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慕姐8265434
TA贡献1813条经验 获得超2个赞
上面给出的levenshtein <= 1的函数不正确 - 它给出了不正确的结果,例如“床”和“出价”。
我在第一个答案中修改了上面给出的“MySQL Levenshtein距离查询”,以接受一个“限制”,这将加快它的速度。基本上,如果你只关心Levenshtein <= 1,将极限设置为“2”,如果它是0或1,函数将返回精确的levenshtein距离; 如果精确的levenshtein距离为2或更大,则为2。
这个mod使它快15%到50% - 搜索词越长,优势越大(因为算法可以提前保释。)例如,搜索200,000个单词以查找距离1的所有匹配项“傻笑”,原版在我的笔记本电脑上花了3分47秒,而“极限”版本需要1:39。当然,这些对于任何实时使用来说都太慢了。
码:
DELIMITER $$CREATE FUNCTION levenshtein_limit_n( s1 VARCHAR(255), s2 VARCHAR(255), n INT) RETURNS INT DETERMINISTIC BEGIN DECLARE s1_len, s2_len, i, j, c, c_temp, cost, c_min INT; DECLARE s1_char CHAR; -- max strlen=255 DECLARE cv0, cv1 VARBINARY(256); SET s1_len = CHAR_LENGTH(s1), s2_len = CHAR_LENGTH(s2), cv1 = 0x00, j = 1, i = 1, c = 0, c_min = 0; IF s1 = s2 THEN RETURN 0; ELSEIF s1_len = 0 THEN RETURN s2_len; ELSEIF s2_len = 0 THEN RETURN s1_len; ELSE WHILE j <= s2_len DO SET cv1 = CONCAT(cv1, UNHEX(HEX(j))), j = j + 1; END WHILE; WHILE i <= s1_len and c_min < n DO -- if actual levenshtein dist >= limit, don't bother computing it SET s1_char = SUBSTRING(s1, i, 1), c = i, c_min = i, cv0 = UNHEX(HEX(i)), j = 1; WHILE j <= s2_len DO SET c = c + 1; IF s1_char = SUBSTRING(s2, j, 1) THEN SET cost = 0; ELSE SET cost = 1; END IF; SET c_temp = CONV(HEX(SUBSTRING(cv1, j, 1)), 16, 10) + cost; IF c > c_temp THEN SET c = c_temp; END IF; SET c_temp = CONV(HEX(SUBSTRING(cv1, j+1, 1)), 16, 10) + 1; IF c > c_temp THEN SET c = c_temp; END IF; SET cv0 = CONCAT(cv0, UNHEX(HEX(c))), j = j + 1; IF c < c_min THEN SET c_min = c; END IF; END WHILE; SET cv1 = cv0, i = i + 1; END WHILE; END IF; IF i <= s1_len THEN -- we didn't finish, limit exceeded SET c = c_min; -- actual distance is >= c_min (i.e., the smallest value in the last computed row of the matrix) END IF; RETURN c; END$$
侃侃尔雅
TA贡献1801条经验 获得超16个赞
damerau-levenshtein距离的实现可以在这里找到: Damerau-Levenshtein算法:具有转置 的Levenshtein对纯Levenshtein距离的改进是考虑字符的交换。我在schnaader链接的评论中找到了它,谢谢!
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