MySQL计算N行的移动平均值

时间:2016-01-16 22:39:20

标签: mysql

我正在尝试计算 N行的移动平均值 单个查询中的所有行。在示例中,我试图计算50行的移动平均值。

SELECT 
    h1.date, 
    h1.security_id, 
    (   SELECT 
            AVG(last50.close)
        FROM (
            SELECT h.close
            FROM history as h
            WHERE h.date <= h1.date AND h.security_id = h1.security_id
            ORDER BY h.date DESC
            LIMIT 50
        ) as last50
    ) as avg50
FROM history as h1

但是,MySQL在运行此查询时出错:

Unknown column 'h1.date' in 'where clause'

我正在尝试这种方法,因为列出的其他解决方案似乎并不适用于我的用例。有一个N 天的移动平均值的解决方案,但由于我的数据集中没有考虑所有日期,我需要N 行的平均值

This solution(如下所示)不起作用,因为AVGSUMCOUNT}不会占LIMIT

SELECT
  t1.data_date
  ( SELECT SUM(t2.price) / COUNT(t2.price)
    FROM t as t2
    WHERE t2.data_date <= t1.data_date
    ORDER BY t2.data_date DESC
    LIMIT 5
  ) AS 'five_row_moving_average_price'
FROM t AS t1
ORDER BY t1.data_date;

This question看起来很有希望,但对我来说有些难以理解。

有什么建议吗? Here's an SQLFiddle来玩。

1 个答案:

答案 0 :(得分:1)

<强>计划

  
      
  • 过去50天的自我加入历史
  •   
  • 按日期和安全ID(当前)进行平均分组
  •   

<强>查询

select curr.date, curr.security_id, avg(prev.close)
from history curr
inner join history prev 
on prev.`date` between date_sub(curr.`date`, interval 49 day) and curr.`date`
and curr.security_id = prev.security_id
group by 1, 2
order by 2, 1
;

<强>输出

+---------------------------+-------------+--------------------+
|           date            | security_id |  avg(prev.close)   |
+---------------------------+-------------+--------------------+
| January, 04 2016 00:00:00 | 1           | 10.770000457763672 |
| January, 05 2016 00:00:00 | 1           | 10.800000190734863 |
| January, 06 2016 00:00:00 | 1           | 10.673333485921225 |
| January, 07 2016 00:00:00 | 1           | 10.59250020980835  |
| January, 08 2016 00:00:00 | 1           | 10.432000160217285 |
| January, 11 2016 00:00:00 | 1           | 10.40166680018107  |
| January, 12 2016 00:00:00 | 1           | 10.344285828726631 |
| January, 13 2016 00:00:00 | 1           | 10.297500133514404 |
| January, 14 2016 00:00:00 | 1           | 10.2877779006958   |
| January, 04 2016 00:00:00 | 2           | 56.15999984741211  |
| January, 05 2016 00:00:00 | 2           | 56.18499946594238  |
| ..                        | ..          | ..                 |
+---------------------------+-------------+--------------------+

sqlfiddle

<强>参考

已修改为使用最后50行

select
rnk_curr.`date`, rnk_curr.security_id, avg(rnk_prev50.close)
from
(
select `date`, security_id,
@row_num := if(@lag = security_id, @row_num + 1,
               if(@lag := security_id, 1, 1)) as row_num
from history 
cross join ( select @row_num := 1, @lag := null ) params
order by security_id, `date`
) rnk_curr
inner join
(
select date, security_id, close,
@row_num := if(@lag = security_id, @row_num + 1,
               if(@lag := security_id, 1, 1)) as row_num
from history 
cross join ( select @row_num := 1, @lag := null ) params
order by security_id, `date`
) rnk_prev50
on  rnk_curr.security_id = rnk_prev50.security_id
and rnk_prev50.row_num between rnk_curr.row_num - 49 and rnk_curr.row_num
group by 1,2
order by 2,1
;

sqlfiddle

注意

if函数是强制正确评估变量的顺序。