使用 dplyr 编程时 ensym 和 enquo 有什么区别？

Question

对 tidy evaluation 比较陌生，虽然我正在使用这些函数，但我想知道为什么使用不同的辅助函数。例如，enquo 和 ensym 有什么区别？在我下面制作的捕捉每日平均线和移动平均线的函数中，它们是可以互换的：

library(dplyr)
library(lubridate)
library(rlang)
library(zoo)

manipulate_for_ma <- function(data, group_var, da_col_name, summary_var, ma_col_name) {
  group_var <- ensym(group_var) 
  summary_var <- enquo(summary_var)
  da_col_name <- ensym(da_col_name) 
  ma_col_name <- enquo(ma_col_name)

  data %>% 
    group_by(!!group_var) %>%
    summarise(!!da_col_name := mean(!!summary_var, na.rm = TRUE)) %>% 
    mutate(!!ma_col_name := rollapply(!!da_col_name,
                                      30,
                                      mean,
                                      na.rm = TRUE,
                                      partial = TRUE,
                                      fill = NA)) %>% 
    rename(date = !!group_var)
}

lakers %>%
 mutate(date = ymd(date)) %>%
 manipulate_for_ma(group_var = date,
                   da_col_name = points_per_play_da,
                   summary_var = points,
                   points_per_play_ma)

# A tibble: 78 x 3
   date       points_per_play_da points_per_play_ma
   <date>                  <dbl>              <dbl>
 1 2008-10-28              0.413              0.458
 2 2008-10-29              0.431              0.459
 3 2008-11-01              0.408              0.456
 4 2008-11-05              0.386              0.457

我读过 enquo here and ensym (here)[https://adv-r.hadley.nz/quasiquotation.html]。 ensym 的区别是限制性更强，只接受字符串或类似字符串的对象吗？

Answer 1

这里有一个例子说明了一个区别（即 enquo 捕获调用环境而 ensym 不捕获）。希望它不言自明：

library(rlang)

f <- function(x) {
  foo <- 42
  print(eval_tidy(quo(!! ensym(x))))
  print(eval_tidy(quo(!! enquo(x))))
}
foo <- 3
f(foo)
# [1] 42
# [1] 3

或者稍微复杂一点的：

library(rlang)

myenv <- new.env()

local(envir = myenv, {
  foo <- 17
  g <- function(x) {
    print(eval_tidy(quo(!! ensym(x))))
    print(eval_tidy(quo(!! enquo(x))))
  }
})

foo <- 123
myenv$g(foo)
#> [1] 17
#> [1] 123

^{由 reprex package (v0.3.0)}

于 2019-09-18 创建

使用 dplyr 时，差异通常不明显，因为它足够简单，总是首先在 .data 参数的上下文中查找名称。

Answer 2

另一个镜头：

library(rlang)
library(dplyr, warn.conflicts = FALSE)

test <- function(x){
  Species <- "bar"
  cat("--- enquo builds a quosure from any expression\n")
  print(enquo(x))
  cat("--- ensym captures a symbol or a literal string as a symbol\n")
  print(ensym(x))
  cat("--- evaltidy will evaluate the quosure in its environment\n")
  print(eval_tidy(enquo(x)))
  cat("--- evaltidy will evaluate a symbol locally\n")
  print(eval_tidy(ensym(x)))
  cat("--- but both work fine where the environment doesn't matter\n")
  identical(select(iris,!!ensym(x)), select(iris,!!enquo(x)))
}

Species = "foo"
test(Species)
#> --- enquo builds a quosure from any expression
#> <quosure>
#> expr: ^Species
#> env:  global
#> --- ensym captures a symbol or a literal string as a symbol
#> Species
#> --- evaltidy will evaluate the quosure in its environment
#> [1] "foo"
#> --- evaltidy will evaluate a symbol locally
#> [1] "bar"
#> --- but both work fine where the environment doesn't matter
#> [1] TRUE

test("Species")
#> --- enquo builds a quosure from any expression
#> <quosure>
#> expr: ^"Species"
#> env:  empty
#> --- ensym captures a symbol or a literal string as a symbol
#> Species
#> --- evaltidy will evaluate the quosure in its environment
#> [1] "Species"
#> --- evaltidy will evaluate a symbol locally
#> [1] "bar"
#> --- but both work fine where the environment doesn't matter
#> [1] TRUE
test(paste0("Spec","ies"))
#> --- enquo builds a quosure from any expression
#> <quosure>
#> expr: ^paste0("Spec", "ies")
#> env:  global
#> --- ensym captures a symbol or a literal string as a symbol
#> Only strings can be converted to symbols

^{由 reprex package (v0.3.0)}

于 2019-09-23 创建

使用 dplyr 编程时 ensym 和 enquo 有什么区别？

What is the difference between ensym and enquo when programming with dplyr?

r

dplyr

tidyeval

rlang