寻求data.table分组的优雅单行解决方案：获取首次出现渠道

Absolutely! You can skip the merge step entirely by using data.table's powerful grouped assignment feature—here's the clean, single-line solution you're after:

dt.master[, first_channel := channel[which.min(date)], by = .(user, target)]

How this works: #

• We're directly creating the first_channel column in your original dt.master table using data.table's j argument.
• For each group defined by .(user, target), which.min(date) finds the row index of the earliest date in that group.
• We pull the channel value from that earliest row, and assign it to every row in the same (user, target) group.

Test it with your data: #

After running the one-liner, your table will match your expected output perfectly:

user target channel       date first_channel
1: 1000  50000       A 2018-02-10             B
2: 1002  50004       B 2018-04-11             C
3: 2008  50501       C 2018-03-14             C
4: 3005  50001       A 2018-03-02             A
5: 1000  50000       B 2018-01-05             B
6: 1002  50000       A 2018-05-08             A
7: 1002  50004       C 2018-03-05             C

Why this beats your original approach: #

• No intermediate tables or merge operations required—everything happens in one readable step.
• It’s more efficient for large datasets, as we avoid the overhead of creating a separate result table and joining it back.
• You can choose to modify the table in-place (as shown) or assign to a new table if you want to preserve the original:

  dt.new <- dt.master[, first_channel := channel[which.min(date)], by = .(user, target)]
  

内容的提问来源于stack exchange，提问作者illanxr