在 R 中，如何在将对象发送给函数后获取对象的名称? [英] In R, how to get an object's name after it is sent to a function?

问题描述

我正在寻找 get() 的反面.

I am looking for the reverse of get().

给定一个对象名称，我希望直接从对象中提取代表该对象的字符串.

Given an object name, I wish to have the character string representing that object extracted directly from the object.

foo 是我正在寻找的函数的占位符的简单示例.

Trivial example with foo being the placeholder for the function I am looking for.

z <- data.frame(x=1:10, y=1:10)

test <- function(a){
  mean.x <- mean(a$x)
  print(foo(a))
  return(mean.x)}

test(z)

会打印:

  "z"

我的解决方法是:

test <- function(a="z"){
  mean.x <- mean(get(a)$x)
  print(a)
  return(mean.x)}

test("z")

推荐答案

旧的 deparse-substitute 技巧:

The old deparse-substitute trick:

a<-data.frame(x=1:10,y=1:10)
test<-function(z){
   mean.x<-mean(z$x)
   nm <-deparse(substitute(z))
   print(nm)
   return(mean.x)}
 
 test(a)
#[1] "a"   ... this is the side-effect of the print() call
#          ... you could have done something useful with that character value
#[1] 5.5   ... this is the result of the function call

用新的测试对象运行它

注意:当一组列表项从第一个参数传递给 lapply 时，这在本地函数中不会成功(当一个对象从给定的列表传递时，它也会失败)for-loop.) 如果它是正在处理的命名向量，您将能够从结构结果中提取.Names"-属性和处理顺序.

Note: this will not succeed inside a local function when a set of list items are passed from the first argument to lapply (and it also fails when an object is passed from a list given to a for-loop.) You would be able to extract the ".Names"-attribute and the order of processing from the structure result, if it were a named vector that were being processed.

> lapply( list(a=4,b=5), function(x) {nm <- deparse(substitute(x)); strsplit(nm, '\[')} )
$a      # This "a" and the next one in the print output are put in after processing
$a[[1]]
[1] "X"    ""     "1L]]"  # Notice that there was no "a"


$b
$b[[1]]
[1] "X"    ""     "2L]]"

> lapply( c(a=4,b=5), function(x) {nm <- deparse(substitute(x)); strsplit(nm, '\[')} )
$a
$a[[1]]   # but it's theoretically possible to extract when its an atomic vector
[1] "structure(c(4, 5), .Names = c("a", "b"))" ""                                            
[3] "1L]]"                                        


$b
$b[[1]]
[1] "structure(c(4, 5), .Names = c("a", "b"))" ""                                            
[3] "2L]]"  

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