search for: evalq

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) ???? ##...

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 example stores "x" in "fenv" on 3.1.2, but throws the error I gave when run on 3.2.1. To give credit, I received the idea for using evalq from SO: http:/...

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)) &gt...

...g eval with an expression-object rather than a call-object seemed to solve the problem when this was posed as a question on StackOverflow, but Dayne was not happy with that solution for other reasons that he is not describing. -- David. > 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))...

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)), evalq, envir=list2env(list(...

...reviously. 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, evalq, envir = fenv) Although the code in the example above is much simpler, it shows the problem. On 3.1.2 the expression is evaluated and x is stored as 10 in the given environment, fenv. In 3.2.1 the code throws an error: Error in eval(substitute(expr), envir, enclos) : object 'X' not found...

...solve the >> problem when this was posed as a question on StackOverflow, but Dayne was >> not happy with that solution for other reasons that he is not describing. >> >> -- >> David. >> >> > 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 'myNum...

...er than a call-object seemed to solve the problem > when this was posed as a question on StackOverflow, but Dayne was not happy > with that solution for other reasons that he is not describing. > > -- > David. > > > 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 > >...

...hen this was posed as a question on StackOverflow, but Dayne was >>> not happy with that solution for other reasons that he is not describing. >>> >>> -- >>> David. >>> >>> > 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) || :...

...uot;l$a",20,env=ref) it creates me a new variable in the ref environment named "l$a" So, I did: eval(l <- alist(), env=ref) but this creates the l list both on the current and on the ref environment. The alternative 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, R...

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(...

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...

...#39;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( subfunc2b()) subfunc2_ function() evalq( outer.var, envir=parent.frame()) #print( subfunc2()) } > test.parent.funcs() [1] 5 [1] 5 If you try this without the hash before the 3rd print statement, it crashes, complaining it can't find outer.var. But AFAICS, subfunc2 and subfunc2b should return the same thing. So, at a minimum,...

...L[[len]],parent.frame()) # =0 MAKES SENSE eval.parent(print(x)) # =1 WHY NOT ZERO ? Somehow this is different from eval.parent(L[[len]]) eval(print(x)) # =1 MAKES SENSE eval(print(x),parent.frame()) # =1 # WHY NOT ZERO ? Somehow this is different from eval(L[[len]],parent.frame) evalq(print(x)) # =1 MAKES SENSE evalq(print(x),parent.frame()) # =1 MAKES SENSE print("====================") x <- 2 eval.parent(L[[len]]) # =0 MAKES SENSE eval(L[[len]]) # =2 MAKES SENSE eval(L[[len]],parent.frame()) # =0 MAKES SENSE eval.parent(print(x)) # =2 WH...

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 Wha...

...never seen that warning on the help page of with() and within(), so I assumed both functions can safely be used in functions and packages. I've now been told that both functions pose the same risk as subset() and transform(). Looking at the source code I've noticed the extra step: e <- evalq(environment(), data, parent) which, at least according to my understanding, should ensure that the functions follow the standard evaluation rules. Could somebody with more knowledge than I have shed a bit of light on this issue? Thank you Joris -- Joris Meys Statistical consultant Ghent Univer...

...$first$b [1] 40 $second [1] 50 and > l$first$b [1] 40 I'm searching for some alternative to solution to define this hierarchy in the designed environment, without needing to use the 'evalq' command..... Is there one? Thanks again, Rita > > One might ask whether you are sure that you need to do this? > I work on very large data sets and copying has pretty much never > been an issue. > > > > > > > So, I did:...

...d regards > Joris > > On Wed, Apr 1, 2015 at 7:55 PM, Duncan Murdoch > <murdoch.duncan at gmail.com <mailto:murdoch.duncan at gmail.com>> wrote: > > 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 > enviro...

...doesn't seem to help for reaching into the namespace of hgu133a.db and inspecting 'hgu133aPFAM', e.g. > library("hgu133a.db") > substitute(hgu133aPFAM, env=ns) Error: hgu133aPFAM is defunct. Please use select() if you need access to PFAM or PROSITE accessions. > evalq(substitute(hgu133aPFAM), envir=ns) Error: hgu133aPFAM is defunct. Please use select() if you need access to PFAM or PROSITE accessions. > evalq(substitute(hgu133aPFAM, env=ns), envir=ns) Error: hgu133aPFAM is defunct. Please use select() if you need access to PFAM or PROSITE accessions. Th...