Question

说我有以下内容

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]

所以我希望通过定义cars1函数对cars2和[.dd进行任意操作，以便使...中执行的所有操作都由cars1执行和cars2使用[ data.table语法，例如

car_list[,.N, by = speed]应该执行以下操作

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

我也想要

car_list[,speed*2]

要做

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

基本上，...中的[.dd必须接受任意代码。

以某种方式我需要捕获...，所以我尝试执行code = rlang::enquos(...)，然后rlang::eval_tidy(quo(cars1[!!!code]))无法正常工作并给出错误消息

[.data.table（cars1，〜，〜.N，by =〜speed）中的错误：参数“ i”丢失，没有默认值

Answer 1

虽然不是rlang类型的咒语，但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，因为它明确说明了所使用的方法。如果看到car_list[, .N, by = speed]，我会期望使用默认的data.table方法。

以下是该方法的一些示例：

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
...

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

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
...

这可与:=运算符一起使用：

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

Answer 2

第一个基本R选项为substitute(...())，后跟do.call：

library(data.table)
cars1 = setDT(copy(cars))
cars2 = setDT(copy(cars))
cars2[, speed := sort(speed, decreasing = TRUE)]

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

`[.dd` <- function(x,...) {
  a <- substitute(...()) #this is an alist
  expr <- quote(x[[i]])
  expr <- c(expr, a)
  res <- list()
  for (i in seq_along(x)) {
    res[[i]] <- do.call(data.table:::`[.data.table`, expr)
  }
  res
}

all.equal(
  car_list[,.N, by = speed],
  list(cars1[,.N, by = speed], cars2[,.N, by = speed])
)
#[1] TRUE

all.equal(
  car_list[, speed*2],
  list(cars1[, speed*2], cars2[, speed*2])
)
#[1] TRUE

第二个基本R选项为match.call，修改该调用，然后求值（您可以在lm中找到此方法）：

`[.dd` <- function(x,...) {
  thecall <- match.call()
  thecall[[1]] <- quote(`[`)
  thecall[[2]] <- quote(x[[i]])
  res <- list()
  for (i in seq_along(x)) {
    res[[i]] <- eval(thecall)
  }
  res
}

all.equal(
  car_list[,.N, by = speed],
  list(cars1[,.N, by = speed], cars2[,.N, by = speed])
)
#[1] TRUE

all.equal(
  car_list[, speed*2],
  list(cars1[, speed*2], cars2[, speed*2])
)
#[1] TRUE

如果您使用:=，我还没有测试过这些方法是否可以复制深层副本。

Answer 3

我的评论中的建议尚未完成。您确实可以使用rlang支持整洁的评估，但是由于data.table本身不直接支持它，您最好使用表达式而不是定量，并且您需要在调用eval_tidy之前构建完整的最终表达式：

`[.dd` <- function(x, ...) {
  code <- rlang::enexprs(...)
  lapply(x, function(dt) {
    ex <- rlang::expr(dt[!!!code])
    rlang::eval_tidy(ex)
  })
}

car_list[, .N, by = speed]
[[1]]
    speed N
 1:     4 2
 2:     7 2
 3:     8 1
 4:     9 1
 5:    10 3
 6:    11 2
 7:    12 4
 8:    13 4
 9:    14 4
10:    15 3
11:    16 2
12:    17 3
13:    18 4
14:    19 3
15:    20 5
16:    22 1
17:    23 1
18:    24 4
19:    25 1

[[2]]
    speed N
 1:     4 2
 2:     7 2
 3:     8 1
 4:     9 1
 5:    10 3
 6:    11 2
 7:    12 4
 8:    13 4
 9:    14 4
10:    15 3
11:    16 2
12:    17 3
13:    18 4
14:    19 3
15:    20 5
16:    22 1
17:    23 1
18:    24 4
19:    25 1

如何使用非标准评估NSE评估data.table上的参数？

3 个答案: