如何在MYSQL中计算两个字符串之间的相似度

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:

和结果是3 /(4 + 4-3)= 0.6

高度赞赏任何想法!

回答:

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

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