Question

为了使自定义函数灵活地接受每个形式参数所接收的一个或多个调用参数，我目前依靠“ ...”：

library(dplyr)

foo <- function(data, ..., dv){
  groups <- enquos(...)
  dv <- enquo(dv)
  data %>% 
    group_by(!!!groups) %>% 
    summarise(group_mean = mean(!!dv))
}

mtcars %>% foo(am, dv = mpg)
mtcars %>% foo(vs, am, dv = mpg)

但是“ ...”掩盖了函数的逻辑，并且不能在具有2个或更多个需要多个调用参数的形式参数的自定义函数中使用。

有没有一种方法可以编写上述函数来利用形式参数（例如“组”）而不是“ ...”，后者可以接收单个向量名称或向量名称向量作为其参数？像这样：

foo <- function(data, groups, dv){
  groups <- enquos(groups)
  dv <- enquo(dv)

  data %>% 
    group_by(!!!groups) %>% 
    summarise(group_mean = mean(!!dv))
}

# Failing code
mtcars %>% foo(groups = c(vs, am), dv = mpg)

请注意，此代码可以工作，但要求用户记住在函数体中使用quos（）：

foo <- function(data, groups, dv){
  dv <- enquo(dv)

  data %>% 
    group_by(!!!groups) %>% 
    summarise(group_mean = mean(!!dv))
}

mtcars %>% foo(groups = quos(vs, am), dv = mpg)

我希望改为依靠函数体中的enquos（）。

Answer 1

我们可以将...放在最后

foo <- function(data,  dv, ...){
   groups <- enquos(...)
   dv <- enquo(dv)
   data %>% 
     group_by(!!!groups) %>% 
     summarise(group_mean = mean(!!dv))
  }

如果我们要传递vector的“组”，则一个选项是group_by_at

foo <- function(data, groups, dv){
  dv <- enquo(dv)

  data %>% 
     group_by_at(vars(groups)) %>% 
     summarise(group_mean = mean(!!dv))
  }

mtcars %>% 
    foo(groups = c("vs", "am"), dv = mpg)
# A tibble: 4 x 3
# Groups:   vs [?]
#     vs    am group_mean
#  <dbl> <dbl>      <dbl>
#1     0     0       15.0
#2     0     1       19.8
#3     1     0       20.7
#4     1     1       28.4

如果我们想用c传递不带引号的表达式，则可以选择将其转换为表达式然后对其求值

foo <- function(data, groups, dv){

 groups <- as.list(rlang::enexpr(groups))[-1]
 dv <- enquo(dv)
   data %>% 
      group_by(!!! groups) %>% 
      summarise(group_mean = mean(!!dv))
 }

mtcars %>% 
      foo(groups = c(vs, am), dv = mpg)
# A tibble: 4 x 3
# Groups:   vs [?]
#     vs    am group_mean
#  <dbl> <dbl>      <dbl>
#1     0     0       15.0
#2     0     1       19.8
#3     1     0       20.7
#4     1     1       28.4

或者如评论中提到的@Joe一样，enquo也应与group_by_at一起使用

foo <- function(data, groups, dv){
   dv <- enquo(dv) 
   groups <- enquos(groups) 
   data %>% 
        group_by_at(vars(!!!groups)) %>% 
        summarise(group_mean = mean(!!dv))
   } 

mtcars %>% 
     foo(groups = c(vs, am), dv = mpg)
# A tibble: 4 x 3
# Groups:   vs [?]
#     vs    am group_mean
#  <dbl> <dbl>      <dbl>
#1     0     0       15.0
#2     0     1       19.8
#3     1     0       20.7
#4     1     1       28.4

将多个调用参数传递给dplyr自定义函数中的形式参数，而不使用“ ...”

1 个答案: