r - 找出给定函数中调用了哪些函数

Answer 1

必须有更好的方法，但这是我的尝试:

listFunctions <- function(function.name, recursive = FALSE, 
                          checked.functions = NULL){

    # Get the function's code:
    function.code <- deparse(get(function.name))

    # break code up into sections preceding left brackets:
    left.brackets <- c(unlist(strsplit(function.code, 
                                       split="[[:space:]]*\\(")))

    called.functions <- unique(c(unlist(sapply(left.brackets, 
                                               function (x) {

        # Split up according to anything that can't be in a function name.
        # split = not alphanumeric, not '_', and not '.'
        words <- c(unlist(strsplit(x, split="[^[:alnum:]_.]")))

        last.word <- tail(words, 1)
        last.word.is.function <- tryCatch(is.function(get(last.word)),
                                      error=function(e) return(FALSE))
        return(last.word[last.word.is.function])
    }))))

    if (recursive){

        # checked.functions: We need to keep track of which functions 
        # we've checked to avoid infinite loops.
        functs.to.check <- called.functions[!(called.functions %in%
                                          checked.functions)]

        called.functions <- unique(c(called.functions,
            do.call(c, lapply(functs.to.check, function(x) {
                listFunctions(x, recursive = T,
                              checked.functions = c(checked.functions,          
                                                    called.functions))
                }))))
    }
    return(called.functions)
}

结果:

> listFunctions("listFunctions", recursive = FALSE)
 [1] "function"      "deparse"       "get"           "c"            
 [5] "unlist"        "strsplit"      "unique"        "sapply"       
 [9] "tail"          "tryCatch"      "is.function"   "return"       
[13] "if"            "do.call"       "lapply"        "listFunctions"

> system.time(all.functions <- listFunctions("listFunctions", recursive = TRUE))
   user  system elapsed 
  92.31    0.08   93.49 

> length(all.functions)
  [1] 518

如您所见，递归版本返回许多函数。这样做的问题是，它会在过程中调用的每个函数都返回，这显然随着您的执行而累加。无论如何，我希望您可以使用它(或对其进行修改)以满足您的需求。