C#和SQL实现的字符串相似度计算代码分享

C#实现：

#region 计算字符串相似度

        /// <summary>

        /// 计算字符串相似度

        /// </summary>

        /// <param name="str1">字符串1</param>

        /// <param name="str2">字符串2</param>

        /// <returns>相似度</returns>

        public static float Levenshtein(string str1, string str2)

        {

            //计算两个字符串的长度。  

            int len1 = str1.Length;

            int len2 = str2.Length;

            //比字符长度大一个空间  

            int[,] dif = new int[len1 + 1, len2 + 1];

            //赋初值，步骤B。  

            for (int a = 0; a <= len1; a++)

            {

                dif[a, 0] = a;

            }

            for (int a = 0; a <= len2; a++)

            {

                dif[0, a] = a;

            }

            //计算两个字符是否一样，计算左上的值  

            int temp;

            for (int i = 1; i <= len1; i++)

            {

                for (int j = 1; j <= len2; j++)

                {

                    if (str1.Substring(i - 1, 1) == str2.Substring(j - 1, 1))

                    {

                        temp = 0;

                    }

                    else

                    {

                        temp = 1;

                    }

                    //取三个值中最小的  

                    dif[i, j] = Min(dif[i - 1, j - 1] + temp, dif[i, j - 1] + 1, dif[i - 1, j] + 1);

                }

            }

            return 1 - (float)dif[len1, len2] / Math.Max(str1.Length, str2.Length);

        }

        #endregion

        //比较3个数字得到最小值  

        private static int Min(int i, int j, int k)

        {

            return i < j ? (i < k ? i : k) : (j < k ? j : k);

        } 

SQL实现：

CREATE   function get_semblance_By_2words 

( 

@word1 varchar(50), 

@word2 varchar(50)   

) 

returns nvarchar(4000) 

as 

begin 

declare @re int 

declare @maxLenth int 

declare @i int,@l int 

declare @tb1 table(child varchar(50)) 

declare @tb2 table(child varchar(50)) 

set @i=1 

set @l=2 

set @maxLenth=len(@word1) 

if len(@word1)<len(@word2)  

begin 

set @maxLenth=len(@word2) 

end 

while @l<=len(@word1)  

begin 

while @i<len(@word1)-1 

begin 

insert @tb1 (child) values( SUBSTRING(@word1,@i,@l) )  

set @i=@i+1 

end 

set @i=1 

set @l=@l+1 

end 

set @i=1 

set @l=2 

while @l<=len(@word2)  

begin 

while @i<len(@word2)-1 

begin 

insert @tb2 (child) values( SUBSTRING(@word2,@i,@l) )  

set @i=@i+1 

end 

set @i=1 

set @l=@l+1 

end   

select @re=isnull(max( len(a.child)*100/  @maxLenth ) ,0) from @tb1 a, @tb2 b where a.child=b.child 

return @re 

end 

GO 

  

--测试 

--select dbo.get_semblance_By_2words('我是谁','我是谁啊')  

--75 

--相似度