Question

我有下表：

+----+----------+----------+----------+
| id |  Column1 |  Column2 |  Column3 |
+----+----------+----------+----------+
| 1  |  1       |  2014    |  0.2     | 
| 2  |  1       |  2013    |  0.5     | 
| 3  |  2       |  2014    |  1.9     | 
| 4  |  2       |  2013    |  1.4     | 
| 5  |  2       |  2012    |  1       | 
| 6  |  2       |  2011    |  0.4     | 
| 7  |  3       |  2016    |  1.4     | 
| 8  |  3       |  2015    |  1.2     | 
| 9  |  3       |  2014    |  0.7     | 
| 10 |  4       |  2015    |  0.5     | 
+----+----------+----------+----------+

我需要的是以下内容我想平均具有相同Column1值的行，但最新数据应该乘以0.6，其余的则乘以0.3

所以例如
其中Column1 = 1, it should output the value of 0.2*0.6+0.5*0.3 Column1 = 2, 1.9*0.6+((1.4+1+0.4)/3)*0.3 Column1 = 3, 1.4*0.6+((1.2+0.7)/2)*0.3 Column1 = 4, 0.5

编辑：如果对于一个查询来说这太复杂了，我也很高兴也可以这样做。

Answer 1

请在此处查看：sqlFiddle

SELECT 
    c1, 
    avg(c3), -- this here is the average per weight
    weight,  -- this is the weight
    avg(c3)*weight as weighted_avg -- product between the two
FROM
(
    SELECT
        table1.*,
        if(no_of_lines is null, 
           0.3,                   -- the default weight for >1 lines
           if(no_of_lines = 1 , 
              1,                  -- the weight if there's only 1 line
              0.6                 -- the weight for the 1st line if there are more
           )
        ) as weight
    FROM
        table1
    Left join
    (
        select min(id) as id, count(id) as no_of_lines ,c1
        from table1
        group by c1
    ) tmp on tmp.id = table1.id
) final
group by c1, weight
order by c1 ASC, weight DESC

将输出：

c1 | avg(c3) | weight | weighted_avg
------------------------------------
1  | 0.2     |    0.6 | 0.12
1  | 0.5     |    0.3 | 0.15
2  | 1.9     |    0.6 | 1.14
2  | 0.9333  |    0.3 | 0.279
3  | 1.4     |    0.6 | 0.84
3  | 0.95    |    0.3 | 0.285
4  | 0.5     |      1 | 0.5

您现在需要做的就是：

SELECT c1, sum(weighted_avg) FROM `that_select`
GROUP by c1

声明：
1）这可能会简化一点，但这是另一个故事 2）删除评论 - 可能会给你错误

MySQL具有特殊权重

1 个答案: