为什么我得到的Levenshtein距离不正确?

时间:2019-05-28 14:17:06

标签: c# levenshtein-distance

我正在尝试在C#中实现Levenshtein距离算法(出于实践考虑,因为它很方便)。我使用了Wikipedia page的实现,但由于某种原因,我在一组单词上的距离不正确。这是代码(来自LinqPad):

void Main()
{
   var ld = new LevenshteinDistance();
   int dist = ld.LevenshteinDistanceCalc("sitting","kitten");
   dist.Dump();
}

// Define other methods and classes here
public class LevenshteinDistance
{
   private int[,] distance;

   public int LevenshteinDistanceCalc(string source, string target)
   {
      int sourceSize = source.Length, targetSize = target.Length;
      distance = new int[sourceSize, targetSize];
      for (int sIndex = 0; sIndex < sourceSize; sIndex++)
      {
         distance[sIndex, 0] = sIndex;
      }
      for (int tIndex = 0; tIndex < targetSize; tIndex++)
      {
         distance[0,tIndex] = tIndex;
      }
      //      for j from 1 to n:
      //      for i from 1 to m:
      //          if s[i] = t[j]:
      //            substitutionCost:= 0
      //          else:
      //            substitutionCost:= 1
      //          d[i, j] := minimum(d[i - 1, j] + 1,                   // deletion
      //                             d[i, j - 1] + 1,                   // insertion
      //                             d[i - 1, j - 1] + substitutionCost)  // substitution
      //
      //
      //      return d[m, n]

      for (int tIndex = 1; tIndex < targetSize; tIndex++)
      {
         for (int sIndex = 1; sIndex < sourceSize; sIndex++)
         {
            int substitutionCost = source[sIndex] == target[tIndex] ? 0 : 1;
            int deletion = distance[sIndex-1, tIndex]+1;
            int insertion = distance[sIndex,tIndex-1]+1;
            int substitution = distance[sIndex-1, tIndex-1] + substitutionCost;

            distance[sIndex, tIndex] = leastOfThree(deletion, insertion, substitution);

         }
      }
      return distance[sourceSize-1,targetSize-1];
   }

   private int leastOfThree(int a, int b, int c)
   {
      return Math.Min(a,(Math.Min(b,c)));
   }
}

当我尝试“坐着”和“小猫”时,得到的LD为2(应为3)。然而,当我尝试“星期六”和“星期天”时,我得到的LD为3(这是正确的)。我知道出了点问题,但是我无法弄清楚我所缺少的。

2 个答案:

答案 0 :(得分:2)

维基百科上的示例使用基于1的字符串。在C#中,我们使用基于0的字符串。

在它们的矩阵中确实存在0行和0列。因此,它们的矩阵大小为[source.Length + 1,source.Length + 1]在您的代码中不存在。

public int LevenshteinDistanceCalc(string source, string target)
{
  int sourceSize = source.Length, targetSize = target.Length;
  distance = new int[sourceSize + 1, targetSize + 1];
  for (int sIndex = 1; sIndex <= sourceSize; sIndex++)
    distance[sIndex, 0] = sIndex;

  for (int tIndex = 1; tIndex <= targetSize; tIndex++)
    distance[0, tIndex] = tIndex;

  for (int tIndex = 1; tIndex <= targetSize; tIndex++)
  {
    for (int sIndex = 1; sIndex <= sourceSize; sIndex++)
    {
      int substitutionCost = source[sIndex-1] == target[tIndex-1] ? 0 : 1;
      int deletion = distance[sIndex - 1, tIndex] + 1;
      int insertion = distance[sIndex, tIndex - 1] + 1;
      int substitution = distance[sIndex - 1, tIndex - 1] + substitutionCost;

      distance[sIndex, tIndex] = leastOfThree(deletion, insertion, substitution);
    }
  }
  return distance[sourceSize, targetSize];
}

答案 1 :(得分:1)

您的矩阵不够大。

在伪代码中,st的长度分别为mnchar s[1..m], char t[1..n])。然而,矩阵具有尺寸[0..m, 0..n]-即比每个方向上的字符串长度多一。您可以在伪代码下方的表格中看到这一点。

所以“坐着”和“小猫”的矩阵是7x8,但是您的矩阵只有6x7。

您还会错误地索引字符串,因为伪代码中的字符串是1索引的,而C#的字符串是0索引的。

修复这些问题后,您将获得此代码,该代码可与“坐着”和“小猫”一起使用:

public static class LevenshteinDistance
{
    public static int LevenshteinDistanceCalc(string source, string target)
    {
        int sourceSize = source.Length + 1, targetSize = target.Length + 1;
        int[,] distance = new int[sourceSize, targetSize];
        for (int sIndex = 0; sIndex < sourceSize; sIndex++)
        {
            distance[sIndex, 0] = sIndex;
        }
        for (int tIndex = 0; tIndex < targetSize; tIndex++)
        {
            distance[0, tIndex] = tIndex;
        }
        //      for j from 1 to n:
        //      for i from 1 to m:
        //          if s[i] = t[j]:
        //            substitutionCost:= 0
        //          else:
        //            substitutionCost:= 1
        //          d[i, j] := minimum(d[i - 1, j] + 1,                   // deletion
        //                             d[i, j - 1] + 1,                   // insertion
        //                             d[i - 1, j - 1] + substitutionCost)  // substitution
        //
        //
        //      return d[m, n]

        for (int tIndex = 1; tIndex < targetSize; tIndex++)
        {
            for (int sIndex = 1; sIndex < sourceSize; sIndex++)
            {
                int substitutionCost = source[sIndex - 1] == target[tIndex - 1] ? 0 : 1;
                int deletion = distance[sIndex - 1, tIndex] + 1;
                int insertion = distance[sIndex, tIndex - 1] + 1;
                int substitution = distance[sIndex - 1, tIndex - 1] + substitutionCost;

                distance[sIndex, tIndex] = leastOfThree(deletion, insertion, substitution);
            }
        }
        return distance[sourceSize - 1, targetSize - 1];
    }

    private static int leastOfThree(int a, int b, int c)
    {
        return Math.Min(a, (Math.Min(b, c)));
    }
}

(我也很自由地将distance设置为局部变量,因为不需要将其作为一个字段(它只会使您的类成为非线程安全的),并使其变为静态以避免不必要的实例化)。

要调试此问题,我在return distance[sourceSize - 1, targetSize - 1]上放置了一个断点,并将distance与Wikipedia上的表进行了比较。很明显,它太小了。