R中的度量值，如何使用！运算符(整洁评估) [英] Quosures in R, how to use the !! operator (tidy-evaluation)

问题描述

我试图了解R中的整洁评估.

I'm trying to understand tidy evaluation in R.

grouped_mean <- function(data, group_var, summary_var) {
    group_var <- enquo(group_var)
    summary_var <- enquo(summary_var)

    data %>%
        group_by(!!group_var) %>%
        summarise(mean = mean(!!summary_var))
}

我了解为什么以及如何使用它，但我猜不出实际发生的情况.

I understand why and how to use it but not what actually happens I guess.

var <- "test"
var <- enquo(var)
!!var 

Error in is_quosure(e2) : argument "e2" is missing, with no default

这给了我一个错误，但我希望它也可以在 dplyr 之外运行.为什么它不起作用，我该如何解决?

This gives me an error while I expected it to work outside dplyr too. Why does it not work and how can I fix it?

推荐答案

!! 是一个不带引号的运算符，仅在带引号的上下文中起作用，即在dplyr动词的参数中起作用.您通过 !! quo(foo)看到的错误消息是当前CRAN版本中的错误.在开发版本中，现在是:

!! is an unquoting operator that only works in quoting context, i.e. in arguments to dplyr verbs. The error message you're seeing with !!quo(foo) is a bug in the current CRAN release. With the development version, it is now:

Error: Quosures can only be unquoted within a quasiquotation context.

  # Bad:
  list(!!myquosure)

  # Good:
  dplyr::mutate(data, !!myquosure)

最后，请注意， enquo()仅应用于引用函数参数.由于与R编译器有关的技术原因，它仍然可以在其他对象上运行，但不会执行您期望的操作.您只能在函数内使用它，并且只能与该函数的参数名称一起使用.

Finally, note that enquo() should only be used to quote function arguments. For technical reasons having to do with the R compiler, it still works on other objects, but won't do what you expect. You should only use it within a function, and only with argument names of that function.

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