示例问题:使用dplyr编程

Question

参考:http://dplyr.tidyverse.org/articles/programming.html

此代码工作正常:

df <- tibble(
  g1 = c(1, 1, 2, 2, 2),
  g2 = c(1, 2, 1, 2, 1),
  a = sample(5), 
  b = sample(5)
)


my_summarise <- function(df, group_by) {
  group_by <- enquo(group_by)
  print(group_by)

  df %>%
    group_by(!!group_by) %>%
    summarise(a = mean(a))
}

my_summarise(df, g1)

但是,如果我们将此函数包装在另一个函数中并进行调用,则它将无效.是因为名称只传递了一个级别？

wrapped_my_Summarize <- function(wdf, w_group_by){
  my_summarise(wdf, w_group_by)
}

wrapped_my_Summarize(df, g1)

总的来说,我觉得上面的例子是一个冒险的例子

Answer 1

将它从符号转换为quosure enquo,然后计算函数参数(!!)my_summarise

wrapped_my_Summarize <- function(wdf, w_group_by){
  w_group_by <- enquo(w_group_by) 
  my_summarise(wdf, !! w_group_by)
}

wrapped_my_Summarize(df, g1)
# A tibble: 2 x 2
#     g1     a
#   <dbl> <dbl>
#1  1.00  2.00
#2  2.00  3.67

identical(wrapped_my_Summarize(df, g1), my_summarise(df, g1))
#[1] TRUE