SQL查询/根据日期和ID计算值的等级

Question

  

对于每个日期和ID，我们必须按降序对值进行排名。

     

下面是输入表。

>input data
+------------+----+-------+
|    date    | id | value |
+------------+----+-------+
| 01-01-2018 | A  |    20 |
| 01-01-2018 | A  |    50 |
| 01-01-2018 | C  |    40 |
| 01-01-2018 | B  |    40 |
| 02-01-2018 | A  |    30 |
| 03-01-2018 | C  |    20 |
| 03-01-2018 | C  |    40 |
| 04-01-2018 | B  |     0 |
| 04-01-2018 | B  |    40 |
| 05-01-2018 | B  |    70 |
+------------+----+-------+

  

输出应如下所示：

>output data
+------------+----+-------+------+
|    date    | id | value | rank |
+------------+----+-------+------+
| 01-01-2018 | A  |    50 |    1 |
| 01-01-2018 | A  |    20 |    2 |
| 01-01-2018 | B  |    40 |    1 |
| 01-01-2018 | C  |    40 |    1 |
| 02-01-2018 | A  |    30 |    1 |
| 03-01-2018 | C  |    40 |    1 |
| 03-01-2018 | C  |    20 |    2 |
| 04-01-2018 | B  |    40 |    1 |
| 04-01-2018 | B  |     0 |    2 |
| 05-01-2018 | B  |    70 |    1 |
+------------+----+-------+------+

Answer 1

您可以使用from g2p_en import G2p texts = ["using g2p"] g2p = G2p() for text in texts: out = g2p(text) print(out) ，在RANK()和date上进行分区，并按id降序排列：

value

输出：

SELECT *,
       RANK() OVER (PARTITION BY date, id ORDER BY value DESC) AS ranking
FROM data

Demo on SQLFiddle

此查询将在标记了问题的所有DBMS上运行。