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C#通過編輯距離計算兩個字符串的相似度的代碼

int != else dha reg ndis namespace i++ 過程

將開發過程中較好的一些代碼段備份一下,下面的代碼是關於C#通過編輯距離計算兩個字符串的相似度的代碼,應該能對碼農們有些幫助。

using System;
using System.Text.RegularExpressions;
using System.Threading.Tasks;

namespace Levenshtein
{
    public delegate void AnalyzerCompletedHander(double sim);

    public class LevenshteinDistance:IDisposable
    {
        private string str1;
        private string str2;
        private int[,] index;
        int k;
        Task<double> task;

        public event AnalyzerCompletedHander AnalyzerCompleted;

        public string Str1
        {
            get { return str1; }
            set
            {
                str1 = Format(value);
                index = new int[str1.Length, str2.Length];
            }
        }

        public string Str2
        {
            get { return str2; }
            set
            {
                str2 = Format(value);
                index = new int[str1.Length, str2.Length];
            }
        }

        public int TotalTimes
        {
        }

        public bool IsCompleted
        {
            get { return task.IsCompleted; }
        }

        public LevenshteinDistance(string str1, string str2)
        {
            this.str1 = Format(str1);
            this.str2 = Format(str2);
            index = new int[str1.Length, str2.Length];
        }

        public LevenshteinDistance()
        {
        }

        public void Start()
        {
            task = new Task<double>(Analyzer);
            task.Start();
            task.ContinueWith(o => Completed(o.Result));
        }

        public double StartAyns()
        {
            task = new Task<double>(Analyzer);
            task.Start();
            task.Wait();
            return task.Result;
        }

        private void Completed(double s)
        {
            if (AnalyzerCompleted != null)
            {
                AnalyzerCompleted(s);
            }
        }

        private double Analyzer()
        {
            if (str1.Length == 0 || str2.Length == 0)
                return 0;
            for (int i = 0; i < str1.Length; i++)
            {
                for (int j = 0; j < str2.Length; j++)
                {
                    k = str1[i] == str2[j] ? 0 : 1;
                    if (i == 0&&j==0)
                    {
                        continue;
                    }
                    else if (i == 0)
                    {
                        index[i, j] = k + index[i, j - 1];
                        continue;
                    }
                    else if (j == 0)
                    {
                        index[i, j] = k + index[i - 1, j];
                        continue;
                    }
                    int temp = Min(index[i, j - 1],
                        index[i - 1, j], 
                        index[i - 1, j - 1]);
                    index[i, j] = temp + k;
                }
            }
            float similarty = 1 - (float)index[str1.Length - 1, str2.Length - 1] 
                / (str1.Length > str2.Length ? str1.Length : str2.Length);
            return similarty;
        }

        private string Format(string str)
        {
            str = Regex.Replace(str, @"[^a-zA-Z0-9u4e00-u9fa5s]", "");
            return str;
        }

        private int Min(int a, int b, int c)
        {
            int temp = a < b ? a : b;
            temp = temp < c ? temp : c;
            return temp;
        }

        public void Dispose()
        {
            task.Dispose();
        }
    }
}

C#通過編輯距離計算兩個字符串的相似度的代碼