Hive：根据年份和员工获得最小值？

Question

我有一个示例 table 下面：

+--------------+------------------+-----------+--+
|  orderdate   |     employee     | minprice  |
+--------------+------------------+-----------+--+
| 1992-01-13   | Clerk#943        | 7328.08   |
| 1992-02-21   | Clerk#328        | 33818.37  |
| 1992-02-22   | Clerk#328        | 914.01    |
| 1992-03-03   | Clerk#943        | 10010.11  |
| 1992-03-19   | Clerk#158        | 2712.00   |
| 1992-03-20   | Clerk#328        | 23920.52  |
| 1992-04-05   | Clerk#158        | 919.01    |
| 1993-01-04   | Clerk#943        | 24786.27  |
| 1993-01-29   | Clerk#158        | 11856.13  |
| 1993-01-30   | Clerk#943        | 2712.00   |
| 1993-02-17   | Clerk#328        | 42958.47  |
| 1993-02-25   | Clerk#328        | 2703.00   |

如何根据年份获得员工的最小值？预期输出：

+--------------+------------------+-----------+--+
|  orderdate   |     employee     | minprice  |
+--------------+------------------+-----------+--+
| 1992-01-13   | Clerk#943        | 7328.08   |
| 1992-02-22   | Clerk#328        | 914.01    |
| 1992-04-05   | Clerk#158        | 919.01    |
| 1993-01-30   | Clerk#943        | 2712.00   |
| 1993-01-29   | Clerk#158        | 11856.13  |
| 1993-02-25   | Clerk#328        | 2703.00   |

我目前拥有的：

SELECT o_orderdate, o_employee, min(sales) AS minprice
FROM orders
INNER JOIN sales 
ON o_orderkey = s_orderkey
GROUP BY o_orderdate, o_employee
GROUPING SETS ((o_orderdate, o_employee));

查询运行但没有按年份和基于员工的筛选器。我找不到很多关于如何在 Hive 和分组集中执行此操作的文档。

感谢任何形式的帮助。

Answer 1

您似乎只需要一个 window 函数，而不是聚合：

SELECT o_orderdate, o_employee, sale
FROM (SELECT o.o_orderdate, o.o_employee, s.sales,
             ROW_NUMBER() OVER (PARTITION BY o.o_employee, YEAR(o.order_date) ORDER BY s.sales) as seqnum
      FROM orders o JOIN
           sales s
           ON o.o_orderkey = s.s_orderkey
     ) os
WHERE seqnum = 1;

Hive：根据年份和员工获得最小值？

Hive: get minimum value based on year and employee?

sql

group-by

hive

hiveql