Question

我们只想说我有两列，DATE和VALUE。日期不一定是规则间隔。如何在过去N天内创建VALUE的移动平均线？我正在使用postgres，但mysql解决方案也会有所帮助。

数据：

DATE    VALUE
2012-11-05 10
2012-10-29 31
2012-10-22 108
2012-10-17 3654
2012-10-16 1187
2012-10-15 12033
2012-10-09 41
2012-10-01 85
2012-09-25 20
2012-09-24 285
2012-09-17 20
2012-09-10 20
2012-09-04 41
2012-08-27 63
2012-08-20 52
2012-08-13 160

Answer 1

下面的MySQL示例包含一个滑动的7天窗口：

select t1.`DATE`, AVG(t2.`VALUE`) as MV_AVG
from MyTable t1
left outer join MyTable t2 
    on t2.`DATE` between DATE_ADD(t1.`DATE`, INTERVAL -6 DAY) 
        and t1.`DATE`
group by t1.`DATE`

SQL Fiddle Example

<强>输出：

|                             DATE |    MV_AVG |
------------------------------------------------
|    August, 12 2012 20:00:00+0000 |       160 |
|    August, 19 2012 20:00:00+0000 |        52 |
|    August, 26 2012 20:00:00+0000 |        63 |
| September, 03 2012 20:00:00+0000 |        41 |
| September, 09 2012 20:00:00+0000 |      30.5 |
| September, 16 2012 20:00:00+0000 |        20 |
| September, 23 2012 20:00:00+0000 |       285 |
| September, 24 2012 20:00:00+0000 |     152.5 |
| September, 30 2012 20:00:00+0000 |      52.5 |
|   October, 08 2012 20:00:00+0000 |        41 |
|   October, 14 2012 20:00:00+0000 |      6037 |
|   October, 15 2012 20:00:00+0000 |      6610 |
|   October, 16 2012 20:00:00+0000 | 5624.6667 |
|   October, 21 2012 20:00:00+0000 | 1649.6667 |
|   October, 28 2012 20:00:00+0000 |        31 |
|  November, 04 2012 19:00:00+0000 |        10 |

Answer 2

RedFilter的优秀answer使用非均匀间隔的存储桶。如果你想要均匀分布的桶，可以这样做：

SELECT
DATE_FORMAT(CURDATE() - INTERVAL ((i.id - 0) * 7) - 1 DAY, '%Y-%m-%d') AS start,
DATE_FORMAT(CURDATE() - INTERVAL ((i.id - 1) * 7) DAY, '%Y-%m-%d') AS end,
AVG(d.value) average
FROM
ids i
LEFT OUTER JOIN
dates d ON
d.date BETWEEN 
  (CURDATE() - INTERVAL ((i.id - 0) * 7) - 1 DAY) AND
  (CURDATE() - INTERVAL ((i.id - 1) * 7) DAY)
WHERE
i.id BETWEEN 0 AND 20
GROUP BY
1
ORDER BY
1 DESC;

提供了这个：

+------------+------------+-----------+
| start      | end        | average   |
+------------+------------+-----------+
| 2012-11-01 | 2012-11-07 |   10.0000 |
| 2012-10-25 | 2012-10-31 |   31.0000 |
| 2012-10-18 | 2012-10-24 |  108.0000 |
| 2012-10-11 | 2012-10-17 | 5624.6667 |
| 2012-10-04 | 2012-10-10 |   41.0000 |
| 2012-09-27 | 2012-10-03 |   85.0000 |
| 2012-09-20 | 2012-09-26 |  152.5000 |
| 2012-09-13 | 2012-09-19 |   20.0000 |
| 2012-09-06 | 2012-09-12 |   20.0000 |
| 2012-08-30 | 2012-09-05 |   41.0000 |
| 2012-08-23 | 2012-08-29 |   63.0000 |
| 2012-08-16 | 2012-08-22 |   52.0000 |
| 2012-08-09 | 2012-08-15 |  160.0000 |
| 2012-08-02 | 2012-08-08 |      NULL |
| 2012-07-26 | 2012-08-01 |      NULL |
| 2012-07-19 | 2012-07-25 |      NULL |
| 2012-07-12 | 2012-07-18 |      NULL |
| 2012-07-05 | 2012-07-11 |      NULL |
| 2012-06-28 | 2012-07-04 |      NULL |
| 2012-06-21 | 2012-06-27 |      NULL |
+------------+------------+-----------+
20 rows in set (0.00 sec)

请参阅http://sqlfiddle.com/#!2/78c52/39了解没有CURDATE（）的工作示例。

如何在SQL中的某个日期范围内执行移动平均线？

2 个答案: