Question

我有以下实现，但我想添加一个阈值，所以如果结果大于它，就停止计算并返回。

我该怎么做？

编辑：这是我当前的代码，threshold尚未使用...目标是使用它

    public static int DamerauLevenshteinDistance(string string1, string string2, int threshold)
    {
        // Return trivial case - where they are equal
        if (string1.Equals(string2))
            return 0;

        // Return trivial case - where one is empty
        if (String.IsNullOrEmpty(string1) || String.IsNullOrEmpty(string2))
            return (string1 ?? "").Length + (string2 ?? "").Length;


        // Ensure string2 (inner cycle) is longer
        if (string1.Length > string2.Length)
        {
            var tmp = string1;
            string1 = string2;
            string2 = tmp;
        }

        // Return trivial case - where string1 is contained within string2
        if (string2.Contains(string1))
            return string2.Length - string1.Length;

        var length1 = string1.Length;
        var length2 = string2.Length;

        var d = new int[length1 + 1, length2 + 1];

        for (var i = 0; i <= d.GetUpperBound(0); i++)
            d[i, 0] = i;

        for (var i = 0; i <= d.GetUpperBound(1); i++)
            d[0, i] = i;

        for (var i = 1; i <= d.GetUpperBound(0); i++)
        {
            for (var j = 1; j <= d.GetUpperBound(1); j++)
            {
                var cost = string1[i - 1] == string2[j - 1] ? 0 : 1;

                var del = d[i - 1, j] + 1;
                var ins = d[i, j - 1] + 1;
                var sub = d[i - 1, j - 1] + cost;

                d[i, j] = Math.Min(del, Math.Min(ins, sub));

                if (i > 1 && j > 1 && string1[i - 1] == string2[j - 2] && string1[i - 2] == string2[j - 1])
                    d[i, j] = Math.Min(d[i, j], d[i - 2, j - 2] + cost);
            }
        }

        return d[d.GetUpperBound(0), d.GetUpperBound(1)];
    }
}

Answer 1

这是关于你的答案：Damerau - Levenshtein Distance, adding a threshold （抱歉无法评论，因为我还没有50名代表）

我认为你在这里犯了一个错误。你初始化了：

var minDistance = threshold;

你的更新规则是：

if (d[i, j] < minDistance)
   minDistance = d[i, j];

此外，您提前退出的标准是：

if (minDistance > threshold)
   return int.MaxValue;

现在，请注意上面的if条件永远不会成立！您应该将minDistance初始化为int.MaxValue

Answer 2

这是我能想到的最优雅的方式。设置d的每个索引后，查看它是否超过了您的阈值。评估是恒定时间，因此与整体算法的理论N ^ 2复杂度相比，它是一个下降：

public static int DamerauLevenshteinDistance(string string1, string string2, int threshold)
{
    ...

    for (var i = 1; i <= d.GetUpperBound(0); i++)
    {
        for (var j = 1; j <= d.GetUpperBound(1); j++)
        {
            ...

            var temp = d[i,j] = Math.Min(del, Math.Min(ins, sub));

            if (i > 1 && j > 1 && string1[i - 1] == string2[j - 2] && string1[i - 2] == string2[j - 1])
                temp = d[i,j] = Math.Min(temp, d[i - 2, j - 2] + cost);

            //Does this value exceed your threshold? if so, get out now
            if(temp > threshold) 
              return temp;
        }
    }

    return d[d.GetUpperBound(0), d.GetUpperBound(1)];
}

Answer 3

您还将此问题称为SQL CLR UDF问题，因此我将在该特定背景下回答：您最好的选择不会来自优化Levenshtein距离，而是来自减少您比较的对数。是的，一个更快的Levenshtein算法将改善一些事情，但几乎不会减少从N平方（数百万行中的N）到N *某些因子的比较次数。我的建议是仅比较长度差异在可容忍的delta内的元素。在您的大表上，您在LEN(Data)上添加一个持久的计算列，然后使用包含数据在其上创建一个索引：

ALTER TABLE Table ADD LenData AS LEN(Data) PERSISTED;
CREATE INDEX ndxTableLenData on Table(LenData) INCLUDE (Data);

现在，您可以通过加长长度上的最大差异来限制纯粹的问题空间（例如说5），如果您的数据LEN(Data)显着变化：

SELECT a.Data, b.Data, dbo.Levenshtein(a.Data, b.Data)
FROM Table A
JOIN Table B ON B.DataLen BETWEEN A.DataLen - 5 AND A.DataLen+5

Answer 4

终于明白了......虽然它没有我希望的那样有益

    public static int DamerauLevenshteinDistance(string string1, string string2, int threshold)
    {
        // Return trivial case - where they are equal
        if (string1.Equals(string2))
            return 0;

        // Return trivial case - where one is empty
        if (String.IsNullOrEmpty(string1) || String.IsNullOrEmpty(string2))
            return (string1 ?? "").Length + (string2 ?? "").Length;


        // Ensure string2 (inner cycle) is longer
        if (string1.Length > string2.Length)
        {
            var tmp = string1;
            string1 = string2;
            string2 = tmp;
        }

        // Return trivial case - where string1 is contained within string2
        if (string2.Contains(string1))
            return string2.Length - string1.Length;

        var length1 = string1.Length;
        var length2 = string2.Length;

        var d = new int[length1 + 1, length2 + 1];

        for (var i = 0; i <= d.GetUpperBound(0); i++)
            d[i, 0] = i;

        for (var i = 0; i <= d.GetUpperBound(1); i++)
            d[0, i] = i;

        for (var i = 1; i <= d.GetUpperBound(0); i++)
        {
            var im1 = i - 1;
            var im2 = i - 2;
            var minDistance = threshold;

            for (var j = 1; j <= d.GetUpperBound(1); j++)
            {
                var jm1 = j - 1;
                var jm2 = j - 2;
                var cost = string1[im1] == string2[jm1] ? 0 : 1;

                var del = d[im1, j] + 1;
                var ins = d[i, jm1] + 1;
                var sub = d[im1, jm1] + cost;

                //Math.Min is slower than native code
                //d[i, j] = Math.Min(del, Math.Min(ins, sub));
                d[i, j] = del <= ins && del <= sub ? del : ins <= sub ? ins : sub;

                if (i > 1 && j > 1 && string1[im1] == string2[jm2] && string1[im2] == string2[jm1])
                    d[i, j] = Math.Min(d[i, j], d[im2, jm2] + cost);

                if (d[i, j] < minDistance)
                    minDistance = d[i, j];
            }

            if (minDistance > threshold)
                return int.MaxValue;
        }

        return d[d.GetUpperBound(0), d.GetUpperBound(1)] > threshold 
            ? int.MaxValue 
            : d[d.GetUpperBound(0), d.GetUpperBound(1)];
    }

Damerau - Levenshtein距离，增加一个门槛

4 个答案: