在表的每个分区中标记行的前90％

Question

我有一张下表。我想派生一列Flag，以便每个分区的前90％的行将有TypeA，其余的10％ 行中将有TypeB作为标志。

+------+----+
| City | id |
+------+----+
| A    | 1A |
| A    | 2A |
| A    | 3A |
| A    | 4A |
| A    | 5A |
| B    | 1B |
| B    | 2B |
| B    | 3B |
| B    | 4B |
| B    | 5B |
| B    | 6B |
| D    | 1D |
| D    | 2D |
| D    | 3D |
| D    | 4D |
| D    | 5D |
| D    | 6D |
| D    | 7D |
| D    | 8D |
+------+----+

所需结果

+------+----+-------+
| City | id | Flag  |
+------+----+-------+
| A    | 1A | TypeA |
| A    | 2A | TypeA |
| A    | 3A | TypeA |
| A    | 4A | TypeA | // Approximately Top 90% of rows for City A: Flag Type A
| A    | 5A | TypeB | // Approximately below 10% of rows for City A: Flag Type B
| B    | 1B | TypeA |
| B    | 2B | TypeA |
| B    | 3B | TypeA |
| B    | 4B | TypeA |// Approximately Top 90% of rows for City B: Flag Type A
| B    | 5B | TypeB |// Approximately below 10% of rows for City B: Flag Type B
| B    | 6B | TypeB |
| D    | 1D | TypeA |
| D    | 2D | TypeA |
| D    | 3D | TypeA |
| D    | 4D | TypeA |
| D    | 5D | TypeA |
| D    | 6D | TypeA |
| D    | 7D | TypeA |
| D    | 8D | TypeB |
+------+----+-------+

任何帮助将不胜感激。

SQL Fiddle

Answer 1

一种方法是进行显式计数：

select t.*,
       (case when row_number() over (partition by city order by id) <=
                  0.9 * count(*) over (partition by city)
             then 'TypeA'
             else 'TypeB'
        end) as flag
from t

Answer 2

这是一个使用COUNT作为分析函数的选项：

SELECT
    City,
    id,
    CASE WHEN COUNT(*) OVER (PARTITION BY City ORDER BY id) /
        COUNT(*) OVER (PARTITION BY City) <= 0.9
         THEN 'TypeA'
         ELSE 'TypeB' END AS Flag
FROM yourTable
ORDER BY
    City,
    Id;

Demo

对COUNT的第一次调用将按照Id的顺序计算每个城市分区中直到当前行的元素数量。然后，我们通过每个城市的记录总数对其进行归一化，然后将其与0.9进行比较，以确定要分配的标志。

Answer 3

SQL Server具有percent_rank()窗口函数，可直接计算所需的数字，而无需自己执行：

SELECT City, id
     , CASE
        WHEN percent_rank() OVER (PARTITION BY City ORDER BY id) <= 0.9 THEN 'TypeA'
        ELSE 'TypeB'
       END AS Flag
FROM table1
ORDER BY City, id;

Fiddle。