我有一个data.table，看起来像这样：
df
seller_id buyer_id hs10 year mean_ln_c_hit
1: 1 5 1200000000 5 4.407456
2: 1 24 1500000000 2 4.173422
3: 1 8 1500000000 3 4.532695
4: 1 8 1500000000 5 4.830106
5: 1 16 1500000000 5 4.830106
6: 1 17 1800000000 3 3.920435
7: 1 20 1800000000 5 4.357196
8: 1 27 1800000000 5 4.357196
9: 1 1 1900000000 1 4.835762
10: 1 2 1900000000 1 4.835762
11: 1 3 1900000000 1 4.835762
12: 1 4 1900000000 1 4.835762
13: 1 5 1900000000 1 4.835762
14: 1 6 1900000000 1 4.835762
15: 1 7 1900000000 1 4.835762
16: 1 8 1900000000 1 4.835762
17: 1 9 1900000000 1 4.835762
18: 1 11 1900000000 1 4.835762
19: 1 12 1900000000 1 4.835762
20: 1 13 1900000000 1 4.835762
21: 1 14 1900000000 1 4.835762
22: 1 15 1900000000 1 4.835762
23: 1 16 1900000000 1 4.835762
24: 1 17 1900000000 1 4.835762
25: 1 18 1900000000 1 4.835762
26: 1 19 1900000000 1 4.835762
27: 1 20 1900000000 1 4.835762
28: 1 21 1900000000 1 4.835762
29: 1 22 1900000000 1 4.835762
30: 1 23 1900000000 1 4.835762
31: 1 24 1900000000 1 4.835762
32: 1 25 1900000000 1 4.835762
33: 1 26 1900000000 1 4.835762
34: 1 27 1900000000 1 4.835762
35: 1 28 1900000000 1 4.835762
36: 1 29 1900000000 1 4.835762
37: 1 30 1900000000 1 4.835762
38: 1 1 1900000000 2 4.409253
39: 1 2 1900000000 2 4.409253
40: 1 3 1900000000 2 4.409253
41: 1 4 1900000000 2 4.409253
42: 1 5 1900000000 2 4.409253
43: 1 6 1900000000 2 4.409253
44: 1 7 1900000000 2 4.409253
45: 1 8 1900000000 2 4.409253
46: 1 9 1900000000 2 4.409253
47: 1 10 1900000000 2 4.409253
48: 1 11 1900000000 2 4.409253
49: 1 12 1900000000 2 4.409253
50: 1 13 1900000000 2 4.409253
51: 1 14 1900000000 2 4.409253
52: 1 15 1900000000 2 4.409253
53: 1 16 1900000000 2 4.409253
54: 1 17 1900000000 2 4.409253
55: 1 18 1900000000 2 4.409253
56: 1 19 1900000000 2 4.409253
57: 1 20 1900000000 2 4.409253
58: 1 21 1900000000 2 4.409253
59: 1 22 1900000000 2 4.409253
60: 1 23 1900000000 2 4.409253
61: 1 25 1900000000 2 4.409253
62: 1 26 1900000000 2 4.409253
63: 1 27 1900000000 2 4.409253
64: 1 28 1900000000 2 4.409253
65: 1 29 1900000000 2 4.409253
66: 1 30 1900000000 2 4.409253
67: 1 1 1900000000 3 4.514642
68: 1 3 1900000000 3 4.514642
69: 1 4 1900000000 3 4.514642
70: 1 5 1900000000 3 4.514642
71: 1 6 1900000000 3 4.514642
72: 1 7 1900000000 3 4.514642
73: 1 9 1900000000 3 4.514642
74: 1 11 1900000000 3 4.514642
75: 1 12 1900000000 3 4.514642
76: 1 13 1900000000 3 4.514642
77: 1 14 1900000000 3 4.514642
78: 1 15 1900000000 3 4.514642
79: 1 16 1900000000 3 4.514642
80: 1 18 1900000000 3 4.514642
81: 1 19 1900000000 3 4.514642
82: 1 20 1900000000 3 4.514642
83: 1 21 1900000000 3 4.514642
84: 1 22 1900000000 3 4.514642
85: 1 23 1900000000 3 4.514642
86: 1 24 1900000000 3 4.514642
87: 1 25 1900000000 3 4.514642
88: 1 26 1900000000 3 4.514642
89: 1 27 1900000000 3 4.514642
90: 1 28 1900000000 3 4.514642
91: 1 29 1900000000 3 4.514642
92: 1 30 1900000000 3 4.514642
93: 1 2 1900000000 5 4.698335
94: 1 3 1900000000 5 4.698335
95: 1 4 1900000000 5 4.698335
96: 1 6 1900000000 5 4.698335
97: 1 7 1900000000 5 4.698335
98: 1 9 1900000000 5 4.698335
99: 1 11 1900000000 5 4.698335
100: 1 12 1900000000 5 4.698335
我想使用data.table功能创建一个名为lag_mean_ln_c_hit的新列，其中包含lag列中值的一年滞后mean_ln_c_hit，以seller_id和hs10为条件。我最好的尝试是：
df[!is.na(year), lag_mean_ln_c_hit:=
(.SD[.(seller_id = seller_id, hs10 = hs10, year = year - 1), mean_ln_c_hit, on = c("seller_id", "hs10", "year"), allow.cartesian = TRUE ]) ]
何时显示NAs正确显示哪种图片（即year = 1是第一年，因此第一year中所有观测值的滞后应为NA），但无法捕获正确的滞后。对于除4.835762以外的任何其他组合，它始终报告NA（尽管一旦报告4.173422，则显示此信息）。
每个seller_id具有相同buyer_id值的多个mean_ln_c_hit条目。因此，我有重复。运行代码，我得到以下Warning as well
Warning message:
In `[.data.table`(df, !is.na(year), `:=`(lag_mean_ln_c_hit, (.SD[.(seller_id = seller_id, :
Supplied 1640 items to be assigned to 100 items of column 'lag_mean_ln_c_hit' (1540 unused)
没有allow.cartesian = TRUE，代码将无法运行。
关于如何解决此问题的任何想法？它应该很简单，但是我无法理解。
seller <- unique(df, by=c("seller_id", "hs10", "year"))
df[, oneyearago := year - 1L]
df[, lag_mean_ln_c_hit := seller[.SD, on=.(seller_id, hs10, year=oneyearago), x.mean_ln_c_hit]]
或者如果至少在一年之前，则可以使用非等额联接并选择最后一个匹配项：
seller <- unique(df, by=c("seller_id", "hs10", "year")) df[, lag_mean_ln_c_hit := seller[.SD, on=.(seller_id, hs10, year<year), x.mean_ln_c_hit, mult="last"]]
数据：
library(data.table) df <- fread("seller_id buyer_id hs10 year mean_ln_c_hit 1 5 1200000000 5 4.407456 1 24 1500000000 2 4.173422 1 8 1500000000 3 4.532695 1 8 1500000000 5 4.830106 1 16 1500000000 5 4.830106 1 17 1800000000 3 3.920435 1 20 1800000000 5 4.357196 1 27 1800000000 5 4.357196 1 1 1900000000 1 4.835762 1 2 1900000000 1 4.835762 1 3 1900000000 1 4.835762 1 4 1900000000 1 4.835762 1 5 1900000000 1 4.835762 1 6 1900000000 1 4.835762 1 7 1900000000 1 4.835762 1 8 1900000000 1 4.835762 1 9 1900000000 1 4.835762 1 11 1900000000 1 4.835762 1 12 1900000000 1 4.835762 1 13 1900000000 1 4.835762 1 14 1900000000 1 4.835762 1 15 1900000000 1 4.835762 1 16 1900000000 1 4.835762 1 17 1900000000 1 4.835762 1 18 1900000000 1 4.835762 1 19 1900000000 1 4.835762 1 20 1900000000 1 4.835762 1 21 1900000000 1 4.835762 1 22 1900000000 1 4.835762 1 23 1900000000 1 4.835762 1 24 1900000000 1 4.835762 1 25 1900000000 1 4.835762 1 26 1900000000 1 4.835762 1 27 1900000000 1 4.835762 1 28 1900000000 1 4.835762 1 29 1900000000 1 4.835762 1 30 1900000000 1 4.835762 1 1 1900000000 2 4.409253 1 2 1900000000 2 4.409253 1 3 1900000000 2 4.409253 1 4 1900000000 2 4.409253 1 5 1900000000 2 4.409253 1 6 1900000000 2 4.409253 1 7 1900000000 2 4.409253 1 8 1900000000 2 4.409253 1 9 1900000000 2 4.409253 1 10 1900000000 2 4.409253 1 11 1900000000 2 4.409253 1 12 1900000000 2 4.409253 1 13 1900000000 2 4.409253 1 14 1900000000 2 4.409253 1 15 1900000000 2 4.409253 1 16 1900000000 2 4.409253 1 17 1900000000 2 4.409253 1 18 1900000000 2 4.409253 1 19 1900000000 2 4.409253 1 20 1900000000 2 4.409253 1 21 1900000000 2 4.409253 1 22 1900000000 2 4.409253 1 23 1900000000 2 4.409253 1 25 1900000000 2 4.409253 1 26 1900000000 2 4.409253 1 27 1900000000 2 4.409253 1 28 1900000000 2 4.409253 1 29 1900000000 2 4.409253 1 30 1900000000 2 4.409253 1 1 1900000000 3 4.514642 1 3 1900000000 3 4.514642 1 4 1900000000 3 4.514642 1 5 1900000000 3 4.514642 1 6 1900000000 3 4.514642 1 7 1900000000 3 4.514642 1 9 1900000000 3 4.514642 1 11 1900000000 3 4.514642 1 12 1900000000 3 4.514642 1 13 1900000000 3 4.514642 1 14 1900000000 3 4.514642 1 15 1900000000 3 4.514642 1 16 1900000000 3 4.514642 1 18 1900000000 3 4.514642 1 19 1900000000 3 4.514642 1 20 1900000000 3 4.514642 1 21 1900000000 3 4.514642 1 22 1900000000 3 4.514642 1 23 1900000000 3 4.514642 1 24 1900000000 3 4.514642 1 25 1900000000 3 4.514642 1 26 1900000000 3 4.514642 1 27 1900000000 3 4.514642 1 28 1900000000 3 4.514642 1 29 1900000000 3 4.514642 1 30 1900000000 3 4.514642 1 2 1900000000 5 4.698335 1 3 1900000000 5 4.698335 1 4 1900000000 5 4.698335 1 6 1900000000 5 4.698335 1 7 1900000000 5 4.698335 1 9 1900000000 5 4.698335 1 11 1900000000 5 4.698335 1 12 1900000000 5 4.698335")