similar to: names of return value of median

Dear R-devel, Is there a reason that lapply(NULL, ...) returns the empty list, rather than NULL? It seems intuitive to expect the latter, and rather counterintuitive that lapply(list(), ... ) returns the same value as lapply(NULL, ...). > lapply(list(), function(x) 1) list() > lapply(NULL, function(x) 1) list() > version _ platform i386-pc-mingw32 arch

Dear R-users, Suppose I have an expression: expr = expression(a>0) and now I want to modify it to expression(a>0 & b>0). The following doesn't work: expr = expression(expr & b>0) What would be a good way of doing this? Thanks, Vadim ________________________________ Note: This email is for the confidential use of the named addressee(s) only and may contain

Dear R-users, I am trying to write a wrapper function around save() that will report the file which is being saved to. So I thought that the followintg would do the trick, but it doesn't. I understand that 'y' is somehow not visible inside save.verbose, but don't know how to fix this. save.verbose <- function(..., file) { cat("save.verbose:", file, "\n")

Dear R Users, Given two vectors, say a = seq(2) and b = seq(3), I want to make an 2*3 array, where (i,j) element is list(a=a[i], b=b[j]). I tried the outer() function but it generates an error message that I don't understand, see below. What do I do wrong? The expan.grid function is not good enough since I need a solution that works when a and b are not atomic, say a=list(list(x=1,

Hi, One way of doing object-oriented programming in R is to use function environment to hold object's data, see for example @Article{Rnews:Chambers+Lang:2001a, author = {John M. Chambers and Duncan Temple Lang}, title = {Object-Oriented Programming in {R}}, journal = {R News}, year = 2001, volume = 1, number = 3, pages = {17--19},

Dear R-devel, When a character vector is used to subscript a list and when some of the subscripts are not present in the list names R returns NULL for those subscripts and generate NA names for each of them: > list(b=1)[c('a','b')] $<NA> <<-- generated name NULL $b [1] 1 Wouldn't it be more intuitive to use the subscript name rather than to generate an NA?

Dear R-developers, I am trying to figure out a way to call browser() when an error occur, and naturally I want the browser() to be called in the environment of the error. I tried something simple in vain: > f <- function() { x <- 1; stop('ok') } > tryCatch(f(), error=browser()) Called from: tryCatch(f(), error = browser()) ## if browser() was called in the local environment

Dear R-help, When the result of a matrix subscription degenerates to a scalar the names implied by the dimnames are discarded. > x <- matrix(0, 1, 1, dimnames=list('a', 'x')) ## below I expected result to have names='x', it's not > x[1,] [1] 0 ## below I expected result to have names='a', it's not > x[,1] [1] 0 This is probably a side effect

Dear R users, Please disregard my previous post "converting result of substitute to 'ordidnary' expression". The problem I have has nothing to do with substitute. Consider: > dat <- data.frame(x=1:10, y=1:10) > subsetexp <- expression(5<x) > ## this does work > subset(dat, eval(subsetexp)) x y 6 6 6 7 7 7 8 8 8 9 9 9 10 10 10 > ##

Hi, I seem to have a problem when passing named parameters to R via Rscript (R2.5.1, bash shell). As soon as I name elements of a list Rscript generates an error. I will appreciate if someone could point to me a correct way of doing this. Thanks, Vadim ## This works bash-3.2$ Rscript.exe -e 'list(1)' [[1]] [1] 1 # and these do not work bash-3.2$ Rscript.exe -e

Hi, I ran into a problem of converting a text representation of an expression into parsed expression to be further evaluated inside lm (). > n <- 100 > data <- data.frame(x= rnorm (n), y= rnorm (n)) > data. lm <- lm (y ~ x, data=data) > > ## this works > update(data. lm , subset=x<0) Call: lm (formula = y ~ x, data = data, subset = x < 0)

Hi, I have few high-level questions about the Snow and Rmpi packages . I understand that Snow uses Rmpi as one of possible transport layers, yet my questions about user experience, not technical details: 1. Does Snow install and work well in Windows? 2. Interruptibility. I understand that currently it is impossible to interrupt a running top-level command in Snow ( Ctl-c or the likes), the

Dear R-users, I am somewhat puzzled by how R treats data frames with nested data frames. Below are a couple of examples, maybe someone could help explain what the guiding logic here is. ## construct plain data frame > z <- data.frame(x=1) ## add a data frame member > z$y <- data.frame(a=1,b=2) ## puzzle 1: z is apparently different from a straightforward construction of the

Hi, I am debugging intermittent crashes of R that seem to happen when multiple R sessions nearly summaltaneously load same dll-based library. I have R and my libraries installed on a network drive (everything is Windows). The drive is visible from a farm of servers. I have an R script, foo.R, that just loads a dll-based library (to be precise it loads a library that requires a dll-based

Hi, When using evalq to evaluate expressions within a say data.frame context I often wish there was a 'subset' argument, much like in lm() or any ather advanced regression model. I would be grateful for a tip how to do this. Here is an illustration of what I want: n <- 100 data <- data.frame(x=rnorm(n), y=rnorm(y), z=rnorm(z)) # this works evalq({ i <- 0<x;

Dear R users, As substitute() help page points out: Substituting and quoting often causes confusion when the argument is 'expression(...)'. The result is a call to the 'expression' constructor function and needs to be evaluated with 'eval' to give the actual expression object. And indeed I am confused. Consider: > dat <- data.frame(x=1:10,

Dear R users, As substitute() help page points out: Substituting and quoting often causes confusion when the argument is 'expression(...)'. The result is a call to the 'expression' constructor function and needs to be evaluated with 'eval' to give the actual expression object. And indeed I am confused. Consider: > dat <- data.frame(x=1:10,

Dear R-users, It is my understanding that cat(shQuote(a.string)) should print the origintal a.string. Is this right? I am not sure cat() correctly prints strings which are generated by triple-shQuote(): > shQuote(shQuote("a")) [1] "\"\\\"a\\\"\"" > cat(shQuote(shQuote(shQuote("a"))), '\n')

Dear R-users, It is my understanding that cat(shQuote(a.string)) should print the origintal a.string. Is this right? I am not sure cat() correctly prints strings which are generated by triple-shQuote(): > shQuote(shQuote("a")) [1] "\"\\\"a\\\"\"" > cat(shQuote(shQuote(shQuote("a"))), '\n')

Dear R-users, Could someone please explain why the message printed by function stopifnot2, see below, is different from that of stopifnot itself? Thank you for your help, Vadim > stopifnot2 <- function(...) stopifnot(...) > stopifnot(F) Error: F is not TRUE > stopifnot2(F) Error: ..1 is not TRUE > version _ platform i386-pc-mingw32 arch i386 os