Question

我有一张这样的表，

Date    User
15-06-2018  A
16-06-2018  A
15-06-2018  B
14-06-2018  C
16-06-2018  C

我想得到像这样的输出，

Date    User
16-06-2018  A
15-06-2018  B
16-06-2018  C

我尝试了Select Max(date),User from Table group by User

Answer 1

根据您的评论，我假设当您group by时，这80个列中有重复的结果。假设是这样的，这里有一个选项，row_number总是为每个用户返回1行：

select *
from (
    select *, row_number() over (partition by user order by date desc) rn
    from yourtable
) t
where rn = 1

Answer 2

您可以使用 correlation subquery：

select t.*
from table t
where date = (select max(t1.date) 
              from table t1 
              where t1.user = t.user
             );

但是，我还建议row_number()：

select top (1) with ties *
from table t
order by row_number() over (partition by user order by date desc);

Answer 3

您还可以使用排名功能

SELECT User, Date
  FROM 
    (
      SELECT User, Date
           , Row_id = Row_Number() OVER (Partition by User,  ORDER BY User, Date desc)
        FROM table
    )q
WHERE Row_Id = 1

Answer 4

我建议你这个

dplyr::filter

选择单个用户的最长日期时间

4 个答案: