用 NULL 替换重复列值的优化解决方案

Question

我有一张表，其结构和数据如下：

Time	Company	AttriA	AttriB	AttriC
12/4/2020 13:21	TestA	12	1	2
12/4/2020 13:21	TestB	1	2	44
12/4/2020 13:22	TestB	34	3	44
12/4/2020 13:22	TestA	12	4	5
12/4/2020 13:21	TestC	15	5	4

我想要达到的目标如下：

• 如果对于特定公司，在 AttriX（X = A、B 或 C）列中有 连续、及时的重复值，那么我想用 NULL 替换这些重复值。
• 数据量很大，大约几百万，查询也必须是高性能的。

按照上述逻辑，我尝试创建的视图应返回如下数据：

Time	Company	AttriA	AttriB	AttriC
12/4/2020 13:21	TestA	NULL	1	2
12/4/2020 13:21	TestB	1	2	NULL
12/4/2020 13:22	TestB	34	3	NULL
12/4/2020 13:22	TestA	NULL	4	5
12/4/2020 13:21	TestC	15	5	4

我遇到了这个问题。谁能帮帮我？

Answer 1

您可以使用lead() 和lag()：

select t.*,
       (case when lead(attria) over (partition by company order by time) = attria or
                  lag(attria) over (partition by company order by time) = attria
             then null
             else attria
        end),
       (case when lead(attrib) over (partition by company order by time) = attrib or
                  lag(attrib) over (partition by company order by time) = attrib
             then null
             else attrib
        end),
       (case when lead(attric) over (partition by company order by time) = attric or
                  lag(attric) over (partition by company order by time) = attric
             then null
             else attric
        end)
from t