[英]Count Number of Consecutive Occurrence of values in Table
我有下表
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 |
+------+------------+
我尝试过的最好的事情是:
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
一种方法是行号的差异:
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。
您可以使用LAG
和 running total 等窗口函数:
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;
第一个 cte 返回组号:
+-----+-------+-----+
| 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 |
+-------+-----+
我使用了递归 CTE 并尽量减少了 row_number 的使用,也避免了 count(*)。
我认为它会表现得更好,但在现实世界中,它取决于您放置的其他过滤器以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
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