如何在函数中使用 dplyr::group_by

How to use dplyr::group_by in a function

我想创建一个函数来生成 table,该函数具有基于一个或多个分组变量的计数。我发现这个 post 如果我向函数传递一个变量名

就可以工作
library(dplyr)
l <- c("a", "b", "c", "e", "f", "g")
animal <- c("dog", "cat", "dog", "dog", "cat", "fish")
sex <- c("m", "f", "f", "m", "f", "unknown")
n <- rep(1, length(animal))
theTibble <- tibble(l, animal, sex, n)


countString <- function(things) {
  theTibble %>% group_by(!! enquo(things)) %>% count()
}

countString(animal)
countString(sex)

效果很好,但我不知道如何将两个变量传递给函数。 这类作品:

countString(paste(animal, sex))

它给了我正确的计数,但返回的 table 将动物和性别变量合并为一个变量。

# A tibble: 4 x 2
# Groups:   paste(animal, sex) [4]
  `paste(animal, sex)`    nn
  <chr>                <int>
1 cat f                    2
2 dog f                    1
3 dog m                    2
4 fish unknown             1

向函数传递以逗号分隔的两个单词的语法是什么?我想得到这个结果:

# A tibble: 4 x 3
# Groups:   animal, sex [4]
  animal sex        nn
  <chr>  <chr>   <int>
1 cat    f           2
2 dog    f           1
3 dog    m           2
4 fish   unknown     1

您可以使用 group_by_at 和列索引,例如:

countString <- function(things) {
  index <- which(colnames(theTibble) %in% things)
  theTibble %>% 
       group_by_at(index) %>% 
       count()
}

countString(c("animal", "sex"))

## A tibble: 4 x 3
## Groups:   animal, sex [4]
#  animal sex        nn
#  <chr>  <chr>   <int>
#1 cat    f           2
#2 dog    f           1
#3 dog    m           2
#4 fish   unknown     1

对于多个参数,我们将 'things' 替换为 ...,类似地,对于多个参数,将 enquos 替换为 !!!。用 count

删除了 group_by
countString <- function(...) {
  grps <- enquos(...)
  theTibble %>%
       count(!!! grps) 
}


countString(sex)
# A tibble: 3 x 2
#  sex        nn
#  <chr>   <int>
#1 f           3
#2 m           2
#3 unknown     1

countString(animal)
# A tibble: 3 x 2
#  animal    nn
#  <chr>  <int>
#1 cat        2
#2 dog        3
#3 fish       1

countString(animal, sex)
# A tibble: 4 x 3
#  animal sex        nn
#  <chr>  <chr>   <int>
#1 cat    f           2
#2 dog    f           1
#3 dog    m           2
#4 fish   unknown     1