similar to: fast NA elimination ?

I ran into an interesting oddity of R, if (0) { print(1); } else { print(2); } is a syntax error, while if (0) { print(1); } else { print(2); } or if (0) { print(1); } else { print(2); } is not. I presume it has to do with the duality of the newline functioning as an end of command (;) character, though it still seems a bit odd, and it took me a while to figure out

Why does stop("we are done") print "Error in eval.with.vis(expr, envir, enclos) :" ? It would seem to me that a plain stop() is not an error, and that it would make more sense to have an error() function that is different from a stop(). Is there a rationale here that I am missing? sincerely, /iaw

dear wizards: FYI: gentoo is a linux meta distribution, which compiles all packages. Once running, gentoo is stable on most applications. (it has some problems with system tools, such as grub.) the compiler is gcc 3.3.3. I do not expect anyone to track down for me why R fails on the gentoo amd64+gcc3.3.3 system, but I thought that it would be good to put it on the record to save

dear wizards: FYI: gentoo is a linux meta distribution, which compiles all packages. Once running, gentoo is stable on most applications. (it has some problems with system tools, such as grub.) the compiler is gcc 3.3.3. I do not expect anyone to track down for me why R fails on the gentoo amd64+gcc3.3.3 system, but I thought that it would be good to put it on the record to save

hi: I searched the last 2 hours for a way to enter a data frame directly in my program. (I know how to read from a file.) that is, I would like to say something like d <- this.is.a.data.frame( c("obs1name", 0.2, 0.3), c("obs2name", 0.4, 1.0), c("obs3name", 0.6, 2.0) ,

Dear R wizards: is there a clean way to assign to elements in a list? what I would like to do, in pseudo R+perl notation is f <- function(a,b) list(a+b,a-b) (c,d) <- f(1,2) and have c be assigned 1+2 and d be assigned 1-2. right now, I use the clunky x <- f(1,2) c <- x[[1]] d <- x[[2]] rm(x) which seems awful. is there a nicer syntax? regards, /iaw ---- Ivo Welch

hi: apologies for taking up everyone's time. my problem is probably documented somewhere, but I again cannot find it. (which reminds me: I cannot find a search engine that allows me to search the archives of this very useful mailing list.) * it seems that lines(lowess()) fails to plot certain line segments. (and, what does it do at the x-min and x-max of a data set?) Rather than

dear R experts: has someone written a function that returns the results of by() as a data frame? ??of course, this can work only if the output of the function that is an argument to by() is a numerical vector. presumably, what is now names(byobject) would become a column in the data frame, and the by object's list elements would become columns. it's a little bit like flattening the by()

I think many people share your view and are aghast at the reception that some well-intentioned posts receive. There have been past discussions on this and many people feel the way you and I do. Just to head off another round, let me acknowledge that there appears to be multiple viewpoints and although hard to believe by myself, there actually is a contingent that views what I see as

Dear R experts--- after many years, I am planning to give in and write my first R package. I want to combine my collection of collected useful utility routines. as my guide, I am planning to use Friedrich Leisch's "Creating R Packages: A Tutorial" from Sep 2009. Is there a newer or better tutorial? this one is 4 years old. I also plan on one change---given that the

Dear R Wizards---is it possible to get R to show its current call stack (sys.calls()) upon an error abort? I don't use ESS for execution, and it is often not obvious how to locate how I triggered an error in an R internal function. Seeing the call stack would make this easier. (right now, I sprinkle "cat" statements everywhere, just to locate the line where the error appears.) Of

dear R wizards: I have a large data frame, a million rows, 40 columns. In this data frame, there are some (about 100,000) rows which I want to recompute (update), while I want to leave others just as is. this is based on a condition that I need to compute, based on what is in a few of the columns. what is the right R way to do this? I could subset out the rows that I want to recompute into a

Ivo, Let me address your points in reverse order: 1. There is a `wishlist' category for bug reports, which I guess you've overlooked. 2. There is also a `Contributed Documentation' section on the R web site, which you can submit your contribution. As well, there are a few introductory level documents there already that you might be interested. 3. I must repectfully disagree about

dear R experts--- I am experimenting with multicore processing, so far with pretty disappointing results. Here is my simple example: A <- 100000 randvalues <- abs(rnorm(A)) minfn <- function( x, i ) { log(abs(x))+x^3+i/A+randvalues[i] } ?## an arbitrary function ARGV <- commandArgs(trailingOnly=TRUE) if (ARGV[1] == "do-onecore") { ?library(foreach) ?discard <-

dear wizards: my .Rprofile has just one command for testing, loadhistory("~/.Rhistory") but this gives me an error on R startup: Error: couldn't find function "loadhistory" Invoking loadhistory() as the first interactive command works fine; incidentally, I believe loadhistory() in the .Rprofile worked in earlier or other platform R releases, too. Is the .Rprofile

dear R wizards---more ignorance on my part, exacerbated by too few examples in the function documentations. > d <- data.frame( id=rep(1:3,3), x=rnorm(9), y=rnorm(9)) Question 1: how do I work with the output of "by"? for example, > b <- by( d, d$id, function(x) coef(lm( y ~ x, data=x ) )) > b d$id: 1 (Intercept) x 0.2303 0.3618

dear R wizards:? I am wrestling with reshape.? I have a long data set that I want to convert into a wide data set, in which rows are firms and columns are years. > summary(rin) firm fyear sim1 Min. :1004.00 Min. :1964.0 Min. : -1.00000 1st Qu.:1010.00 1st Qu.:1979.0 1st Qu.: -0.14334 Median :1016.00 Median :1986.0 Median : 0.00116 Mean

dear r experts---Is there a multicore equivalent of by(), just like mclapply() is the multicore equivalent of lapply()? if not, is there a fast way to convert a data.table into a list based on a column that lapply and mclapply can consume? advice appreciated...as always. regards, /iaw ---- Ivo Welch (ivo.welch at gmail.com)

The treatment of dates seems to be a little inconsistent in R 2.11.1 (2010-05-31): [1] The choice of origins? > as.integer(as.Date("1970-01-01")) works and assumes as origin 1970-01-01. However, > as.Date(1) does not work. It requires an origin (as.Date(1, origin="1970-01-01")). If we set a default origin in the former, it should probably work when the input

is it possible to throw a stop() that is so hard that it will escape even tryCatch? /iaw ---- Ivo Welch (ivo.welch at gmail.com)