如何使用非标准评估 NSE 来评估 data.table 上的参数？

假设我有以下内容

library(data.table)
cars1 = setDT(copy(cars))
cars2 = setDT(copy(cars))

car_list = list(cars1, cars2)
class(car_list) <- "dd"

`[.dd` <- function(x,...) {
  code = rlang::enquos(...)
  cars1 = x[[1]]
  rlang::eval_tidy(quo(cars1[!!!code]))
}

car_list[,.N, by = speed]

所以我希望执行任意操作cars1 and cars2通过定义[.dd功能这样无论我放入什么...被执行cars1 and cars2使用[data.table 语法例如

car_list[,.N, by = speed]应执行以下操作

cars1[,.N, by = speed]
cars2[,.N, by = speed]

我也想要

car_list[,speed*2]

to do

cars1[,speed*2]
cars2[,speed*2]

基本上，... in [.dd必须接受任意代码。

不知何故我需要捕捉...所以我试着做code = rlang::enquos(...)进而rlang::eval_tidy(quo(cars1[!!!code]))不起作用并给出错误

错误于[.data.table（汽车1，〜，〜.N，按=〜速度）： 缺少参数“i”，没有默认值

虽然不在下面rlang类型的咒语，这种方法似乎效果很好：lapply(dt_list, '[', ...)该代码对我来说更具可读性，因为它明确说明了正在使用的方法。如果我看到car_list[, .N, by = speed]我希望默认data.table方法。

将其作为一个函数可以让您两全其美：

class(car_list) <- "dd"

`[.dd` <- function(x,...) {
 lapply(x, '[', ...)
}

car_list[, .N, speed]
car_list[, speed * 2]
car_list[, .(.N, max(dist)), speed]
car_list[, `:=` (more_speed = speed+5)]

以下是该方法的一些示例：

car_list[, .N, speed]
# lapply(car_list, '[', j = .N, by = speed)
# or
# lapply(car_list, '[', , .N, speed)
[[1]]
    speed N
 1:     4 2
 2:     7 2
 3:     8 1
 4:     9 1
 5:    10 3
...
[[2]]
    speed N
 1:     4 2
 2:     7 2
 3:     8 1
 4:     9 1
 5:    10 3
...
car_list[, speed * 2]
# lapply(car_list, '[', j = speed*2)
# or
# lapply(car_list, '[', , speed*2)
[[1]]
 [1]  8  8 14 14 16 18 20 20 20 22 22 24 24 24 24 26 26
[18] 26 26 28 28 28 28 30 30 30 32 32 34 34 34 36 36 36
[35] 36 38 38 38 40 40 40 40 40 44 46 48 48 48 48 50

[[2]]
 [1]  8  8 14 14 16 18 20 20 20 22 22 24 24 24 24 26 26
[18] 26 26 28 28 28 28 30 30 30 32 32 34 34 34 36 36 36
[35] 36 38 38 38 40 40 40 40 40 44 46 48 48 48 48 50

car_list[, .(.N, max(dist)), speed]
# lapply(car_list, '[', j = list(.N, max(dist)), by = speed)
# or 
# lapply(car_list, '[', ,.(.N, max(dist)), speed)

[[1]]
    speed N  V2
 1:     4 2  10
 2:     7 2  22
 3:     8 1  16
 4:     9 1  10
 5:    10 3  34
...

[[2]]
    speed N  V2
 1:     4 2  10
 2:     7 2  22
 3:     8 1  16
 4:     9 1  10
 5:    10 3  34
...

这适用于:=操作员：

car_list[, `:=` (more_speed = speed+5)]
# or
# lapply(car_list, '[', , `:=` (more_speed = speed+5))

car_list
[[1]]
    speed dist more_speed
 1:     4    2          9
 2:     4   10          9
 3:     7    4         12
 4:     7   22         12
 5:     8   16         13
...

[[2]]
    speed dist more_speed
 1:     4    2          9
 2:     4   10          9
 3:     7    4         12
 4:     7   22         12
 5:     8   16         13