rlang: Error: Can't convert a function to a string - r

I created a function to convert a function name to string. Version 1 func_to_string1 works well, but version 2 func_to_string2 doesn't work.
func_to_string1 <- function(fun){
print(rlang::as_string(rlang::enexpr(fun)))
}
func_to_string2 <- function(fun){
is.function(fun)
print(rlang::as_string(rlang::enexpr(fun)))
}
func_to_string1 works:
> func_to_string1(sum)
[1] "sum"
func_to_string2 doesn't work.
> func_to_string2(sum)
Error: Can't convert a primitive function to a string
Call `rlang::last_error()` to see a backtrace
My guess is that by calling the fun before converting it to a string, it gets evaluated inside function and hence throw the error message. But why does this happen since I didn't do any assignments?
My questions are why does it happen and is there a better way to convert function name to string?
Any help is appreciated, thanks!

This isn't a complete answer, but I don't think it fits in a comment.
R has a mechanism called pass-by-promise,
whereby a function's formal arguments are lazy objects (promises) that only get evaluated when they are used.
Even if you didn't perform any assignment,
the call to is.function uses the argument,
so the promise is "replaced" by the result of evaluating it.
Nevertheless, in my opinion, this seems like an inconsistency in rlang*,
especially given cory's answer,
which implies that R can still find the promise object even after a given parameter has been used;
the mechanism to do so might not be part of R's public API though.
*EDIT: see coments.
Regardless, you could treat enexpr/enquo/ensym like base::missing,
in the sense that you should only use them with parameters you haven't used at all in the function's body.

Maybe use this instead?
func_to_string2 <- function(fun){
is.function(fun)
deparse(substitute(fun))
#print(rlang::as_string(rlang::enexpr(fun)))
}
> func_to_string2(sum)
[1] "sum"

This question brings up an interesting point on lazy evaluations.
R arguments are lazily evaluated, meaning the arguments are not evaluated until its required.
This is best understood in the Advanced R book which has the following example,
f <- function(x) {
10
}
f(stop("This is an error!"))
the result is 10, which is surprising because x is never called and hence never evaluated. We can force x to be evaluated by using force()
f <- function(x) {
force(x)
10
}
f(stop("This is an error!"))
This behaves as expected. In fact we dont even need force() (Although it is good to be explicit).
f <- function(x) {
x
10
}
f(stop("This is an error!"))
This what is happening with your call here. The function sum which is a symbol initially is being evaluated with no arguments when is.function() is being called. In fact, even this will fail.
func_to_string2 <- function(fun){
fun
print(rlang::as_string(rlang::ensym(fun)))
}
Overall, I think its best to use enexpr() at the very beginning of the function.
Source:
http://adv-r.had.co.nz/Functions.html

Related

How an R function can be used before being defined

I came across this spinet of code where the function rval_top_ingredients() was used to render a D3wordcloud before it was defined. I think that would throw an error in case of Python as the script is executed from top to bottom. Why did it work in R then? Thankyou.
output$wc_ingredients <- d3wordcloud::renderD3wordcloud({
ingredients_df <- rval_top_ingredients()
d3wordcloud(ingredients_df$ingredient, ingredients_df$nb_recipes, tooltip = TRUE)
})
rval_top_ingredients <- reactive({
recipes_enriched %>%
filter(cuisine == input$cuisine) %>%
arrange(desc(tf_idf)) %>%
head(input$nb_ingredients) %>%
mutate(ingredient = forcats::fct_reorder(ingredient, tf_idf))
})
R doesn’t differ from Python here: you can’t use a function before it’s defined. But, despite appearances to the contrary, this also isn’t happening here.
d3wordcloud::renderD3wordcloud is a special function call which doesn’t evaluate its arguments immediately. In fact, the argument is stored internally as an unevaluated expression and is only evaluated later after a certain trigger. By that time, rval_top_ingredients has been defined.
This is a pervasive pattern in Shiny, but you can harness this behaviour yourself. Consider the following:
f = function (expr) {}
f(g())
g = function () { stop('oh no!') }
This code works, since f never uses its argument, and since R uses lazy evaluation for function arguments: unlike most other languages, a function argument only gets evaluated once it is used. Arguments that are never used are never evaluated.
So, despite the fact that f(g()) appears to use g before it’s defined, the actual call to f never evaluates its arguments so there’s no issue. The only constraint is that the argument needs to be syntactically valid.
Here’s a slightly more meaningful example which does something useful (it creates a function that creates a log message before evaluating an expression:
make_verbose = function (expr) {
function () {
message(sprintf('Evaluating %s', deparse(substitute(expr))))
expr
}
}
verbose_g = make_verbose(g())
g = function () {
message('g was called!')
}
verbose_g()
Python doesn’t quite support this, since Python doesn’t have lazy and non-standard evaluation. But a similar situation still exists in Python:
def f():
g()
def g():
print('g()')
f()
Here, g() is seemingly used before it was defined; but this is only true if we’re reading the code textually from top top bottom without paying attention to scope. In reality, g() is only ever called after it was defined. The same is true in the R code you’ve posted.

Creating a list of functions using a loop in R

In this introduction to functional programming, the author Hadley Wickham creates the following function factory:
power <- function(exponent) {
function(x) {
x ^ exponent
}
}
He then shows how this function can be used to define other functions, such as
square <- power(2)
cube <- power(3)
Now suppose I wanted to create these functions simultaneously via the following loop:
ftns <- lapply(2:3, power)
This doesn't seem to work, as 3 gets assigned to the exponent for all entries of the list:
as.list(environment(ftns[[1]]))
$exponent
[1] 3
Can someone please help me understand what's wrong with this code?
Thanks!
What you're seeing is a consequence of R's use of promises to implement lazy argument evaluation. See Promise objects.
The problem is that in the power() function the exponent argument is never evaluated, meaning the underlying promise is never called (at least not until the generated function is evaluated).
You can force the promise to be evaluated like this:
power <- function(exponent) { exponent; function(x) x^exponent; };
ftns <- lapply(2:3,power);
sapply(ftns,function(ftn) environment(ftn)$exponent);
## [1] 2 3
Without the exponent; statement to force evaluation of the promise, we see the issue:
power <- function(exponent) { function(x) x^exponent; };
ftns <- lapply(2:3,power);
sapply(ftns,function(ftn) environment(ftn)$exponent);
## [1] 3 3
Aha! The R changelog shows that this behavior was changed in 3.2.0:
* Higher order functions such as the apply functions and Reduce()
now force arguments to the functions they apply in order to
eliminate undesirable interactions between lazy evaluation and
variable capture in closures. This resolves PR#16093.
It's classified as a new feature.

Confused by ...()?

In another question, sapply(substitute(...()), as.character) was used inside a function to obtain the names passed to the function. The as.character part sounds fine, but what on earth does ...() do?
It's not valid code outside of substitute:
> test <- function(...) ...()
> test(T,F)
Error in test(T, F) : could not find function "..."
Some more test cases:
> test <- function(...) substitute(...())
> test(T,F)
[[1]]
T
[[2]]
F
> test <- function(...) substitute(...)
> test(T,F)
T
Here's a sketch of why ...() works the way it does. I'll fill in with more details and references later, but this touches on the key points.
Before performing substitution on any of its components, substitute() first parses an R statement.
...() parses to a call object, whereas ... parses to a name object.
... is a special object, intended only to be used in function calls. As a consequence, the C code that implements substitution takes special measures to handle ... when it is found in a call object. Similar precautions are not taken when ... occurs as a symbol. (The relevant code is in the functions do_substitute, substitute, and substituteList (especially the latter two) in R_SRCDIR/src/main/coerce.c.)
So, the role of the () in ...() is to cause the statement to be parsed as a call (aka language) object, so that substitution will return the fully expanded value of the dots. It may seem surprising that ... gets substituted for even when it's on the outside of the (), but: (a) calls are stored internally as list-like objects and (b) the relevant C code seems to make no distinction between the first element of that list and the subsequent ones.
Just a side note: for examining behavior of substitute or the classes of various objects, I find it useful to set up a little sandbox, like this:
f <- function(...) browser()
f(a = 4, 77, B = "char")
## Then play around within the browser
class(quote(...)) ## quote() parses without substituting
class(quote(...()))
substitute({...})
substitute(...(..., X, ...))
substitute(2 <- (makes * list(no - sense))(...))

find all functions in a package that use a function

I would like to find all functions in a package that use a function. By functionB "using" functionA I mean that there exists a set of parameters such that functionA is called when functionB is given those parameters.
Also, it would be nice to be able to control the level at which the results are reported. For example, if I have the following:
outer_fn <- function(a,b,c) {
inner_fn <- function(a,b) {
my_arg <- function(a) {
a^2
}
my_arg(a)
}
inner_fn(a,b)
}
I might or might not care to have inner_fn reported. Probably in most cases not, but I think this might be difficult to do.
Can someone give me some direction on this?
Thanks
A small step to find uses of functions is to find where the function name is used. Here's a small example of how to do that:
findRefs <- function(pkg, fn) {
ns <- getNamespace(pkg)
found <- vapply(ls(ns, all.names=TRUE), function(n) {
f <- get(n, ns)
is.function(f) && fn %in% all.names(body(f))
}, logical(1))
names(found[found])
}
findRefs('stats', 'lm.fit')
#[1] "add1.lm" "aov" "drop1.lm" "lm" "promax"
...To go further you'd need to analyze the body to ensure it is a function call or the FUN argument to an apply-like function or the f argument to Map etc... - so in the general case, it is nearly impossible to find all legal references...
Then you should really also check that getting the name from that function's environment returns the same function you are looking for (it might use a different function with the same name)... This would actually handle your "inner function" case.
(Upgraded from a comment.) There is a very nice foodweb function in Mark Bravington's mvbutils package with a lot of this capability, including graphical representations of the resulting call graphs. This blog post gives a brief description.

Can an R function access its own name?

Can you write a function that prints out its own name?
(without hard-coding it in, obviously)
You sure can.
fun <- function(x, y, z) deparse(match.call()[[1]])
fun(1,2,3)
# [1] "fun"
You can, but just in case it's because you want to call the function recursively see ?Recall which is robust to name changes and avoids the need to otherwise process to get the name.
Recall package:base R Documentation
Recursive Calling
Description:
‘Recall’ is used as a placeholder for the name of the function in
which it is called. It allows the definition of recursive
functions which still work after being renamed, see example below.
As you've seen in the other great answers here, the answer seems to be "yes"...
However, the correct answer is actually "yes, but not always". What you can get is actually the name (or expression!) that was used to call the function.
First, using sys.call is probably the most direct way of finding the name, but then you need to coerce it into a string. deparse is more robust for that.
myfunc <- function(x, y=42) deparse(sys.call()[[1]])
myfunc (3) # "myfunc"
...but you can call a function in many ways:
lapply(1:2, myfunc) # "FUN"
Map(myfunc, 1:2) # (the whole function definition!)
x<-myfunc; x(3) # "x"
get("myfunc")(3) # "get(\"myfunc\")"
The basic issue is that a function doesn't have a name - it's just that you typically assign the function to a variable name. Not that you have to - you can have anonymous functions - or assign many variable names to the same function (the x case above).

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