search for: evalqs

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)

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 <-

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")

I think rapply() was changed to act like lapply() in this respect. In R-3.1.3 we got rapply(list(quote(1+myNumber)), evalq, envir=list2env(list(myNumber=17))) #[1] 18 rapply(list(quote(1+myNumber)), eval, envir=list2env(list(myNumber=17))) #Error in (function (expr, envir = parent.frame(), enclos = if (is.list(envir) || : object 'myNumber' not found lapply(list(quote(1+myNumber)),

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,

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,

Another aspect of the change is (using TERR's RinR package): > options(REvaluators=list(makeREvaluator("R-3.1.3"), makeREvaluator("R-3.2.0"))) > RCompare(rapply(list(quote(function(x)x),list(quote(pi),quote(7-4))), function(arg)typeof(arg))) R version 3.1.3 (2015-03-09) R version 3.2.0 (2015-04-16) [1,] [1]

Bill, Is your conclusion to just update the code and enforce using the most recent version of R? Dayne On Wed, Jul 15, 2015 at 4:44 PM, Dayne Filer <dayne.filer at gmail.com> wrote: > 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.

...lternative solution that I found out was: evalq(l<-alist(),env=ref) and then evalq(ind <-c("a","b"), env=ref) evalq(l[ind] <- as.list(c(20,40)), env=ref) I would like to know if there is another possible solution, instead of doing these 'evalqs' along the program code. Thanks, Rita -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body&quo...

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;

I am curious why you used evalq instead of eval in this code. Bill Dunlap TIBCO Software wdunlap tibco.com On Wed, Jul 15, 2015 at 11:49 AM, Dayne Filer <dayne.filer at gmail.com> wrote: > 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

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(

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

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

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?

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

...> > and then > > > > evalq(ind <-c("a","b"), env=ref) > > evalq(l[ind] <- as.list(c(20,40)), env=ref) > > > > > > I would like to know if there is another possible solution, instead of > > doing these 'evalqs' along the program code. > > > > > > Thanks, > > > > Rita > > > > > > > > -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- > > r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.h...

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