在dplyr中使用group_by时处理缺少的字符串值到函数 [英] Handling a missing string value to a function when using group_by in dplyr

问题描述

我希望创建一个可以接受多个字符串输入的函数（在此示例中为2个），并使用group_by返回即使返回一个字符串也返回结果。我知道我可以创建if语句来解决只将一个字符串传递给函数的情况，但是group_by是否有更好的方法可以在不构建条件语言的情况下仍然产生输出（即，使用多个输入变得更加麻烦）。

I'm looking to create a function that can take multiple string inputs (2 in this example), and using group_by, return results even if only one string is input. I know I could create if statements to get around the case when only one string is passed to the function, but is there a better way for group_by to still produce output without building in conditional language (i.e., gets more cumbersome with multiple inputs).

可复制的示例

library(dplyr)

# Create simple function
car_fx <- function(df, grp1, grp2) {
  output <- df %>% 
    group_by(.data[[grp1]], .data[[grp2]]) %>% 
    summarize(mean_hp = mean(hp, na.rm = TRUE))
}



# String inputs
grp1 <- "cyl"
grp2 <- "carb"



# Run and print function output
(car_fx(mtcars, grp1, grp2))

# works fine
# A tibble: 9 x 3
# Groups:   cyl [3]
    cyl  carb mean_hp
  <dbl> <dbl>   <dbl>
1     4     1    77.4
2     4     2    87  
3     6     1   108. 
4     6     4   116. 
5     6     6   175  
6     8     2   162. 
7     8     3   180  
8     8     4   234  
9     8     8   335 

如果仅传递一个变量，则该函数将引发错误。我想做的是让函数的行为就像我仅传递单个变量一样，并且可以在可能创建3个或更多变量输入的函数中使用。

If I only pass one variable, the function throws an error. What I'd like to do is have the function behave as if I only passed the single variable, and be able to use in function where I might create 3 or more variable inputs.

# Try with just one group, including with NA.  Throws error.
(car_fx(mtcars, grp1))
(car_fx(mtcars, grp1, NA))

推荐答案

您可以使用省略号 ... 将任意数量的参数传递给函数。在这种情况下，要在group_by函数中使用的任何列名称。

You can use ellipsis ... to pass an arbitrary number of arguments to a function. In this case any column names you want to use in the group_by function.

# Create simple function
car_fx <- function(df, ...) {
  output <- df %>% 
    group_by_at(c(...)) %>% 
    summarize(mean_hp = mean(hp, na.rm = TRUE))
}

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