如何在 MySQL 中对具有不同平均值的三个变量进行分组?
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【中文标题】如何在 MySQL 中对具有不同平均值的三个变量进行分组?【英文标题】:How to group three variable with different averages in MySQL? 【发布时间】:2016-03-30 17:41:39 【问题描述】:+-----------------+---------------------+
| date_time | vale |
+-----------------+---------------------+
| 12/13/2015 0:00 | 56.75 |
| 12/13/2015 0:15 | 208.75 |
| 12/13/2015 0:30 | 58.8 |
| 12/13/2015 0:45 | 61.79 |
| 12/13/2015 1:00 | 288.65 |
| 12/13/2015 1:15 | 89.1 |
| 12/13/2015 1:30 | 28.9 |
| 12/13/2015 1:45 | 57.04 |
| 12/14/2015 1:00 | 63.87 |
| 12/14/2015 1:15 | 219.83 |
| 12/14/2015 1:30 | 64.95 |
| 12/14/2015 1:45 | 65.24 |
| 12/14/2015 2:00 | 55.67 |
| 12/14/2015 2:15 | 21.63 |
| 12/14/2015 2:30 | 56.75 |
| 12/14/2015 2:45 | 57.04 |
+-----------------+---------------------+
我有 date_time 及其各自的值,现在如何根据天、周和小时取平均值,如下所示:
+-----------------+-----------------+-----------+----------+
| date_time | hour_avg | day_avg | week_avg |
+-----------------+-----------------+-----------+----------+
| 12/13/2015 0:00 | 96.52 | 106.2 | 90.9 |
| 12/13/2015 1:00 | 115.9 | 106.2 | 90.9 |
| 12/14/2015 1:00 | 103.4 | 75.6 | 90.9 |
| 12/14/2015 2:00 | 47.7 | 75.6 | 90.9 |
+-----------------+-----------------+-----------+----------+
【问题讨论】:
如果你想在date_time
列中显示小时,那么你必须计算day_avg
和week_avg
也考虑时间......而且,顺便说一句,你想如何计算那些@ 987654327@和week_avg
???
它没有回答您的确切问题,但另一种方法(具有不同的输出风格)将使用“WITH ROLLUP”-dev.mysql.com/doc/refman/5.7/en/group-by-modifiers.html
【参考方案1】:
实现它的一种方法是使用 GROUP BY
日期和小时 + 整天/周的相关子查询:
SELECT
DATE_ADD(CAST(date_time AS DATE), INTERVAL HOUR(date_time) HOUR) AS date_time
,ROUND(AVG(vale),1) AS hour_avg
,ROUND((SELECT AVG(vale) FROM tab t2 WHERE DATE(t2.date_time) = DATE(t.date_time) GROUP BY DATE(date_time)),1) AS day_avg
,ROUND((SELECT AVG(vale) FROM tab t2 WHERE WEEK(t2.date_time) = WEEK(t.date_time) AND YEAR(t.date_time) = YEAR(t2.date_time) GROUP BY WEEK(date_time)),1) AS week_avg
FROM tab t
GROUP BY DATE(date_time), HOUR(date_time);
SqlFiddleDemo
输出:
╔═════════════════════════════╦═══════════╦══════════╦══════════╗
║ date_time ║ hour_avg ║ day_avg ║ week_avg ║
╠═════════════════════════════╬═══════════╬══════════╬══════════╣
║ December, 13 2015 00:00:00 ║ 96.5 ║ 106.2 ║ 90.9 ║
║ December, 13 2015 01:00:00 ║ 115.9 ║ 106.2 ║ 90.9 ║
║ December, 14 2015 01:00:00 ║ 103.5 ║ 75.6 ║ 90.9 ║
║ December, 14 2015 02:00:00 ║ 47.8 ║ 75.6 ║ 90.9 ║
╚═════════════════════════════╩═══════════╩══════════╩══════════╝
【讨论】:
【参考方案2】:计划
计算每个分组粒度的平均值 在小时级别将谷物连接在一起
查询
select ha.grain, ha.hour_avg, da.day_avg, wa.week_avg
from
(
select date(date_time) + interval hour(date_time) hour as grain, avg(vale) hour_avg
from temperature
group by date(date_time), hour(date_time)
) ha
inner join
(
select date(date_time) as day, avg(vale) as day_avg
from temperature
group by date(date_time)
) da
on date(grain) = da.day
inner join
(
select year(date_time) as year, week(date_time) as week, avg(vale) as week_avg
from temperature
group by year(date_time), week(date_time)
) wa
on wa.year = year(ha.grain)
and wa.week = week(ha.grain)
;
输出
+----------------------------+----------+----------+----------+
| grain | hour_avg | day_avg | week_avg |
+----------------------------+----------+----------+----------+
| December, 13 2015 00:00:00 | 96.5225 | 106.2225 | 90.9225 |
| December, 13 2015 01:00:00 | 115.9225 | 106.2225 | 90.9225 |
| December, 14 2015 01:00:00 | 103.4725 | 75.6225 | 90.9225 |
| December, 14 2015 02:00:00 | 47.7725 | 75.6225 | 90.9225 |
+----------------------------+----------+----------+----------+
sqlfiddle
【讨论】:
您应该将YEAR
和 WEEK
加入到 wa
子查询中。当表包含多年的数据时,仅一周会产生不正确的结果。 Demo以上是关于如何在 MySQL 中对具有不同平均值的三个变量进行分组?的主要内容,如果未能解决你的问题,请参考以下文章