从值获取R表达式(类似于enquote) [英] Getting an R expression from a value (similar to enquote)

问题描述

假设我有一个值x，它是某些(未知)类型(尤其是:标量，向量或列表).我想获得代表该值的R表达式.如果为x == 1，则此函数应简单地返回expression(1).对于x == c(1,2))，此函数应返回expression(c(1,2)). enquote函数与我想要的功能非常接近，但不完全相同.

Assume I have a value x which is of some (unknown) type (especially: scalar, vector or list). I would like to get the R expression representing this value. If x == 1 then this function should simply return expression(1). For x == c(1,2)) this function should return expression(c(1,2)). The enquote function is quite near to that what I want, but not exactly.

通过玩耍，我找到了解决问题的以下解决方案":

By some playing around I found the following "solution" to my problem:

get_expr <- function(val) {
  tmp_expr <- enquote(val)
  tmp_expr[1] <- quote(expression())
  return(eval(tmp_expr))
}

get_expr(1) # returns expression(1)
get_expr(c(1, 2)) # returns expression(c(1, 2))
get_expr(list(x = 1)) # returns expression(list(x = 1))

但是我认为我的get_expr函数是某种hack.从逻辑上讲，不需要评估.

But I think my get_expr function is some kind of hack. Logically, the evaluation should not be necessary.

是否有一些更优雅的方法可以做到这一点?据我所知，substitute对我来说并不真正有用，因为我的get_expr函数的参数可能是求值的结果(而substitute(eval(expr))并不求值).

Is there some more elegant way to do this? As far as I see, substitute does not really work for me, because the parameter of my get_expr function may be the result of an evaluation (and substitute(eval(expr)) does not do the evaluation).

我通过parse(text = deparse(val))找到了另一种方法，但这更是一个糟糕的hack ...

I found another way via parse(text = deparse(val)), but this is even more a bad hack...

推荐答案

as.expression(list(...))似乎可以做到:

> get_expr <- function(val) as.expression(list(val))
> str(get_expr(1))
  expression(1)
> str(get_expr(c(1, 2)))
  expression(c(1, 2))
> str(get_expr(list(x=1)))
  expression(list(x = 1))
> val <- list(x=1, y=2)
> str(get_expr(val))
  expression(list(x = 1, y = 2))

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