计算值在 Table 中连续出现的次数

Question

我有以下table

create table #t (Id int, Name char)

insert into #t values
(1, 'A'),
(2, 'A'),
(3, 'B'),
(4, 'B'),
(5, 'B'),
(6, 'B'),
(7, 'C'),
(8, 'B'),
(9, 'B')

我想计算名称列中的连续值

+------+------------+
| Name | Repetition |
+------+------------+
| A    |          2 |
| B    |          4 |
| C    |          1 |
| B    |          2 |
+------+------------+

我尝试过的最好的方法是：

select Name
, COUNT(*) over (partition by Name order by Id) AS Repetition
from #t
order by Id

但它没有给我预期的结果

Answer 1

一种方法是行号的不同：

select name, count(*) 
from (select t.*,
             (row_number() over (order by id) -
              row_number() over (partition by name order by id)
             ) as grp
      from t
     ) t
group by grp, name;

如果您运行子查询并分别查看每个行号的值，然后查看差异，则逻辑最容易理解。

Answer 2

我已经使用递归 CTE 并尽量减少 row_number 的使用，同时避免使用 count(*)。

我认为它会表现得更好，但在现实世界中，这取决于您放置了什么过滤器来最小化受影响的行数。

如果 ID 具有离散值，则将使用一个额外的 CTE 来生成连续的 ID。

 ;With CTE2 as
(
select ROW_NUMBER()over(order by id) id, name,1 Repetition ,1 Marker  from @t
)
, CTE as
(
select top 1 cast(id as int) id, name,1 Repetition ,1 Marker  from CTE2 order by id

union all

select a.id, a.name
, case when a.name=c.name then Repetition +1 else 1 end  
, case when a.name=c.name then c.Marker else  Marker+1 end
from @t a
inner join CTE c on a.id=c.id+1

)
,CTE1 as
(select *,ROW_NUMBER()over(partition by marker order by id desc)rn from cte c
)
select Name,Repetition from cte1 where rn=1

Answer 3

您可以使用 LAG 和运行等窗口函数总计：

WITH cte AS (
 SELECT Id, Name, grp = SUM(CASE WHEN Name = prev THEN 0 ELSE 1 END) OVER(ORDER BY id)
 FROM (SELECT *, prev = LAG(Name) OVER(ORDER BY id) FROM t) s
)
SELECT name, cnt = COUNT(*)
FROM cte
GROUP BY grp,name
ORDER BY grp;

db<>fiddle demo

第一个ctereturns群号：

+-----+-------+-----+
| Id  | Name  | grp |
+-----+-------+-----+
|  1  | A     |   1 |
|  2  | A     |   1 |
|  3  | B     |   2 |
|  4  | B     |   2 |
|  5  | B     |   2 |
|  6  | B     |   2 |
|  7  | C     |   3 |
|  8  | B     |   4 |
|  9  | B     |   4 |
+-----+-------+-----+

主查询根据之前计算的 grp 列对其进行分组：

+-------+-----+
| name  | cnt |
+-------+-----+
| A     |   2 |
| B     |   4 |
| C     |   1 |
| B     |   2 |
+-------+-----+

计算值在 Table 中连续出现的次数

Count Number of Consecutive Occurrence of values in Table

sql

tsql

sql-server

aggregation

sql-server-2012