R data.table包中点函数“.（）”的实现

从

？data.table:：data.table

：

表达式“（）”是list（）的简写别名；他们都是说 同样的

但是，该功能在任何地方都找不到：

data.table:::.

get（name，envir=asNamespace（pkg），inherits=FALSE）中出错：
找不到对象“”

所以我假设输入是以某种方式解析的，它是如何完成的？我想在我自己的软件包中使用相同的功能

以下几点效果不错：

test  <- function(x) {
  eval(substitute(
    eval.parent(substitute(x, list(.=list)))
    ))
}

foo <- "bar"
test(.(foo))
# [[1]]
# [1] "bar"
identical(test(.(foo)), list(foo))
# [1] TRUE

---

data.table

包使用这一位代码完成它

replace_dot_alias <- function(e) {
  # we don't just simply alias .=list because i) list is a primitive (faster to iterate) and ii) we test for use
  # of "list" in several places so it saves having to remember to write "." || "list" in those places
  if (is.call(e)) {
    # . alias also used within bquote, #1912
    if (e[[1L]] == 'bquote') return(e)
    if (e[[1L]] == ".") e[[1L]] = quote(list)
    for (i in seq_along(e)[-1L]) if (!is.null(e[[i]])) e[[i]] = replace_dot_alias(e[[i]])
  }
  e
}

---

…太棒了！这个函数将产生我期望的输出：

精确地测试！这是几天前问的同一个
composer
软件包吗？的确，我对此很感兴趣，这让我震惊了
[[
对于向量来说是多余的，它可以用于很多有趣的事情。我想，
dplyr:：n
也使用了类似的手法，它存在，但只包含
中止（“不应该直接调用此函数”）
我确信这是正确的
# [[1]]
# [1] "bar"
# 
# [[2]]
# [1] "baz"

replace_dot_alias <- function(e) {
  # we don't just simply alias .=list because i) list is a primitive (faster to iterate) and ii) we test for use
  # of "list" in several places so it saves having to remember to write "." || "list" in those places
  if (is.call(e)) {
    # . alias also used within bquote, #1912
    if (e[[1L]] == 'bquote') return(e)
    if (e[[1L]] == ".") e[[1L]] = quote(list)
    for (i in seq_along(e)[-1L]) if (!is.null(e[[i]])) e[[i]] = replace_dot_alias(e[[i]])
  }
  e
}

if (!missing(j)) {
    jsub = replace_dot_alias(substitute(j))
    root = if (is.call(jsub)) as.character(jsub[[1L]])[1L] else ""