similar to: evaluation revisited

I ran into an interesting issue with `evalq` (and also `eval(quote(...))`): ???? f <- function() { ?????? list( ???????? sys.parent(1), ???????? evalq(sys.parent(1)), ???????? evalq((function() sys.parent(2))()),? # add an anon fun layer ???????? evalq((function() sys.parent(1))()) ?????? ) ???? } ???? res <- f() ???? str(res) ???? ## List of 4 ???? ##? $ : int 0???????? # sys.parent(1)

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

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

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(

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

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 <-

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,

Hi, Why can't I pass an expression to `within' by way of textual input to the 'parse' function? e.g., > x <- data.frame(a=1:5,b=LETTERS[1:5]) > x a b 1 1 A 2 2 B 3 3 C 4 4 D 5 5 E > within(x, parse(text="a<-a*10; b<-2:6")) a b 1 1 A 2 2 B 3 3 C 4 4 D 5 5 E > within(x, parse(text="a<-a*10; b<-2:6")[[1]]) a b 1 1 A 2 2 B 3

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?

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

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 <-

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 would expect that expression to be evaluated in the data frame first. I

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 downloaded R v2.1.1 earlier this morning to compile under Fedora Core 4. It compiled without incident, but 'make check' failed. Below is the relevant part of its report. Is this a known problem? I used a locally compiled version of GCC v4.0.0 that reports [kent at d89h102 R-2.1.1]$ gcc -v Using built-in specs. Target: i686-pc-linux-gnu Configured with: ../gcc-4.0.0/configure

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". Note that that is is only generating e, the environment that expr will be evaluated within in the next line (the call to eval). This means that expr is evaluated in an environment which is inside the environment defined

I've been looking at the help page for eval for a while, but I can't make sense of why this example does not work. show.a <- function() { a } init.env <- function() { a <- 200 environment() } my.env <- init.env() ls(envir=my.env) # returns this: # > ls(envir=my.env) # [1] "a" # but this does not work: eval(expression(show.a()),envir=my.env) # >

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

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

I maintain the package affylmGUI. It works when installed on many previous versions of R. I have today tested exactly the same code under R-2.15.3, R-3.0.2, R-3.1.0, R-3.1.1, R-3.1.2 and R-devel. I have also tested the versions of affylmGUI downloaded by biocLite for each version of R and the same result applies. I have no errors under 2.15.3, 3.0.2, 3.1.0 and 3.1.1. The following error