R data.table 通过 id、使用 shift 等更新参数的最后值以及一些动态

问题描述 投票：0回答：1

这是一个 data.table，其中每个 id 都有一些参数，按一些定期的季度日期排列。 它最初是随机打乱的，但是，首先，比方说，它是按 fab_date 和 id 排序的。

set.seed(1)
dt_to_fun <- data.table(fab_date = structure(c(18993, 19174, 19358, 19539, 
                            18993, 19174, 19358, 19539, 18993, 19174, 19358, 19539, 18993, 
                            19174, 19358, 19539, 18993, 19174, 19358, 19539), class = "Date"), 
     id = c("n_01", "n_01", "n_01", "n_01", "n_02", "n_02", "n_02", 
                 "n_02", "n_03", "n_03", "n_03", "n_03", "n_04", "n_04", "n_04", 
                 "n_04", "n_05", "n_05", "n_05", "n_05"), 
     param_01 = sample(c(10,20, 30, 40, 50), 20, replace = TRUE), 
     param_02 = sample(c(10,20, 30, 40, 50), 20, replace = TRUE), 
     param_03 = sample(c(10,20, 30, 40, 50), 20, replace = TRUE))

dt_to_fun

       fab_date   id param_01 param_02 param_03
 1: 2022-01-01 n_01       10       50       30
 2: 2022-07-01 n_01       40       20       20
 3: 2023-01-01 n_01       10       20       20
 4: 2023-07-01 n_01       20       10       50
 5: 2022-01-01 n_02       50       40       20
 6: 2022-07-01 n_02       30       10       10
 7: 2023-01-01 n_02       20       40       30
 8: 2023-07-01 n_02       30       30       30
 9: 2022-01-01 n_03       30       20       40
10: 2022-07-01 n_03       10       20       30
11: 2023-01-01 n_03       50       40       10
12: 2023-07-01 n_03       50       40       40
13: 2022-01-01 n_04       20       40       50
14: 2022-07-01 n_04       20       20       10
15: 2023-01-01 n_04       10       40       10
16: 2023-07-01 n_04       50       10       40
17: 2022-01-01 n_05       50       10       50
18: 2022-07-01 n_05       10       40       50
19: 2023-01-01 n_05       10       10       40
20: 2023-07-01 n_05       50       20       50
>

目标是：对于每个 id 替换（最后日期）参数值，如下所示： param_01(最后日期) = param_01(最后日期) + param_01(-1 周期) - param_01(-2 周期) 例如，对于 id n_05，param_02 的最后一个值为 20，前一个值为 10，前两次值为 40， 所以结果应该是 20 + 10 - 40 = -10。 所有 id 和所有 param 列都相同。最后日期是“2023-07-01”，因此仅应更新具有该日期的行中的参数。

我设法完成了这个计算，但我的方式适用于所有日期，但它需要 仅针对每个 ID 的最后日期执行此操作。 这是函数：

quarterly_process_fun <- function(dt) {
  param_cols <- c("param_01", "param_02", "param_03")
  dt[, (param_cols) := lapply(
    .SD,
    \(x) (x + data.table::shift(x, n = 1L, fill = x[1L], type = "lag") - data.table::shift(x, n = 2L, fill = x[1L], type = "lag")
    )),
    by = .(id),
    .SDcols = param_cols
  ]
  return(dt)
} 
quarterly_process_fun(dt_to_fun)

结果：

dt_to_fun
        fab_date   id param_01 param_02 param_03
     1: 2022-01-01 n_01       10       50       30
     2: 2022-07-01 n_01       40       20       20
     3: 2023-01-01 n_01       40      -10       10
     4: 2023-07-01 n_01      -10       10       50
     5: 2022-01-01 n_02       50       40       20
     6: 2022-07-01 n_02       30       10       10
     7: 2023-01-01 n_02        0       10       20
     8: 2023-07-01 n_02       20       60       50
     9: 2022-01-01 n_03       30       20       40
    10: 2022-07-01 n_03       10       20       30
    11: 2023-01-01 n_03       30       40        0
    12: 2023-07-01 n_03       90       60       20
    13: 2022-01-01 n_04       20       40       50
    14: 2022-07-01 n_04       20       20       10
    15: 2023-01-01 n_04       10       20      -30
    16: 2023-07-01 n_04       40       30       40
    17: 2022-01-01 n_05       50       10       50
    18: 2022-07-01 n_05       10       40       50
    19: 2023-01-01 n_05      -30       40       40
    20: 2023-07-01 n_05       50      -10       40

那么，我该如何调整它，以便它只计算和替换每个 id 最后日期的差异？

另一个问题是，这是否可以在打乱的数据上进行管理？

    set.seed(1)
    dt_to_fun <- data.table(fab_date = structure(c(18993, 19174, 19358, 19539, 
                                18993, 19174, 19358, 19539, 18993, 19174, 19358, 19539, 18993, 
                                19174, 19358, 19539, 18993, 19174, 19358, 19539), class = "Date"), 
         id = c("n_01", "n_01", "n_01", "n_01", "n_02", "n_02", "n_02", 
                     "n_02", "n_03", "n_03", "n_03", "n_03", "n_04", "n_04", "n_04", 
                     "n_04", "n_05", "n_05", "n_05", "n_05"), 
         param_01 = sample(c(10,20, 30, 40, 50), 20, replace = TRUE), 
         param_02 = sample(c(10,20, 30, 40, 50), 20, replace = TRUE), 
         param_03 = sample(c(10,20, 30, 40, 50), 20, replace = TRUE))
    
#shuffle rows
dt_to_fun <- dt_to_fun[sample(nrow(dt_to_fun)),]
     fab_date   id param_01 param_02 param_03
 1: 2023-07-01 n_03       50       40       40
 2: 2023-07-01 n_04       50       10       40
 3: 2022-01-01 n_01       10       50       30
 4: 2022-01-01 n_04       20       40       50
 5: 2022-01-01 n_02       50       40       20
 6: 2023-01-01 n_04       10       40       10
 7: 2022-07-01 n_02       30       10       10
 8: 2022-07-01 n_05       10       40       50
 9: 2022-01-01 n_03       30       20       40
10: 2023-01-01 n_02       20       40       30
11: 2023-01-01 n_03       50       40       10
12: 2023-01-01 n_01       10       20       20
13: 2022-07-01 n_04       20       20       10
14: 2022-07-01 n_01       40       20       20
15: 2022-07-01 n_03       10       20       30
16: 2023-07-01 n_05       50       20       50
17: 2023-07-01 n_01       20       10       50
18: 2023-07-01 n_02       30       30       30
19: 2023-01-01 n_05       10       10       40
20: 2022-01-01 n_05       50       10       50
r data.table self-join
1个回答
0
投票

我们可以稍微调整您的功能： 首先我们 

sort
by 
id
和 
fab_date
，然后 我们使用 
.I
中的 
data.table
（行索引）功能来识别每个 
id
的最后日期。

library(data.table)

quarterly_process_fun <- function(dt) {
  # Sort by id and fab_date
  setorder(dt, id, fab_date)
  
  # last row for each id
  last_row <- dt[, .I[fab_date == max(fab_date)], by = id]$V1
  
  dt[last_row, c("param_01", "param_02", "param_03") := lapply(
    .SD,
    function(x) (x + shift(x, n = 1L, fill = x[1L], type = "lag") - shift(x, n = 2L, fill = x[1L], type = "lag"))
  ), .SDcols = c("param_01", "param_02", "param_03")]
  
  return(dt)
}

quarterly_process_fun(dt_to_fun)
dt_to_fun

  fab_date   id param_01 param_02 param_03
 1: 2022-01-01 n_01       10       50       30
 2: 2022-07-01 n_01       40       20       20
 3: 2023-01-01 n_01       10       20       20
 4: 2023-07-01 n_01       20       10       50
 5: 2022-01-01 n_02       50       40       20
 6: 2022-07-01 n_02       30       10       10
 7: 2023-01-01 n_02       20       40       30
 8: 2023-07-01 n_02       30       30       30
 9: 2022-01-01 n_03       30       20       40
10: 2022-07-01 n_03       10       20       30
11: 2023-01-01 n_03       50       40       10
12: 2023-07-01 n_03       60       60       20
13: 2022-01-01 n_04       20       40       50
14: 2022-07-01 n_04       20       20       10
15: 2023-01-01 n_04       10       40       10
16: 2023-07-01 n_04       70       20       50
17: 2022-01-01 n_05       50       10       50
18: 2022-07-01 n_05       10       40       50
19: 2023-01-01 n_05       10       10       40
20: 2023-07-01 n_05       50      -10       50
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