计算日期范围内的前 N ​​条记录 Postgres

Question

我有一个如下所示的表格：

+---------+---------+----------+
|id       |title    |date      |
+---------+---------+----------+
|1        |A        |2017-05-01|
|2        |C        |2017-05-01|
|3        |A        |2017-05-02|
|4        |A        |2017-05-01|
|5        |A        |2017-05-03|
|6        |B        |2017-05-01|
|7        |A        |2017-05-01|
|8        |D        |2017-05-03|
|9        |A        |2017-05-02|
|10       |B        |2017-05-01|
|11       |A        |2017-05-01|
|12       |C        |2017-05-02|
|13       |A        |2017-05-01|
|14       |B        |2017-05-01|
|15       |B        |2017-05-04|
|16       |A        |2017-05-03|
|17       |B        |2017-05-01|
|18       |A        |2017-05-01|
+---------+---------+----------+

我想查询这个表并按日期返回一个计数（前 2 个）。所以，我希望它返回以下内容：

+---------+---------+----------+
|count    |title    |date      |
+---------+---------+----------+
|6        |A        |2017-05-01|
|4        |A        |2017-05-02|
|1        |A        |2017-05-03|
|4        |B        |2017-05-01|
|5        |B        |2017-05-02|
|6        |B        |2017-05-03|
+---------+---------+----------+

（结果可能与我在示例表中的结果不匹配），但我认为重点可能就在那里。

Answer 1

我认为您的问题需要两个聚合，第一个找到每个标题/日期组合的计数，第二个保留每个日期的两个最大计数。

WITH cte1 AS (
    SELECT title, date, COUNT(*) AS t_count
    FROM yourTable
    GROUP BY title, date
),
cte2 AS (
    SELECT title, date, t_count,
           ROW_NUMBER() OVER (PARTITION BY date ORDER BY t_count DESC) rn
    FROM cte1
)
SELECT title, date, t_count
FROM cte2
WHERE rn <= 2

Answer 2

您可以一步组合窗口函数和聚合函数：

select cnt, title, date
from (select t.date, t.title, count(*) as cnt,
             row_number() over (partition by date order by count(*) desc) as seqnum
      from t
      group by t.date, t.title
     ) t
where seqnum <= 2;