比较单词相似度的一个算法(c#)
正好这几天在关注这块,看到了别人的一个java实现(http://www.360doc.com/content/090201/10/96202_2430993.html),边改成了c#版本的,使用了一下,还可以,符合我的要求。
using System;
using System.Collections.Generic;
using System.Text;
namespace test
{
public class Similarity
{
private int min(int one, int two, int three)
{
int min = one;
if (two < min)
{
min = two;
}
if (three < min)
{
min = three;
}
return min;
}
public int levDistance(String str1, String str2)
{
int n = str1.Length;
int m = str2.Length;
int i; //遍历str1的
int j; //遍历str2的
char ch1; //str1的
char ch2; //str2的
int temp; //记录相同字符,在某个矩阵位置值的增量,不是0就是1
if (n == 0)
{
return m;
}
if (m == 0)
{
return n;
}
int[][] d = new int[n + 1][]; //矩阵
for (i = 0; i <= n; i++)
{ //初始化第一列
d[i] = new int[m + 1];
d[i][0] = i;
}
for (j = 0; j <= m; j++)
{ //初始化第一行
d[0][j] = j;
}
for (i = 1; i <= n; i++)
{ //遍历str1
ch1 = str1[i - 1];
//去匹配str2
for (j = 1; j <= m; j++)
{
ch2 = str2[j - 1];
if (ch1 == ch2)
{
temp = 0;
}
else
{
temp = 1;
}
//左边+1,上边+1, 左上角+temp取最小
d[i][j] = min(d[i - 1][j] + 1, d[i][j - 1] + 1, d[i - 1][j - 1] + temp);
}
}
return d[n][m];
}
public double similarity(String str1, String str2)
{
int ld = levDistance(str1, str2);
return 1 - (double)ld / Math.Max(str1.Length, str2.Length);
}
static void Main(string[] args)
{
Similarity s = new Similarity();
String str11 = "expert";
String str22 = "except";
Console.WriteLine("ld=" + s.levDistance(str11, str22));
Console.WriteLine("sim=" + s.similarity(str11, str22));
}
}
}