在c＃中打印LevenshteinDistance矩阵＃

Question

我已经在c＃中实现了LevenshteinDistance算法，如下所示。此代码运行完美。但出于调试目的，我想打印该矩阵，但无法确定应该在哪里放置Print语句。

using System; 
using System.Collections.Generic; 
using System.Linq; 
using System.Text; 

namespace _Levenshtein_ 
{ 
    class Program 
    { 
     public static void Print(int[,] data) 
     { 
      for (int i = 0; i < data.GetUpperBound(0); i++) 
      { 
       for (int j = 0; j < data.GetUpperBound(1); j++) 
       { 
        Console.Write(data[i, j] + " "); 
       } 
      } 
      Console.WriteLine(); 
     } 
     public static int LevenshteinDistance(string source, string target) 
     { 
      if (String.IsNullOrEmpty(source)) 
      { 
       if (String.IsNullOrEmpty(target)) return 0; 
       { 

        return target.Length; 
       } 
      } 
      if (String.IsNullOrEmpty(target)) return source.Length; 

      if (source.Length > target.Length) 
      { 
       var temp = target; 
       target = source; 
       source = temp; 
      } 

      var m = target.Length; 
      var n = source.Length; 
      var distance = new int[2, m + 1]; 
      // Initialize the distance 'matrix' 
      for (var j = 1; j <= m; j++) distance[0, j] = j; 


       Console.Write(target + " "); 

      var currentRow = 0; 
      for (var i = 1; i <= n; ++i) 
      { 
       currentRow = i & 1; 
       distance[currentRow, 0] = i; 
       var previousRow = currentRow^1; 
       Console.WriteLine(source[i-1] + " "); 
       for (var j = 1; j <= m; j++) 
       { 
        var cost = (target[j - 1] == source[i - 1] ? 0 : 1); 

        distance[currentRow, j] = Math.Min(Math.Min(distance[previousRow, j] + 1,distance[currentRow, j - 1] + 1),distance[previousRow, j - 1] + cost); 
        Print(distance);   
       } 
       Console.WriteLine(); 
      } 
      return distance[currentRow, m]; 
     } 
     static void Main(string[] args) 
     { 
      LevenshteinDistance("Sunday", "Saturday"); 
     } 
    } 
} 

Answer 1

我在打印距离矩阵中添加了注释。

// `target` string in first ROW, each char in 4 width 
// for (var j = 0; j <=target.Length; j++) 
// Console.Write(target[j] + " ") 
for (var i = 1; i <= n; ++i) 
{ 
    currentRow = i & 1; 
    distance[currentRow, 0] = i; 
    var previousRow = currentRow^1; 
    // print: `source[i]` ith char only one Console.Write(source[i] + " ") 
    // Console.Write(source[i] + " ") 
    for (var j = 1; j <= m; j++) 
    { 
     var cost = (target[j - 1] == source[i - 1] ? 0 : 1); 
     distance[currentRow, j] = Math.M........); 
     // write distance in 3 width 
      //Console.Write(distance[currentRow, j] + " ") 
    }  

     // Console.Write("\n") 
} 

编辑：

长度 “星期日”= 6
长度 “星期六”= 8

if (source.Length > target.Length){ // target is large . 
    // swap means 
} 

所以目标是 “星期六” 的字符串。 （横纹）
和“星期日”是垂直

你会得到像下面的图输出：

这里就Codepade是我工作的C代码，可以打印编辑距离矩阵，如：

 S a t u r d a y 

S  0 1 2 3 4 5 6 7 
u  1 1 2 2 3 4 5 6 
n  2 2 2 3 3 4 5 6 
d  3 3 3 3 4 3 4 5 
a  4 3 4 4 4 4 3 4 
y  5 4 4 5 5 5 4 3