3

我想制作一个简单的函数，该函数采用数据框和用户为该数据框中的两列提供的名称。目的是允许它轻松地与dplyr管道一起工作。它将生成粘合字符串的字符向量：

func <- function(data, last, first) {
  last <- rlang::enquo(last)
  first <- rlang::enquo(first)
  
  glue::glue_data(data, "{!!last}, {!!first}")
}

理想情况下，我希望用户能够调用：

df %>% func(lastName, firstName)

这将产生一个包含多个值的字符向量，格式为Smith, John.

我的功能目前不起作用，因为 bang-bang 运算符在glue_data. 如何在仍然使用 NSE 的同时解决这个问题？我收到的错误是：

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

代表：

df <- data.frame(lastName = c("Smith", "Bond", "Trump"), firstName = c("John","James","Donald"))

> df
  lastName firstName
1    Smith      John
2     Bond     James
3    Trump    Donald

预期输出

> glue::glue_data(df, "{lastName}, {firstName}")

Smith, John
Bond, James
Trump, Donald

但是，我希望能够通过使用我的函数并调用来实现预期的输出：

df %>% func(lastName, firstName)

以上是我实际用例的简化版本，我实际上将glue在后续函数中将该语句作为参数调用：

biggerfn <- function(data, subject, first, last) {
  subject <- rlang::enquo(subject)
  first <- rlang::enquo(first)
  last <- rlang::enquo(last)
  
  data %>%
    dplyr::distinct(!!subject, !!first, !!last) %>%
    smallerfunc(!!subject, glue::glue_data(data, "{!!last}, {!!first}"))
}
4

2 回答 2

5

我不知道您是否致力于glue，但这可以使用以下方法轻松完成tidyr::unite：

func <- function(data, last, first) {
    data %>%
        tidyr::unite(result, {{last}}, {{first}}, sep=", ")
}

df %>% func(lastName, firstName)
#           result
#  1   Smith, John
#  2   Bond, James
#  3 Trump, Donald

# Optionally, follow up with: 
#   %>% dplyr::pull(result)
# to retrieve the column

这里，{{x}}是 的简写!!enquo(x)。

于 2020-11-18T16:27:12.837 回答
1

我认为您不想强制评估firstand last。您只需要名称来构造字符串"{lastName}, {firstName}"。

library(dplyr)
library(rlang)
library(glue)

func <- function(data, last, first) {
  first <- as_name(enquo(first))
  last <- as_name(enquo(last))
  
  glue_data(
    data,
    glue("{[last]}, {[first]}", .open = "[", .close = "]")
  )
}

df %>% func(lastName, firstName)
#> Smith, John
#> Bond, James
#> Trump, Donald

对于您的新示例，请as_name在内部调用中glue调用。

df <- data.frame(
  lastName = c("Smith", "Bond", "Trump"),
  firstName = c("John","James","Donald"),
  other = 1:3
)

biggerfn <- function(data, subject, first, last) {
  subject <- enquo(subject)
  first <- enquo(first)
  last <- enquo(last)
  
  data %>%
    distinct(!!subject, !!first, !!last) %>%
    transmute(
      !!subject := glue_data(
        data,
        glue("{[as_name(last)]}, {[as_name(first)]}", .open = "[", .close = "]")
      )
    )
}

biggerfn(df, other, firstName, lastName)
#>           other
#> 1   Smith, John
#> 2   Bond, James
#> 3 Trump, Donald
于 2020-11-18T06:21:49.837 回答