如何在SQL Server中的滑動窗口上聚合（計算不同的項目）？

Question

我目前正在使用此查詢（在SQL Server中）每天計算唯一項目的數量：

SELECT Date, COUNT(DISTINCT item) 
FROM myTable 
GROUP BY Date 
ORDER BY Date

如何對此進行轉換以獲取過去3天 （包括當天）中每個日期的唯一商品數量 ？

輸出應該是一個包含2列的表：一列包含原始表中的所有日期。 在第二列，我們有每個日期的唯一項目數。

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

根據我上面的查詢，我目前獲得每天的唯一計數：

Date        count  
01/01/2018  2  
02/01/2018  1  
03/01/2018  1  
04/01/2018  1

我希望得到3天滾動窗口的獨特計數：

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)    

Answer 1

使用apply提供了一種形成滑動窗口的便捷方式

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])  
;    

查詢 ：

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

結果 ：

|              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 |

通過在使用apply之前減少date列可能會有一些性能提升，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
;

您可以使用group by而不是在子查詢中使用select distinct ，但執行計划將保持不變。

在SQL Fiddle演示

Answer 2

最直接的解決方案是根據日期加入表格：

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

輸出：

| 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 |

Answer 3

GROUP BY應該比DISTINCT快（確保在Date列上有索引）

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];

結果

Date        CountedItems    ItemCount
2018-01-01  AB              2
2018-01-02  ABC             3
2018-01-03  ABC             3
2018-01-04  C               1

Answer 4

該解決方案與其他解決方案不同。 你可以通過與其他答案的比較來檢查這個查詢在真實數據上的表現嗎？

基本思想是每行可以在其自己的日期，后一天或后一天參與窗口。 因此，首先將行擴展為三行，並附加不同的日期，然后它可以在計算日期使用常規COUNT(DISTINCT)聚合。 HAVING子句只是為了避免返回單獨計算並且不存在於基礎數據中的日期的結果。

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)

輸出：

|        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   |

如果您有許多重復行可能會更快：

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)

Answer 5

使用GETDATE()函數獲取當前日期，使用DATEADD()獲取最近3天

 SELECT Date, count(DISTINCT item) 
 FROM myTable 
 WHERE [Date] >= DATEADD(day,-3, GETDATE())
 GROUP BY Date 
 ORDER BY Date

Answer 6

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

Answer 7

由於不支持COUNT(DISTINCT item) OVER (PARTITION BY [Date])您可以使用dense_rank來模擬：

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 

需要注意的一點是， dense_rank將計為null，而COUNT則不計算。

請參閱此職位的更多細節。

Answer 8

這是一個簡單的解決方案，它使用myTable本身作為分組日期的來源（為SQLServer dateadd編輯）。 請注意，此查詢假定myTable中每個日期至少會有一條記錄; 如果沒有任何日期，即使前兩天有記錄，它也不會出現在查詢結果中：

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

Answer 9

我用Math解決了這個問題。

z（任何一天）= 3x + y（y是模式3值）我需要從3 *（x - 1）+ y + 1到3 *（x - 1）+ y + 3

3 *（x-1）+ y + 1 = 3 *（z / 3-1）+ z％3 + 1

在這種情況下; 我可以使用group by（在3 *（z / 3 - 1）+ z％3 + 1和z之間）

    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)

如果你需要其他日組，你可以使用;

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)