similar to: eval and Calling Frames

Hi Consider the following example: f <- function(expr) g(expr) g <- function(expr) { ? h(expr) } h <- function(expr) { ? expr # evaluation happens here ? i(expr) } i <- function(expr) { ? expr # already evaluated, no costs here ? invisible() } rprof <- tempfile() Rprof(rprof) f(replicate(1e2, sample.int(1e4))) Rprof(NULL) cat(readLines(rprof), sep = "\n") #>

Dear mailing list, I have found some strange behaviour which I think relates to parent frames and eval. Can anyone explain what's going on here? First example: > test.parent.funcs_ function() { outer.var_ 5 subfunc1_ function() substitute( outer.var, envir=parent.frame()) print( subfunc1()) subfunc2b_ function() eval( quote( outer.var), envir=parent.frame()) print(

On Jul 15, 2015, at 12:51 PM, William Dunlap wrote: > I think rapply() was changed to act like lapply() in this respect. > When I looked at the source of the difference, it was that typeof() returned 'language' in 3.2.1, while it returned 'list' in the earlier version of R. The first check in rapply's code in both version was: if (typeof(object) != "list")

David, If you are referring to the solution that would be: rapply(list(test), eval, envir = fenv) I thought I explained in the question that the above code does not work. It does not throw an error, but the behavior is no different (at least in the output or result). Using the above code still results in the x object not being stored in fenv on 3.1.2. Dayne On Wed, Jul 15, 2015 at 4:40 PM,

In 3.1.2 eval does not store the result of the bquote-generated call in the given environment. Interestingly, in 3.2.1 eval does store the result of the bquote-generated call in the given environment. In other words if I run the given example with eval rather than evalq, on 3.1.2 "x" is never stored in "fenv," but it is when I run the same code on 3.2.1. However, the given

The help page of eval says: The 'evalq' form is equivalent to 'eval(quote(expr), ...)'. But the following is not equivalent. Can anyone give me some explaination? Thanks very much. > f1 <- function(x,digits=5) lapply(x, f2) > f2 <- function(x) eval(quote(print(x+1,digits=digits)),list(x=x),parent.frame(2)) > f1(list(x1=1)) [1] 2 $x1 [1] 2 > > f1 <-

Hello, I upgraded from 3.1.2 to 3.2.1 and am receiving errors on code that worked as I intended previously. Briefly, I am using bquote to generate expressions to modify data.table objects within a function, so I need the changes to actually be stored in the given environment. Previously, I used code like the following: test <- list(bquote(x <- 10)) fenv <- environment() rapply(test,

I'm still going over old emails and trying to get my head around evaluation so I'm persistent if nothing else. A while back , an expert sent me below as an exercise in understanding and I only got around to it tonight. I understand some of the output but not all of it and I put "Why not Zero ?" next to the ones that I don't understand based on my reading of the various

Hi, In order to avoid deep copies by passing large arguments to functions or returning values, I'm trying to do the assignment of variables in a given environment. The problem is when I try to assign a structure: a list for example. If I have: ind <- c("a","b") my idea is doing something like l <- alist() l[ind] <- as.list(c(20,40)) in a given

Sorry to intervene. Argument passed to 'eval' is evaluated first. So, eval(x <- 2) is effectively like { x <- 2; eval(2) } , which is effectively { x <- 2; 2 } . The result is visible. eval(expression(x <- 2)) or eval(quote(x <- 2)) or evalq(x <- 2) gives the same effect as x <- 2 . The result is invisible. In function 'eval2', res <-

Given > D = data.frame(o=gl(2,1,4)) this works as I expected: > evalq(o, D) [1] 1 2 1 2 Levels: 1 2 but neither of these does: > f <- function(x, dat) evalq(x, dat) > f(o, D) Error in eval(expr, envir, enclos) : object "o" not found > g <- function(x, dat) eval(x, dat) > g(o, D) Error in eval(x, dat) : object "o" not found What am I doing wrong?

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;

On 01/04/2015 1:35 PM, Gabriel Becker wrote: > Joris, > > > The second argument to evalq is envir, so that line says, roughly, "call > environment() to generate me a new environment within the environment > defined by data". I think that's not quite right. environment() returns the current environment, it doesn't create a new one. It is evalq() that created

Greetings all, Recently ran into a seg fault the keeps reoccurring whenever R-java is used (I believe). Not sure if this is an R issue or an openjdk issue so I'm trying to cover all bases with this report. I tried downgrading R version from 3.4.0 to 3.3.3 with the same results. Also tried downgrading java version from 1.8 to 1.7 with no luck. Running on a fully updated CentOS 7 x86_64,

On 01/04/2015 2:33 PM, Joris Meys wrote: > Thank you for the insights. I understood as much from the code, but I > can't really see how this can cause a problem when using with() or > within() within a package or a function. The environments behave like > I would expect, as does the evaluation of the arguments. The second > argument is supposed to be an expression, so I

On Mon, Jan 26, 2015 at 12:24 PM, Hadley Wickham <h.wickham at gmail.com> wrote: > If it was any other environment than the global, you could use substitute: > > e <- new.env() > delayedAssign("foo", stop("Hey!"), assign.env = e) > substitute(foo, e) > > delayedAssign("foo", stop("Hey!")) > substitute(foo) Hmm... interesting

Dear list members, I'm a bit confused about the evaluation of expressions using with() or within() versus subset() and transform(). I always teach my students to use with() and within() because of the warning mentioned in the helppages of subset() and transform(). Both functions use nonstandard evaluation and are to be used only interactively. I've never seen that warning on the help

Hi, I've just created: newEnvEval <- function(..., hash=FALSE, parent=parent.frame(), size=29L) { envir <- new.env(hash=hash, parent=parent, size=size); evalq(..., envir=envir); envir; } # newEnvEval() so that I can create an environment and assign objects to it in one go, e.g. env <- newEnvEval({ a <- 1; b <- 2; }); print(env$a); Does this already exists somewhere?

Greetings, Say I have defined mp <- function(a) function(x) x^a f2 <- mp(2) and I would like to retrieve the "a" which is local to f2. Two options come to mind; get("a", envir=environment(f2)) eval(substitute(a), environment(f2)) I'm curious if one of these is preferred over the other in terms of efficiency, robustness, aesthetics, etc. Or perhaps

Hi all, I'm not sure if I'm not supposed to do the following (the dyn.unload part, I mean) or this could be a bug (in R or Rcpp): ``` Rcpp::sourceCpp(code=' #include <Rcpp.h> class Object {}; //[[Rcpp::export]] SEXP new_object() { return Rcpp::XPtr<Object>(new Object()); }' ) new_object() dyn.unload(list.files(tempdir(), ".(so|dll)$",