R bquote的反向行为 [英] R reverse behaviour of bquote

问题描述

bquote函数允许计算表达式的部分，这些部分包装在.()调用中.例如

bquote function allows to evaluate parts of the expression which are wrapped in .() call. For example,

a <- 2
b <- 100
bquote(.(2 * a) * x + .(log10(b)))

会返回

4 * x + 2

我想重写此函数以评估除.()调用内的内容以外的所有内容.这是所需的行为:

I want to rewrite this function to evaluate everything except for things inside .() call. This is the desired behavior:

a <- 2
b <- 100
bquote(2 * a * .(x) + log10(b))

> 4 * x + 2

我知道要这样做，我必须仔细阅读提取语法树，并在未调用.()的情况下评估早午餐，但我无法处理所有此类递归.

I understand that to do so I have to go over the abstract syntax tree and evaluate brunches without .() call in them but I couldn't handle all this recursion.

您能帮我编写这样的功能吗?

Could you help me to write such a function?

推荐答案

subst将替换.(...)中的所有变量，而simplify函数将简化没有变量的子树-如果不需要简化，则省略简化部分.不使用任何软件包.

subst will substitute all variables except those within .(...) and the simplify function will simplify sub-trees that have no variables -- omit the simplify part if simplification is not needed. No packages are used.

subst <- function(e) {
   if (typeof(e) == "language") {
      if (identical(e[[1]], as.name("."))) e[[2]]
      else {
        if (length(e) > 1) e[-1] <- lapply(as.list(e[-1]), subst)
        e
      }
   } else {
      eval(e)
   }
}

simplify <- function(e) {
  if (typeof(e) == "language") {
     if (length(all.vars(e))) {
         if (length(e) > 1) {
           e[-1] <- lapply(as.list(e[-1]), simplify)
           e
         } else e
     } else eval(e)
  } else e
}

inverse_bquote <- function(x, SIMPLIFY = TRUE) {
   result <- subst(substitute(x))
   if (SIMPLIFY) simplify(result) else result
}

现在进行测试.

a <- 2
b <- 100

inverse_bquote(2 * a * .(x) + log10(b))
## 4 * x + 2

# without simplification

inverse_bquote(2 * a * .(x) + log10(b), SIMPLIFY = FALSE)
## 2 * 2 * x + log10(100)

更新:添加了简化功能.使其成为可选.

Update: Added simplification. Made it optional.

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