Levenshtein距离c#计数错误类型

本文关键字：错误 类型 距离 Levenshtein | 更新日期: 2023-09-27 18:05:17

我找到了这段代码，用来计算答案和猜测之间的距离:

int CheckErrors(string Answer, string Guess)
{
    int[,] d = new int[Answer.Length + 1, Guess.Length + 1];
    for (int i = 0; i <= Answer.Length; i++)
        d[i, 0] = i;
    for (int j = 0; j <= Guess.Length; j++)
        d[0, j] = j;
    for (int j = 1; j <= Guess.Length; j++)
        for (int i = 1; i <= Answer.Length; i++)
            if (Answer[i - 1] == Guess[j - 1])
                d[i, j] = d[i - 1, j - 1];  //no operation
            else
                d[i, j] = Math.Min(Math.Min(
                    d[i - 1, j] + 1,    //a deletion
                    d[i, j - 1] + 1),   //an insertion
                    d[i - 1, j - 1] + 1 //a substitution
                );
    return d[Answer.Length, Guess.Length];
}

但是我需要一种方法来计算每个错误发生的次数。有没有简单的方法来实现?

Levenshtein距离c#计数错误类型

似乎可以为每个操作添加计数器:

                if (Answer[i - 1] == Guess[j - 1])
                    d[i, j] = d[i - 1, j - 1];  //no operation
                else
                {
                    int del = d[i-1, j] + 1;
                    int ins = d[i, j-1] + 1;
                    int sub = d[i-1, j-1] + 1;
                    int op = Math.Min(Math.Min(del, ins), sub);
                    d[i, j] = op;
                    if (i == j)
                    {
                        if (op == del)
                            ++deletions;
                        else if (op == ins)
                            ++insertions;
                        else
                            ++substitutions;
                    }
                }
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