volume max) and each of those delivers volumes in steps of 50 µL, 5 µL or 1µL, respectively Since the above values would eventually become c(1800, 195, 3) it is then necessary to recalculate all the final concentrations of A, B and C, because the volumes are changed. I know that in most spreadsheets (Calc in Open Office, Gnumeric, Excel and so on) there's a function such as mround(num; num) that give the results I need, but I want to learn more on R functions. I played a little with R functions such as round, signif, ceiling, trunc, and floor but without success. Any hint to solve this problem ? Thanks a lot http://www.openofficetips.com/blog/archives/2005/04/rounding_to_the.html http://www.gnome.org/projects/gnumeric/doc/gnumeric-MROUND.shtml -- Ottorino-Luca Pantani, Università di Firenze Dip. Scienza del Suolo e Nutrizione della Pianta P.zle Cascine 28 50144 Firenze Italia Tel 39 055 3288 202 (348 lab) Fax 39 055 333 273 [EMAIL PROTECTED] http://www4.unifi.it/dssnp/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - Yasir H. Kaheil, Ph.D. Catchment Research Facility The University of Western Ontario -- View this message in context: http://www.nabble.com/Is-there-in-R-a-function-equivalent-to-the-mround%2C-as-found-in-most-spreadsheets--tp17086856p17087399.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Negative memory size: -1012.854
Dear all, With R.2.7.0 (on windows XP) I have encountered what seems to be a negative memory size. If I use gc() afterwards the R goes down. I use R_alloc for *most* memory allocations in my C-routines: memory.size() [1] 11.08132 dyn.load(rconipm.dll) dyn.unload(rconipm.dll) memory.size() [1] -1012.854 Has anybody encountered similar problems? (It all worked well under R.2.6.2, but I am of course not blaming R for failing). Regards Søren __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] trouble installing Rmpi on 64-bit Ubuntu 8.04 with openmpi
Subject pretty much says it all. I am running 64-bit Ubuntu 8.04, i.e. Hardy Heron, have openmpi installed, and get the following error message with attempted install of Rmpi. sessionInfo() follows. Mark checking for ANSI C header files... yes checking for sys/types.h... yes checking for sys/stat.h... yes checking for stdlib.h... yes checking for string.h... yes checking for memory.h... yes checking for strings.h... yes checking for inttypes.h... yes checking for stdint.h... yes checking for unistd.h... yes checking mpi.h usability... no checking mpi.h presence... no checking for mpi.h... no Try to find libmpi.so or libmpich.a checking for main in -lmpi... yes checking for openpty in -lutil... yes checking for main in -lpthread... yes configure: creating ./config.status config.status: creating src/Makevars ** libs gcc -std=gnu99 -I/home/mkimpel/R_HOME/R-patched/R-build/lib64/R/include -DPACKAGE_NAME=\\ -DPACKAGE_TARNAME=\\ -DPACKAGE_VERSION=\\ -DPACKAGE_STRING=\\ -DPACKAGE_BUGREPORT=\\ -DSTDC_HEADERS=1 -DHAVE_SYS_TYPES_H=1 -DHAVE_SYS_STAT_H=1 -DHAVE_STDLIB_H=1 -DHAVE_STRING_H=1 -DHAVE_MEMORY_H=1 -DHAVE_STRINGS_H=1 -DHAVE_INTTYPES_H=1 -DHAVE_STDINT_H=1 -DHAVE_UNISTD_H=1 -DUNKNOWN -fPIC -I/usr/local/include-fpic -g -O2 -c conversion.c -o conversion.o In file included from conversion.c:18: Rmpi.h:1:17: error: mpi.h: No such file or directory In file included from conversion.c:18: Rmpi.h:14: error: expected '=', ',', ';', 'asm' or '__attribute__' before 'mpitype' make: *** [conversion.o] Error 1 chmod: cannot access `/home/mkimpel/R_HOME/site-library-2.7.0/Rmpi/libs/*': No such file or directory ERROR: compilation failed for package 'Rmpi' ** Removing '/home/mkimpel/R_HOME/site-library-2.7.0/Rmpi' The downloaded packages are in /tmp/RtmppcK0FI/downloaded_packages Warning message: sessionInfo() R version 2.7.0 Patched (2008-05-04 r45620) x86_64-unknown-linux-gnu locale: LC_CTYPE=en_US.UTF-8;LC_NUMERIC=C;LC_TIME=en_US.UTF-8;LC_COLLATE=en_US.UTF-8;LC_MONETARY=C;LC_MESSAGES=en_US.UTF-8;LC_PAPER=en_US.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_US.UTF-8;LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] graph_1.18.0 loaded via a namespace (and not attached): [1] cluster_1.11.10 tcltk_2.7.0 tools_2.7.0 -- Mark W. Kimpel MD ** Neuroinformatics ** Dept. of Psychiatry Indiana University School of Medicine 15032 Hunter Court, Westfield, IN 46074 (317) 490-5129 Work, Mobile VoiceMail (317) 663-0513 Home (no voice mail please) ** [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] General Plotting Question
cast the two vectors as.matrix-- see here: plot(as.matrix(f[2,4:26]), as.matrix(f[1,4:26])) y stephen sefick wrote: f - (structure(list(X = structure(96:97, .Label = c(119DAmm, 119DN, 119DNN, 119DO, 119DOC, 119Flow, 119Nit, 119ON, 119OPhos, 119OrgP, 119Phos, 119TKN, 119TOC, 148DAmm, 148DN, 148DNN, 148DO, 148DOC, 148Flow, 148Nit, 148ON, 148OPhos, 148OrgP, 148Phos, 148TKN, 148TOC, 179DAmm, 179DN, 179DNN, 179DO, 179DOC, 179Flow, 179Nit, 179ON, 179OPhos, 179OrgP, 179Phos, 179TKN, 179TOC, 185DAmm, 185DN, 185DNN, 185DO, 185DOC, 185Flow, 185Nit, 185ON, 185OPhos, 185OrgP, 185Phos, 185TKN, 185TOC, 190DAmm, 190DN, 190DNN, 190DO, 190DOC, 190Flow, 190Nit, 190ON, 190OPhos, 190OrgP, 190Phos, 190TKN, 190TOC, 198DAmm, 198DN, 198DNN, 198DO, 198DOC, 198Flow, 198Nit, 198ON, 198OPhos, 198OrgP, 198Phos, 198TKN, 198TOC, 202DAmm, 202DN, 202DNN, 202DO, 202DOC, 202Flow, 202Nit, 202ON, 202OPhos, 202OrgP, 202Phos, 202TKN, 202TOC, 215DAmm, 215DN, 215DNN, 215DO, 215DOC, 215Flow, 215Nit, 215ON, 215OPhos, 215OrgP, 215Phos, 215TKN, 215TOC, 61DAmm, 61DN, 61DNN, 61DO, 61DOC, 61Flow, 61Nit, 61ON, 61OPhos, 61OrgP, 61Phos, 61TKN, 61TOC, BCOPhos, BCOrgP, BCPhos, BCTKN, BCTOC, HCDAmm, HCDN, HCDNN, HCDO, HCDOC, HCFlow, HCNit, HCON, HCOPhos, HCOrgP, HCPhos, HCTKN, HCTOC, SCDAmm, SCDN, SCDNN, SCDO, SCDOC, SCFlow, SCNit, SCON, SCOPhos, SCOrgP, SCPhos, SCTKN, SCTOC), class = factor), RiverMile = structure(c(8L, 8L), .Label = c(119, 148, 179, 185, 190, 198, 202, 215, 61, BC, HC, SC), class = factor), Constituent = structure(5:6, .Label = c(DAmm, DN, DNN, DO, DOC, Flow, Nit, ON, OPhos, OrgP, Phos, TKN, TOC), class = factor), X2.1.06 = c(2.5, 2494), X4.1.06 = c(2.4, NA), X5.1.06 = c(2.3, NA), X6.1.06 = c(2.1, NA), X7.1.06 = c(2, NA), X8.1.06 = c(1.9, NA), X9.1.06 = c(2, NA), X10.1.06 = c(2.1, NA), X11.1.06 = c(2.9, NA), X12.1.06 = c(2.6, NA), X1.1.07 = c(2.1, 4229.55), X2.1.07 = c(2.1, 1823.5), X3.1.07 = c(2.8, 3617.5), X4.1.07 = c(2.3, NA), X5.1.07 = c(3.5, NA), X6.1.07 = c(2.8, 10974.5), X7.1.07 = c(2.5, 9652), X8.1.07 = c(2.4, 6700.5), X9.1.07 = c(2.2, 11438.5), X10.1.07 = structure(c(34L, 33L), .Label = c(-0.0214, -0.3660, #VALUE!, 0., 0.0010, 0.0040, 0.0060, 0.0080, 0.0086, 0.0090, 0.0100, 0.0110, 0.0120, 0.0140, 0.0220, 0.0230, 0.0240, 0.0280, 0.0300, 0.0720, 0.0890, 0.1000, 0.1200, 0.1280, 0.1400, 0.1500, 0.1880, 0.1910, 0.2000, 0.2080, 0.2200, 0.3900, 1141., 2.2000, 2.3000, 2.5000, 2.6000, 2606., 3., 3.4000, 4.5000, 4.6000, 4710., 4720., 4729., 6.9768, 7.7065, 7.7338, 8.0097), class = factor), X11.1.07 = c(2.2, 10968), X12.1.07 = c(2.2, 7361.5), X1.1.08 = c(2, 6797)), .Names = c(X, RiverMile, Constituent, X2.1.06, X4.1.06, X5.1.06, X6.1.06, X7.1.06, X8.1.06, X9.1.06, X10.1.06, X11.1.06, X12.1.06, X1.1.07, X2.1.07, X3.1.07, X4.1.07, X5.1.07, X6.1.07, X7.1.07, X8.1.07, X9.1.07, X10.1.07, X11.1.07, X12.1.07, X1.1.08), row.names = c(NA, -2L), class = data.frame)) plot(f[2,4:26], f[1,4:26]) #Error in plot.window(...) : need finite 'xlim' values #In addition: Warning messages: #1: In min(x) : no non-missing arguments to min; returning Inf #2: In max(x) : no non-missing arguments to max; returning -Inf #I want a scatterplot like what I thought would happen with the plot command and the rows and columns that contain the data. If there is a NA in either case there is no point . plotted. This is how plot works with the data in columns, I think. This is a small example of a much larger data set are there any suggestions. What am I missing? thanks stephen -- Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - Yasir H. Kaheil, Ph.D. Catchment Research Facility The University of Western Ontario -- View this message in context: http://www.nabble.com/General-Plotting-Question-tp17087346p17087502.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] rggobi is crashing R-2.7.0
I am running 64-bit Ubuntu 8.04 and when I invoke rggobi the interactive graph displays but R crashes. See my sessionInfo() and a short example below. Ggobi and rggobi installed without complaints. Mark sessionInfo() R version 2.7.0 Patched (2008-05-04 r45620) x86_64-unknown-linux-gnu locale: LC_CTYPE=en_US.UTF-8;LC_NUMERIC=C;LC_TIME=en_US.UTF-8;LC_COLLATE=en_US.UTF-8;LC_MONETARY=C;LC_MESSAGES=en_US.UTF-8;LC_PAPER=en_US.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_US.UTF-8;LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] rggobi_2.1.9 RGtk2_2.12.5-3 graph_1.18.0 loaded via a namespace (and not attached): [1] cluster_1.11.10 tools_2.7.0 a - matrix(rnorm(1000), nrow = 10) g - ggobi(a) ** (R:25146): CRITICAL **: Error on loading plugin library plugins/GraphLayout/plugin.la: libgvc.so.3: cannot open shared object file: No such file or directory ** (R:25146): CRITICAL **: Error on loading plugin library plugins/GraphLayout/plugin.la: libgvc.so.3: cannot open shared object file: No such file or directory ** (R:25146): CRITICAL **: can't locate required plugin routine addToToolsMenu in GraphLayout -- Mark W. Kimpel MD ** Neuroinformatics ** Dept. of Psychiatry Indiana University School of Medicine 15032 Hunter Court, Westfield, IN 46074 (317) 490-5129 Work, Mobile VoiceMail (317) 663-0513 Home (no voice mail please) ** [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Plotting separate populations in scatterplot3d
Hi, I'm having trouble plotting populations as separate colors and points in the 3d scatterplot package. I have a column with 4 different population names and 3 columns with my data. I want to plot each population with a different color and pch. In addition, I want to use the type=h in my plotting so that I can see the points clearly in my graph( for publication purposes). I've also attached an example dataset. I simply want to show that each population clusters separately in space. I've attached a visual representation of what I want my data to look like. The scripts for this graph is as follows: data(trees) s3d - scatterplot3d(trees, type = h, color = blue, angle = 55, scale.y = 0.7, pch = 16, main = Adding elements) For some reason, R will not read my header, so I had to use the data.frame function to get R to read my data. This is a real pain because I have tons of data. I've tried the following to get R to recognize each population type separately: PopulationType.LK1-PopulationType[LK1] PopulationType.LK2-PopulationType[LK2] points((PopulationType.LK1, col=red, pch=19), (PopulationType.LK2, col=blue, pch=22), .) When I tweak the example tree code from above, R will plot all of my data no problem (with the rgl 3d graphs, not scatterplot3d), but only with one color and plotting character. It seems pretty straightforward, but I've spent 4 days on this already. Any help would be greatly appreciated. Regards, EJ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plotting separate populations in scatterplot3d
#Install library rgl #here is the function which you need to run first: rgl.plot3d-function(z, x, y, cols=red,axes=T,new=T) {xr-range(x) x01-(x-xr[1])/(xr[2]-xr[1]) yr-range(y) y01-(y-yr[1])/(yr[2]-yr[1]) zr-range(z) z01-(z-zr[1])/(zr[2]-zr[1]) if(new) rgl.clear() if(axes) {xlab-pretty(x) ylab-pretty(y) zlab-pretty(z) xat-(xlab-xr[1])/(xr[2]-xr[1]) yat-(ylab-yr[1])/(yr[2]-yr[1]) zat-(zlab-zr[1])/(zr[2]-zr[1]) rgl.lines(c(0,1.1),0,0) rgl.lines(0,c(0,1.1),0) rgl.lines(0,0,c(0,1.1)) rgl.texts(xat,-.05,-.05,xlab) rgl.texts(-.05,yat,-.05,ylab) rgl.texts(-.05,-.05,zat,zlab) rgl.texts(c(0.5,-.15,-.15),c(-.15,.5,-.15),c(-.15,-.15,.5), c(deparse(substitute(x)),deparse(substitute(y)),deparse(substitute(z } rgl.spheres(x01,y01,z01,.01,color=cols) } #and here is how you call it library(rgl) data(iris) iris.pc-prcomp(iris[,1:4],scale=T) rgl.plot3d(iris.pc$x[,1],iris.pc$x[,2],iris.pc$x[,3]) # different colors rgl.plot3d(iris.pc$x[,1],iris.pc$x[,2],iris.pc$x[,3],col=unclass(iris[,5])+1) Refer to: http://www.nabble.com/interactive-rotatable-3d-scatterplot-td17030023.html#a17030164 thanks y Eleca Dunham wrote: Hi, I'm having trouble plotting populations as separate colors and points in the 3d scatterplot package. I have a column with 4 different population names and 3 columns with my data. I want to plot each population with a different color and pch. In addition, I want to use the type=h in my plotting so that I can see the points clearly in my graph( for publication purposes). I've also attached an example dataset. I simply want to show that each population clusters separately in space. I've attached a visual representation of what I want my data to look like. The scripts for this graph is as follows: data(trees) s3d - scatterplot3d(trees, type = h, color = blue, angle = 55, scale.y = 0.7, pch = 16, main = Adding elements) For some reason, R will not read my header, so I had to use the data.frame function to get R to read my data. This is a real pain because I have tons of data. I've tried the following to get R to recognize each population type separately: PopulationType.LK1-PopulationType[LK1] PopulationType.LK2-PopulationType[LK2] points((PopulationType.LK1, col=red, pch=19), (PopulationType.LK2, col=blue, pch=22), .) When I tweak the example tree code from above, R will plot all of my data no problem (with the rgl 3d graphs, not scatterplot3d), but only with one color and plotting character. It seems pretty straightforward, but I've spent 4 days on this already. Any help would be greatly appreciated. Regards, EJ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - Yasir H. Kaheil, Ph.D. Catchment Research Facility The University of Western Ontario -- View this message in context: http://www.nabble.com/Plotting-separate-populations-in-scatterplot3d-tp17088785p17088937.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] General Plotting Question
Thanks for the help- It worked just fine: This will cast my ignorance across the table, but here it goes. Why do I need to make them a matrix? because they are in a row? if d - as.matrix(f) f.t - t(d) f.m - as.data.frame(f.t) now I could use just the following plot(f.m[c(rownumber), column], f.m[c(rownumber), column]) ? I am sure that I need to brush up on underlying data structures, but I am trying to analyze data as we speak and R along with all of your help is an invaluable tool. Am I missing something easy, do I not understand the logic/philosophy, or...? thanks Stephen On Tue, May 6, 2008 at 12:53 PM, Yasir Kaheil [EMAIL PROTECTED] wrote: cast the two vectors as.matrix-- see here: plot(as.matrix(f[2,4:26]), as.matrix(f[1,4:26])) y stephen sefick wrote: f - (structure(list(X = structure(96:97, .Label = c(119DAmm, 119DN, 119DNN, 119DO, 119DOC, 119Flow, 119Nit, 119ON, 119OPhos, 119OrgP, 119Phos, 119TKN, 119TOC, 148DAmm, 148DN, 148DNN, 148DO, 148DOC, 148Flow, 148Nit, 148ON, 148OPhos, 148OrgP, 148Phos, 148TKN, 148TOC, 179DAmm, 179DN, 179DNN, 179DO, 179DOC, 179Flow, 179Nit, 179ON, 179OPhos, 179OrgP, 179Phos, 179TKN, 179TOC, 185DAmm, 185DN, 185DNN, 185DO, 185DOC, 185Flow, 185Nit, 185ON, 185OPhos, 185OrgP, 185Phos, 185TKN, 185TOC, 190DAmm, 190DN, 190DNN, 190DO, 190DOC, 190Flow, 190Nit, 190ON, 190OPhos, 190OrgP, 190Phos, 190TKN, 190TOC, 198DAmm, 198DN, 198DNN, 198DO, 198DOC, 198Flow, 198Nit, 198ON, 198OPhos, 198OrgP, 198Phos, 198TKN, 198TOC, 202DAmm, 202DN, 202DNN, 202DO, 202DOC, 202Flow, 202Nit, 202ON, 202OPhos, 202OrgP, 202Phos, 202TKN, 202TOC, 215DAmm, 215DN, 215DNN, 215DO, 215DOC, 215Flow, 215Nit, 215ON, 215OPhos, 215OrgP, 215Phos, 215TKN, 215TOC, 61DAmm, 61DN, 61DNN, 61DO, 61DOC, 61Flow, 61Nit, 61ON, 61OPhos, 61OrgP, 61Phos, 61TKN, 61TOC, BCOPhos, BCOrgP, BCPhos, BCTKN, BCTOC, HCDAmm, HCDN, HCDNN, HCDO, HCDOC, HCFlow, HCNit, HCON, HCOPhos, HCOrgP, HCPhos, HCTKN, HCTOC, SCDAmm, SCDN, SCDNN, SCDO, SCDOC, SCFlow, SCNit, SCON, SCOPhos, SCOrgP, SCPhos, SCTKN, SCTOC), class = factor), RiverMile = structure(c(8L, 8L), .Label = c(119, 148, 179, 185, 190, 198, 202, 215, 61, BC, HC, SC), class = factor), Constituent = structure(5:6, .Label = c(DAmm, DN, DNN, DO, DOC, Flow, Nit, ON, OPhos, OrgP, Phos, TKN, TOC), class = factor), X2.1.06 = c(2.5, 2494), X4.1.06 = c(2.4, NA), X5.1.06 = c(2.3, NA), X6.1.06 = c(2.1, NA), X7.1.06 = c(2, NA), X8.1.06 = c(1.9, NA), X9.1.06 = c(2, NA), X10.1.06 = c(2.1, NA), X11.1.06 = c(2.9, NA), X12.1.06 = c(2.6, NA), X1.1.07 = c(2.1, 4229.55), X2.1.07 = c(2.1, 1823.5), X3.1.07 = c(2.8, 3617.5), X4.1.07 = c(2.3, NA), X5.1.07 = c(3.5, NA), X6.1.07 = c(2.8, 10974.5), X7.1.07 = c(2.5, 9652), X8.1.07 = c(2.4, 6700.5), X9.1.07 = c(2.2, 11438.5), X10.1.07 = structure(c(34L, 33L), .Label = c(-0.0214, -0.3660, #VALUE!, 0., 0.0010, 0.0040, 0.0060, 0.0080, 0.0086, 0.0090, 0.0100, 0.0110, 0.0120, 0.0140, 0.0220, 0.0230, 0.0240, 0.0280, 0.0300, 0.0720, 0.0890, 0.1000, 0.1200, 0.1280, 0.1400, 0.1500, 0.1880, 0.1910, 0.2000, 0.2080, 0.2200, 0.3900, 1141., 2.2000, 2.3000, 2.5000, 2.6000, 2606., 3., 3.4000, 4.5000, 4.6000, 4710., 4720., 4729., 6.9768, 7.7065, 7.7338, 8.0097), class = factor), X11.1.07 = c(2.2, 10968), X12.1.07 = c(2.2, 7361.5), X1.1.08 = c(2, 6797)), .Names = c(X, RiverMile, Constituent, X2.1.06, X4.1.06, X5.1.06, X6.1.06, X7.1.06, X8.1.06, X9.1.06, X10.1.06, X11.1.06, X12.1.06, X1.1.07, X2.1.07, X3.1.07, X4.1.07, X5.1.07, X6.1.07, X7.1.07, X8.1.07, X9.1.07, X10.1.07, X11.1.07, X12.1.07, X1.1.08), row.names = c(NA, -2L), class = data.frame)) plot(f[2,4:26], f[1,4:26]) #Error in plot.window(...) : need finite 'xlim' values #In addition: Warning messages: #1: In min(x) : no non-missing arguments to min; returning Inf #2: In max(x) : no non-missing arguments to max; returning -Inf #I want a scatterplot like what I thought would happen with the plot command and the rows and columns that contain the data. If there is a NA in either case there is no point . plotted. This is how plot works with the data in columns, I think. This is a small example of a much larger data set are there any suggestions. What am I missing? thanks stephen -- Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal,
Re: [R] substring
?gsub() Match the start (_) followed by anything (.*) and replace by gsub(_.*,,Name) Weidong Gu, Department of Medicine University of Alabama, Birmingham 1900 University Blvd., Birmingham, Alabama 35294 PH: (205)-975-9053 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Qian R Sent: Tuesday, May 06, 2008 12:56 PM To: R-help@r-project.org Subject: [R] substring I have a data frame such as Name DISC1_5916 DABC2_789 ABCD_1234 I want to substring column Name so that DISC1 DABC2 ABCD ... Thanks. - [[elided Yahoo spam]] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] General Plotting Question
the data frames work as follows: columns= variables and rows= records. when you plot with a data frame it's logical to plot one variable against another not one record against another. with casting into a matrix and rotating then casting back into a dataframe, you made your records variables.. this is why it worked. I hope this makes sense to you. Thanks stephen sefick wrote: Thanks for the help- It worked just fine: This will cast my ignorance across the table, but here it goes. Why do I need to make them a matrix? because they are in a row? if d - as.matrix(f) f.t - t(d) f.m - as.data.frame(f.t) now I could use just the following plot(f.m[c(rownumber), column], f.m[c(rownumber), column]) ? I am sure that I need to brush up on underlying data structures, but I am trying to analyze data as we speak and R along with all of your help is an invaluable tool. Am I missing something easy, do I not understand the logic/philosophy, or...? thanks Stephen On Tue, May 6, 2008 at 12:53 PM, Yasir Kaheil [EMAIL PROTECTED] wrote: cast the two vectors as.matrix-- see here: plot(as.matrix(f[2,4:26]), as.matrix(f[1,4:26])) y stephen sefick wrote: f - (structure(list(X = structure(96:97, .Label = c(119DAmm, 119DN, 119DNN, 119DO, 119DOC, 119Flow, 119Nit, 119ON, 119OPhos, 119OrgP, 119Phos, 119TKN, 119TOC, 148DAmm, 148DN, 148DNN, 148DO, 148DOC, 148Flow, 148Nit, 148ON, 148OPhos, 148OrgP, 148Phos, 148TKN, 148TOC, 179DAmm, 179DN, 179DNN, 179DO, 179DOC, 179Flow, 179Nit, 179ON, 179OPhos, 179OrgP, 179Phos, 179TKN, 179TOC, 185DAmm, 185DN, 185DNN, 185DO, 185DOC, 185Flow, 185Nit, 185ON, 185OPhos, 185OrgP, 185Phos, 185TKN, 185TOC, 190DAmm, 190DN, 190DNN, 190DO, 190DOC, 190Flow, 190Nit, 190ON, 190OPhos, 190OrgP, 190Phos, 190TKN, 190TOC, 198DAmm, 198DN, 198DNN, 198DO, 198DOC, 198Flow, 198Nit, 198ON, 198OPhos, 198OrgP, 198Phos, 198TKN, 198TOC, 202DAmm, 202DN, 202DNN, 202DO, 202DOC, 202Flow, 202Nit, 202ON, 202OPhos, 202OrgP, 202Phos, 202TKN, 202TOC, 215DAmm, 215DN, 215DNN, 215DO, 215DOC, 215Flow, 215Nit, 215ON, 215OPhos, 215OrgP, 215Phos, 215TKN, 215TOC, 61DAmm, 61DN, 61DNN, 61DO, 61DOC, 61Flow, 61Nit, 61ON, 61OPhos, 61OrgP, 61Phos, 61TKN, 61TOC, BCOPhos, BCOrgP, BCPhos, BCTKN, BCTOC, HCDAmm, HCDN, HCDNN, HCDO, HCDOC, HCFlow, HCNit, HCON, HCOPhos, HCOrgP, HCPhos, HCTKN, HCTOC, SCDAmm, SCDN, SCDNN, SCDO, SCDOC, SCFlow, SCNit, SCON, SCOPhos, SCOrgP, SCPhos, SCTKN, SCTOC), class = factor), RiverMile = structure(c(8L, 8L), .Label = c(119, 148, 179, 185, 190, 198, 202, 215, 61, BC, HC, SC), class = factor), Constituent = structure(5:6, .Label = c(DAmm, DN, DNN, DO, DOC, Flow, Nit, ON, OPhos, OrgP, Phos, TKN, TOC), class = factor), X2.1.06 = c(2.5, 2494), X4.1.06 = c(2.4, NA), X5.1.06 = c(2.3, NA), X6.1.06 = c(2.1, NA), X7.1.06 = c(2, NA), X8.1.06 = c(1.9, NA), X9.1.06 = c(2, NA), X10.1.06 = c(2.1, NA), X11.1.06 = c(2.9, NA), X12.1.06 = c(2.6, NA), X1.1.07 = c(2.1, 4229.55), X2.1.07 = c(2.1, 1823.5), X3.1.07 = c(2.8, 3617.5), X4.1.07 = c(2.3, NA), X5.1.07 = c(3.5, NA), X6.1.07 = c(2.8, 10974.5), X7.1.07 = c(2.5, 9652), X8.1.07 = c(2.4, 6700.5), X9.1.07 = c(2.2, 11438.5), X10.1.07 = structure(c(34L, 33L), .Label = c(-0.0214, -0.3660, #VALUE!, 0., 0.0010, 0.0040, 0.0060, 0.0080, 0.0086, 0.0090, 0.0100, 0.0110, 0.0120, 0.0140, 0.0220, 0.0230, 0.0240, 0.0280, 0.0300, 0.0720, 0.0890, 0.1000, 0.1200, 0.1280, 0.1400, 0.1500, 0.1880, 0.1910, 0.2000, 0.2080, 0.2200, 0.3900, 1141., 2.2000, 2.3000, 2.5000, 2.6000, 2606., 3., 3.4000, 4.5000, 4.6000, 4710., 4720., 4729., 6.9768, 7.7065, 7.7338, 8.0097), class = factor), X11.1.07 = c(2.2, 10968), X12.1.07 = c(2.2, 7361.5), X1.1.08 = c(2, 6797)), .Names = c(X, RiverMile, Constituent, X2.1.06, X4.1.06, X5.1.06, X6.1.06, X7.1.06, X8.1.06, X9.1.06, X10.1.06, X11.1.06, X12.1.06, X1.1.07, X2.1.07, X3.1.07, X4.1.07, X5.1.07, X6.1.07, X7.1.07, X8.1.07, X9.1.07, X10.1.07, X11.1.07, X12.1.07, X1.1.08), row.names = c(NA, -2L), class = data.frame)) plot(f[2,4:26], f[1,4:26]) #Error in plot.window(...) : need finite 'xlim' values #In addition: Warning messages: #1: In min(x) : no non-missing arguments to min; returning Inf #2: In max(x) : no non-missing arguments to max; returning -Inf #I want a scatterplot like what I thought would happen with the plot command and the rows and columns that contain the data. If there is a NA in either case there is no point . plotted. This is how plot works with the data in columns, I think. This is a small example of a much larger data set are there any suggestions. What am I missing? thanks stephen -- Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel
[R] Type I or III SS with mixed model function lme
Hello, I have come across a result that I cannot explain, and am hoping that someone else can provide an answer. A student fitted a mixed model using the lme function: out- lme(fixed=Y~A+B+A:B, random=~1|Site). Y is a continuous variable while A and B are factors. The data set is balanced with the same number of observations in each combination of A and B. There are two hierarchical levels: Site and plots nested in site. He tried two different ways of getting theANOVA table: anova(out) and anova(out, type=marginal). Since the data were balanced, these two ways should (I think) give the same output since they correspond to Type I and III sums of squares in the SAS terminology. At least, this is the case with normal (i.e. not mixed) linear models. However, he finds very different results of these two types of ANOVA tables. Why? Bill Shipley North American Editor, Annals of Botany Département de biologie Université de Sherbrooke Sherbrooke (Québec) J1K 2R1 Canada (819) 821-8000, poste 62079 (819) 821-8049 FAX http://pages.usherbrooke.ca/jshipley/recherche/ http://pages.usherbrooke.ca/jshipley/recherche/ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Concatenate a vector into a string, only using distinct component
I'm trying to use combine c('a','b','c','a','c') into 'a, b, c', the order does not matter. paste(c('a','b','c','a','c'), collapse=', ') yields 'a, b, c, a, c'. Any idea? -- Regards, Anh Tran [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] list manipulation
Hello, I have a set of one-liners (many thanks to previous responses from this list) that I use to look at newly imported data sets with functions like dim(), names(), str(), etc. within lapply(). Generally, these commands work for me but, I am apparently still missing some aspect of list manipulation. I don't understand why I get a set of NULL list elements at the end of each output as demonstrated below. How can I generate this (and similar) result(s) without all the trailing NULLs? lapply(ls(pattern='bn'), function(x) cat(x, dim(get(x)), \t, names(get(x)), \n)) bn1993 2885 11 oplt rplt rsiz tree bd ht oaz odst raz rdst spr bn1994 3158 7oplt tree bd ht spr stat dam bn1995 734 7 oplt tree bd ht spr stat dam bn1996 293 7 oplt tree bd ht spr stat dam bn1997 264 7 oplt tree bd ht spr stat dam bn1998 768 7 oplt tree bd ht spr stat dam bn1999 654 7 oplt tree bd ht dbh stat dam bn2003 1407 9oplt tree bd94 ht94 ht99 ht02 ht03 stat dam [[1]] NULL [[2]] NULL [[3]] NULL [[4]] NULL [[5]] NULL [[6]] NULL [[7]] NULL [[8]] NULL Thanx, DaveT. * Silviculture Data Analyst Ontario Forest Research Institute Ontario Ministry of Natural Resources [EMAIL PROTECTED] http://ofri.mnr.gov.on.ca __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 write a sum of matrixes??
3) Bill Venables offered this about a week ago in this list: -- This is probably as good a way as any way for this kind of problem. First define a binary operator: %^% - function(x, n) with(eigen(x), vectors %*% (values^n * t(vectors))) This example only works for _diagonalizable_ matrices. It crashes, for example, in cases like: m - rbind(c(1,1,0), c(0,1,1), c(0,0,1)) m %^% 2 m %*% m Alberto Monteiro __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Concatenate a vector into a string, only using distinct component
Hi Anh, Try this, x=c('a','b','c','a','c') paste(unique(x),collapse=, ) [1] a, b, c HTH, Jorge On Tue, May 6, 2008 at 2:22 PM, Anh Tran [EMAIL PROTECTED] wrote: I'm trying to use combine c('a','b','c','a','c') into 'a, b, c', the order does not matter. paste(c('a','b','c','a','c'), collapse=', ') yields 'a, b, c, a, c'. Any idea? -- Regards, Anh Tran [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How do I write a sum of matrixes??
Hi Pascal, I think the function could be better but try this: # Function: M is your matrix and n MUST be an integer0 mat.pow-function(M,n) { result-M if(n1){ for ( iter in 2:n) result-M%*%result result } else {result} result } # The matrix m - rbind(c(1,1,0), c(0,1,1), c(0,0,1)) # Goal m^2 = m x m goal=m%*%m # matpow res=mat.pow(m,2) # Check point all.equal(goal,res) See RSiteSearch('nth step transition matrices') HTH, Jorge On Tue, May 6, 2008 at 1:37 AM, pascal vrolijk [EMAIL PROTECTED] wrote: Hello best helpers, I am a new user and I have been struggling for hours. So finally I decide to ask you: If I have a matrix P, and P.2 = P%*%P, and P.3=P.2%*%P is there a function to calculate the power of a matrix?? if not how can i do: for (i in 1:10) {P.i=P^i} after this I need to sum them up and my problem is to combine P and i to P.i can anyone help me please??? Thanks and have a nice day, Pascal. _ [[elided Hotmail spam]] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] list manipulation
Thanks Mark, Your suggestion led me to this: !is.null(lapply(ls(pattern='bn'), function(y) cat(y, dim(get(y)), \t, names(get(y)), \n))) bn1993 2885 11 oplt rplt rsiz tree bd ht oaz odst raz rdst spr bn1994 3158 7oplt tree bd ht spr stat dam bn1995 734 7 oplt tree bd ht spr stat dam bn1996 293 7 oplt tree bd ht spr stat dam bn1997 264 7 oplt tree bd ht spr stat dam bn1998 768 7 oplt tree bd ht spr stat dam bn1999 654 7 oplt tree bd ht dbh stat dam bn2003 1407 9oplt tree bd94 ht94 ht99 ht02 ht03 stat dam [1] TRUE Which is far cleaner and more compact (which was probably what I was _actually_ looking for) than what I had started with. Thanx, DaveT. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Sent: May 6, 2008 02:47 PM To: Thompson, David (MNR) Subject: RE: [R] list manipulation Hi: i would guess that you are sending objects into the lapply that dim and names don't work on so you get NULL in those cases. you can probably figure out what's happening by doing ls(pattern='bn') and seeing what gets returned ? There could be some other reason but that's my best guess. if you don't care why and you just want to get rid of them you can do initialoutput-lapply( that stuff) finaloutput-initialoutput[!sapply(initialoutput,is.null] print(finaloutput) i'm no R expert by any stretch but, if you want to get better at R , the best way is to live on this list if possible and watch the questions and solutions. for me, it beats any book out there. good luck and let me know if anything above is unclear or doesn't work. On Tue, May 6, 2008 at 2:35 PM, Thompson, David (MNR) wrote: Hello, I have a set of one-liners (many thanks to previous responses from this list) that I use to look at newly imported data sets with functions like dim(), names(), str(), etc. within lapply(). Generally, these commands work for me but, I am apparently still missing some aspect of list manipulation. I don't understand why I get a set of NULL list elements at the end of each output as demonstrated below. How can I generate this (and similar) result(s) without all the trailing NULLs? lapply(ls(pattern='bn'), function(x) cat(x, dim(get(x)), \t, names(get(x)), \n)) bn1993 2885 11 oplt rplt rsiz tree bd ht oaz odst raz rdst spr bn1994 3158 7oplt tree bd ht spr stat dam bn1995 734 7 oplt tree bd ht spr stat dam bn1996 293 7 oplt tree bd ht spr stat dam bn1997 264 7 oplt tree bd ht spr stat dam bn1998 768 7 oplt tree bd ht spr stat dam bn1999 654 7 oplt tree bd ht dbh stat dam bn2003 1407 9oplt tree bd94 ht94 ht99 ht02 ht03 stat dam [[1]] NULL [[2]] NULL [[3]] NULL [[4]] NULL [[5]] NULL [[6]] NULL [[7]] NULL [[8]] NULL Thanx, DaveT. * Silviculture Data Analyst Ontario Forest Research Institute Ontario Ministry of Natural Resources [EMAIL PROTECTED] http://ofri.mnr.gov.on.ca __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Concatenate a vector into a string, only using distinct component
probably you want: stg - c('a', 'b', 'c', 'a', 'c') paste(unique(stg), collapse = , ) I hope it helps. 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 Quoting Anh Tran [EMAIL PROTECTED]: I'm trying to use combine c('a','b','c','a','c') into 'a, b, c', the order does not matter. paste(c('a','b','c','a','c'), collapse=', ') yields 'a, b, c, a, c'. Any idea? -- Regards, Anh Tran [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] building a formula string bit by bit ..
Hello, Still a newbie with R, though I have learned a lot from reading this list. I'm hoping someone can help with this question: I have two vectors, one for variables, and one for bits. I want to build a string (really a formula) based on the values in my vector of 1s and 0s in bits. If I have a one, I want to include the corresponding entry in the vars vector, otherwise ignore it. Of course the bits vector will not stay the same. So for example: vars=c('X.1', 'X.2', 'X.3', 'X.4', 'X.5') bits=c(1, 0, 1, 1, 0) ones=which(bits==1) should yield: X.1 + X.3 + X.4 where as bits=c(1, 1, 0, 0, 0) would yield X.1 + X.2 the which operator gives me the index values, is there an easy and *efficient* way to build this string so that I can pass it on to glm? I can probably hack some ugly code together to do this, but it won't be efficient, and it won't be elegant :-) Can anyone help? Thanks! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] building a formula string bit by bit ..
Hi Esmail, Try this: vars=c('X.1', 'X.2', 'X.3', 'X.4', 'X.5') bits=c(1, 0, 1, 1, 0) paste(vars[which(bits==1)],collapse=+) HTH, Jorge On Tue, May 6, 2008 at 3:06 PM, Esmail Bonakdarian [EMAIL PROTECTED] wrote: Hello, Still a newbie with R, though I have learned a lot from reading this list. I'm hoping someone can help with this question: I have two vectors, one for variables, and one for bits. I want to build a string (really a formula) based on the values in my vector of 1s and 0s in bits. If I have a one, I want to include the corresponding entry in the vars vector, otherwise ignore it. Of course the bits vector will not stay the same. So for example: vars=c('X.1', 'X.2', 'X.3', 'X.4', 'X.5') bits=c(1, 0, 1, 1, 0) ones=which(bits==1) should yield: X.1 + X.3 + X.4 where as bits=c(1, 1, 0, 0, 0) would yield X.1 + X.2 the which operator gives me the index values, is there an easy and *efficient* way to build this string so that I can pass it on to glm? I can probably hack some ugly code together to do this, but it won't be efficient, and it won't be elegant :-) Can anyone help? Thanks! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] building a formula string bit by bit ..
Esmail Bonakdarian wrote: Hello, Still a newbie with R, though I have learned a lot from reading this list. I'm hoping someone can help with this question: I have two vectors, one for variables, and one for bits. I want to build a string (really a formula) based on the values in my vector of 1s and 0s in bits. If I have a one, I want to include the corresponding entry in the vars vector, otherwise ignore it. Of course the bits vector will not stay the same. So for example: vars=c('X.1', 'X.2', 'X.3', 'X.4', 'X.5') bits=c(1, 0, 1, 1, 0) ones=which(bits==1) should yield: X.1 + X.3 + X.4 where as bits=c(1, 1, 0, 0, 0) would yield X.1 + X.2 the which operator gives me the index values, is there an easy and *efficient* way to build this string so that I can pass it on to glm? I can probably hack some ugly code together to do this, but it won't be efficient, and it won't be elegant :-) Can anyone help? Thanks! Depending upon other factors, there may be a better and more general approach to what you are trying to do relative to selecting subsets of IV's. However, given the information you have provided: vars - c('X.1', 'X.2', 'X.3', 'X.4', 'X.5') bits - c(1, 0, 1, 1, 0) paste(vars[bits == 1], collapse = + ) [1] X.1 + X.3 + X.4 bits - c(1, 1, 0, 0, 0) paste(vars[bits == 1], collapse = + ) [1] X.1 + X.2 You will then need to use as.formula() to coerce the resultant vectors before passing them to the model function. HTH, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] building a formula string bit by bit ..
I would actually go with this: bits=c(1, 0, 1, 1, 0) paste(X, which(bits==1), sep=.,collapse=+) No need for the vars variable. Though admittedly it breaks down if bits is identically 0. Haris Skiadas Department of Mathematics and Computer Science Hanover College On May 6, 2008, at 3:11 PM, Jorge Ivan Velez wrote: Hi Esmail, Try this: vars=c('X.1', 'X.2', 'X.3', 'X.4', 'X.5') bits=c(1, 0, 1, 1, 0) paste(vars[which(bits==1)],collapse=+) HTH, Jorge On Tue, May 6, 2008 at 3:06 PM, Esmail Bonakdarian [EMAIL PROTECTED] wrote: Hello, Still a newbie with R, though I have learned a lot from reading this list. I'm hoping someone can help with this question: I have two vectors, one for variables, and one for bits. I want to build a string (really a formula) based on the values in my vector of 1s and 0s in bits. If I have a one, I want to include the corresponding entry in the vars vector, otherwise ignore it. Of course the bits vector will not stay the same. So for example: vars=c('X.1', 'X.2', 'X.3', 'X.4', 'X.5') bits=c(1, 0, 1, 1, 0) ones=which(bits==1) should yield: X.1 + X.3 + X.4 where as bits=c(1, 1, 0, 0, 0) would yield X.1 + X.2 the which operator gives me the index values, is there an easy and *efficient* way to build this string so that I can pass it on to glm? I can probably hack some ugly code together to do this, but it won't be efficient, and it won't be elegant :-) Can anyone help? Thanks! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Installing A2R in Windows
Hi there, I've tried to install the A2R package using the files from http://addictedtor.free.fr/packages/A2R/lastVersion/ This is the error I get when trying to load the library: library(A2R) Error in library(A2R) : 'A2R' is not a valid package -- installed 2.0.0? Can anyone please help? Thanks. Kate [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] building a formula string bit by bit ..
Marc Schwartz wrote: Esmail Bonakdarian wrote: Hello, Still a newbie with R, though I have learned a lot from reading this list. I'm hoping someone can help with this question: I have two vectors, one for variables, and one for bits. I want to build a string (really a formula) based on the values in my vector of 1s and 0s in bits. If I have a one, I want to include the corresponding entry in the vars vector, otherwise ignore it. Of course the bits vector will not stay the same. So for example: vars=c('X.1', 'X.2', 'X.3', 'X.4', 'X.5') bits=c(1, 0, 1, 1, 0) ones=which(bits==1) should yield: X.1 + X.3 + X.4 where as bits=c(1, 1, 0, 0, 0) would yield X.1 + X.2 the which operator gives me the index values, is there an easy and *efficient* way to build this string so that I can pass it on to glm? I can probably hack some ugly code together to do this, but it won't be efficient, and it won't be elegant :-) Can anyone help? Thanks! Depending upon other factors, there may be a better and more general approach to what you are trying to do relative to selecting subsets of IV's. However, given the information you have provided: vars - c('X.1', 'X.2', 'X.3', 'X.4', 'X.5') bits - c(1, 0, 1, 1, 0) paste(vars[bits == 1], collapse = + ) [1] X.1 + X.3 + X.4 bits - c(1, 1, 0, 0, 0) paste(vars[bits == 1], collapse = + ) [1] X.1 + X.2 You will then need to use as.formula() to coerce the resultant vectors before passing them to the model function. One other tweak here, if you use TRUE/FALSE rather than 1/0 to define the positions and this gets rid of the need for 'vars': bits - c(TRUE, FALSE, TRUE, TRUE, FALSE) paste(X, which(bits), collapse = + , sep = .) [1] X.1 + X.3 + X.4 HTH, Marc __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] categorical data analysis
Fisher's exact test works with small cells. See ?fisher.test -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of raymond chiruka Sent: Tuesday, May 06, 2008 10:34 AM To: r Subject: [R] categorical data analysis hie all i am trying to carry out a categorical data analysis but my problem is that when in i use the chi squared test some of my expected values are less than 5. is there a test that can handle this situation. the data is not a 2*2 table. its more from the social sciences where you have from strongly agree to strongly disagree. i know i can collapse vthe tables but there is a loss of information . is the a test vthat i can for this kind of data. thanks in advancde. ray - [[elided Yahoo spam]] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 write a sum of matrixes??
Le mar. 06 mai à 14:23, Alberto Monteiro a écrit : 3) Bill Venables offered this about a week ago in this list: -- This is probably as good a way as any way for this kind of problem. First define a binary operator: %^% - function(x, n) with(eigen(x), vectors %*% (values^n * t(vectors))) This example only works for _diagonalizable_ matrices. It crashes, for example, in cases like: m - rbind(c(1,1,0), c(0,1,1), c(0,0,1)) m %^% 2 m %*% m Then the %^% operator defined in package expm [1] will work for any kind of (square) matrix. The actual work is done in C, so it is pretty fast. [1] On R-Forge: http://r-forge.r-project.org/projects/expm/ --- Vincent Goulet, Associate Professor École d'actuariat Université Laval, Québec [EMAIL PROTECTED] http://vgoulet.act.ulaval.ca __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] ridge regression
Thanks to all of you that helped me with the issues of bootstrapping and downloading packages to a local disk. As an starter I'm in the lower side of the learning curve, but this R software is awesome. What I like most is this kind of forums when people share their problems and we can find solutions. My new inquiry is short: do any of you have some example (besides the one that appears in help) about a simple ridge regression application in R? I'm trying to figure out how the program estimates the constant value that is most appropiate to run regression and how can I see that value in the plot. My operative system: windows XP, running R 2.6.2 Thanks to all __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] building a formula string bit by bit ..
Jorge Ivan Velez wrote: Hi Esmail, Try this: vars=c('X.1', 'X.2', 'X.3', 'X.4', 'X.5') bits=c(1, 0, 1, 1, 0) paste(vars[which(bits==1)],collapse=+) HTH, Jorge Wow .. that is beautiful :-) .. and exactly what I was looking for (and suspected existed). I ended up doing this: eqn=(paste(vars[which(bits==1)],collapse= + )) eqn=paste(c(Y.data ~), eqn) GLM.9 - glm(as.formula(eqn) , family=gaussian(identity), data=simpleData) For some reason I couldn't collapse the two eqn/paste statements into one statement but this seems to work. Thank again Jorge and everyone else .. this group is a big help. Esmail __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Print table data on to a plot
Hi, Is there away to print a short table out along side with a plot? I'm thinking about doing a par(mfrow = c(1,2)) Then, the plot is on one side, summary result on the other. Is there any quick way to print out a data.frame in table format? Thanks -- Regards, Anh Tran [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Print table data on to a plot
Hi Anh, Take a look at http://finzi.psych.upenn.edu/R/Rhelp02a/archive/128041.html HTH, Jorge On Tue, May 6, 2008 at 3:46 PM, Anh Tran [EMAIL PROTECTED] wrote: Hi, Is there away to print a short table out along side with a plot? I'm thinking about doing a par(mfrow = c(1,2)) Then, the plot is on one side, summary result on the other. Is there any quick way to print out a data.frame in table format? Thanks -- Regards, Anh Tran [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Print table data on to a plot
Look at the addtable2plot function in the plotrix package. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Anh Tran Sent: Tuesday, May 06, 2008 1:47 PM To: R-help@r-project.org Subject: [R] Print table data on to a plot Hi, Is there away to print a short table out along side with a plot? I'm thinking about doing a par(mfrow = c(1,2)) Then, the plot is on one side, summary result on the other. Is there any quick way to print out a data.frame in table format? Thanks -- Regards, Anh Tran [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Count data in random Forest
Hi Birgit, I'm not sure that I understand your question. I'll try to answer anyways. Regression trees and therefore also RandomForests are invariant to monotonic transformations in the independent variables. There are no distributional assumptions for the independent variables. The dependent variable, however, is used to calculate the variances within the two groups of cases that result from a split. Therefore, it would make sense to have the dependent variable follow the typical distributional requirements of least-squares driven models such as homoscedasity, symmetrical distribution etc. For count data a square root transformation is often appropriate. HTH Volker Birgit Lemcke wrote: div class=moz-text-flowed style=font-family: -moz-fixedHello R-user! I am running R 2.7.0 on a Power Book (Tiger). (I am still R and statistics beginner) I try to find the most important variables to divide my dataset as given in a categorical variable using randomForest. Is randomForest() able to deal with count data? Or is there no difference because only the ranks are used in the trees? Thanks in advance Birgit Birgit Lemcke Institut für Systematische Botanik Zollikerstrasse 107 CH-8008 Zürich Switzerland Ph: +41 (0)44 634 8351 [EMAIL PROTECTED] 175 Jahre UZH «staunen.erleben.begreifen. Naturwissenschaft zum Anfassen.» MNF-Jubiläumsevent für gross und klein. 19. April 2008, 10.00 Uhr bis 02.00 Uhr Campus Irchel, Winterthurerstrasse 190, 8057 Zürich Weitere Informationen http://www.175jahre.uzh.ch/naturwissenschaft /div __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] ggplo2: x_discrete labels size/direction
Hi Xavier, Changing the grid settings has worked! Thanks a lot! Mikhail PS Perhaps it'd still be really useful to be able to change text direction in labels. Hadley, what do you think? -- View this message in context: http://www.nabble.com/ggplo2%3A-x_discrete-labels-size-direction-tp17077479p17091078.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] rggobi is crashing R-2.7.0
On Tue, May 6, 2008 at 10:32 AM, Mark Kimpel [EMAIL PROTECTED] wrote: I am running 64-bit Ubuntu 8.04 and when I invoke rggobi the interactive graph displays but R crashes. See my sessionInfo() and a short example below. Ggobi and rggobi installed without complaints. Mark sessionInfo() R version 2.7.0 Patched (2008-05-04 r45620) x86_64-unknown-linux-gnu locale: LC_CTYPE=en_US.UTF-8;LC_NUMERIC=C;LC_TIME=en_US.UTF-8;LC_COLLATE=en_US.UTF-8;LC_MONETARY=C;LC_MESSAGES=en_US.UTF-8;LC_PAPER=en_US.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_US.UTF-8;LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] rggobi_2.1.9 RGtk2_2.12.5-3 graph_1.18.0 loaded via a namespace (and not attached): [1] cluster_1.11.10 tools_2.7.0 a - matrix(rnorm(1000), nrow = 10) g - ggobi(a) ** (R:25146): CRITICAL **: Error on loading plugin library plugins/GraphLayout/plugin.la: libgvc.so.3: cannot open shared object file: No such file or directory ** (R:25146): CRITICAL **: Error on loading plugin library plugins/GraphLayout/plugin.la: libgvc.so.3: cannot open shared object file: No such file or directory ** (R:25146): CRITICAL **: can't locate required plugin routine addToToolsMenu in GraphLayout It's not clear to me - did R crash or did you just receive these warnings? These warnings are due to a missing graphviz, so the GraphLayout plugin fails to load. -- Mark W. Kimpel MD ** Neuroinformatics ** Dept. of Psychiatry Indiana University School of Medicine 15032 Hunter Court, Westfield, IN 46074 (317) 490-5129 Work, Mobile VoiceMail (317) 663-0513 Home (no voice mail please) ** [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] NLS plinear question
Hi All. I've run into a problem with the plinear algorithm in nls that is confusing me. Assume the following reaction time data over 15 trials for a single unit. Trials are coded from 0-14 so that the intercept represents reaction time in the first trial. trl RT 01132.0 1 630.5 21371.5 3 704.0 4 488.5 5 575.5 6 613.0 7 824.5 8 509.0 9 791.0 10 492.5 11 515.5 12 467.0 13 556.5 14 456.0 Now fit a power function to this data using nls with the plinear algorithm fit.pw -nls(RT ~ cbind(1,trl, trl^p), start = c(p = -.2), algorithm = plinear, data=df.one) Yields the following error message Error in numericDeriv(form[[3]], names(ind), env) : Missing value or an infinity produced when evaluating the model Now, recode trial from 1-15 and run the same model. fit.pw -nls(RT ~ cbind(1,trl, trl^p), start = c(p = -.2), algorithm = plinear, data=df.one) Seems to work fine now... Nonlinear regression model model: RT ~ cbind(1, trl, trl^p) data: df.one p .lin1.lin.trl .lin3 -0.2845 200.3230-8.9467 904.7582 residual sum-of-squares: 555915 Number of iterations to convergence: 11 Any idea why having a zero for the first value of X causes this problem? Thanks in advance, Rick DeShon [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Installing A2R in Windows
How did you install it? You need to get A2R_0.0-4.tar.gz and do R CMD INSTALL A2R_0.0-4.tar.gz, after reading the 'R Installation and Administration manual, especially the sections for your OS. The package in A2R/lastVersion/ was built for Unix on R 2.2.1. An expert might be able to get it working, but we don't even know your OS and R version. On Tue, 6 May 2008, Kate wrote: Hi there, I've tried to install the A2R package using the files from http://addictedtor.free.fr/packages/A2R/lastVersion/ This is the error I get when trying to load the library: library(A2R) Error in library(A2R) : 'A2R' is not a valid package -- installed 2.0.0? Can anyone please help? Thanks. Kate [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. PLEASE do read it. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] unable to use functions require DLL from package base
I suspect you have more than one version of R installed and are mixing them up. Those symbols have been in package stats for quite a while. Try starting R with --vanilla, and if that works, clean out your startup files (see ?Startup). If not, remove all your R installations and reinstall R 2.7.0 (or R-patched). On Tue, 6 May 2008, A.N. wrote: Hi all, I have issues using some basic functions in R such as these ones : pp.test(R) (where is a vector of returns) Error in .C(R_approx, as.double(x), as.double(y), as.integer(nx), xout = as.double(xout), : C symbol name R_approx not in DLL for package base boxcox(reg,plotit=T) (where reg is an lm object) Error in .C(spline_coef, method = as.integer(method), n = as.integer(nx), : C symbol name spline_coef not in DLL for package base as I do miss some symbol names. How can I overcome this serious problem ? -- View this message in context: http://www.nabble.com/unable-to-use-functions-require-DLL-from-package-%22base%22-tp17086783p17086783.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] NLS plinear question
recall that 0 ^{-.2} = 1/0^{.2}, and that dividing by 0 gives Inf. so when 0 is in trl, part of your model for RT is Inf: trl - 0:14 p - -.2 cbind(1,trl, trl^p) trl [1,] 1 0 Inf [2,] 1 1 1.000 [3,] 1 2 0.8705506 [4,] 1 3 0.8027416 [5,] 1 4 0.7578583 [6,] 1 5 0.7247797 [7,] 1 6 0.6988271 [8,] 1 7 0.6776109 [9,] 1 8 0.6597540 [10,] 1 9 0.6443940 [11,] 1 10 0.6309573 [12,] 1 11 0.6190439 [13,] 1 12 0.6083643 [14,] 1 13 0.5987029 [15,] 1 14 0.5898946 On Tue, 6 May 2008, Rick DeShon wrote: Hi All. I've run into a problem with the plinear algorithm in nls that is confusing me. Assume the following reaction time data over 15 trials for a single unit. Trials are coded from 0-14 so that the intercept represents reaction time in the first trial. trl RT 01132.0 1 630.5 21371.5 3 704.0 4 488.5 5 575.5 6 613.0 7 824.5 8 509.0 9 791.0 10 492.5 11 515.5 12 467.0 13 556.5 14 456.0 Now fit a power function to this data using nls with the plinear algorithm fit.pw -nls(RT ~ cbind(1,trl, trl^p), start = c(p = -.2), algorithm = plinear, data=df.one) Yields the following error message Error in numericDeriv(form[[3]], names(ind), env) : Missing value or an infinity produced when evaluating the model Now, recode trial from 1-15 and run the same model. fit.pw -nls(RT ~ cbind(1,trl, trl^p), start = c(p = -.2), algorithm = plinear, data=df.one) Seems to work fine now... Nonlinear regression model model: RT ~ cbind(1, trl, trl^p) data: df.one p .lin1.lin.trl .lin3 -0.2845 200.3230-8.9467 904.7582 residual sum-of-squares: 555915 Number of iterations to convergence: 11 Any idea why having a zero for the first value of X causes this problem? Thanks in advance, Rick DeShon [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Dendrogram label size
Alex -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Alex Reynolds Sent: Wednesday, 7 May 2008 12:26 a.m. To: r-help@r-project.org Subject: [R] Dendrogram label size Is it possible to resize the labels in a dendrogram without applying circles and triangles to edges? I tried cex.labels: plot(scoreDendogramObj, horiz=TRUE, axes=FALSE, cex.labels=0.8) but that didn't have any effect. One way is to get at the object itself to find where the labels are located and then use text() to draw them. This has the advantage of allowing some labels to be emphasised by different colours/font sizes or whatever. HTH ... Peter Alspach The contents of this e-mail are privileged and/or confidential to the named recipient and are not to be used by any other person and/or organisation. If you have received this e-mail in error, please notify the sender and delete all material pertaining to this e-mail. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] NLS plinear question
0^(-0.2) = Inf, so you started with an infinite prediction for your first point and hence an infinite sum of squares. On Tue, 6 May 2008, Rick DeShon wrote: Hi All. I've run into a problem with the plinear algorithm in nls that is confusing me. Assume the following reaction time data over 15 trials for a single unit. Trials are coded from 0-14 so that the intercept represents reaction time in the first trial. trl RT 01132.0 1 630.5 21371.5 3 704.0 4 488.5 5 575.5 6 613.0 7 824.5 8 509.0 9 791.0 10 492.5 11 515.5 12 467.0 13 556.5 14 456.0 Now fit a power function to this data using nls with the plinear algorithm fit.pw -nls(RT ~ cbind(1,trl, trl^p), start = c(p = -.2), algorithm = plinear, data=df.one) Yields the following error message Error in numericDeriv(form[[3]], names(ind), env) : Missing value or an infinity produced when evaluating the model Now, recode trial from 1-15 and run the same model. fit.pw -nls(RT ~ cbind(1,trl, trl^p), start = c(p = -.2), algorithm = plinear, data=df.one) Seems to work fine now... Nonlinear regression model model: RT ~ cbind(1, trl, trl^p) data: df.one p .lin1.lin.trl .lin3 -0.2845 200.3230-8.9467 904.7582 residual sum-of-squares: 555915 Number of iterations to convergence: 11 Any idea why having a zero for the first value of X causes this problem? Thanks in advance, Rick DeShon [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] ridge regression
Thanks to all of you that helped me with the issues of bootstrapping and downloading packages to a local disk. As an starter I'm in the lower side of the learning curve, but this R software is awesome. What I like most is this kind of forums when people share their problems and we can find solutions. My new inquiry is short: do any of you have some example (besides the one that appears in help) about a simple ridge regression application in R? I'm trying to figure out how the program estimates the constant value that is most appropiate to run regression and how can I see that value in the plot. My operative system: windows XP, running R 2.6.2 Thanks to all __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 Case/Control R Beginner
I am looking for a way to simulate genotypes of cases and control at a disease locus in R. I am supposed to set the allele frequency as control/cases. for each of the column below simulate 200 snp dataset. I am looking at treesim function from popgen to stimulate the genotypes in R. Here is what i have size case control 20000.60.15 25000.10.17 what am i missing in the treesim parameter and how do i calculate the p-value of trend test? I am a bit confused with the parameter. N-2000 Sn-200 theta-0 a-treesim(N, Sn, theta, rho.vec, mutations=FALSE, sample=FALSE) mat-LDmat(.) Here is the documentation. http://cran.r-project.org/web/packages/popgen/popgen.pdf any help will really be helpful. thank you, Claire -- View this message in context: http://www.nabble.com/Simulation-Case-Control-R-Beginner-tp17092885p17092885.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Type I or III SS with mixed model function lme
Dear Bill, I expect that the problem is in the contrasts that your student used for A and B, though I haven't thought specifically about the context of a mixed model. If he or she used the default contr.treatment(), then the contrasts for different factors (and the interaction) are not orthogonal in the row basis of the model matrix and hence are not orthogonal, even for balanced data. Using, e.g., contr.sum() should provide A, B, and A:B contrasts that are orthogonal to each other. I hope this helps, John -- John Fox, Professor Department of Sociology McMaster University Hamilton, Ontario, Canada web: socserv.mcmaster.ca/jfox -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Bill Shipley Sent: May-06-08 2:13 PM To: R help list Subject: [R] Type I or III SS with mixed model function lme Hello, I have come across a result that I cannot explain, and am hoping that someone else can provide an answer. A student fitted a mixed model using the lme function: out- lme(fixed=Y~A+B+A:B, random=~1|Site). Y is a continuous variable while A and B are factors. The data set is balanced with the same number of observations in each combination of A and B. There are two hierarchical levels: Site and plots nested in site. He tried two different ways of getting theANOVA table: anova(out) and anova(out, type=marginal). Since the data were balanced, these two ways should (I think) give the same output since they correspond to Type I and III sums of squares in the SAS terminology. At least, this is the case with normal (i.e. not mixed) linear models. However, he finds very different results of these two types of ANOVA tables. Why? Bill Shipley North American Editor, Annals of Botany Dipartement de biologie Universiti de Sherbrooke Sherbrooke (Quibec) J1K 2R1 Canada (819) 821-8000, poste 62079 (819) 821-8049 FAX http://pages.usherbrooke.ca/jshipley/recherche/ http://pages.usherbrooke.ca/jshipley/recherche/ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Spatial join between two datasets using x and y co-ordinates
Hi R users I am trying to create a spatial join between two datasets. The first data set is large and contains descriptive data including x and y co-ordinates. The second dataset is small and has been selected spatially. The only data contained within the second dataset is the x and y coordinates only i.e. no descriptive data. The aim of a join made between the two datasets is to select those points (and hence all the descriptive data) from dataset one that has the same x and y co-ordinates as dataset two. x1-c(1824615,1823650,1821910) y1-c(5980732,5983220,5990931) descript-c(cat, dog, horse) dat1-data.frame(x1,y1,descript) x2-c(1824615,1823650) y2-c(5980732,5983220) dat2-data.frame(x2,y2) dat1 dat2 The aim of the join is to produce: x1 y1 descript 1 1824615 5980732 cat 2 1823650 5983220 dog And therefore limit the data in dataset 1 to that which has the same x and y co-ordinates as dataset 2. Any suggestions? Regards Andy Andrew McFadden MVS BVSc Incursion Investigator Investigation Diagnostic Centres - Wallaceville Biosecurity New Zealand Ministry of Agriculture and Forestry Phone 04 894 5600 Fax 04 894 4973 Mobile 029 894 5611 Postal address: Investigation and Diagnostic Centre- Wallaceville Box 40742 Ward St Upper Hutt This email message and any attachment(s) is intended solely for the addressee(s) named above. The information it contains is confidential and may be legally privileged. Unauthorised use of the message, or the information it contains, may be unlawful. If you have received this message by mistake please call the sender immediately on 64 4 8940100 or notify us by return email and erase the original message and attachments. Thank you. The Ministry of Agriculture and Forestry accepts no responsibility for changes made to this email or to any attachments after transmission from the office. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is there in R a function equivalent to the mround, as found in most spreadsheets?
I believe this function matches the description in OOO: mround - function(number, multiple) multiple * round(number/multiple) I've implemented a slightly more general form in the reshape package: round_any - function (x, accuracy, f = round) { f(x/accuracy) * accuracy } Hadley -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Installing A2R in Windows
Thanks. I used the command you mention and it works fine in Linux. But I need to get it to work for Windows XP as well (currently running R-2.7.0). Any idea if it's possible? On Tue, May 6, 2008 at 5:18 PM, Prof Brian Ripley [EMAIL PROTECTED] wrote: How did you install it? You need to get A2R_0.0-4.tar.gz and do R CMD INSTALL A2R_0.0-4.tar.gz, after reading the 'R Installation and Administration manual, especially the sections for your OS. The package in A2R/lastVersion/ was built for Unix on R 2.2.1. An expert might be able to get it working, but we don't even know your OS and R version. On Tue, 6 May 2008, Kate wrote: Hi there, I've tried to install the A2R package using the files from http://addictedtor.free.fr/packages/A2R/lastVersion/ This is the error I get when trying to load the library: library(A2R) Error in library(A2R) : 'A2R' is not a valid package -- installed 2.0.0? Can anyone please help? Thanks. Kate [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. PLEASE do read it. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/http://www.stats.ox.ac.uk/%7Eripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Spatial join between two datasets using x and y co-ordinates
Andrew ?merge HTH .. Peter Alspach -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Andrew McFadden Sent: Wednesday, 7 May 2008 9:23 a.m. To: r-help@r-project.org Subject: [R] Spatial join between two datasets using x and y co-ordinates Hi R users I am trying to create a spatial join between two datasets. The first data set is large and contains descriptive data including x and y co-ordinates. The second dataset is small and has been selected spatially. The only data contained within the second dataset is the x and y coordinates only i.e. no descriptive data. The aim of a join made between the two datasets is to select those points (and hence all the descriptive data) from dataset one that has the same x and y co-ordinates as dataset two. x1-c(1824615,1823650,1821910) y1-c(5980732,5983220,5990931) descript-c(cat, dog, horse) dat1-data.frame(x1,y1,descript) x2-c(1824615,1823650) y2-c(5980732,5983220) dat2-data.frame(x2,y2) dat1 dat2 The aim of the join is to produce: x1 y1 descript 1 1824615 5980732 cat 2 1823650 5983220 dog And therefore limit the data in dataset 1 to that which has the same x and y co-ordinates as dataset 2. Any suggestions? Regards Andy Andrew McFadden MVS BVSc Incursion Investigator Investigation Diagnostic Centres - Wallaceville Biosecurity New Zealand Ministry of Agriculture and Forestry Phone 04 894 5600 Fax 04 894 4973 Mobile 029 894 5611 Postal address: Investigation and Diagnostic Centre- Wallaceville Box 40742 Ward St Upper Hutt ## ## This email message and any attachment(s) is intended solely for the addressee(s) named above. The information it contains is confidential and may be legally privileged. Unauthorised use of the message, or the information it contains, may be unlawful. If you have received this message by mistake please call the sender immediately on 64 4 8940100 or notify us by return email and erase the original message and attachments. Thank you. The Ministry of Agriculture and Forestry accepts no responsibility for changes made to this email or to any attachments after transmission from the office. ## ## [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. The contents of this e-mail are privileged and/or confidential to the named recipient and are not to be used by any other person and/or organisation. If you have received this e-mail in error, please notify the sender and delete all material pertaining to this e-mail. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] mgcv::gam shrinkage of smooths
In Dr. Wood's book on GAM, he suggests in section 4.1.6 that it might be useful to shrink a single smooth by adding S=S+epsilon*I to the penalty matrix S. The context was the need to be able to shrink the term to zero if appropriate. I'd like to do this in order to shrink the coefficients towards zero (irrespective of the penalty for wiggliness) - but not necessarily all the way to zero. IE, my informal prior is to keep the contribution of a specific term small. 1) Is adding eps*I to the penalty matrix an effective way to achieve this goal? 2) How do I accomplish this in practice using mgcv::gam? Thanks. -- View this message in context: http://www.nabble.com/mgcv%3A%3Agam-shrinkage-of-smooths-tp17093645p17093645.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Spatial join between two datasets using x and y co-ordinates
vector dat1.select is the selected records from dat1 by dat2. dat1.select- dat1$x1 %in% dat2$x2 dat1$y1 %in% dat2$y2 dat1[dat1.select,] x1 y1 descript 1 1824615 5980732 cat 2 1823650 5983220 dog Andrew McFadden wrote: Hi R users I am trying to create a spatial join between two datasets. The first data set is large and contains descriptive data including x and y co-ordinates. The second dataset is small and has been selected spatially. The only data contained within the second dataset is the x and y coordinates only i.e. no descriptive data. The aim of a join made between the two datasets is to select those points (and hence all the descriptive data) from dataset one that has the same x and y co-ordinates as dataset two. x1-c(1824615,1823650,1821910) y1-c(5980732,5983220,5990931) descript-c(cat, dog, horse) dat1-data.frame(x1,y1,descript) x2-c(1824615,1823650) y2-c(5980732,5983220) dat2-data.frame(x2,y2) dat1 dat2 The aim of the join is to produce: x1 y1 descript 1 1824615 5980732 cat 2 1823650 5983220 dog And therefore limit the data in dataset 1 to that which has the same x and y co-ordinates as dataset 2. Any suggestions? Regards Andy Andrew McFadden MVS BVSc Incursion Investigator Investigation Diagnostic Centres - Wallaceville Biosecurity New Zealand Ministry of Agriculture and Forestry Phone 04 894 5600 Fax 04 894 4973 Mobile 029 894 5611 Postal address: Investigation and Diagnostic Centre- Wallaceville Box 40742 Ward St Upper Hutt This email message and any attachment(s) is intended solely for the addressee(s) named above. The information it contains is confidential and may be legally privileged. Unauthorised use of the message, or the information it contains, may be unlawful. If you have received this message by mistake please call the sender immediately on 64 4 8940100 or notify us by return email and erase the original message and attachments. Thank you. The Ministry of Agriculture and Forestry accepts no responsibility for changes made to this email or to any attachments after transmission from the office. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - Yasir H. Kaheil, Ph.D. Catchment Research Facility The University of Western Ontario -- View this message in context: http://www.nabble.com/Spatial-join-between-two-datasets-using-x-and-y-co-ordinates-tp17093486p17093656.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Spatial join between two datasets using x and y co-ordinates
Hi Andrew, Try also: x1-c(1824615,1823650,1821910) y1-c(5980732,5983220,5990931) descript-c(cat, dog, horse) dat1-data.frame(x1,y1,descript) x2-c(1824615,1823650) y2-c(5980732,5983220) dat2-data.frame(x2,y2) colnames(dat2)=c('x1','y1') merge(dat1,dat2,by=c('x1','y1')) HTH, Jorge On Tue, May 6, 2008 at 5:22 PM, Andrew McFadden [EMAIL PROTECTED] wrote: Hi R users I am trying to create a spatial join between two datasets. The first data set is large and contains descriptive data including x and y co-ordinates. The second dataset is small and has been selected spatially. The only data contained within the second dataset is the x and y coordinates only i.e. no descriptive data. The aim of a join made between the two datasets is to select those points (and hence all the descriptive data) from dataset one that has the same x and y co-ordinates as dataset two. x1-c(1824615,1823650,1821910) y1-c(5980732,5983220,5990931) descript-c(cat, dog, horse) dat1-data.frame(x1,y1,descript) x2-c(1824615,1823650) y2-c(5980732,5983220) dat2-data.frame(x2,y2) dat1 dat2 The aim of the join is to produce: x1 y1 descript 1 1824615 5980732 cat 2 1823650 5983220 dog And therefore limit the data in dataset 1 to that which has the same x and y co-ordinates as dataset 2. Any suggestions? Regards Andy Andrew McFadden MVS BVSc Incursion Investigator Investigation Diagnostic Centres - Wallaceville Biosecurity New Zealand Ministry of Agriculture and Forestry Phone 04 894 5600 Fax 04 894 4973 Mobile 029 894 5611 Postal address: Investigation and Diagnostic Centre- Wallaceville Box 40742 Ward St Upper Hutt This email message and any attachment(s) is intended solely for the addressee(s) named above. The information it contains is confidential and may be legally privileged. Unauthorised use of the message, or the information it contains, may be unlawful. If you have received this message by mistake please call the sender immediately on 64 4 8940100 or notify us by return email and erase the original message and attachments. Thank you. The Ministry of Agriculture and Forestry accepts no responsibility for changes made to this email or to any attachments after transmission from the office. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Estimating QAIC using glm with the quasibinomial family
Hello R-list. I am a long time listener - first time caller who has been using R in research and graduate teaching for over 5 years. I hope that my question is simple but not too foolish. I've looked through the FAQ and searched the R site mail list with some close hits but no direct answers, so... I would like to estimate QAIC (and QAICc) for a glm fit using the quasibinomial family. I found a general reference suggesting a simple solution: we calculated QAICc adjusting for overdispersion by dividing the residual deviance (i.e. -2 loglikelihood) with the overdispersion parameter calculated from the most complex model as the sum of squares Pearson residuals divided by the number of degrees of freedom (Burnham Anderson, 2002). - Mystrud et al. 2007. Animal Conservation. 10:77-87. My question is: Will this calculation be valid with the residual deviance returned by the glm() function using the quasibinomial family as reported in R? I thought I should ask to be certain that there is no dispersion correction applied to the reported deviance, as encouraged by Burnham and Anderson, 2nd ed., 2002 on p.69: When data are overdispersed and c 1, the proper likelihood is log(L)/c. Regards, Darren Gillis Department of Biological Sciences Faculty of Science University of Manitoba Winnipeg, MB Canada, R3T 2N2 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] loess model with different percentile
Hi all, I'm looking for a way to use loess model (?loess) with a different percentile (instead of median). I hope I made myself clear enough. Here's and example: a-data.frame(x=seq(1:200),y=200*rnorm(200)) lss-loess(y~x,a) predict(lss, 120) will give a local median of that dataset. How about a local 99 percentile? Thanks.- -- Regards, Anh Tran [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] To preserve the class Matrix
Shubha Vishwanath Karanth [EMAIL PROTECTED] wrote in news:[EMAIL PROTECTED]: Thank you very much Mark! That worked Just a question, ?[ does give an error to me...how do I find it? Try: ?[ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Spatial join between two datasets using x and y co-ordinates
-Original Message- From: [EMAIL PROTECTED] on behalf of Yasir Kaheil Sent: Tue 5/6/2008 4:24 PM To: r-help@r-project.org Subject: Re: [R] Spatial join between two datasets using x and y co-ordinates vector dat1.select is the selected records from dat1 by dat2. dat1.select- dat1$x1 %in% dat2$x2 dat1$y1 %in% dat2$y2 dat1[dat1.select,] x1 y1 descript 1 1824615 5980732 cat 2 1823650 5983220 dog This won't always work, for example if another observation in dat1 had the x coordinate of one of the dat2 entries and the y coordinate of another: x1-c(1824615,1823650,1821910,1823650) y1-c(5980732,5983220,5990931,5980732) descript-c(cat, dog, horse, cow) dat1-data.frame(x1,y1,descript) x2-c(1824615,1823650) y2-c(5980732,5983220) dat2-data.frame(x2,y2) dat1 x1 y1 descript 1 1824615 5980732 cat 2 1823650 5983220 dog 3 1821910 5990931horse 4 1823650 5980732 cow dat2 x2 y2 1 1824615 5980732 2 1823650 5983220 dat1.select- dat1$x1 %in% dat2$x2 dat1$y1 %in% dat2$y2 dat1[dat1.select,] x1 y1 descript 1 1824615 5980732 cat 2 1823650 5983220 dog 4 1823650 5980732 cow merge() uses the key variables pasted together, to form a unique key, e.g. dat1[match(paste(dat1$x1, dat1$y1, sep = \r), paste(dat2$x2, dat2$y2, sep = \r), nomatch = 0),] x1 y1 descript 1 1824615 5980732 cat 2 1823650 5983220 dog see e.g. merge.data.frame Steven McKinney Statistician Molecular Oncology and Breast Cancer Program British Columbia Cancer Research Centre email: smckinney +at+ bccrc +dot+ ca tel: 604-675-8000 x7561 BCCRC Molecular Oncology 675 West 10th Ave, Floor 4 Vancouver B.C. V5Z 1L3 Canada Andrew McFadden wrote: Hi R users I am trying to create a spatial join between two datasets. The first data set is large and contains descriptive data including x and y co-ordinates. The second dataset is small and has been selected spatially. The only data contained within the second dataset is the x and y coordinates only i.e. no descriptive data. The aim of a join made between the two datasets is to select those points (and hence all the descriptive data) from dataset one that has the same x and y co-ordinates as dataset two. x1-c(1824615,1823650,1821910) y1-c(5980732,5983220,5990931) descript-c(cat, dog, horse) dat1-data.frame(x1,y1,descript) x2-c(1824615,1823650) y2-c(5980732,5983220) dat2-data.frame(x2,y2) dat1 dat2 The aim of the join is to produce: x1 y1 descript 1 1824615 5980732 cat 2 1823650 5983220 dog And therefore limit the data in dataset 1 to that which has the same x and y co-ordinates as dataset 2. Any suggestions? Regards Andy Andrew McFadden MVS BVSc Incursion Investigator Investigation Diagnostic Centres - Wallaceville Biosecurity New Zealand Ministry of Agriculture and Forestry Phone 04 894 5600 Fax 04 894 4973 Mobile 029 894 5611 Postal address: Investigation and Diagnostic Centre- Wallaceville Box 40742 Ward St Upper Hutt This email message and any attachment(s) is intended solely for the addressee(s) named above. The information it contains is confidential and may be legally privileged. Unauthorised use of the message, or the information it contains, may be unlawful. If you have received this message by mistake please call the sender immediately on 64 4 8940100 or notify us by return email and erase the original message and attachments. Thank you. The Ministry of Agriculture and Forestry accepts no responsibility for changes made to this email or to any attachments after transmission from the office. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - Yasir H. Kaheil, Ph.D. Catchment Research Facility The University of Western Ontario -- View this message in context: http://www.nabble.com/Spatial-join-between-two-datasets-using-x-and-y-co-ordinates-tp17093486p17093656.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list