MYSQL 日平均值统计

Question

用例:我有一个 cron每 5 分钟检查一些统计信息并将其插入数据库表stats

**Structure**

`time` as DATETIME   (index)

`skey` as VARCHAR(50)  (index)

`value` as BIGINT

Primary (time and skey)

现在我想创建一个图表来显示一天中正在进行的每日平均值 - 所以即游戏用户图表:

from 0-1 i have 10 playing users (avg value from 0-1 is now 10)
from 1-2 i have 6 playing users (avg value is now 8 =>  (10+6) / 2)
from 2-3 i have 14 playing users (avg value is no 10 => (10+6+14) / 3

第二天从头开始

我已经在运行查询，但运行时间超过 3.5 秒

第一次尝试:

SELECT *  
     , (SELECT AVG(value) 
          FROM stats as b 
         WHERE b.skey = stats.skey 
           AND b.time <= stats.time 
           AND DATE(b.time) = DATE(stats.time)) 
  FROM stats 
 ORDER 
    BY stats.time DESC

第二次尝试:

SELECT *
     , (SELECT AVG(b.value) 
          FROM stats as b 
         WHERE b.skey = stats.skey 
           AND DATE(b.time) = DATE(stats.time) 
           AND b.time <= stats.time) as avg 
  FROM stats 
 WHERE skey = 'playingUsers' 
 GROUP 
    BY HOUR(stats.time)
     , DATE(stats.time)

第一次尝试是获取每个条目并计算平均值

第二次尝试是按小时分组(就像我的例子)

无论如何，这不会改变任何性能

是否有办法提高 mysql 的性能，或者我是否必须更改其背后的完整逻辑？

数据库 fiddle : https://www.db-fiddle.com/f/krFmR1yPsmnPny2zi5NJGv/4

Answer 1

我建议将每小时平均值的计算与天平均值的计算分开，并通过分组每小时仅计算一次这些值。

如果您使用的是 MySQL 8，我建议按如下方式使用 CTE:

with HOURLY AS (
    SELECT distinct
                    DATE_,
                    HOUR_,
                    AVG(b.value) as avg_per_hour
    FROM (SELECT s.value, DATE(s.time) DATE_, HOUR(s.time) HOUR_
          FROM stats s
          where skey = 'playingUsers'
         ) b
    GROUP BY b.DATE_, b.HOUR_
    ORDER BY b.DATE_ DESC, b.HOUR_ DESC
)
SELECT *
     , (SELECT AVG(b.avg_per_hour)
        FROM HOURLY as b
        WHERE b.DATE_ = HOURLY.DATE_
          AND b.HOUR_ <= HOURLY.HOUR_) as avg
FROM HOURLY

该语句在给定的 fiddle 中持续 < 300 毫秒。

计算对应于您在上表中描述的算法。

但是，结果与所提供的陈述不同。