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MySQL top-N排名并对同一组的其余部分求和

转载 作者:行者123 更新时间:2023-11-29 04:14:57 24 4
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我大部分时间都在研究这个主题,但是对于使用 sum() 对其余部分进行分组和聚合的 MySQL 表进行排名(前 3 名),我无法得到有效和完美的答案。

数据如下:

TS         | Name     | Count
=============================
1552286160 | Apple | 7
1552286160 | Orange | 8
1552286160 | Grape | 8
1552286160 | Pear | 9
1552286160 | Kiwi | 10
...
1552286100 | Apple | 10
1552286100 | Orange | 12
1552286100 | Grape | 14
1552286100 | Pear | 16
1552286100 | Kiwi | 9
...
1552286040 | Apple | 4
1552286040 | Orange | 2
1552286040 | Grape | 3
1552286040 | Pear | 7
1552286040 | Kiwi | 9
...

有了这个数据集,我想按每个 TS 组形成前 3 名,并且 1 行包含该组其余部分的总和(计数),如下所示:

TS         | Name     | Count
=============================
1552286160 | Kiwi | 10
1552286160 | Pear | 9
1552286160 | Grape | 8
1552286160 | Other | 8 + 7
...
1552286100 | Pear | 16
1552286100 | Grape | 14
1552286100 | Orange | 12
1552286100 | Other | 10 + 9
...
1552286040 | Kiwi | 9
1552286040 | Pear | 7
1552286040 | Apple | 4
1552286040 | Other | 3 + 2
...

最接近的提示实际上是通过 http://www.silota.com/docs/recipes/sql-top-n-aggregate-rest-other.html 提供的但是,该解决方案仅适用于单个组。

我准备的 SQL Fiddle 位于此处:http://sqlfiddle.com/#!9/3cedd0/10

感谢您是否有任何解决方案。

最佳答案

DROP TABLE IF EXISTS my_table;

CREATE TABLE my_table
(ts INT NOT NULL
,name VARCHAR(12) NOT NULL
,count INT NOT NULL
,PRIMARY KEY(ts,name)
);

INSERT INTO my_table VALUES
(1552286160,'Apple' , 7),
(1552286160,'Orange', 8),
(1552286160,'Grape' , 8),
(1552286160,'Pear' , 9),
(1552286160,'Kiwi' ,10),
(1552286100,'Apple' ,10),
(1552286100,'Orange',12),
(1552286100,'Grape' ,14),
(1552286100,'Pear' ,16),
(1552286100,'Kiwi' , 9),
(1552286040,'Apple' , 4),
(1552286040,'Orange', 2),
(1552286040,'Grape' , 3),
(1552286040,'Pear' , 7),
(1552286040,'Kiwi' , 9);

SELECT ts
, CASE WHEN i>3 THEN 'other' ELSE name END name
, SUM(count) count
FROM
( SELECT x.*
, CASE WHEN @prev=ts THEN @i:=@i+1 ELSE @i:=1 END i
, @prev:=ts
FROM my_table x
, (SELECT @prev:=null,@i:=0) vars
ORDER
BY ts
, count DESC
, name
) a
GROUP
BY ts
, CASE WHEN i>3 THEN 'other' ELSE name END;

+------------+--------+-------+
| ts | name | count |
+------------+--------+-------+
| 1552286040 | Apple | 4 |
| 1552286040 | Kiwi | 9 |
| 1552286040 | other | 5 |
| 1552286040 | Pear | 7 |
| 1552286100 | Grape | 14 |
| 1552286100 | Orange | 12 |
| 1552286100 | other | 19 |
| 1552286100 | Pear | 16 |
| 1552286160 | Grape | 8 |
| 1552286160 | Kiwi | 10 |
| 1552286160 | other | 15 |
| 1552286160 | Pear | 9 |
+------------+--------+-------+

关于MySQL top-N排名并对同一组的其余部分求和,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/55098157/

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