挣扎着一个非常简单的领先/滞后，希望你们能帮忙。假设我有：
set.seed(1234)
foo2 <- data.table(data.frame(Wk = rep(c(1:5), 4), CODE = c(rep('a',5),rep('b',5),rep('c',5),rep('d',5)), METRIC = rnorm(20)))[order(-Wk,CODE)]
currentWeek <- max(foo2[["Wk"]])
foo2：
Wk CODE METRIC
1: 5 a 0.85023226
2: 5 b -1.19452788
3: 5 c -0.49558344
4: 5 d 2.12111711
5: 4 a -0.17378717
6: 4 b -0.19159377
7: 4 c 1.00151325
8: 4 d 0.97291675
9: 3 a -1.37230189
10: 3 b -0.40273198
11: 3 c 1.70596401
12: 3 d 0.87820363
13: 2 a -0.16999408
14: 2 b 0.54999735
15: 2 c 0.25519600
16: 2 d -1.13460804
17: 1 a -0.17778996
18: 1 b 0.69760871
19: 1 c -0.05315882
20: 1 d 0.35555030
我如何将上述内容更改为：
Wk CODE METRIC METRIC_currentWeekMinus1 METRIC_currentWeekMinus2
1: 5 a 0.85023226 -0.17378717 -1.37230189
2: 5 b -1.19452788 -0.19159377 -0.40273198
3: 5 c -0.49558344 1.00151325 1.70596401
4: 5 d 2.12111711 0.97291675 0.87820363
5: 4 a -0.17378717 . .
6: 4 b -0.19159377 . .
7: 4 c 1.00151325 . .
8: 4 d 0.97291675 . .
9: 3 a -1.37230189 . .
10: 3 b -0.40273198 . .
11: 3 c 1.70596401 . .
12: 3 d 0.87820363 . .
13: 2 a -0.16999408 . NA
14: 2 b 0.54999735 . NA
15: 2 c 0.25519600 . NA
16: 2 d -1.13460804 . NA
17: 1 a -0.17778996 NA NA
18: 1 b 0.69760871 NA NA
19: 1 c -0.05315882 NA NA
20: 1 d 0.35555030 NA NA
有什么想法吗？提前谢谢！
答案 0 :(得分：3)
使用dplyr：
library(dplyr)
foo2 %>% group_by(CODE) %>% mutate(last1 = lead(METRIC), last2 = lead(METRIC, 2))
和data.table，版本v1.9.5（从github安装）：
foo2[ , c("last1", "last2") := shift(METRIC, 1:2, type = "lead"), by = CODE]