Use character string as function argument - string

Question

I'm sure this is simple, but I cannot find a solution ...
I would like to use a variable containing a character string as argument for a function.
x <- c(1:10)
myoptions <- "trim=0, na.rm=FALSE"
Now, something like
foo <- mean(x, myoptions)
should be the same as
foo <- mean(x, trim=0, na.rm=FALSE)
Thanks in advance!

Answer 1

You can use eval and parse:
foo <- eval(parse(text = paste("mean(x,", myoptions, ")")))

Answer 2

A more natural way to do what you want is to use do.call. For example,
R> l[["trim"]] = 0
R> l[["na.rm"]] = FALSE
R> l[["x"]] = 1:10
##Or l <- list(trim = 0, na.rm = FALSE, x = 1:10)
R> do.call(mean, l)
[1] 5.5
If for some reason you really want to use a myoptions string, you could always use strsplit to coarce it into a list form. For example,
R> y = "trim=0, na.rm=FALSE"
R> strsplit(y, ", ")
[[1]]
[1] "trim=0" "na.rm=FALSE"
R> strsplit(y, ", ")[[1]][1]
[1] "trim=0"

Answer 3

Here's a third answer that both uses parse, alist and do.call. My motivation for this new answer, is in the case where arguments are passed interactively from a client-side as chars. Then I guess, there is no good way around not using parse. Suggested solution with strsplit, cannot understand the context whether a comma , means next argument or next argument within an argument. strsplit does not understand context as strsplit is not a parser.
here arguments can be passed as "a=c(2,4), b=3,5" or list("c(a=(2,4)","b=3","5")
#' convert and evaluate a list of char args to a list of arguments
#'
#' #param listOfCharArgs a list of chars
#'
#' #return
#' #export
#'
#' #examples
#' myCharArgs = list('x=c(1:3,NA)',"trim=0","TRUE")
#' myArgs = callMeMaybe(myCharArgs)
#' do.call(mean,myArgs)
callMeMaybe2 = function(listOfCharArgs) {
CharArgs = unlist(listOfCharArgs)
if(is.null(CharArgs)) return(alist())
.out = eval(parse(text = paste0("alist(",
paste(parse(text=CharArgs),collapse = ","),")")))
}
myCharArgs = list('x=c(1:3,NA)',"trim=0","TRUE")
myArgs = callMeMaybe2(myCharArgs)
do.call(mean,myArgs)
[1] 2

Answer 4

Using all of do.call, eval and parse (combining kohske's and csgillespie's answers, and also WoDoSc's answer to 'Pass a comma separated string as a list'):
x <- c(1:10)
myoptions <- "trim = 0, na.rm = FALSE"
do.call(
what = mean,
args = append(list(x = x), eval(parse(text = paste0("list(", myoptions, ")"))))
)
This solution can be quite resilient in a more complex case, such as shown below.
myfn <- function(x, y = 0, z = 0, ...) {
print(paste("x:", x))
print(paste("y:", y))
print(paste("z:", z))
if (length(list(...)) > 0) {
print("other:")
print(list(...))
}
}
myextraargs <- paste(
"y = c(11, 14), z = 47,",
"t = data.frame(p = c('apple', 'plum'), j = c(7, 2), k = c(3, 21))"
)
do.call(
what = myfn,
args = append(
list(x = 7),
eval(parse(text = paste0("list(", myextraargs, ")")))
)
)
results in:
[1] "x: 7"
[1] "y: 11" "y: 14"
[1] "z: 47"
[1] "other:"
$t
p j k
1 apple 7 3
2 plum 2 21
...and...
myextraargs <- NULL
do.call(
what = myfn,
args = append(
list(x = 7),
eval(parse(text = paste0("list(", myextraargs, ")")))
)
)
results in
[1] "x: 7"
[1] "y: 0"
[1] "z: 0"