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).
Related
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
Let consider a typical deparse(substitute( R call:
f1 <-function(u,x,y)
{print(deparse(substitute(x)))}
varU='vu'
varX='vx'
varY='vy'
f1(u=varU,x=varX,y=varY)
That results in
[1] "varX"
which is what we expect and we want.
Then, comes the trouble, I try to get a similar behaviour using the ... arguments i.e.
f2 <- function(...)
{ l <- list(...)
x=l$x
print(deparse(substitute(x))) ### this cannot work but I would like something like that
}
That, not surprisingly, does not work :
f2(u=varU,x=varX,y=varY)
[1] "\"vx\"" ### wrong ! I would like "varX"
I tried to get the expected behaviour using a different combination of solutions but none provides me what expected and it seems that I am still not clear enough on lazy eval to find myself the how-to in a resonable amount of time.
You can get the list of all unevaluated arguments by doing
match.call(expand.dots = FALSE)$...
Or, if you only have dot arguments, via
as.list(match.call()[-1L])
This will give you a named list, similarly to list(...), but in its unevaluated form (similarly to what substitute does on a single argument).
An alternative is using rlang::quos(...) if you’re willing to use the {rlang} package, which returns a similar result in a slightly different form.
Is there a way to call a function when the name of the function is decided at the runtime? For example, calling pdf would look like:
pdf("myfile.pdf")
but is there a way, I could do something like:
media_type = "pdf"
media_type("myfile.pdf")
1) do.call Use do.call
do.call(media_type, list("myfile.pdf"))
2) match.fun Another approach is match.fun
fun <- match.fun(media_type)
fun("myfile.pdf")
3) switch Another approach is the following where an argument to switch would be added for each media type. stop is the default. It generates an error when called.
fun <- switch(media_type, pdf = pdf, stop)
fun("myfile.pdf")
4) eval/call This also works although the use of eval is generally frowned upon:
eval(call(media_type, "myfile.pdf"))
I'm not sure if this is what you're wanting, but given the example in your question, it is possible. Assuming you have code that determines which function you want to call, you can use do.call to pass in the string based function name. I had to wrap the input in a list to make it happy, but that's not a big deal most of the time.
f = "mean"
d = c(1,2,3)
do.call(f,list(d))
#> 2
consumeSingleRequest <- function(api_key, URL, columnNames, globalParam="", ...)
consumeSingleRequest <- function(api_key, URL, columnNames, valuesList, globalParam="")
I am trying to overload a function like this, that takes in multiple lists in the first function and combines them into one list of lists. However, I don't seem to be able to skip passing in globalParam and pass in oly the multiple lists in the ...
Does anyone know how to do that?
I've heard S3 methods could be used for that? Does anyone know how?
R doesn't support a concept of overloading functions. It supports function calls with variable number of arguments. So you can declare a function with any number of arguments, but supply only a subset of those when actually calling a function. Take vector function as an example:
> vector
function (mode = "logical", length = 0L)
.Internal(vector(mode, length))
<bytecode: 0x103b89070>
<environment: namespace:base>
It supports up to 2 parameters, but can be called with none or some subset(in that case default values are used) :
> vector()
logical(0)
> vector(mode='numeric')
numeric(0)
So you only need a second declaration:
consumeSingleRequest <- function(api_key, URL, columnNames, valuesList, globalParam="")
And supply just supply the needed parameters when actually calling the function
consumeSingleRequest(api_key=..., valueList=...)
P.S. A good explanation can be found in Advanced R Book.
I'd like to give a params argument to a function and then attach it so that I can use a instead of params$a everytime I refer to the list element a.
run.simulation<-function(model,params){
attach(params)
#
# Use elements of params as parameters in a simulation
detach(params)
}
Is there a problem with this? If I have defined a global variable named c and have also defined an element named c of the list "params" , whose value would be used after the attach command?
Noah has already pointed out that using attach is a bad idea, even though you see it in some examples and books. There is a way around. You can use "local attach" that's called with. In Noah's dummy example, this would look like
with(params, print(a))
which will yield identical result, but is tidier.
Another possibility is:
run.simulation <- function(model, params){
# Assume params is a list of parameters from
# "params <- list(name1=value1, name2=value2, etc.)"
for (v in 1:length(params)) assign(names(params)[v], params[[v]])
# Use elements of params as parameters in a simulation
}
Easiest way to solve scope problems like this is usually to try something simple out:
a = 1
params = c()
params$a = 2
myfun <- function(params) {
attach(params)
print(a)
detach(params)
}
myfun(params)
The following object(s) are masked _by_ .GlobalEnv:
a
# [1] 1
As you can see, R is picking up the global attribute a here.
It's almost always a good idea to avoid using attach and detach wherever possible -- scope ends up being tricky to handle (incidentally, it's also best to avoid naming variables c -- R will often figure out what you're referring to, but there are so many other letters out there, why risk it?). In addition, I find code using attach/detach almost impossible to decipher.
Jean-Luc's answer helped me immensely for a case that I had a data.frame Dat instead of the list as specified in the OP:
for (v in 1:ncol(Dat)) assign(names(Dat)[v], Dat[,v])