similar to: convert text to exprission good for lm arguments

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,

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

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 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

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-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-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

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-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?

Hi, I thought it would be convenient if the check.names argument to read.table, which currently can only be TRUE/FALSE, could take a function value as well. If the function is supplied it should be used instead of the default make.names. Here is an example where it can come in handy. I tend to keep my data in coma-separated files with a header line. The header line is prefixed with a comment

Hi, Is it possible to modify an object in the parent.env (as opposed to re-bind)? Here is what I tried: > x = 1:3 # try to modify the first element of x from within a new environment > local(get("x", parent.env(environment()))[1] <- NA) Error in eval(expr, envir, enclos) : Target of assignment expands to non-language object # On the other hand retrieval works just fine >

Hi, This is not an R question per ce, but I feel like this is a right community to ask it. As a part of our work we run a lot of non-interactive computational jobs. To increase the throughput we would like to distribute the load over the entire network and we are looking at Linux network as a platform. Ideally we would like to be able to submit a job to the network, rather than to a computer, and

Hi, As was discussed earlier in another thread and as documented in R-exts .Call() should not be interruptible by Ctrl-C. However the following code, which spends most of its time inside mkChar, turned out to be interruptible on RH-7.3 R-1.8.1 gcc-2.96: #include <Rinternals.h> #include <R.h> SEXP foo0(const SEXP nSexp) { int i, n; SEXP resSexp; if (!isInteger(nSexp))

Hi, I tried to write the dim method for the list class, but R doesn't seem to dispatch to it: > dim.list = function(x) c(length(x[[1]]), length(x)) > dim(list(1)) NULL > dim.list(list(1)) [1] 1 1 What is the correct way of registering dim.list with .Primitive("dim")? Thanks, Vadim [[alternative HTML version deleted]]

Hi, I am looking for an efficient way of skipping big chunks of lines on a connection (not necessarily at the beginning of the file). One way is to use read lines, e.g. readLines(1e6), but a) this incurs the overhead of construction of the return char vector and b) has a (fairly remote) potential to blow up the memory. Another way would be to use scan(), e.g. scan(con, skip=1e6, nmax=0)

Hi, I am trying to figure a way to allocate a string SEXP so that gc() won't ever collect it. Here is a little bit of a background. Suppose I want to write a .Call-callable function that upon each call returns the same value, say mkChar("foo"): SEXP getFoo() { return mkChar("foo"); } The above implementation doesn't take advantage of the fact that

Hi, I am having a problem to understand why as.data.frame method doesn't dispatch properly on my class: > setClass("Foo", "character") [1] "Foo" > as.data.frame(list(foo=new("Foo", .Data="a"))) Error in as.data.frame.default(x[[i]], optional = TRUE) : can't coerce Foo into a data.frame I was expecting that this would call

Hi, What is the reason for seq.Date to require the 'by' argument and not to default it to 1 in the example below? > seq(from=as.Date("1996-01-01"), to=as.Date("1996-12-01")) Error in seq.Date(from = as.Date("1996-01-01"), to = as.Date("1996-12-01")) : exactly two of `to', `by' and `length.out' / `along.with' must be specified

Dear R-Users, I need to find a smooth function f() and coefficients a_i that give the best fit to y ~ a_0 + a_1*f(x_1) + a_2*f(x_2) Note that it is the same non-linear transformation f() that is applied to both x_1 and x_2. So my first question is how can I do it in R? A more general question is this: suppose I have a utility function U(a_i, f()), where f() is say a spline. Is there a general