R如何将一个函数作为字符串传递给另一个函数

Question

如能就此小难题提供任何帮助，我们将不胜感激。

我正在尝试将参数从 tidyquant 包传递给 tq_transmute 函数；参数的值是一个函数，但是我想将它作为字符串传递（在下面示例的范围之外，我将通过 Shiny selectInput 传递它）。

我已经尝试了所有我能想到的方法来将字符串 'apply.quarterly' 转换为 mutate_fun 参数接受的对象 apply.quarterly。注释行是我失败的尝试。

最后，我想将这个概念扩展到其他参数，即 FUN = max 到 FUN = ‘max’。

library(tidyverse)
library(tidyquant)
library(rlang)

FANG %>%
  group_by(symbol) %>%
  tq_transmute(select     = adjusted, 
               mutate_fun = apply.quarterly,
               # mutate_fun = sym('apply.quarterly'),
               # mutate_fun = syms('apply.quarterly'),
               # mutate_fun = !!sym('apply.quarterly'),
               # mutate_fun = !!!sym('apply.quarterly'),
               # mutate_fun = eval(parse('apply.quarterly')),
               # mutate_fun = eval(substitute('apply.quarterly')),
               # mutate_fun = enquo('apply.quarterly'),
               # mutate_fun = expr(!!enquo('apply.quarterly')),
               # mutate_fun = quo('apply.quarterly'),
               # mutate_fun = enquos('apply.quarterly'),
               # mutate_fun = enquote('apply.quarterly'),
               # mutate_fun = quote('apply.quarterly'),
               # mutate_fun = substitute('apply.quarterly'),
               # mutate_fun = parse('apply.quarterly'),
               # mutate_fun = parse('apply.quarterly'),
               # mutate_fun = ensym('apply.quarterly'),
               # mutate_fun = rlang::as_function('apply.quarterly'),
               # mutate_fun = rlang::as_closure('apply.quarterly'),
               # mutate_fun = rlang::as_quosure('apply.quarterly'),
               # mutate_fun = rlang::as_quosure('apply.quarterly'),
               # mutate_fun = enexpr('apply.quarterly'),
               # mutate_fun = enexprs('apply.quarterly'),
               # mutate_fun = ensym('apply.quarterly'),
               # mutate_fun = ensyms('apply.quarterly'),
               # mutate_fun = eval_tidy('apply.quarterly'),
               # mutate_fun = exprs('apply.quarterly'),
               # mutate_fun = expr_deparse('apply.quarterly'),
               # mutate_fun = expr_label('apply.quarterly'),
               # mutate_fun = expr_label(substitute('apply.quarterly')),
               # mutate_fun = expr_label(quote('apply.quarterly')),
               # mutate_fun = parse_expr('apply.quarterly'),
               # mutate_fun = quasiquotation('apply.quarterly'),
               # mutate_fun = quotation('apply.quarterly'),
               # mutate_fun = quotation('apply.quarterly'),
               FUN        = max, 
               col_rename = "max.close")

Answer 1

可以通过get获取函数名字符串中的函数。以下作品为例：

s <- get("sum")
s(c(1,2,3))

Answer 2

由于某种原因，功能似乎有点挑剔。一种方法是更改调用然后对其进行评估。例如

myfun <- "apply.quarterly"
bquote(FANG %>%
  group_by(symbol) %>%
  tq_transmute(select     = adjusted, 
               mutate_fun = .(as.name(myfun)),
               FUN        = max, 
               col_rename = "max.close")) %>% 
  eval()

或者如果您更喜欢 rlang 语法

myfun <- "apply.quarterly"
quo(FANG %>%
         group_by(symbol) %>%
         tq_transmute(select     = adjusted, 
                      mutate_fun = !!sym(myfun),
                      FUN        = max, 
                      col_rename = "max.close")) %>% 
  eval_tidy()

请注意，我们必须将整个表达式视为 rlang quosure。除非 tq_transmute 函数是专门为处理 !! 之类的 rlang 特性而编写的，否则默认情况下它们将不起作用。

Answer 3

您可以使用 blast() 进行即时评估的准报价

blast <- function(expr, env = caller_env()) {
  eval_bare(enexpr(expr), env)
}

blast(list(!!!1:3))
#> [[1]]
#> [1] 1
#>
#> [[2]]
#> [1] 2
#>
#> [[3]]
#> [1] 3

然后

myfun <- "apply.quarterly"

blast(
  FANG %>%
    group_by(symbol) %>%
    tq_transmute(
      select = adjusted,
      mutate_fun = !!sym(myfun),
      FUN = max,
      col_rename = "max.close"
    )
)

R如何将一个函数作为字符串传递给另一个函数

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