Google BigQuery SQL：独立排序两列

Question

假设我有一些数据，例如：

grp   v1   v2
---   --   --
 2    5    7
 2    4    9
 3    10   2
 3    11   1

我想创建独立于表顺序的新列 - 使两列具有独立的顺序，即按 v1 独立于 v2 排序，同时按 grp 分区。

结果（独立排序，按 grp 分区）将是：

grp   v1   v2  v1_ordered v2_ordered
---   --   --  ---------- ----------
 2    5    7       4          7
 2    4    9       5          9
 3    10   2      10          1
 3    11   1      11          2

一种方法是创建两个表并进行交叉连接。但是，我正在处理太多的数据行，以至于计算上难以处理 - 有没有办法在没有 JOIN 的单个查询中做到这一点？

基本上，我想编写如下 SQL：

SELECT
  *,
  v1 OVER (PARTITION BY grp ORDER BY v1 ASC) as v1_ordered,
  v2 OVER (PARTITION BY grp ORDER BY v2 ASC) as v2_ordered
FROM [example_table]

这打破了表格行的含义，但它是许多应用程序的必要功能 - 例如计算两个字段之间的有序相关性CORR(v1_ordered, v2_ordered).

这可能吗？

Answer 1

我认为你的方向是正确的！您只需要使用适当的窗口功能。 Row_number() 在这种情况下。它应该可以工作！

根据@cgn 请求添加工作示例：
我认为没有办法完全避免使用 JOIN。
同时，下面的示例在其他答案中仅使用 ONE JOIN 与 TWO JOIN：

SELECT 
  a.grp AS grp, 
  a.v1 AS v1, 
  a.v2 AS v2, 
  a.v1 AS v1_ordered, 
  b.v2 AS v2_ordered 
FROM (
  SELECT grp, v1, v2, ROW_NUMBER() OVER(PARTITION BY grp ORDER BY v1) AS v1_order
  FROM [example_table]
) AS a
JOIN (
  SELECT grp, v1, v2, ROW_NUMBER() OVER(PARTITION BY grp ORDER BY v2) AS v2_order
  FROM [example_table]
) AS b
ON a.grp = b.grp AND a.v1_order = b.v2_order 

结果如预期：

grp v1  v2  v1_ordered  v2_ordered   
2    4   9           4           7   
2    5   7           5           9   
3   10   2          10           1   
3   11   1          11           2   

现在您可以使用如下所示的 CORR()

SELECT grp, CORR(v1_ordered, v2_ordered) AS [corr]
FROM (
  SELECT 
    a.grp AS grp, 
    a.v1 AS v1, 
    a.v2 AS v2, 
    a.v1 AS v1_ordered, 
    b.v2 AS v2_ordered 
  FROM (
    SELECT grp, v1, v2, ROW_NUMBER() OVER(PARTITION BY grp ORDER BY v1) AS v1_order
    FROM [example_table]
  ) AS a
  JOIN (
    SELECT grp, v1, v2, ROW_NUMBER() OVER(PARTITION BY grp ORDER BY v2) AS v2_order
    FROM [example_table]
  ) AS b
  ON a.grp = b.grp AND a.v1_order = b.v2_order
)
GROUP BY grp

Answer 2

这对你有用。

SQLFiddle Demo in SQL Server

注意：您在示例中提到的序列，对于如何从数据库返回行不是必需的。在我的情况下，对于v1，我得到了4,5,10,11，不像你的5,4,10,11。但是，您的输出将与您想要的相同。

Select t.grp,t.v1,t.v2,
v1.v1 as v1_ordered,v2.v2 as v2_ordered
From
(
    select t1.*,
    row_number() over (partition by grp
                   Order by v1) v1o
    ,
    row_number() over (partition by grp
                   Order by v2) v2o
    from table1 t1
) t
Inner join
(
    Select t.*,
    row_number() over (partition by grp
                   Order by v1) v1o
    From table1 t
) v1
On t.grp=v1.grp
And t.v1o=v1.v1o
Inner join
(
    Select t.*,
    row_number() over (partition by grp
                   Order by v2) v2o
    From table1 t
) v2
On t.grp=v2.grp
And t.v1o=v2.v2o

输出：

+------+-----+-----+-------------+------------+
| grp  | v1  | v2  | v1_ordered  | v2_ordered |
+------+-----+-----+-------------+------------+
|   2  |  4  |  9  |          4  |          7 |
|   2  |  5  |  7  |          5  |          9 |
|   3  | 10  |  2  |         10  |          1 |
|   3  | 11  |  1  |         11  |          2 |
+------+-----+-----+-------------+------------+

Answer 3

AI 不能 100% 确定这在 BigQuery 中是否有效，但情况如下：

select e.*, ev1.v1, ev2.v2
from (select e.*,
             row_number() over (partition by grp order by v1) as seqnum_v1,
             row_number() over (partition by grp order by v2) as seqnum_v2
      from example e
     ) e join
     (select e.*, row_number() over (partition by grp order by v1) as seqnum_v1
      from example e
     ) ev1
     on ev1.grp = e.grp and ev1.seqnum_v1 = e.seqnum_v1 join
     (select e.*, row_number() over (partition by grp order by v2) as seqnum_v2
      from example e
     ) ev2
     on ev2.grp = e.grp and ev2.seqnum_v2 = e.seqnum_v2;

这个想法是为每一列分配一个独立的顺序。然后连接回原始表以获取实际值。