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...ion 1.4-9 o Upated Russian translation (courtesy of Alexey Shipunov). o Fixed bug in BreuschPaganTest() that prevented dialog from functioning (reported by Michel Pouchain). Version 1.4-10 o Corrected problem in paste in Command window in non-Windows systems (reported by Liviu Andronic). o Fixed handling of X11 warnings which was disrupted by showData(), and added check for "Warning in structure" (problems reported by Liviu Andronic). o Prefixes used when output is directed to the R console are no longer hard-coded, but can be set by the "prefixes...

Dear R users I would like to do some spreadsheet style expansion of dates. For example, I would need to obtain a vector of months. I approached in an obviously wrong way: > paste(01:12) [1] "1" "2" "3" "4" "5" "6" "7" "8" "9" "10" "11" "12" > as.Date(paste(01:12),

Dear all, Given the following vector: > (z <- c('R project', 'hello world', 'something Else')) [1] "R project" "hello world" "something Else" I know how to obtain all capitals or all lower case letters: > tolower(z) [1] "r project" "hello world" "something else" > toupper(z) [1] "R

Dear all Is there an easier way to retrieve the name of an object? For example, > tmp <- 1:10 > as.character(quote(tmp)) [1] "tmp" > as.character(quote(mtcars$cyl)) [1] "$" "mtcars" "cyl" > as.character(quote(mtcars$cyl))[3] [1] "cyl" The last call more than anything seems a hack. Is there a better way? Thank you Liviu

Dear all, I'm trying to gsub() "%" with "\%" with no obvious success. > temp1 <- c("mean", "sd", "0%", "25%", "50%", "75%", "100%") > temp1 [1] "mean" "sd" "0%" "25%" "50%" "75%" "100%" > gsub("%",

Dear all How does one convert a "non-symmetric" list to a vector? See below: > x <- list() > x[[1]] <- letters[1:5] > x[[2]] <- letters[6:10] > x[[3]] <- letters[11:12] > x [[1]] [1] "a" "b" "c" "d" "e" [[2]] [1] "f" "g" "h" "i" "j" [[3]] [1] "k"

Dear all Is there a simpler method to achieve the following: When I obtain an empty data.frame after subsetting, I need for it to contain one line of NAs. Here's a dummy example: > (.xb <- iris[ iris$Species=='zz', ]) [1] Sepal.Length Sepal.Width Petal.Length Petal.Width Species <0 rows> (or 0-length row.names) > dim(.xb) [1] 0 5 > (.xa <-

Dear all I'd like to make a beeping sound in R, but alarm() doesn't beep? I checked ?alarm but I couldn't find any pointers to system configuration. Any ideas? Regards Liviu > sessionInfo() R version 2.14.2 (2012-02-29) Platform: x86_64-pc-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C [3] LC_TIME=en_US.UTF-8 LC_COLLATE=en_US.UTF-8 [5]

Dear all This may be obvious, but I cannot get it working. I'm trying to subset & sort a data frame in one go. x <- iris x$Species1 <- as.character(x$Species) ##subsetting alone works fine with(x, x[Sepal.Length==6.7,]) ##sorting alone works fine with(x, x[order(Sepal.Length, rev(sort(Species1))),]) ##gets subsetted, but not sorted as expected with(x, x[(Sepal.Length==6.7) &

Dear all I'm using Xubuntu Lucid and I keep getting the same random numbers whenever I start a new session of R. For example, I keep getting > sample(1:1000, 1) [1] 87 or > rnorm(1:10) [1] -1.3618103 0.4241701 1.0720076 0.2208145 -0.5375314 -0.4846588 [7] 0.7576768 0.6527407 -0.6868786 0.8718527 I expected that some set.seed() instruction woudl be present in a config file in

Why does R CMD INSTALL work for some packages (e.g., lme4) but not others (e.g., nlme)? Thanks, Gang

Dear all, I'm looking for an alternative way to replicate the "2," string for an x number of times, and end up with one string containing "2," x times. I can partly achieve this using replicate(). > y <- rep("2,", times=3) > y [1] "2," "2," "2," The output that I am looking for is, however, "2,2,2,". I also tried

Dear all, The other day I stumbled on this article, "A critique of R and S-PLUS" [1], and got curious on whether the points outlined are (still) valid. The article is quite old, dating 2004, but was updated several times. Regards, Liviu [1] http://fluff.info/blog/arch/00000041.htm -- Do you know how to read? http://www.alienetworks.com/srtest.cfm Do you know how to write?

Dear all Is there an easy way to display only one half (top-right or bottom-left) of a correlation matrix? > require(Hmisc) > rcorr(as.matrix(mtcars[ , 1:4])) mpg cyl disp hp mpg 1.00 -0.85 -0.85 -0.78 cyl -0.85 1.00 0.90 0.83 disp -0.85 0.90 1.00 0.79 hp -0.78 0.83 0.79 1.00 n= 32 P mpg cyl disp hp mpg 0 0 0 cyl 0 0 0 disp 0 0

...0 & 4267 & 6553 & 179774 \\ \hline \end{tabular} \end{center} \end{table} Regards, Liviu ---------- Forwarded message ---------- From: Romain Francois <rfrancois at mango-solutions.com> Date: Dec 5, 2007 2:10 PM Subject: RE: [R] alternatives to latex() or xtable() ? To: Liviu Andronic <landronimirc at gmail.com> You need to look at the digits argument of xtable that would allow you to control this i think. > xtable( numSummary( iris[,1:4] ) , digit = c( 0, 0, 2,2,2,2,2,2,0) ) % latex table generated in R 2.6.0 by xtable 1.5-2 package % Wed Dec 05 13:07:47 2007...

hi, i am running R with JGR and Deducer under Ubuntu Mint Lucid. since i updated to R 2.12 i can no longer make graphs - R just crashes. the full error report was posted to launchpad (https://bugs.launchpad.net/ubuntu/+source/openjdk-6/+bug/675905) without any success so far. anyone with a similar problem or helpful suggestions? thanks! kat

Hi R friends and users, Just for fun (or concern): I received a R spam mail. Perhaps the first in history... Subject: R Courses and Consulting From: R Training33 <rtrainers33a at gmail.com> > R Courses and Consulting > Dear Sir, > > We are working on our 2010 R training schedule and would like to know > if you are interested in attending R courses this year. >

Dear all How does print.htest display the p-value in scientific notation? > (x <- cor.test(iris[[1]], iris[[3]])) Pearson's product-moment correlation data: iris[[1]] and iris[[3]] t = 21.65, df = 148, p-value < 2.2e-16 alternative hypothesis: true correlation is not equal to 0 95 percent confidence interval: 0.8270 0.9055 sample estimates: cor 0.8718 Above the p-value comes

Dear all I have a work-flow issue with lm(). When I use > lm(y1~x1, anscombe) Call: lm(formula = y1 ~ x1, data = anscombe) Coefficients: (Intercept) x1 3.0001 0.5001 I get as expected the formula, "y1 ~ x1", in the print()ed results or summary(). However, if I pass through a formula object > (form <- formula(y1~x1)) y1 ~ x1 > lm(form, anscombe) Call:

Hello everyone, As some may know, today Google unveiled its 2001 search index [1]. I was curious to see how was R like at that time, and was not disappointed. Compared to today's main page [2], seven years ago the page looked [3] a bit rudimentary, especially the graphic. (It is wort noting that structurally the pages are very similar.) What definitely changed is the `Contributed