通过带有无引号的元素的显式参数指定多个变量进行分组 [英] Specifying multiple variables to group by via explicit argument with unquoted elements

问题描述

基于的部分 nofollow noreferrer>使用dplyr编程，我试图指定

1. 多个变量按 dplyr :: group_by

而不依赖 ... ，但使用显式列表参数 group_vars

without relying on ... but using an explicit list argument group_vars instead

，而无需引用列表元素arg group_vars

without needing to quote the list elements in arg group_vars

示例数据

df <- tibble::tribble(
  ~a,   ~b,  ~c,
  "A",  "a", 10,
  "A",  "a", 20,
  "A",  "b", 1000,
  "B",  "a", 5,
  "B",  "b", 1
)

基于<$ c $的方法c> ... 来自编程与dplyr

# Approach 1 -----
my_summarise <- function(df, ...) {
  group_vars <- dplyr::enquos(...)

  df %>%
    dplyr::group_by(!!!group_vars) %>%
    dplyr::summarise(x = mean(c))
}

my_summarise(df, a, b)
#> # A tibble: 4 x 3
#> # Groups:   a [2]
#>   a     b         x
#>   <chr> <chr> <dbl>
#> 1 A     a        15
#> 2 A     b      1000
#> 3 B     a         5
#> 4 B     b         1

基于带引号元素的列表参数的方法：

Approach based on list argument with quoted elements:

# Approach 2 -----
my_summarise_2 <- function(df, group_vars = c("a", "b")) {
  group_vars <- dplyr::syms(group_vars)

  df %>%
    dplyr::group_by(!!!group_vars) %>%
    dplyr::summarise(x = mean(c))
}

my_summarise_2(df)
#> # A tibble: 4 x 3
#> # Groups:   a [2]
#>   a     b         x
#>   <chr> <chr> <dbl>
#> 1 A     a        15
#> 2 A     b      1000
#> 3 B     a         5
#> 4 B     b         1

my_summarise_2(df, group_vars = "a")
#> # A tibble: 2 x 2
#>   a         x
#>   <chr> <dbl>
#> 1 A      343.
#> 2 B        3

我找不到让我提供未加引号的列名的方法：

I can't find an approach that lets me supply unquoted column names:

# Approach 3 -----
my_summarise_3 <- function(df, group_vars = list(a, b)) {
  group_vars <- dplyr::enquos(group_vars)

  df %>%
    dplyr::group_by(!!!group_vars) %>%
    dplyr::summarise(x = mean(c))
}

my_summarise_3(df)
#> Error: Column `list(a, b)` must be length 5 (the number of rows) or one, not 2

我想关键的是要在调用 group_vars<-dplyr :: enquos（...）相同的列表结构。 c $ c>：

I guess the crucial thing is to end up with an identical list structure as the one after calling group_vars <- dplyr::enquos(...):

<list_of<quosure>>

[[1]]
<quosure>
expr: ^a
env:  global

[[2]]
<quosure>
expr: ^b
env:  global

我试图用 group_vars％>％purrr :: map（dplyr :: enquo），但是R当然会抱怨 a 和 b ，因为需要对其进行评估。

I tried to tackle it with group_vars %>% purrr::map(dplyr::enquo), but of course R complains about a and b as they need to be evaluated.

推荐答案

主要问题是 list（a，b）不能捕获未评估的表达式 a 和 b ，而是计算这些表达式并创建一个包含结果的两元素列表。您基本上有两个选择：

The main issue is that list(a, b) does not capture unevaluated expressions a and b, but instead evaluates those expressions and creates a two-element list with results. You basically have two options:

解决方案一：使用 rlang :: exprs（）捕获实际表达式。由于表达式已经过计算，因此您不再需要在函数内部使用 enquos ，而只需将其变为

Solution one: Use rlang::exprs() to capture the actual expressions. Since the expressions are already unevaluated, you no longer need an enquos inside your function, which simply becomes

my_summarise_3 <- function(df, group_vars = rlang::exprs(a, b)) {
  df %>%
    dplyr::group_by(!!!group_vars) %>%
    dplyr::summarise(x = mean(c))
}

my_summarise_3(df)
# # A tibble: 4 x 3
# # Groups:   a [2]
#   a     b         x
#   <chr> <chr> <dbl>
# 1 A     a        15
# 2 A     b      1000
# 3 B     a         5
# 4 B     b         1

此界面的缺点是用户现在负责引用（即捕获其表达式）参数：

The down side of this interface is that the user is now responsible for quoting (i.e, capturing the expressions of) the arguments:

# Note that it can be done using quote() from base R
my_summarise_3(df, group_vars=quote(a))
# # A tibble: 2 x 2
#   a         x
#   <chr> <dbl>
# 1 A      343.
# 2 B        3 

解决方案两个：完整捕获未评估的表达式 list（a，b）并手动解析。

Solution two: Capture the unevaluated expression list(a,b) in its entirety and parse it by hand.

## Helper function to recursively construct an abstract syntax tree
getAST <- function( ee ) { as.list(ee) %>% map_if(is.call, getAST) }

my_summarise_3 <- function(df, group_vars = list(a,b)) {
  ## Capture the expression and parse it
  ast <- rlang::enexpr(group_vars) %>% getAST()

  ## Identify symbols present in the data
  gvars <- unlist(ast) %>% map_chr(deparse) %>%
      intersect(names(df)) %>% rlang::syms()

  df %>%
      dplyr::group_by(!!!gvars) %>%
      dplyr::summarise(x = mean(c))
}

my_summarise_3(df, list(a,b))
# # A tibble: 4 x 3
# # Groups:   a [2]
#   a     b         x
#   <chr> <chr> <dbl>
# 1 A     a        15
# 2 A     b      1000
# 3 B     a         5
# 4 B     b         1

my_summarise_3(df, b)
# # A tibble: 2 x 2
#   b         x
#   <chr> <dbl>
# 1 a      11.7
# 2 b     500. 

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