Question

我有下表：

Row   Column   Type
1     1        =
1     2        =
1     3        O
1     4        =
1     5        =
1     6        O
2     1        =

我需要得到类似的东西

Row   Start_Column   End_Column   Type
1     1              2           =
1     3              3           O
1     4              5           =
1     6              6           O
2     1              1           =

我尝试对其进行分组，以使用 ROW_NUMBER、RANK 进行操作，但没有运气

有人知道如何做到这一点吗？

Answer 1

您可以使用 LAG() 和 SUM() 窗口函数来创建您想要的组然后聚合：

SELECT [Row], 
       MIN([Column]) Start_Column, 
       MAX([Column]) End_Column, 
       MAX([Type]) [Type]
FROM (
  SELECT *, SUM(flag) OVER (PARTITION BY [Row] ORDER BY [Column]) grp
  FROM (
    SELECT *, 
      CASE WHEN [Type] = LAG([Type]) OVER (PARTITION BY [Row] ORDER BY [Column]) THEN 0 ELSE 1 END flag
    FROM tablename
  ) t
) t
GROUP BY [Row], grp
ORDER BY [Row], grp

参见demo。
结果：

Row | Start_Column | End_Column | Type
--- | ------------ | ---------- | ----
  1 |            1 |          2 | =   
  1 |            3 |          3 | O   
  1 |            4 |          5 | =   
  1 |            6 |          6 | O   
  2 |            1 |          1 | =

Answer 2

这是一种间隙和孤岛问题。这种情况下，最简单的方法大概就是行号差了：

select row, type, min(column), max(column)
from (select t.*,
             row_number() over (partition by row, type order by column) as seqnum_2,
             row_number() over (partition by row order by column) as seqnum
      from t
     ) t
group by row, type, (seqnum - seqnum_2)
order by row, min(column);

如果 column 是连续的且没有间隙，您可以进一步简化：

select row, type, min(column), max(column)
from (select t.*,
             row_number() over (partition by row, type order by column) as seqnum_2
      from t
     ) t
group by row, type, (column - seqnum_2)
order by row, min(column);

为什么会这样？好吧，如果你从 column 中减去一个递增的序列，那么结果是常数——当类型相同时。

Here 是一个 db<>fiddle。

分组数据以创建范围列

2 个答案: