sql - 将订单量四舍五入至限制总数

Question

昨天工作中遇到一个有趣的小问题。这是一个关于算术的问题，也是一个关于 SQL 的问题。假设您有很多订单，并且订单数量有限制(在本例中为全部 20 个):

if object_id('tempdb..#OMAX') is not null drop table #OMAX
create table #OMAX
    (
    OrderId int primary key,
    MaxVol decimal(15,3)
    )
insert into #OMAX(OrderId, MaxVol) values (1, 20), (2, 20), (3, 20)

以下是您的订单行项目及其当前、建议的数量:

if object_id('tempdb..#OLI') is not null drop table #OLI
create table #OLI
    (
    OrderId int,
    ProposedVolume decimal(15,3)
    )

insert into #OLI(OrderId, ProposedVolume)
values
    (1, 11.6),
    (1, 5.4),
    (2, 9.744),
    (2, 16.254),
    (2, 9.556),
    (3, 7.1),
    (3, 7.23),
    (3, 7.45)

您还希望将结果四舍五入到特定的精确度，暂时假设为 1.0(整数):

declare @nOrderRoundAmt decimal(15,3) = 1.0;

问题:对于当前总计大于 OMAX.MaxVol 的订单，您能否编写一条 SQL 语句来缩小 ProposeVolumes，以便订单行的新总计等于 MaxVol ？它必须等于，而不是小于(原因:这里的业务案例是订单 2 的建议总数量为 35.554，但我们说允许的最大数量是 20，因此当我们减少订单时，我们需要减少它到 20，不能少于 20，否则不合理)。

复杂性:一个订单可以有 1..N 个行项目。不要认为这是一组详尽的测试数据，我怀疑还有其他棘手的情况。

在这种情况下，除了四舍五入之外，阶数 1 应保持不变，阶数 2 和阶数 3 应减少并四舍五入为 20。

这是我迄今为止所做的最大努力:

; with OrderTotals as
    (
    select OrderId, sum(ProposedVolume) as TotalVolume
    from #OLI
    group by OrderId
    )
select
    OLI.*, 
    Ratio.Ratio,
    Scaled.Vol as SVol,
    ScaledAndRounded.Vol as SRVol
from
    #OLI OLI
    join OrderTotals OT on OLI.OrderId = OT.OrderId
    join #OMAX OMAX on OLI.OrderId = OMAX.OrderId
    cross apply
        (
        -- Don't reduce orders that are already below the max.
        select
            case when OMAX.MaxVol / OT.TotalVolume > 1 then 1
            else OMAX.MaxVol / OT.TotalVolume
            end as Ratio
        ) Ratio
    cross apply (select OLI.ProposedVolume * Ratio.Ratio as Vol) Scaled
    -- Rounds to nearest.
    cross apply (select round(Scaled.Vol / @nOrderRoundAmt, 0) * @nOrderRoundAmt as Vol) ScaledAndRounded
    -- Rounds down.
    -- cast(Scaled.Vol / @nOrderRoundAmt as bigint) * @nOrderRoundAmt as ScaledAndRoundedDown,

这说明了两个问题:订单 2 的总数为 19，订单 3 的总数为 21。您可以通过始终向下舍入来阻止订单 3 超过 20，但是您可能会遇到这样的情况:订单总数为 18。

那么这可以用一个语句来实现吗？到目前为止，我最好的解决方案是应用上述逻辑(使用向下舍入)，然后在游标中应用第二步处理以添加差异，直到我们回到 20 的总数。

您能证明您的解决方案适用于所有情况吗？

以下用于生成测试随机订单的代码可能有用:

declare @OrderId int = 0, @NumLineItems int;

while @OrderId < 1000 begin
    set @NumLineItems = cast(rand() * 5 as int) + 1

    insert into #OLI(OrderId, ProposedVolume)
    select top (@NumLineItems) @OrderId, rand(cast(newId() as varbinary)) * 15
    from sys.objects

    set @OrderId = @OrderId + 1
end

解决方案

如果有人对我根据戈登的答案制定的最终解决方案感兴趣，就在这里。它有点冗长，返回的列远多于实际需要的列，但这有助于调试/理解。尝试将舍入程度设置为 0.1 或 0.01。如果任何行项目的建议数量为 0，则该解决方案很容易出现被零除错误，但它们很容易被预先过滤掉。它还可以生成一些四舍五入为零的行项目，事后需要将其排除。

declare @nOrderRoundAmt decimal(15,3) = 0.1;  -- Degree of rounding required.
if object_id('tempdb..#Results') is not null drop table #Results

select
    T.*,
    row_number() over (partition by OrderId order by Remainder desc) as seqnum,
    case
        when NeedsAdjustment = 0 then ProposedVolumeRounded
        else
            (case when row_number() over (partition by OrderId order by Remainder desc) <= LeftOver
            then AppliedVolInt + 1
            else AppliedVolInt
            end)
    end * @nOrderRoundAmt as NewVolume
--into #Results
from
    (
    select
        T.*,
        floor(T.AppliedVol) as AppliedVolInt,
        (T.AppliedVol - 1.000 * floor(T.AppliedVol)) as Remainder,
        T.MaxVol * 1.0 - sum(floor(T.AppliedVol)) over (partition by T.OrderId) as LeftOver
    from
        (
        select
            OLI.OrderId,
            OMAX.MaxVol as OrigMaxVol,
            MaxVol.Vol as MaxVol,
            OLI.ProposedVolume as OrigProposedVolume,
            ProposedVolume.Vol as ProposedVolume,
            ProposedVolumeRounded.Vol as ProposedVolumeRounded,
            sum(ProposedVolume.Vol) over (partition by OLI.OrderId) as SumProposedVolume,
            sum(ProposedVolumeRounded.Vol) over (partition by OLI.OrderId) as SumProposedVolumeRounded, -- Round, THEN sum.
            case
                -- when SumProposedVolumeRounded > MaxVol, i.e. the sum of the rounded line items would be
                -- greater than the order limit, then scale, else take the original.
                when sum(ProposedVolumeRounded.Vol) over (partition by OLI.OrderId) > MaxVol.Vol then 1
                else 0
            end as NeedsAdjustment,
            case
                -- when SumProposedVolumeRounded > MaxVol, i.e. the sum of the rounded line items would be
                -- greater than the order limit, then scale, else take the original.
                when sum(ProposedVolumeRounded.Vol) over (partition by OLI.OrderId) > MaxVol.Vol then MaxVol.Vol * (ProposedVolume.Vol / sum(ProposedVolume.Vol) over (partition by OLI.OrderId))
                else ProposedVolume.Vol
            end as AppliedVol
        from
            ##OLI OLI
            join ##OMax OMAX on OLI.OrderId = OMAX.OrderId
            cross apply (select OLI.ProposedVolume / @nOrderRoundAmt as Vol) ProposedVolume
            cross apply (select OMAX.MaxVol / @nOrderRoundAmt as Vol) MaxVol
            cross apply (select round(ProposedVolume.Vol, 0) as Vol) ProposedVolumeRounded
        ) T
    ) T

Answer 1

这是一个分区问题，您试图使结果为整数(或等效地，整数的某个固定倍数)。策略是将所有内容都计算为整数，找到余数，然后将余数分配给各个项目。

以下是计算概述:

1. 以 float 形式计算订单中每个条目的新交易量
2. 将此体积中的整数部分与分数分开。
3. 计算最大体积减去整数比例之和。差异在于您必须弥补的金额。
4. 从最大到最小枚举分数。
5. 将最终金额计算为整数金额加上 1 或 0。当枚举小于或等于要补足的金额时，请使用 1。其他为 0。

以下 SQL 执行此操作:

select t.*, row_number() over (partition by orderid order by remainder desc) as seqnum,
       (case when row_number() over (partition by orderid order by remainder desc) <= LeftOver
             then AppliedVolInt + 1
             else AppliedVolInt
        end) as NewVolume
from (select t.*, floor(AppliedVol) as AppliedVolInt,
             (AppliedVol - 1.000*floor(AppliedVol)) as Remainder,
             maxvol*1.0 - sum(floor(AppliedVol)) over (partition by orderid) as LeftOver
      from (select oli.orderid, oli.ProposedVolume, omax.MaxVol,
                   sum(proposedVolume) over (partition by oli.orderid) as sumProposed,
                   omax.maxvol * (oli.ProposedVolume / sum(proposedVolume) over (partition by oli.orderid)) as AppliedVol
            from #OLI oli join
                 #OMax omax
                 on oli.orderid = omax.orderid
           ) t
     ) t

如果你没有整数，算术会稍微复杂一些(因为使用了从(4)到(5)的枚举。我的建议是将所有数字乘以一个常数，然后将其转换为整数问题或将(4)中的枚举乘以因子。

而且，是的，我已经在您的测试数据上对此进行了测试。它不仅在逻辑上有效，而且在实践中也有效。