分组，每组最多选择一个分组

Question

这是一个ClickHouse请求，但我认为我缺少一些核心sql解决方案。我想计算每天每分钟的最大行数：

SELECT date_date,
       formatDateTime(date_time,'%F %H:%M:00', 'Asia/Dubai') AS Max_TPS_Time,
       count(1) AS Max_TPS
FROM TEST
GROUP BY Max_TPS_Time,
         date_date

哪个结果

date_date   Max_TPS_Time    Max_TPS
19/11/19    2019-11-19 11:37:00 4
19/11/19    2019-11-19 11:15:00 2
19/11/19    2019-11-19 11:40:00 5
19/11/18    2019-11-18 12:37:00 4
19/11/18    2019-11-18 12:15:00 2
19/11/18    2019-11-18 12:40:00 6

最后，我想在适当的日期时间看到每天的最大值。

date_date   Max_TPS_Time    Max_TPS
19/11/19    2019-11-19 11:40:00 5
19/11/18    2019-11-18 12:40:00 6

Answer 1

您正在从汇总查询的结果中查找每组的前1条记录。

据我所知，clickhouse不支持窗口功能（这会使此任务更加容易）。

但是它支持常用的表表达式：因此您可以将现有查询转换为CTE，然后使用自相关子查询进行过滤：

WITH cte as (
    SELECT 
        date_date,
        formatDateTime(date_time,'%F %H:%M:00', 'Asia/Dubai') AS Max_TPS_Time,
        count(*) AS Max_TPS
    FROM TEST
    GROUP BY Max_TPS_Time, date_date
)
SELECT c.*
FROM cte c
WHERE c.Max_TPS = (SELECT MAX(c1.Max_TPS) FROM cte c1 WHERE c1.date_date = c.date_date)

[请注意，如果您每天只需要Max_TPS，但不需要相应的Max_TPS_Time，那么它很简单，您可以在当前查询中添加另一级别的汇总：

SELECT date_date, MAX(Max_TPS) Max_TPS
FROM (
    SELECT 
        date_date,
        formatDateTime(date_time,'%F %H:%M:00', 'Asia/Dubai') AS Max_TPS_Time,
        count(*) AS Max_TPS
    FROM TEST
    GROUP BY Max_TPS_Time, date_date
) t
GROUP BY date_date

Answer 2

SELECT date_date,
       formatDateTime(date_time,'%F %H:%M:00', 'Asia/Dubai') AS Max_TPS_Time,
       count() AS Max_TPS
FROM TEST GROUP BY Max_TPS_Time, date_date
order by Max_TPS_Time, date_date, Max_TPS desc 
limit 1 by Max_TPS_Time, date_date


select argMax(date_date, Max_TPS) date_date,  argMax(Max_TPS_Time, Max_TPS) Max_TPS_Time
from (
SELECT date_date,
       formatDateTime(date_time,'%F %H:%M:00', 'Asia/Dubai') AS Max_TPS_Time,
       count() AS Max_TPS
FROM TEST GROUP BY Max_TPS_Time, date_date)