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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