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C#通过编辑距离算法实现字符串相似度比较
编辑距离:通过插入、删除、替换一个字符(和交换相邻字符)的操作,使得字符串A和字符串B相同,而最少的操作次数就是编辑距离。
如字符串abcd和aca的距离是2
编辑距离:通过插入、删除、替换一个字符(和交换相邻字符)的操作,使得字符串A和字符串B相同,而最少的操作次数就是编辑距离。
如字符串abcd和aca的距离是2
public class LevenshteinDistance { private static LevenshteinDistance _instance=null; public static LevenshteinDistance Instance { get { if (_instance == null) { return new LevenshteinDistance(); } return _instance; } } /// <summary> /// 取最小的一位数 /// </summary> /// <param name="first"></param> /// <param name="second"></param> /// <param name="third"></param> /// <returns></returns> public int LowerOfThree(int first, int second, int third) { int min = first; if (second < min) min = second; if (third < min) min = third; return min; } public int Levenshtein_Distance(string str1, string str2) { int[,] Matrix; int n=str1.Length; int m=str2.Length; int temp = 0; char ch1; char ch2; int i = 0; int j = 0; if (n ==0) { return m; } if (m == 0) { return n; } Matrix=new int[n+1,m+1]; for (i = 0; i <= n; i++) { //初始化第一列 Matrix[i,0] = i; } for (j = 0; j <= m; j++) { //初始化第一行 Matrix[0, j] = j; } for (i = 1; i <= n; i++) { ch1 = str1[i-1]; for (j = 1; j <= m; j++) { ch2 = str2[j-1]; if (ch1.Equals(ch2)) { temp = 0; } else { temp = 1; } Matrix[i,j] = LowerOfThree(Matrix[i - 1,j] + 1, Matrix[i,j - 1] + 1, Matrix[i - 1,j - 1] + temp); } } for (i = 0; i <= n; i++) { for (j = 0; j <= m; j++) { Console.Write(" {0} ", Matrix[i, j]); } Console.WriteLine(""); } return Matrix[n, m]; } /// <summary> /// 计算字符串相似度 /// </summary> /// <param name="str1"></param> /// <param name="str2"></param> /// <returns></returns> public decimal LevenshteinDistancePercent(string str1,string str2) { int maxLenth = str1.Length > str2.Length ? str1.Length : str2.Length; int val = Levenshtein_Distance(str1, str2); return 1 - (decimal)val / maxLenth; } } class Program { static void Main(string[] args) { string str1 = "你好蒂蒂"; string str2="你好蒂芬"; Console.WriteLine("字符串1 {0}", str1); Console.WriteLine("字符串2 {0}", str2); Console.WriteLine("相似度 {0} %", LevenshteinDistance.Instance.LevenshteinDistancePercent(str1, str2)*100); Console.ReadLine(); } }