r - 确定传递给函数的值是否是一个变量

Question

简短版本:函数如何区分接收这两个输入？

value <- "asdf"
"asdf"

一个可重现的例子: 假设我有一个寻找邪恶的函数，或者至少寻找字符串 "evil"

library(stringr)

Find_Evil <- function(x) {
  x <- str_to_lower(x)
  if(str_detect(x, "evil")) {
    message("Evil detected")
  } else {
    return(x)
  }
}

该函数可以很好地检测其输入中是否存在 "evil"。

Find_Evil("this has some evil in it")
Evil detected

如果我还想检测变量名中是否包含 evil 怎么办？这在不应该通过的时候通过了。

sneaky_evil <- "sounds good but isn't"
Find_Evil(sneaky_evil)
[1] "sounds good but isn't"

我可以使用 deparse(substitute()) 检查变量名，但我如何首先检查某物是否为变量？我想要的是符合以下元代码的内容(不起作用):

Find_Evil <- function(x) {
  x <- str_to_lower(x)
  if (x is a variable) { # this is the part I need help on
    x_name <- str_to_lower(deparse(substitute(sneaky_evil))) # convert value name to string
    evil_name <- str_detect(x_name, "evil") # check name for evil
    evil_x <- str_detect(x, "evil") # check contents for evil
    if (sum(evil_name, evil_x) != 2) { # if either name or contents contains evil
      message("evil detected")
    } else {
      return(x)
    }
  } else {
    if (str_detect(x, "evil")) { # if x isn't a variable just check x (no name)
      message("Evil detected")
    } else {
      return(x)
    }
  }

Answer 1

可能，这样的事情会有所帮助:

Find_Evil <- function(x) {
   var_type <- substitute(x)
   var_name <- deparse(substitute(x))
   if(grepl('evil', var_name) & is.name(var_type)) {
     return("evil detected in variable")
   }
   x <- tolower(x)
   if(grepl('evil', x)) {
     return("Evil detected in string")
   }
   else return(x)
}

sneaky_evil <- "sounds good but isn't"
Find_Evil(sneaky_evil)
#[1] "evil detected in variable"

sneaky_only <- "sounds good but isn't"
Find_Evil(sneaky_only)
#[1] "sounds good but isn't"

sneaky_only <- "sounds good but evil"
Find_Evil(sneaky_only)
#[1] "Evil detected in string"

Find_Evil("sounds good but isn't")
#[1] "sounds good but isn't"

Find_Evil("sounds good but evil")
#[1] "Evil detected in string"