一尘不染

如何在MYSQL中计算两个字符串之间的相似度

mysql

如果我在mysql中有两个字符串:

@ a =“欢迎使用堆栈溢出”
@ b =“你好,堆栈溢出”;

有没有办法使用MYSQL获得这两个字符串之间的相似性百分比?例如,这里有3个单词是相似的,因此相似度应为:
count(@a和@b之间的相似单词)/(count(@a)+ count(@b)-count(intersection))
和结果是3 /(4 + 4-3)= 0.6
高度赞赏任何想法!


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2020-05-17

共1个答案

一尘不染

您可以使用此功能(从http://www.artfulsoftware.com/infotree/queries.php#552改编的cop
^ H ^ H ^ ):

CREATE FUNCTION `levenshtein`( s1 text, s2 text) RETURNS int(11)
    DETERMINISTIC
BEGIN 
    DECLARE s1_len, s2_len, i, j, c, c_temp, cost INT; 
    DECLARE s1_char CHAR; 
    DECLARE cv0, cv1 text; 
    SET s1_len = CHAR_LENGTH(s1), s2_len = CHAR_LENGTH(s2), cv1 = 0x00, j = 1, i = 1, c = 0; 
    IF s1 = s2 THEN 
      RETURN 0; 
    ELSEIF s1_len = 0 THEN 
      RETURN s2_len; 
    ELSEIF s2_len = 0 THEN 
      RETURN s1_len; 
    ELSE 
      WHILE j <= s2_len DO 
        SET cv1 = CONCAT(cv1, UNHEX(HEX(j))), j = j + 1; 
      END WHILE; 
      WHILE i <= s1_len DO 
        SET s1_char = SUBSTRING(s1, i, 1), c = i, cv0 = UNHEX(HEX(i)), j = 1; 
        WHILE j <= s2_len DO 
          SET c = c + 1; 
          IF s1_char = SUBSTRING(s2, j, 1) THEN  
            SET cost = 0; ELSE SET cost = 1; 
          END IF; 
          SET c_temp = CONV(HEX(SUBSTRING(cv1, j, 1)), 16, 10) + cost; 
          IF c > c_temp THEN SET c = c_temp; END IF; 
            SET c_temp = CONV(HEX(SUBSTRING(cv1, j+1, 1)), 16, 10) + 1; 
            IF c > c_temp THEN  
              SET c = c_temp;  
            END IF; 
            SET cv0 = CONCAT(cv0, UNHEX(HEX(c))), j = j + 1; 
        END WHILE; 
        SET cv1 = cv0, i = i + 1; 
      END WHILE; 
    END IF; 
    RETURN c; 
  END

并以XX%的价格使用此功能

CREATE FUNCTION `levenshtein_ratio`( s1 text, s2 text ) RETURNS int(11)
    DETERMINISTIC
BEGIN 
    DECLARE s1_len, s2_len, max_len INT; 
    SET s1_len = LENGTH(s1), s2_len = LENGTH(s2); 
    IF s1_len > s2_len THEN  
      SET max_len = s1_len;  
    ELSE  
      SET max_len = s2_len;  
    END IF; 
    RETURN ROUND((1 - LEVENSHTEIN(s1, s2) / max_len) * 100); 
  END
2020-05-17