similar to: Handling NA values

Can anyone help me how to read xls file into R. I have tried following library(gdata) xlsfile <- file.path(.path.package('gdata'),'xls','iris.xls') read.xls(xlsfile) I got following error: Converting xls file to csv file... Error in system(cmd, intern = !verbose) : perl not found Error in file.exists(tfn) : invalid 'file' argument Question *1) What is the way

Dear all, Does R use double precision for calculation as default? If not, how to enforce double precision calculation in R for my current calculation session? I Use R-2.14.0 with windows XP. Thanks,

Tony Plate <tplate <at> acm.org> writes: > > Since the virtue and reliability of Wikis was brought up, I created a R > Wiki page for this at > http://www.sciviews.org/_rgui/wiki/doku.php?id=beginners:surprises:emptysetfuncs > > > Anyone: please correct errors and improve it! > > Tony Plate > OK, now I have another question: I see a wiki at

Tony Plate <tplate <at> acm.org> writes: > > Since the virtue and reliability of Wikis was brought up, I created a R > Wiki page for this at > http://www.sciviews.org/_rgui/wiki/doku.php?id=beginners:surprises:emptysetfuncs > > > Anyone: please correct errors and improve it! > > Tony Plate > OK, now I have another question: I see a wiki at

Dear R-help, I first thought that the empty set (for a vector) would be NULL. x <- c() x However, the documentation seems to make clear that there _many_ empty sets depending on the vector's mode, namely, numeric(0), character(0), logical(0), etc. This is borne out by y <- letters[1:3] z <- letters[4:6] intersect(y,z) which, of course, is non-NULL: is.null(character(0)) #

I happened to see these: > round(.5, 0) [1] 0 > round(1.5, 0) [1] 2 > round(2.5, 0) [1] 2 > round(3.5, 0) [1] 4 > round(4.5, 0) [1] 4 What is the rule here? Should not round(.5, 0) = 1, round(2.5, 0) = 3 etc? Thanks and regards,

Hello R-users, Checking the archives, I recently came across this topic: "export tables to Excel files" (http://r.789695.n4.nabble.com/export-tables-to-Excel-files-td1565679.html#a1565679), and the following interesting references have been proposed: http://rwiki.sciviews.org/doku.php?id=tips:data-io:ms_windows

I need a picture like this: http://rwiki.sciviews.org/doku.php?id=graph_gallery:graph38 http://rwiki.sciviews.org/doku.php?id=graph_gallery:graph38 but when I try compile it require("hexbin") data(NHANES)# pretty large data set! good <- !(is.na(NHANES$Albumin) | is.na(NHANES$Transferin)) NH.vars <- NHANES[good,

Hello again, Let say I have a matrix: Mat <- matrix(1:12, 4, 3) And a vector: Vec <- 5:8 Now I want to do following: Each element of row-i in 'Mat' will be divided by i-th element of Vec Is there any direct way to doing that? Thanks for your help

Hello I tried to install R on my Transformer Prime without root. I downloaded the three files i found on: http://rwiki.sciviews.org/doku.php?id=getting-started:installation:android and i extract all the files, also without root my prime. But now i can't install R with the files i extractet. Is it possible to install R on my Prime without root and how can i do it? Or is root a must, for

Per http://rwiki.sciviews.org/doku.php?id=developers:projects:gsoc2010 -- and http://rwiki.sciviews.org/doku.php?id=developers:projects:gsoc2010:syrfr -- I am applying to mentor the "Symbolic Regression for R" (syrfr) package for the Google Summer of Code 2010. I propose the following test which an applicant would have to pass in order to qualify for the topic: 1. Describe each of the

Dear all, I have a system of simultaneous equations with 2 unknowns as follows: x*y + (1-x) = 0.05 x*(y - .5)^2 + (1-x)*0.6 = 0.56^2 Ofcourse I can do it manually however wondering whether there is any direct way in R available to get the solution of this system? Thanks and regards, [[alternative HTML version deleted]]

I'm trying to do this in several ways but havent had any result. Im asked to install python, or perl.... etc. Can anybody suggest a direct, easy and understandable way? Every help would be appreciated. Thx. -- View this message in context: http://r.789695.n4.nabble.com/Export-R-dataframes-to-excel-tp3330399p3330399.html Sent from the R help mailing list archive at Nabble.com.

  Hola:   Perdonad por la insistencia, pero me gustaría saber si cualquier proyecto en R debe afrontarse desde el uso de clases, o si sólo procede ante determinados casos. De ser así, ¿cuáles serían?.   Gracias.   Eva [[alternative HTML version deleted]]

Dear all, I need to download an excel file from net, on which I have address like http://www.2shared.com/file/MMSMWv4B/MyData.html. Can I somehow directly download this file into my R workbook? Thanks, [[alternative HTML version deleted]]

Hello Rusers, I have worked on a R Wiki page for solutions in exporting R graphs, especially, the often-asked questions: - How can I export R graphs in vectorized format (EMF) for inclusion in MS Word or OpenOffice outside of Windows? - What is the best solution(s) for post-editing/annotating R graphs. The page is at: http://wiki.r-project.org/rwiki/doku.php?id=tips:graphics-misc:export. I

Hola, es la primera lista en la que participo, espero no meter mucho la pata y aprender rápido, se ruega comprensión. Simplemente me gustaría saber la forma de importar datos desde un archivo ".sf3' de Statgraphics. Creo que ese es un programa bastante difundido, pero no lo tengo y en principio no quiero tenerlo porque no es libre. Pero encuentro referencias a todo (piInfo, Minitab,

Dear all, I have a date related question. Suppose I have a character string "March-2009", how I can convert it to a valid date object like as.yearmon("2009-01-03") in the zoo package? Is there any possibility there? Ans secondly is there any R function which will give the names of of all months as "LETTERS" does? Thanks for your time. [[alternative HTML version

Below is my full implementation (tried to make it simple as for demonstration) Lapply_me = function(X = X, FUN = FUN, Apply_MC = FALSE, ...) { if (Apply_MC) { return(mclapply(X, FUN, ...)) } else { if (any(names(list(...)) == 'mc.cores')) { myList = list(...)[!names(list(...)) %in% 'mc.cores'] } return(lapply(X, FUN, myList)) } } Lapply_me(as.list(1:4), function(xx) { if (xx ==

Hi all, please see my code: > library(zoo) > a <- as.yearmon("March-2010", "%B-%Y") > b <- as.yearmon("May-2010", "%B-%Y") > > nn <- (b-a)*12 # number of months in between them > nn [1] 2 > as.integer(nn) [1] 1 What is the correct way to find the number of months between "a" and "b", still