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!
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