Question

我几个月来一直想知道如何在Transact SQL中实现此算法，https://en.wikipedia.org/wiki/Jaro%E2%80%93Winkler_distance

怎么做？

Answer 1

今天我终于偶然发现了this Stack Overflow-leebickmtu的答案，显示了C＃中的一个实现，最初是从Java移植过来的。我冒昧地将它移植到Transact SQL函数，享受！

IF OBJECT_ID (N'dbo.InlineMax', N'FN') IS NOT NULL
    DROP FUNCTION dbo.InlineMax;
GO

CREATE FUNCTION dbo.InlineMax(@valueOne int, @valueTwo int)
    RETURNS FLOAT
AS
BEGIN
    IF @valueOne > @valueTwo
    BEGIN
        RETURN @valueOne
    END

    RETURN ISNULL(@valueTwo, @valueOne)
END;
GO

IF OBJECT_ID (N'dbo.InlineMin', N'FN') IS NOT NULL
    DROP FUNCTION dbo.InlineMin;
GO

CREATE FUNCTION dbo.InlineMin(@valueOne int, @valueTwo int)
    RETURNS FLOAT
AS
BEGIN
    IF @valueOne < @valueTwo
        RETURN @valueOne

    RETURN @valueTwo
END;
GO

IF OBJECT_ID (N'dbo.JaroWinklerDistance', N'FN') IS NOT NULL
    DROP FUNCTION dbo.JaroWinklerDistance;
GO

CREATE FUNCTION dbo.JaroWinklerDistance(@stringOne varchar(MAX), @stringTwo varchar(MAX))
RETURNS FLOAT
WITH EXECUTE AS CALLER
AS
BEGIN
    DECLARE @mWeightThreshold FLOAT; SET @mWeightThreshold = 0.7;
    DECLARE @mNuMChars INT; SET @mNumChars = 4;
    DECLARE @lLen1 int; SET @lLen1 = LEN(@stringOne)
    DECLARE @lLen2 int; SET @lLen2 = LEN(@stringTwo)

    IF @lLen1 = 0
        RETURN CASE WHEN @lLen2 = 0 THEN 1 ELSE 0 END

    DECLARE @lSearchRange int; SET @lSearchRange = dbo.InlineMax(0,dbo.InlineMax(@lLen1, @lLen2)/2 - 1);

    DECLARE @lMatched1 TABLE (position int not null, [status] bit not null)
    DECLARE @lMatched2 TABLE (position int not null, [status] bit not null)

    DECLARE @lNumCommon int; SET @lNumCommon = 0
    DECLARE @i int; SET @i = 1; WHILE(@i <= @lLen1)
    BEGIN
        DECLARE @lStart int; SET @lStart = dbo.InlineMax(1, @i - @lSearchRange)
        DECLARE @lEnd int; SET @lEnd = dbo.InlineMin(@i + @lSearchRange + 1, @lLen2)

        DECLARE @j int; SET @j = @lStart; WHILE(@j <= @lEnd)
        BEGIN
            IF((SELECT [status] FROM @lMatched2 WHERE position = @j) = 1)
            BEGIN
                SET @j = @j + 1
                CONTINUE
            END

            IF (SELECT SUBSTRING(@stringOne, @i, 1)) <> (SELECT SUBSTRING(@stringTwo, @j, 1))
            BEGIN
                SET @j = @j + 1
                CONTINUE
            END

            INSERT INTO @lMatched1 (position, [status]) VALUES(@i, 1)
            INSERT INTO @lMatched2 (position, [status]) VALUES(@j, 1)

            SET @lNumCommon = @lNumCommon + 1
            BREAK
        END

        SET @i = @i + 1
    END

    IF @lNumCommon = 0
    BEGIN
        RETURN 0.0;
    END

    DECLARE @lNumHalfTransposed int; SET @lNumHalfTransposed = 0
    DECLARE @k INT; SET @k = 1;
    DECLARE @stopLoop bit; SET @stopLoop = 0;
    SET @i = 1; WHILE(@i <= @lLen1)
    BEGIN
        IF ((SELECT [status] FROM @lMatched1 WHERE position = @i) = 1)
        BEGIN
            SET @i = @i + 1
            CONTINUE;
        END

        WHILE(@stopLoop = 0)
        BEGIN
            IF((SELECT [status] FROM @lMatched2 WHERE position = @k) = 0)
                SET @k = @k + 1
            ELSE
                BREAK

            IF((SELECT SUBSTRING(@stringOne, @i, 1)) <> (SELECT SUBSTRING(@stringTwo, @k, 1)))
                SET @lNumHalfTransposed = @lNumHalfTransposed + 1

            SET @k = @k + 1
        END

        SET @i = @i + 1
    END

    DECLARE @lNumTransposed INT; SET @lNumTransposed = @lNumHalfTransposed/2;

    DECLARE @lNumCommonD FLOAT; SET @lNumCommonD = @lNumCommon
    DECLARE @lWeight FLOAT; SET @lWeight = (@lNumCommonD / @lLen1 + @lNumCommonD / @lLen2 + (@lNumCommon - @lNumTransposed) / @lNumCommonD) / 3.0;

    IF(@lWeight <= @mWeightThreshold)
        RETURN @lWeight
    DECLARE @lMax INT; SET @lMax = dbo.InlineMin(@mNumChars, dbo.InlineMin(@lLen1, @lLen2))
    DECLARE @lPos INT; SET @lPos = 0
    WHILE(@lPos < @lMax AND (SELECT SUBSTRING(@stringOne, @lPos, 1)) = (SELECT SUBSTRING(@stringTwo, @lPos, 1)))
    BEGIN
        SET @lPos = @lPos + 1
    END

    IF @lPos = 0
        RETURN @lWeight

    RETURN @lWeight + 0.1 * @lPos * (1.0 - @lWeight)
END;
GO

Transact SQL中Jaro Winkler距离算法的实现

1 个答案: