SQL中的组合优化匹配

Question

我在SQL中开发匹配算法时遇到问题.我有一张桌子subjects.这些中的每一个都需要与表中相同数量的行匹配controls(为了这个问题,我们需要为每个主题选择两行或控件).所选控件的位置必须完全匹配,并且所选控件的值应match_field尽可能接近主题.

以下是一些示例数据:

表科目:

id   location    match_field
1    1           190
2    2           2000
3    1           100

表格控件:

id   location    match_field
17    1          70
11    1          180
12    1          220
13    1          240
14    1          500
15    1          600
16    1          600
10    2          30
78    2          1840
79    2          2250

以下是样本数据的最佳结果:

subject_id control_id  location    match_field_diff
1          12          1           30
1          13          1           50
2          78          2           160
2          79          2           250
3          17          1           30
3          11          1           80

它变得棘手,因为例如,控制11是与对象1最接近的匹配.然而,在最佳解决方案中,控制11与对象3匹配.

我相信匈牙利算法接近这个问题的"正确"解决方案.然而,没有相同数量的受试者和对照,也不会使用所有对照(我有几千个受试者和几百万个潜在的对照).

没有必要获得绝对最佳结果; 一个非常好的近似对我来说没问题.

似乎应该有一个很好的基于集合的解决方案来解决这个问题,但我想不出怎么做.以下是一些代码,仅根据位置为每个主题分配相同数量的控件:

select * from (
    select   subject.id, 
             control.id,
             subject.location,
             row_number() over (
                 partition by subject.location
                 order by subject.id, control.id
             ) as rn,
             count(distinct control.id)     over (
                 partition by subject.location
             ) as controls_in_loc
         from subjects
         join controls on control.location = subject.location
    )
    where mod(rn,controls_in_loc+1) = 1

但是,我无法弄清楚如何添加模糊匹配组件.我正在使用DB2,但如果您使用其他东西,可以将算法转换为DB2.

在此先感谢您的帮助!

更新:我基本上确信SQL不适合这项工作.但是,为了确保(并且因为它是一个有趣的问题),我提供了一个赏金,看看是否可以使用有效的SQL解决方案.它需要是一个基于集合的解决方案.它可以使用迭代(多次循环同一个查询以实现结果),但迭代次数需要远远小于大型表的行数.它不应该循环遍历表中的每个元素或使用游标.

Answer 1

虽然匈牙利算法可以运行,但在您的情况下可以使用更简单的算法.隐式成本矩阵是特殊形式的对称矩阵:

ABS(SUBJ.match_field-CTRL.match_field)

因此,可以相对容易地证明,在最佳的分配{SUBJ _我,CTRL _Ĵ }通过有序SUBJ.match_field的值CTRL.match_field将被责令为好.

证明:考虑一个任务 {SUBJ _我,CTRL _Ĵ }通过命令SUBJ.match_field不是由命令CTRL.match_field.然后你至少有一个反转,即一对赋值 {SUBJ _i1,CTRL _j1 }和 {SUBJ _i2,CTRL _j2 }这样

SUBJ.match_field_i1 < SUBJ.match_field_i2,但是

CTRL.match_field_j1 > CTRL.match_field_j2

然后你可以用一个非反转的对替换倒置的对

{SUBJ _i1,CTRL _j2 }和{SUBJ _i2,CTRL _j1 }

成本小于或等于SUBJ.match_field(_i1,_i2)和CTRL.match_field(_j1,_j2)的所有六个相对位置的反向分配的成本(链接到Wolfram Alpha).:证明

有了这个观察结果,很容易证明下面的动态编程算法提出了最佳分配:

• 制作N每个主题的副本; 订购match_field
• 订单控制 match_field
• 准备一个空数组assignments大小的N * subject.SIZE
• 准备一个空的2D阵列mem尺寸的N * subject.SIZE由control.SIZE用于记忆化 ; 将所有元素设置为-1
• 调用Recursive_Assign以下伪代码定义
• 该assignments表现在包含N每个主题i在N*i(包括)和N*(i+1)独占之间的位置的分配.

FUNCTION Recursive_Assign
    // subjects contains each original subj repeated N times
    PARAM subjects : array of int[subjectLength]
    PARAM controls: array of int[controlLength]
    PARAM mem : array of int[subjectLength,controlLength]
    PARAM sp : int // current subject position
    PARAM cp : int // current control position
    PARAM assign : array of int[subjectLength]
BEGIN
    IF sp == subjects.Length THEN RETURN 0 ENDIF
    IF mem[sp, cp] > 0 THEN RETURN mem[sp, cp] ENDIF
    int res = ABS(subjects[sp] - controls[cp])
            + Recursive_Assign(subjects, controls, mem, sp + 1, cp + 1, assign)
    assign[sp] = cp
    IF cp+1+subjects.Length-sp < controls.Length THEN
        int alt = Recursive_Assign(subjects, controls, mem, sp, cp + 1, assign)
        IF alt < res THEN
            res = alt
        ELSE
            assign[sp] = cp
        ENDIF
    ENDIF
    RETURN (mem[sp, cp] = res)
END

这是在ideone上使用C#的上述伪代码的实现.

该算法已准备好在SQL中以基于集合的方式重写.试图将其纳入原始问题设置(按位置分组并制作主题的多个副本)会给已经相当复杂的过程增加不必要的复杂层,因此我将通过使用表来简化相当多的事情SQL Server的评估参数.我不确定DB2是否提供类似的功能,但如果没有,您应该能够用临时表替换它们.

下面的存储过程几乎直接将上述伪代码转换为SQL Server的存储过程语法:

CREATE TYPE SubjTableType AS TABLE (row int, id int, match_field int)
CREATE TYPE ControlTableType AS TABLE (row int, id int, match_field int)
CREATE PROCEDURE RecAssign (
    @subjects SubjTableType READONLY
,   @controls ControlTableType READONLY
,   @sp int
,   @cp int
,   @subjCount int
,   @ctrlCount int
) AS BEGIN
    IF @sp = @subjCount BEGIN
        RETURN 0
    END
    IF 1 = (SELECT COUNT(1) FROM #MemoTable WHERE sRow=@sp AND cRow=@cp) BEGIN
        RETURN (SELECT best FROM #MemoTable WHERE sRow=@sp AND cRow=@cp)
    END
    DECLARE @res int, @spNext int, @cpNext int, @prelim int, @alt int, @diff int, @sId int, @cId int
    SET @spNext = @sp + 1
    SET @cpNext = @cp + 1
    SET @sId = (SELECT id FROM @subjects WHERE row = @sp)
    SET @cId = (SELECT id FROM @controls WHERE row = @cp)
    EXEC @prelim = RecAssign @subjects=@subjects, @controls=@controls, @sp=@spNext, @cp=@cpNext, @subjCount=@subjCount, @ctrlCount=@ctrlCount
    SET @diff = ABS((SELECT match_field FROM @subjects WHERE row=@sp)-(SELECT match_field FROM @controls WHERE row=@cp))
    SET @res = @prelim + @diff
    IF 1 = (SELECT COUNT(1) FROM #Assignments WHERE sRow=@sp) BEGIN
        UPDATE #Assignments SET cId=@cId, sId=@sId, diff=@diff WHERE sRow=@sp
    END
    ELSE BEGIN
        INSERT INTO #Assignments(sRow, sId, cId, diff) VALUES (@sp, @sId, @cId, @diff)
    END
    IF @cp+1+@subjCount-@sp < @ctrlCount BEGIN
        EXEC @alt = RecAssign @subjects=@subjects, @controls=@controls, @sp=@sp, @cp=@cpNext, @subjCount=@subjCount, @ctrlCount=@ctrlCount
        IF @alt < @res BEGIN
            SET @res = @alt
        END
        ELSE BEGIN
            UPDATE #Assignments SET cId=@cId, sId=@sId, diff=@diff WHERE sRow=@sp
        END
    END
    INSERT INTO #MemoTable (sRow, cRow, best) VALUES (@sp, @cp, @res)
    RETURN @res
END

以下是调用此存储过程的方法:

-- The procedure uses a temporary table for memoization:
CREATE TABLE #MemoTable (sRow int, cRow int, best int)
-- The procedure returns a table with assignments:
CREATE TABLE #Assignments (sRow int, sId int, cId int, diff int)

DECLARE @subj as SubjTableType
INSERT INTO @SUBJ (row, id, match_field) SELECT ROW_NUMBER() OVER(ORDER BY match_field ASC)-1 AS row, id, match_field FROM subjects
DECLARE @ctrl as ControlTableType
INSERT INTO @ctrl (row, id, match_field) SELECT ROW_NUMBER() OVER(ORDER BY match_field ASC)-1 AS row, id, match_field FROM controls
DECLARE @subjCount int
SET @subjCount = (SELECT COUNT(1) FROM subjects)
DECLARE @ctrlCount int
SET @ctrlCount = (SELECT COUNT(1) FROM controls)
DECLARE @best int
EXEC @best = RecAssign
    @subjects=@subj
,   @controls=@ctrl
,   @sp=0
,   @cp=0
,   @subjCount=@subjCount
,   @ctrlCount=@ctrlCount
SELECT @best
SELECT sId, cId, diff FROM #Assignments

上面的呼叫假定两个subjects和controls通过位置已被滤除,以及N拷贝subjects已被插入到在进行呼叫之前的表值参数(或DB2的情况下,临时表).

这是sqlfiddle上的运行演示.