使用字符串作为函数参数

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

您可以使用eval和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"