我目前正在使用此查询(在 SQL Server 中(来计算每天唯一项目的数量：
SELECT Date, COUNT(DISTINCT item)
FROM myTable
GROUP BY Date
ORDER BY Date
如何转换它以获取每个日期过去 3 天(包括当天(的唯一项目数？
输出应为包含 2 列的表：一列包含原始表中的所有日期。在第二列中，我们有每个日期的唯一项目数。
例如，如果原始表是：
Date Item
01/01/2018 A
01/01/2018 B
02/01/2018 C
03/01/2018 C
04/01/2018 C
通过上面的查询，我目前可以获得每天的唯一计数：
Date count
01/01/2018 2
02/01/2018 1
03/01/2018 1
04/01/2018 1
我希望获得 3 天滚动窗口的唯一计数：
Date count
01/01/2018 2
02/01/2018 3 (because items ABC on 1st and 2nd Jan)
03/01/2018 3 (because items ABC on 1st,2nd,3rd Jan)
04/01/2018 1 (because only item C on 2nd,3rd,4th Jan)
使用apply提供了一种形成滑动窗口的便捷方法
CREATE TABLE myTable
([DateCol] datetime, [Item] varchar(1))
;
INSERT INTO myTable
([DateCol], [Item])
VALUES
('2018-01-01 00:00:00', 'A'),
('2018-01-01 00:00:00', 'B'),
('2018-01-02 00:00:00', 'C'),
('2018-01-03 00:00:00', 'C'),
('2018-01-04 00:00:00', 'C')
;
CREATE NONCLUSTERED INDEX IX_DateCol
ON MyTable([Date])
;
查询：
select distinct
t1.dateCol
, oa.ItemCount
from myTable t1
outer apply (
select count(distinct t2.item) as ItemCount
from myTable t2
where t2.DateCol between dateadd(day,-2,t1.DateCol) and t1.DateCol
) oa
order by t1.dateCol ASC
结果：
| dateCol | ItemCount |
|----------------------|-----------|
| 2018-01-01T00:00:00Z | 2 |
| 2018-01-02T00:00:00Z | 3 |
| 2018-01-03T00:00:00Z | 3 |
| 2018-01-04T00:00:00Z | 1 |
通过使用apply之前减少date列可能会有一些性能提升，如下所示：
select
d.date
, oa.ItemCount
from (
select distinct t1.date
from myTable t1
) d
outer apply (
select count(distinct t2.item) as ItemCount
from myTable t2
where t2.Date between dateadd(day,-2,d.Date) and d.Date
) oa
order by d.date ASC
;
您可以在该子查询中使用group by而不是使用select distinct，但执行计划将保持不变。
SQL Fiddle 上的演示
最直接的解决方案是根据日期将表与自身联接：
SELECT t1.DateCol, COUNT(DISTINCT t2.Item) AS C
FROM testdata AS t1
LEFT JOIN testdata AS t2 ON t2.DateCol BETWEEN DATEADD(dd, -2, t1.DateCol) AND t1.DateCol
GROUP BY t1.DateCol
ORDER BY t1.DateCol
输出：
| DateCol | C |
|-------------------------|---|
| 2018-01-01 00:00:00.000 | 2 |
| 2018-01-02 00:00:00.000 | 3 |
| 2018-01-03 00:00:00.000 | 3 |
| 2018-01-04 00:00:00.000 | 1 |
GROUP BY应该比DISTINCT更快(确保在Date列上有一个索引(
DECLARE @tbl TABLE([Date] DATE, [Item] VARCHAR(100))
;
INSERT INTO @tbl VALUES
('2018-01-01 00:00:00', 'A'),
('2018-01-01 00:00:00', 'B'),
('2018-01-02 00:00:00', 'C'),
('2018-01-03 00:00:00', 'C'),
('2018-01-04 00:00:00', 'C');
SELECT t.[Date]
--Just for control. You can take this part away
,(SELECT DISTINCT t2.[Item] AS [*]
FROM @tbl AS t2
WHERE t2.[Date]<=t.[Date]
AND t2.[Date]>=DATEADD(DAY,-2,t.[Date]) FOR XML PATH('')) AS CountedItems
--This sub-select comes back with your counts
,(SELECT COUNT(DISTINCT t2.[Item])
FROM @tbl AS t2
WHERE t2.[Date]<=t.[Date]
AND t2.[Date]>=DATEADD(DAY,-2,t.[Date])) AS ItemCount
FROM @tbl AS t
GROUP BY t.[Date];
结果
Date CountedItems ItemCount
2018-01-01 AB 2
2018-01-02 ABC 3
2018-01-03 ABC 3
2018-01-04 C 1
此解决方案与其他解决方案不同。您能否与其他答案进行比较来检查此查询在真实数据上的性能？
基本思想是，每一行都可以参与其自己的日期、后天或后天的窗口。因此，这首先将行扩展为三行，并附加了这些不同的日期，然后它可以使用常规COUNT(DISTINCT)聚合计算的日期。HAVING 子句只是为了避免返回仅计算且未存在于基本数据中的日期的结果。
with cte(Date, Item) as (
select cast(a as datetime), b
from (values
('01/01/2018','A')
,('01/01/2018','B')
,('02/01/2018','C')
,('03/01/2018','C')
,('04/01/2018','C')) t(a,b)
)
select
[Date] = dateadd(dd, n, Date), [Count] = count(distinct Item)
from
cte
cross join (values (0),(1),(2)) t(n)
group by dateadd(dd, n, Date)
having max(iif(n = 0, 1, 0)) = 1
option (force order)
输出：
| Date | Count |
|-------------------------|-------|
| 2018-01-01 00:00:00.000 | 2 |
| 2018-01-02 00:00:00.000 | 3 |
| 2018-01-03 00:00:00.000 | 3 |
| 2018-01-04 00:00:00.000 | 1 |
如果有许多重复的行，则可能会更快：
select
[Date] = dateadd(dd, n, Date), [Count] = count(distinct Item)
from
(select distinct Date, Item from cte) c
cross join (values (0),(1),(2)) t(n)
group by dateadd(dd, n, Date)
having max(iif(n = 0, 1, 0)) = 1
option (force order)
使用GETDATE()函数获取当前日期，DATEADD()获取最近 3 天
SELECT Date, count(DISTINCT item)
FROM myTable
WHERE [Date] >= DATEADD(day,-3, GETDATE())
GROUP BY Date
ORDER BY Date
SQL
SELECT DISTINCT Date,
(SELECT COUNT(DISTINCT item)
FROM myTable t2
WHERE t2.Date BETWEEN DATEADD(day, -2, t1.Date) AND t1.Date) AS count
FROM myTable t1
ORDER BY Date;
演示
Rextester 演示：http://rextester.com/ZRDQ22190
由于不支持COUNT(DISTINCT item) OVER (PARTITION BY [Date])，您可以使用dense_rank来模拟：
SELECT Date, dense_rank() over (partition by [Date] order by [item])
+ dense_rank() over (partition by [Date] order by [item] desc)
- 1 as count_distinct_item
FROM myTable
需要注意的一件事是，dense_rank将计为空，而COUNT则不会。
有关更多详细信息，请参阅此帖子。
这是一个简单的解决方案，它使用 myTable 本身作为分组日期的来源(为 SQLServer dateadd 编辑(。请注意，此查询假定每个日期在 myTable 中至少会有一条记录;如果缺少任何日期，则即使有前 2 天的记录，它也不会显示在查询结果中：
select
date,
(select
count(distinct item)
from (select distinct date, item from myTable) as d2
where
d2.date between dateadd(day,-2,d.date) and d.date
) as count
from (select distinct date from myTable) as d
我用数学解决了这个问题。
Z(任何一天(= 3x + y(y 是模式 3 值(我需要从 3
* (x - 1( + y + 1 到 3 * (x - 1( + y + 33 * (x- 1( + y + 1= 3* (z/3 - 1( + z % 3 + 1
在这种情况下;我可以使用分组依据(在 3* (z/3 - 1( + z % 3 + 1 和 z 之间(
SELECT iif(OrderDate between 3 * (cast(OrderDate as int) / 3 - 1) + (cast(OrderDate as int) % 3) + 1
and orderdate, Orderdate, 0)
, count(sh.SalesOrderID) FROM Sales.SalesOrderDetail shd
JOIN Sales.SalesOrderHeader sh on sh.SalesOrderID = shd.SalesOrderID
group by iif(OrderDate between 3 * (cast(OrderDate as int) / 3 - 1) + (cast(OrderDate as int) % 3) + 1
and orderdate, Orderdate, 0)
order by iif(OrderDate between 3 * (cast(OrderDate as int) / 3 - 1) + (cast(OrderDate as int) % 3) + 1
and orderdate, Orderdate, 0)
如果您需要其他日组，您可以使用;
declare @n int = 4 (another day count)
SELECT iif(OrderDate between @n * (cast(OrderDate as int) / @n - 1) + (cast(OrderDate as int) % @n) + 1
and orderdate, Orderdate, 0)
, count(sh.SalesOrderID) FROM Sales.SalesOrderDetail shd
JOIN Sales.SalesOrderHeader sh on sh.SalesOrderID = shd.SalesOrderID
group by iif(OrderDate between @n * (cast(OrderDate as int) / @n - 1) + (cast(OrderDate as int) % @n) + 1
and orderdate, Orderdate, 0)
order by iif(OrderDate between @n * (cast(OrderDate as int) / @n - 1) + (cast(OrderDate as int) % @n) + 1
and orderdate, Orderdate, 0)