Finding and writing environment - r

Question

I am writing a R equivalent to Pythons 'pop' method. I know 99th percentile has one but I'd prefer my own (practice/understanding/consistency etc).
For reference, pop() takes an object and removes the first item from the object whilt also returning it. So
> l <- c(1,3,5)
> x <- pop(l)
> print(l)
> 3, 5
> print(x)
> 1
I am using assign() to replace the input object with one less the first value and returning said first value from the function.
My question is, how do I get the environment of the input object and use this environment within assign()?
I have tried using pryr::where() which returns 'R_GlobalEnv' but I can't use this value in assign(). Instead the only value I can get to work in assign() is 'globalenv()'.
Posted from mobile so let me know if something doesn't work.

Answer 1

You can implement this in base R, though it's not advised. R is a functional language and functions with side effects are not expected by end-users.
pop <- function(vec)
{
vec_name <- deparse(substitute(vec))
assign(vec_name, vec[-1], envir = parent.frame())
vec[1]
}
a <- c(2, 7, 9)
a
#> [1] 2 7 9
pop(a)
#> [1] 2
a
#> [1] 7 9
pop(a)
#> [1] 7
a
#> [1] 9
Created on 2020-08-15 by the reprex package (v0.3.0)

Answer 2

The following answer is based in this R-Help post, function pop with function getEnvOf from this SO post, both adapted to the question's problem.
getEnvOf <- function(what, which=rev(sys.parents())) {
what <- as.character(substitute(what))
for (frame in which)
if (exists(what, frame=frame, inherits=FALSE))
return(sys.frame(frame))
return(NULL)
}
pop <- function(x){
y <- as.character(substitute(x))
e <- getEnvOf(y)
if(length(x) > 0) {
val <- x[[length(x)]]
assign(y, x[-length(x)], envir = parent.env(e))
val
} else {
msg <- paste(sQuote(y), "length is not > 0")
warning(msg)
NULL
}
}
y <- c(1,3,5)
pop(y)
This also works with lists.
z <- list(1, 2, 5)
pop(z)
w <- list(1, c(2, 4, 6), 5)
pop(w)
#[1] 5
pop(w)
#[1] 2 4 6
pop(w)
#[1] 1
pop(w)
#NULL
#Warning message:
#In pop(w) : ‘w’ length is not > 0

Answer 3

You can do it using pryr::promise_info(l)$env, but it's a very un-R-like thing to do. Functions shouldn't have side effects.
For example,
pop <- function(l) {
info <- pryr::promise_info(l)
if (!is.name(info$code))
stop("Argument expression should be a name.")
result <- l[[1]] # work on lists too
assign(as.character(info$code), l[-1], envir = info$env)
result
}
l <- c(1, 3, 5)
pop(l)
#> Registered S3 method overwritten by 'pryr':
#> method from
#> print.bytes Rcpp
#> [1] 1
l
#> [1] 3 5
Created on 2020-08-15 by the reprex package (v0.3.0)
Edited to add: Interestingly, none of the three answers so far works in complicated situations like this one:
f <- function(x) {
cat("The pop(x) result is", pop(x), "\n")
cat("Now x is ", x, "\n")
cat("Now l is ", l, "\n")
}
l <- c(1, 3, 5)
f(l)
#RuiBarradas's answer gives
The pop(x) result is 5
Now x is 1 3 5
Now l is 1 3 5
(He pops the last value rather than the first which is not a big deal, but neither x nor l is modified.)
#AllanCameron's answer gives
The pop(x) result is 1
Now x is 3 5
Now l is 1 3 5
This is arguably correct (x got popped), but I think it would be nice to have l being popped, and that seems tricky.
My answer dies with this message:
Error in pop(x) : Argument expression should be a name.
which seems like a bug: obviously whether it's getting x or l, it really is a name. The problem seems to be in pryr::promise_info, which returns the compiled code that would return the value of x, rather than just the code for x. If I turn off JIT compiling by compiler::enableJIT(0), I get the same result as #AllanCameron. It's not clear to me how to unwind back the right amount to pop l instead of just x.