Manipulating enclosing environment of a function - r

I'm trying to get a better understanding of closures, in particular details on a function's scope and how to work with its enclosing environment(s)
Based on the Description section of the help page on rlang::fn_env(), I had the understanding, that a function always has access to all variables in its scope and that its enclosing environment belongs to that scope.
But then, why isn't it possible to manipulate the contents of the closure environment "after the fact", i.e. after the function has been created?
By means of R's lexical scoping, shouldn't bar() be able to find x when I put into its enclosing environment?
foo <- function(fun) {
env_closure <- rlang::fn_env(fun)
env_closure$x <- 5
fun()
}
bar <- function(x) x
foo(bar)
#> Error in fun(): argument "x" is missing, with no default

Ah, I think I got it down now.
It has to do with the structure of a function's formal arguments:
If an argument is defined without a default value, R will complain when you call the function without specifiying that even though it might technically be able to look it up in its scope.
One way to kick off lexical scoping even though you don't want to define a default value would be to set the defaults "on the fly" at run time via rlang::fn_fmls().
foo <- function(fun) {
env_enclosing <- rlang::fn_env(fun)
env_enclosing$x <- 5
fun()
}
# No argument at all -> lexical scoping takes over
baz <- function() x
foo(baz)
#> [1] 5
# Set defaults to desired values on the fly at run time of `foo()`
foo <- function(fun) {
env_enclosing <- rlang::fn_env(fun)
env_enclosing$x <- 5
fmls <- rlang::fn_fmls(fun)
fmls$x <- substitute(get("x", envir = env_enclosing, inherits = FALSE))
rlang::fn_fmls(fun) <- fmls
fun()
}
bar <- function(x) x
foo(bar)
#> [1] 5

I can't really follow your example as I am unfamiliar with the rlang library but I think a good example of a closure in R would be:
bucket <- function() {
n <- 1
foo <- function(x) {
assign("n", n+1, envir = parent.env(environment()))
n
}
foo
}
bar <- bucket()
Because bar() is define in the function environment of bucket then its parent environment is bucket and therefore you can carry some data there. Each time you run it you modify the bucket environment:
bar()
[1] 2
bar()
[1] 3
bar()
[1] 4

Related

Why is this simple function not working?

I first defined new variable x, then created function that require x within its body (not as argument). See code below
x <- c(1,2,3)
f1 <- function() {
x^2
}
rm(x)
f2 <- function() {
x <- c(1,2,3)
f1()
}
f(2)
Error in f1() : object 'x' not found
When I removed x, and defined new function f2 that first define x and then execute f1, it shows objects x not found.
I just wanted to know why this is not working and how I can overcome this problem. I do not want x to be name as argument in f1.
Please provide appropriate title because I do not know what kind of problem is this.
You could use a closure to make an f1 with the desired properties:
makeF <- function(){
x <- c(1,2,3)
f1 <- function() {
x^2
}
f1
}
f1 <- makeF()
f1() #returns 1 4 9
There is no x in the global scope but f1 still knows about the x in the environment that it was defined in.
In short: Your are expecting dynamic scoping but are a victim of R's lexical scoping:
dynamic scoping = the enclosing environment of a command is determined during run-time
lexical scoping = the enclosing environment of a command is determined at "compile time"
To understand the lookup path of your variable x in the current and parent environments try this code.
It shows that both functions do not share the environment in with x is defined in f2 so it can't never be found:
# list all parent environments of an environment to show the "search path"
parents <- function(env) {
while (TRUE) {
name <- environmentName(env)
txt <- if (nzchar(name)) name else format(env)
cat(txt, "\n")
if (txt == "R_EmptyEnv") break
env <- parent.env(env)
}
}
x <- c(1,2,3)
f1 <- function() {
print("f1:")
parents(environment())
x^2
}
f1() # works
# [1] "f1:"
# <environment: 0x4ebb8b8>
# R_GlobalEnv
# ...
rm(x)
f2 <- function() {
print("f2:")
parents(environment())
x <- c(1,2,3)
f1()
}
f2() # does not find "x"
# [1] "f2:"
# <environment: 0x47b2d18>
# R_GlobalEnv
# ...
# [1] "f1:"
# <environment: 0x4765828>
# R_GlobalEnv
# ...
Possible solutions:
Declare x in the global environment (bad programming style due to lack of encapsulation)
Use function parameters (this is what functions are made for)
Use a closure if x has always the same value for each call of f1 (not for beginners). See the other answer from #JohnColeman...
I strongly propose using 2. (add x as parameter - why do you want to avoid this?).

Finding the origin environment of the ... (dots) arguments of a call

I want to be able to find the environment from which the ... (dots) arguments of a call originate.
Scenario
For example, consider a function
foo <- function(x, ...) {
# do something
}
We want a function env_dots(), which we invoke from within foo(), that finds the originating environment of the ... in a call to foo(), even when the call to foo() is deeply nested. That is, if we define
foo <- function(x, ...) {
# find the originating environment of '...'
env <- env_dots()
# do something
}
and nest a call to foo, like so,
baz <- function(...) {
a <- "You found the dots"
bar(1, 2)
}
bar <- function(...)
foo(...)
then calling baz() should return the environment in which the ... in the (nested) call to foo(...) originates: this is the environment where the call bar(1, 2) is made, since the 2 (but not the 1) gets passed to the dots of foo. In particular, we should get
baz()$a
#> [1] "You found the dots"
Naive implementation of env_dots()
Update — env_dots(), as defined here, will not work in general, because the final ... may be populated by arguments that are called at multiple levels of the call stack.
Here's one possibility for env_dots():
# mc: match.call() of function from which env_dots() is called
env_dots <- function(mc) {
# Return NULL if initial call invokes no dots
if (!rlang::has_name(mc, "...")) return(NULL)
# Otherwise, climb the call stack until the dots origin is found
stack <- rlang::call_stack()[-1]
l <- length(stack)
i <- 1
while (i <= l && has_dots(stack[[i]]$expr)) i <- i + 1
# return NULL if no dots invoked
if (i <= l) stack[[i + 1]]$env else NULL
}
# Does a call have dots?
has_dots <- function(x) {
if (is.null(x))
return(FALSE)
args <- rlang::lang_tail(x)
any(vapply(args, identical, logical(1), y = quote(...)))
}
This seems to work: with
foo <- function(x, ...)
env_dots(match.call(expand.dots = FALSE))
we get
baz()$a
#> [1] "You found the dots"
bar(1, 2) # 2 gets passed down to the dots of foo()
#> <environment: R_GlobalEnv>
bar(1) # foo() captures no dots
#> NULL
Questions
The above implementation of env_dots() is not very efficient.
Is there are more skillful way to implement env_dots() in rlang and/or base R?
How can I move the match.call() invocation to within env_dots()?
match.call(sys.function(-1), call = sys.call(-1), expand.dots = FALSE) will indeed work.
Remark — One can't infer the origin environment of the dots from rlang::quos(...), because some quosures won't be endowed with the calling environment (e.g., when an expression is a literal object).
I'm sorry to dig up an old question, but I'm not sure the desired behavior is well-defined. ... is not a single expression; it's a list of expressions. In case of rlang quosures, each of those expressions has their own environment. So what should the environment of the list be?
Furthermore, the ... list itself can be modified. Consider the following example, where g takes its ..., prepends it with an (unevaluated) expression x+3 and passes it onto f.
f <- function(...) {rlang::enquos( ... )}
g <- function(...) {
a <- rlang::quo( x + 3 )
l <- rlang::list2( a, ... )
f(!!!l)
}
b <- rlang::quo( 5 * y )
g( b, 10 )
# [[1]]
# <quosure>
# expr: ^x + 3
# env: 0x7ffd1eca16f0
# [[2]]
# <quosure>
# expr: ^5 * y
# env: global
# [[3]]
# <quosure>
# expr: ^10
# env: empty
Notice that each of the three quosures that make it over to f has their own environment. (As you noted in your question, literals like 10 have an empty environment. This is because the value is the same independent of which environment it's evaluated in.)
Given this scenario, what should the hypothetical env_dots() return when called inside f()?

Environments in R, mapply and get

Let x<-2 in the global env:
x <-2
x
[1] 2
Let a be a function that defines another x locally and uses get:
a<-function(){
x<-1
get("x")
}
This function correctly gets x from the local enviroment:
a()
[1] 1
Now let's define a function b as below, that uses mapply with get:
b<-function(){
x<-1
mapply(get,"x")
}
If I call b, it seems that mapply makes get not search the function environment first. Instead, it tries to get x directly form the global enviroment, and if x is not defined in the global env, it gives an error message:
b()
x
2
rm(x)
b()
Error in (function (x, pos = -1L, envir = as.environment(pos), mode = "any", :
object 'x' not found
The solution to this is to explicitly define envir=environment().
c<-function(){
x<-1
mapply(get,"x", MoreArgs = list(envir=environment()))
}
c()
x
1
But I would like to know what exactly is going on here. What is mapplydoing? (And why? is this the expected behavior?) Is this "pitfall" common in other R functions?
The problem is that get looks into the envivornment that its called from but here we are passing get to mapply and then calling get from the local environment within mapply. If x is not found within the mapply local environment then it looks the into the parent environment of that, i.e. into environment(mapply) (which is the lexical environment that mapply was defined in which is the base namespace environment); if it is not there either, it looks into the parent of that, which is the global environment, i.e. your R workspace.
This is because R uses lexical scoping, as opposed to dynamic scoping.
We can show this by getting a variable that exists within mapply.
x <- 2
b2<-function(){
x<-1
mapply(get, "USE.NAMES")
}
b2() # it finds USE.NAMES in mapply
## USE.NAMES
## TRUE
In addition to the workaround involving MoreArgs shown in the question this also works since it causes the search to look into the local environment within b after failing to find it mapply. (This is just for illustrating what is going on and in actual practice we would prefer the workaround shown in the question.)
x <- 2
b3 <-function(){
x<-1
environment(mapply) <- environment()
mapply(get, "x")
}
b3()
## 1
ADDED Expanded explanation. Also note that we can view the chain of environments like this:
> debug(get)
> b()
debugging in: (function (x, pos = -1L, envir = as.environment(pos), mode = "any",
inherits = TRUE)
.Internal(get(x, envir, mode, inherits)))(dots[[1L]][[1L]])
debug: .Internal(get(x, envir, mode, inherits))
Browse[2]> envir
<environment: 0x0000000021ada818>
Browse[2]> ls(envir) ### this shows that envir is the local env in mapply
[1] "dots" "FUN" "MoreArgs" "SIMPLIFY" "USE.NAMES"
Browse[2]> parent.env(envir) ### the parent of envir is the base namespace env
<environment: namespace:base>
Browse[2]> parent.env(parent.env(envir)) ### and grandparent of envir is the global env
<environment: R_GlobalEnv>
Thus, the ancestory of environments potentially followed is this (where arrow points to parent):
local environment within mapply --> environment(mapply) --> .GlobalEnv
where environment(mapply) equals asNamespace("base"), the base namespace environment.
R is lexically scoped, not dynamically scoped, meaning that when you search through parent environments to find a value, you are searching through the lexical parents (as written in the source code), not through the dynamic parents (as invoked). Consider this example:
x <- "Global!"
fun1 <- function() print(x)
fun2 <- function() {
x <- "Local!"
fun1a <- function() print(x)
fun1() # fun2() is dynamic but not lexical parent of fun1()
fun1a() # fun2() is both dynamic and lexical parent of fun1a()
}
fun2()
outputs:
[1] "Global!"
[1] "Local!"
In this case fun2 is the lexical parent of fun1a, but not of fun1. Since mapply is not defined inside your functions, your functions are not the lexical parents of mapply and the xs defined therein are not directly accessible to mapply.
The issue is an interplay with built-in C code. Namely, considering the following:
fx <- function(x) environment()
env <- NULL; fn <- function() { env <<- environment(); mapply(fx, 1)[[1]] }
Then
env2 <- fn()
identical(env2, env)
# [1] FALSE
identical(parent.env(env2), env)
# [1] FALSE
identical(parent.env(env2), globalenv())
# [1] TRUE
More specifically, the problem lies in the underlying C code, which fails to consider executing environment, and hands it off to an as-is underlying C eval call which creates a temp environment branching directly off of R_GlobalEnv.
Note this really is what is going on, since no level of stack nesting fixes the issue:
env <- NULL; fn2 <- function() { env <<- environment(); (function() { mapply(fx, 1)[[1]] })() }
identical(parent.env(fn2()), globalenv())
# [1] TRUE

Convenience function for exporting objects to the global environment

UPDATE: I have added a variant
of Roland's implementation to the kimisc package.
Is there a convenience function for exporting objects to the global environment, which can be called from a function to make objects available globally?
I'm looking for something like
export(obj.a, obj.b)
which would behave like
assign("obj.a", obj.a, .GlobalEnv)
assign("obj.b", obj.b, .GlobalEnv)
Rationale
I am aware of <<- and assign. I need this to refactor oldish code which is simply a concatenation of scripts:
input("script1.R")
input("script2.R")
input("script3.R")
script2.R uses results from script1.R, and script3.R potentially uses results from both 1 and 2. This creates a heavily polluted namespace, and I wanted to change each script
pollute <- the(namespace)
useful <- result
to
(function() {
pollute <- the(namespace)
useful <- result
export(useful)
})()
as a first cheap countermeasure.
Simply write a wrapper:
myexport <- function(...) {
arg.list <- list(...)
names <- all.names(match.call())[-1]
for (i in seq_along(names)) assign(names[i],arg.list[[i]],.GlobalEnv)
}
fun <- function(a) {
ttt <- a+1
ttt2 <- a+2
myexport(ttt,ttt2)
return(a)
}
print(ttt)
#object not found error
fun(2)
#[1] 2
print(ttt)
#[1] 3
print(ttt2)
#[1] 4
Not tested thoroughly and not sure how "safe" that is.
You can create an environment variable and use it within your export function. For example:
env <- .GlobalEnv ## better here to create a new one :new.env()
exportx <- function(x)
{
x <- x+1
env$y <- x
}
exportx(3)
y
[1] 4
For example , If you want to define a global options(emulate the classic R options) in your package ,
my.options <- new.env()
setOption1 <- function(value) my.options$Option1 <- value
EDIT after OP clarification:
You can use evalq which take 2 arguments :
envir the environment in which expr is to be evaluated
enclos where R looks for objects not found in envir.
Here an example:
env.script1 <- new.env()
env.script2 <- new.env()
evalq({
x <- 2
p <- 3
z <- 5
} ,envir = env.script1,enclos=.GlobalEnv)
evalq({
h <- x +2
} ,envir = env.script2,enclos=myenv.script1)`
You can see that all variable are created within the environnment ( like local)
env.script2$h
[1] 4
env.script1$p
[1] 3
> env.script1$x
[1] 2
First, given your use case, I don't see how an export function is any better than using good (?) old-fashioned <<-. You could just do
(function() {
pollute <- the(namespace)
useful <<- result
})()
which will give the same result as what's in your example.
Second, rather than anonymous functions, it seems better form to use local, which allows you to run involved computations without littering your workspace with various temporary objects.
local({
pollute <- the(namespace)
useful <<- result
})
ETA: If it's important for whatever reason to avoid modifying an existing variable called useful, put an exists check in there. The same applies to the other solutions presented.
local({
.....
useful <- result
if(!exists("useful", globalenv())) useful <<- useful
})

R specify function environment

I have a question about function environments in the R language.
I know that everytime a function is called in R, a new environment E
is created in which the function body is executed. The parent link of
E points to the environment in which the function was created.
My question: Is it possible to specify the environment E somehow, i.e., can one
provide a certain environment in which function execution should happen?
A function has an environment that can be changed from outside the function, but not inside the function itself. The environment is a property of the function and can be retrieved/set with environment(). A function has at most one environment, but you can make copies of that function with different environments.
Let's set up some environments with values for x.
x <- 0
a <- new.env(); a$x <- 5
b <- new.env(); b$x <- 10
and a function foo that uses x from the environment
foo <- function(a) {
a + x
}
foo(1)
# [1] 1
Now we can write a helper function that we can use to call a function with any environment.
with_env <- function(f, e=parent.frame()) {
stopifnot(is.function(f))
environment(f) <- e
f
}
This actually returns a new function with a different environment assigned (or it uses the calling environment if unspecified) and we can call that function by just passing parameters. Observe
with_env(foo, a)(1)
# [1] 6
with_env(foo, b)(1)
# [1] 11
foo(1)
# [1] 1
Here's another approach to the problem, taken directly from http://adv-r.had.co.nz/Functional-programming.html
Consider the code
new_counter <- function() {
i <- 0
function() {
i <<- i + 1
i
}
}
(Updated to improve accuracy)
The outer function creates an environment, which is saved as a variable. Calling this variable (a function) effectively calls the inner function, which updates the environment associated with the outer function. (I don't want to directly copy Wickham's entire section on this, but I strongly recommend that anyone interested read the section entitled "Mutable state". I suspect you could get fancier than this. For example, here's a modification with a reset option:
new_counter <- function() {
i <- 0
function(reset = FALSE) {
if(reset) i <<- 0
i <<- i + 1
i
}
}
counter_one <- new_counter()
counter_one()
counter_one()
counter_two <- new_counter()
counter_two()
counter_two()
counter_one(reset = TRUE)
I am not sure I completely track the goal of the question. But one can set the environment that a function executes in, modify the objects in that environment and then reference them from the global environment. Here is an illustrative example, but again I do not know if this answers the questioners question:
e <- new.env()
e$a <- TRUE
testFun <- function(){
print(a)
}
testFun()
Results in: Error in print(a) : object 'a' not found
testFun2 <- function(){
e$a <- !(a)
print(a)
}
environment(testFun2) <- e
testFun2()
Returns: FALSE
e$a
Returns: FALSE

Resources