根据累计值排名

Question

我想根据 UID 的升序对 ID 和值列进行排名。一旦值列的值与先前的值不同，预期输出就必须更改。排名必须在每个新 ID 上重新开始

UID ID  Value Expected Output
1   1   0         1
2   1   0         1
3   1   1         2
4   1   1         2
5   1   1         2
6   1   0         3
7   1   1         4
8   1   0         5
9   1   0         5
10  1   0         5
11  2   1         1
12  2   1         1
13  2   0         2
14  2   0         2
15  2   1         3

这是我创建的示例数据集：

CREATE TABLE [dbo].[Data] (
    [UID] [int] NOT NULL,
    [ID] [int] NULL,
    [Value] [int] NULL
);

INSERT [dbo].[Data] ([UID], [ID], [Value]) VALUES (1, 1, 0);
INSERT [dbo].[Data] ([UID], [ID], [Value]) VALUES (2, 1, 0);
INSERT [dbo].[Data] ([UID], [ID], [Value]) VALUES (3, 1, 1);
INSERT [dbo].[Data] ([UID], [ID], [Value]) VALUES (4, 1, 1);
INSERT [dbo].[Data] ([UID], [ID], [Value]) VALUES (5, 1, 1);
INSERT [dbo].[Data] ([UID], [ID], [Value]) VALUES (6, 1, 0);
INSERT [dbo].[Data] ([UID], [ID], [Value]) VALUES (7, 1, 1);
INSERT [dbo].[Data] ([UID], [ID], [Value]) VALUES (8, 1, 0);
INSERT [dbo].[Data] ([UID], [ID], [Value]) VALUES (9, 1, 0);
INSERT [dbo].[Data] ([UID], [ID], [Value]) VALUES (10, 1, 0);
INSERT [dbo].[Data] ([UID], [ID], [Value]) VALUES (11, 2, 1);
INSERT [dbo].[Data] ([UID], [ID], [Value]) VALUES (12, 2, 1);
INSERT [dbo].[Data] ([UID], [ID], [Value]) VALUES (13, 2, 0);
INSERT [dbo].[Data] ([UID], [ID], [Value]) VALUES (14, 2, 0);
INSERT [dbo].[Data] ([UID], [ID], [Value]) VALUES (15, 2, 1);

Answer 1

我认为解决此 gaps-and-islands 问题的最简单方法是使用 lag() 检索“先前”值，然后每次值更改时递增的 window 总和.

select uid, id, value,
    1 + sum(case when value <> lag_value then 1 else 0 end) 
        over(partition by id order by uid) grp
from (
    select d.*, lag(value, 1, value) over(partition by id order by uid) lag_value
    from data d
) d
order by uid

Demo on DB Fiddle:

uid | id | value | grp
--: | -: | ----: | --:
  1 |  1 |     0 |   1
  2 |  1 |     0 |   1
  3 |  1 |     1 |   2
  4 |  1 |     1 |   2
  5 |  1 |     1 |   2
  6 |  1 |     0 |   3
  7 |  1 |     1 |   4
  8 |  1 |     0 |   5
  9 |  1 |     0 |   5
 10 |  1 |     0 |   5
 11 |  2 |     1 |   1
 12 |  2 |     1 |   1
 13 |  2 |     0 |   2
 14 |  2 |     0 |   2
 15 |  2 |     1 |   3

Answer 2

这是一个缺口和孤岛问题。我认为最简单的方法是使用行号差异方法：

WITH cte AS (
    SELECT *, ROW_NUMBER() OVER (PARTITION BY ID ORDER BY UID) rn1,
        ROW_NUMBER() OVER (PARTITION BY ID, [Value] ORDER BY UID) rn2
    FROM Data
)

SELECT *, DENSE_RANK() OVER (PARTITION BY ID ORDER BY rn1 - rn2, [Value]) AS output
FROM cte
ORDER BY UID;

根据累计值排名

Rank based on cumulative value

sql

sql-server

rank

window-functions

gaps-and-islands

Demo