Question

我有以下data.table，dtgrouped2：

     MonthNo Unique Total
 1:       1    AAA    10
 2:       1    BBB     0
 3:       2    CCC     3
 4:       2    DDD     0
 5:       3    AAA     0
 6:       3    BBB    35
 7:       4    CCC    15
 8:       4    AAA     0
 9:       5    BBB    60
10:       5    CCC     0
11:       6    DDD   100
12:       6    AAA     0

另一张表dt2：

     Unique1 StartDate EndDate Amount1 Amount2
1:     AAA        0        1       7       0
3:     AAA        1        2       5       0
2:     AAA        2        4       3       2

我希望根据以下逻辑将dt2中的Amount1和Amount2插入dtgrouped2;＆＃34;唯一＆＃34;评估了dtgrouped2的每一行：

filter(StartDate< MonthNo & EndDate>=MonthNo)
then MAX(EndDate)
then insert Amount1 as Amount1 and Amount2 as Amount2

因此，您可以看到结果因行而异。这将是预期的输出：

Date    MonthNo Unique  Items   Amounts Amount1 Amount2
Jan       1      AAA    x         10        7   0
Jan       1      BBB    y          2        NA  NA
Feb       2      CCC    x          3        NA  NA
Feb       2      DDD    y         15       NA   NA
March     3      AAA    y         20        3   2
March     3      BBB    x         35       NA   NA
April     4      CCC    x         15       NA   NA
April     4      AAA    y         50       3    2
May       5      BBB    x         60      NA    NA
May       5      CCC    y         70      NA    NA
June      6      DDD    x         100     NA    NA
June      6      AAA    y         20       NA   NA

Answer 1

我建议使用非等联接与mult = "last"结合使用（以便仅捕获最新的EndDate）

dtgrouped2[, c("Amount1", "Amount2") := # Assign the below result to new columns in dtgrouped2
              dt2[dtgrouped2, # join
                  .(Amount1, Amount2), # get the column you need
                  on = .(Unique1 = Unique, # join conditions
                         StartDate < MonthNo, 
                         EndDate >= MonthNo), 
                  mult = "last"]] # get always the latest EndDate
dtgrouped2

#     MonthNo Unique Total Amount1 Amount2
#  1:       1    AAA    10       7       0
#  2:       1    BBB     0      NA      NA
#  3:       2    CCC     3      NA      NA
#  4:       2    DDD     0      NA      NA
#  5:       3    AAA     0       3       2
#  6:       3    BBB    35      NA      NA
#  7:       4    CCC    15      NA      NA
#  8:       4    AAA     0       3       2
#  9:       5    BBB    60      NA      NA
# 10:       5    CCC     0      NA      NA
# 11:       6    DDD   100      NA      NA
# 12:       6    AAA     0      NA      NA

您需要首先加入dt2[dtgrouped]（而不是相反）的原因是因为您希望为dt2中的每个可能值加入dtgrouped，因此允许多个dt2中要加入dtgrouped

的值

有条件地加入data.table？

1 个答案: