Question

我有一个抵达日期01/01/2010，这已经发生了50次，我想使用下面的停留时间加权指南将50个出发日期随机化，因为你可以在2天后将大部分离开，但是我无法弄清楚如何编写代码，你能帮忙。

LengthofStay LengthofStayWeighting
------------ ---------------------
1            1
2            5
3            4
4            3
5            3
6            3
7            3
8            1
9            1
10           1

我已经开始但已经卡住了

SELECT ArrivalDate,RAND(checksum(NEWID())) * LengthOfStay.LengthofStayWeighting AS Expr1, 
ArrivalDate + Expr1 as DepartureDate

FROM Bookings, LengthOfStay
ORDER BY ArrivalDate

Answer 1

您可能需要使用DATEADD

SELECT ArrivalDate, DATEADD(day, RAND(checksum(NEWID())) * LengthOfStay.LengthofStayWeighting, ArrivalDate) AS DepartureDate    
FROM Bookings, LengthOfStay
ORDER BY ArrivalDate

更新：根据您的评论，我认为我误解了这个问题。这是你需要的吗？：

SELECT ArrivalDate, 
DATEADD(day, (select TOP 1 LengthofStayWeighting FROM LengthOfStay group by LengthofStayWeighting ORDER BY LengthofStayWeighting DESC), ArrivalDate) AS DepartureDate    
    FROM Bookings
    ORDER BY ArrivalDate

基本上你需要获得最重复的长度，在你的情况下为“1”。如果是这样，我认为你需要包括一个外国钥匙..

SELECT ArrivalDate, 
DATEADD(day, (select TOP 1 LengthofStayWeighting FROM LengthOfStay l WHERE b.Id = l.BookingId GROUP BY LengthofStayWeighting  ORDER BY LengthofStayWeighting DESC), ArrivalDate) AS DepartureDate    
    FROM Bookings b
    ORDER BY ArrivalDate

Answer 2

您正试图从累积分布中提取数字。这需要生成一个随机数，然后从分布中拉出来。

以下代码给出了一个示例：

with LengthOfStay as (select 1 as LengthOfStay, 1 as LengthOfStayWeighting union all
                   select 2 as LengthOfStay, 5 union all
                   select 3, 4 union all
                   select 4, 4
                  ),
     Bookings as (select cast('2013-01-01' as DATETIME) as ArrivalDate),
     CumeLengthOfStay as
         (select los.*,
                 (select SUM(LengthOfStayWeighting) from LengthOfStay los2 where los2.LengthOfStay <= los.LengthOfStay
                 ) as cumeweighting
          from LengthOfStay los
         ) -- select * from CumeLengthOfStay
SELECT ArrivalDate, clos.LengthOfStay, randnum % sumweighting, sumweighting,
       ArrivalDate + clos.LengthOfStay as DepartureDate
FROM (select b.*, ABS(CAST(NEWID() AS binary(6))+0) as randnum
      from Bookings b
     ) b cross join
     (select SUM(LengthOfStayWeighting) as sumweighting from LengthOfStay) const left outer join
     CumeLengthOfStay clos
     on (b.randnum % const.sumweighting) between clos.cumeweighting - clos.LengthOfStayWeighting  and clos.cumeweighting - 1
ORDER BY ArrivalDate

基本上，您将权重相加，生成小于最高权重的随机数（使用%运算符），然后在权重的累积总和中查找此值。

加权一段时间以获得每次不同的日期

2 个答案: