问题描述
让我们定义:
f <- function(x) deparse(substitute(x))
挑战:找到<something>，以便f(<something>)返回"abc".当然不包括f(abc).
The challenge: find <something> so that f(<something>) returns "abc". Excluding, of course, f(abc).
使用整洁的NSE"(即准引用)，这非常容易.但是，根据NSE引用( 1 ， 2 ， 3 )，这是不可能的，因为substitute是纯引用(而不是准引用)功能.
With "tidy NSE", i.e. quasiquoting, this is very easy. However, according to the NSE references (1, 2, 3), it is impossible since substitute is a pure quoting (as opposed to quasiquoting) function.
我想知道是否有任何晦涩或未记载的东西(不是那么不常见！)允许在substitute中取消报价，因此是一个挑战.
I wonder if there is anything obscure or undocumented (not that uncommon!) that allows to unquote in substitute, hence the challenge.
推荐答案
@Roland是正确的.由于未评估x，因此您无法为f提供不会被逐字转换为字符串的表达式. R中的准引号由bquote()处理，它具有.()机制，其作用类似于rlang的!!:
@Roland is correct. Because x is not evaluated, there is no expression you can provide to f that won't be converted to a string verbatim. Quasiquotation in base R is handled by bquote(), which has a .() mechanism that works similarly to rlang's !!:
# Quasiquotation with base R
f1 <- function(x) bquote( .(substitute(x)) + 5 )
# Quasiquotation with rlang
f2 <- function(x) rlang::expr( !!rlang::enexpr(x) + 5 )
e1 <- f1(y) # y + 5
e2 <- f2(y) # y + 5
identical(e1, e2) # TRUE
eval(e1, list(y=10)) # 15
这篇关于NSE挑战:脱颖而出(substitute(...))的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！