计算SQL中两个字符串之间的相似字数
我计划编写一个包含两个输入字符串和一个单词相似度百分比的TSQL函数作为输出，例如：计算SQL中两个字符串之间的相似字数,sql,sql-server,string,algorithm,tsql,Sql,Sql Server,String,Algorithm,Tsql,我计划编写一个包含两个输入字符串和一个单词相似度百分比的TSQL函数作为输出，例如： SELECT [dbo].[FN_CalcSimilarWords]('Golden horses hotel','Hotel Golden Horses') 返回： 3/3 2/3 或 返回： 3/3 2/3 我在考虑在解析字符串后循环并比较单词，还有其他更好的方法吗？如果您想在SQL中实现这一点，我将采用以下方法 使用拆分例程创建两个临时表，称为Words1和Words2 现在加入表并获得计数，即
SELECT [dbo].[FN_CalcSimilarWords]('Golden horses hotel','Hotel Golden Horses')
3/3
2/3
3/3
2/3
我在考虑在解析字符串后循环并比较单词，还有其他更好的方法吗？如果您想在SQL中实现这一点，我将采用以下方法 使用拆分例程创建两个临时表，称为Words1和Words2 现在加入表并获得计数，即
select count(*)
from Words1 w1
join Words2 w2 on w1.word=w2.word
select count(distinct w1.word) as Matches,
count(distinct w1.word) as FromW1,
count(distinct w2.word) as FromW2
from #Words1 w1
left join #Words2 w2 on w1.word=w2.word
如果您想在SQL中实现这一点，我将采用以下方法 使用拆分例程创建两个临时表，称为Words1和Words2 现在加入表并获得计数，即
select count(*)
from Words1 w1
join Words2 w2 on w1.word=w2.word
select count(distinct w1.word) as Matches,
count(distinct w1.word) as FromW1,
count(distinct w2.word) as FromW2
from #Words1 w1
left join #Words2 w2 on w1.word=w2.word
如果在部署CLR程序集方面没有任何限制，则可以尝试此路由并比较性能。如果在部署CLR程序集方面没有限制，您可以尝试此方法并比较性能。如果您不担心常见项目的精确计数，则可以使用SQL Server全文搜索功能。
ContainsTableFREETEXT如果您不担心常见项目的精确计数，可以使用SQL Server全文搜索功能。
ContainsTableFREETEXTCREATE TABLE StringTable
(
Id INT IDentity,
String varchar(max)
)
INSERT INTO StringTable
VALUES ('xx xx Golden horses Malaysia'),
('xx xx xx xx xx')
WITH StringsCTE
AS
(
SELECT ID,String As StringValue,
CASE CHARINDEX(' ', String)
WHEN 0 THEN String
ELSE LEFT(String, CHARINDEX(' ',String) -1)
END AS Word,
1 as Position,
CASE CHARINDEX(' ',String)
WHEN 0 THEN ''
ELSE RIGHT(String, LEN(String) - CHARINDEX(' ',String))
END AS RestOfLine
FROM StringTable
UNION ALL
SELECT Id,S.StringValue,
CASE CHARINDEX(' ',RestOfLine)
WHEN 0 THEN RestOfLine
ELSE LEFT(RestOfLine, CHARINDEX(' ',RestOfLine) -1)
END,
Position + 1,
CASE CHARINDEX(' ',RestOfLine)
WHEN 0 THEN ''
ELSE RIGHT(RestOfLine, LEN(RestOfLine) - CHARINDEX(' ',RestOfLine))
END
FROM StringsCTE S
WHERE s.RestOfLine != ''
),
WordsPerString
As
(
SELECT S.Id, COUNT(s.Word) As NumberOfWords
FROM StringsCTE S
GROUP BY S.Id
)
SELECT COUNT(*) As Matches, (SELECT MAX(NumberOfWords) FROM WordsPerString) as Total
FROM StringsCTE S1
INNER JOIN StringsCTE S2
ON S1.Word = S2.Word AND S1.Id <> S2.Id
WHERE S1.Id = 1 AND
NOT EXISTS -- Not already matched
(SELECT * FROM StringsCTE S3 WHERE S3.Word = S2.Word AND S3.Id <> S1.ID AND S3.Position < S2.Position)
| MATCHES | TOTAL |
|---------|-------|
| 2 | 5 |
CREATE TABLE StringTable
(
Id INT IDentity,
String varchar(max)
)
INSERT INTO StringTable
VALUES ('xx xx Golden horses Malaysia'),
('xx xx xx xx xx')
WITH StringsCTE
AS
(
SELECT ID,String As StringValue,
CASE CHARINDEX(' ', String)
WHEN 0 THEN String
ELSE LEFT(String, CHARINDEX(' ',String) -1)
END AS Word,
1 as Position,
CASE CHARINDEX(' ',String)
WHEN 0 THEN ''
ELSE RIGHT(String, LEN(String) - CHARINDEX(' ',String))
END AS RestOfLine
FROM StringTable
UNION ALL
SELECT Id,S.StringValue,
CASE CHARINDEX(' ',RestOfLine)
WHEN 0 THEN RestOfLine
ELSE LEFT(RestOfLine, CHARINDEX(' ',RestOfLine) -1)
END,
Position + 1,
CASE CHARINDEX(' ',RestOfLine)
WHEN 0 THEN ''
ELSE RIGHT(RestOfLine, LEN(RestOfLine) - CHARINDEX(' ',RestOfLine))
END
FROM StringsCTE S
WHERE s.RestOfLine != ''
),
WordsPerString
As
(
SELECT S.Id, COUNT(s.Word) As NumberOfWords
FROM StringsCTE S
GROUP BY S.Id
)
SELECT COUNT(*) As Matches, (SELECT MAX(NumberOfWords) FROM WordsPerString) as Total
FROM StringsCTE S1
INNER JOIN StringsCTE S2
ON S1.Word = S2.Word AND S1.Id <> S2.Id
WHERE S1.Id = 1 AND
NOT EXISTS -- Not already matched
(SELECT * FROM StringsCTE S3 WHERE S3.Word = S2.Word AND S3.Id <> S1.ID AND S3.Position < S2.Position)
| MATCHES | TOTAL |
|---------|-------|
| 2 | 5 |
CREATE TABLE StringTable
(
Id INT IDentity,
String varchar(max)
)
INSERT INTO StringTable
VALUES ('xx xx Golden horses Malaysia'),
('xx xx xx xx xx')
WITH StringsCTE
AS
(
SELECT ID,String As StringValue,
CASE CHARINDEX(' ', String)
WHEN 0 THEN String
ELSE LEFT(String, CHARINDEX(' ',String) -1)
END AS Word,
1 as Position,
CASE CHARINDEX(' ',String)
WHEN 0 THEN ''
ELSE RIGHT(String, LEN(String) - CHARINDEX(' ',String))
END AS RestOfLine
FROM StringTable
UNION ALL
SELECT Id,S.StringValue,
CASE CHARINDEX(' ',RestOfLine)
WHEN 0 THEN RestOfLine
ELSE LEFT(RestOfLine, CHARINDEX(' ',RestOfLine) -1)
END,
Position + 1,
CASE CHARINDEX(' ',RestOfLine)
WHEN 0 THEN ''
ELSE RIGHT(RestOfLine, LEN(RestOfLine) - CHARINDEX(' ',RestOfLine))
END
FROM StringsCTE S
WHERE s.RestOfLine != ''
),
WordsPerString
As
(
SELECT S.Id, COUNT(s.Word) As NumberOfWords
FROM StringsCTE S
GROUP BY S.Id
)
SELECT COUNT(*) As Matches, (SELECT MAX(NumberOfWords) FROM WordsPerString) as Total
FROM StringsCTE S1
INNER JOIN StringsCTE S2
ON S1.Word = S2.Word AND S1.Id <> S2.Id
WHERE S1.Id = 1 AND
NOT EXISTS -- Not already matched
(SELECT * FROM StringsCTE S3 WHERE S3.Word = S2.Word AND S3.Id <> S1.ID AND S3.Position < S2.Position)
| MATCHES | TOTAL |
|---------|-------|
| 2 | 5 |
CREATE TABLE StringTable
(
Id INT IDentity,
String varchar(max)
)
INSERT INTO StringTable
VALUES ('xx xx Golden horses Malaysia'),
('xx xx xx xx xx')
WITH StringsCTE
AS
(
SELECT ID,String As StringValue,
CASE CHARINDEX(' ', String)
WHEN 0 THEN String
ELSE LEFT(String, CHARINDEX(' ',String) -1)
END AS Word,
1 as Position,
CASE CHARINDEX(' ',String)
WHEN 0 THEN ''
ELSE RIGHT(String, LEN(String) - CHARINDEX(' ',String))
END AS RestOfLine
FROM StringTable
UNION ALL
SELECT Id,S.StringValue,
CASE CHARINDEX(' ',RestOfLine)
WHEN 0 THEN RestOfLine
ELSE LEFT(RestOfLine, CHARINDEX(' ',RestOfLine) -1)
END,
Position + 1,
CASE CHARINDEX(' ',RestOfLine)
WHEN 0 THEN ''
ELSE RIGHT(RestOfLine, LEN(RestOfLine) - CHARINDEX(' ',RestOfLine))
END
FROM StringsCTE S
WHERE s.RestOfLine != ''
),
WordsPerString
As
(
SELECT S.Id, COUNT(s.Word) As NumberOfWords
FROM StringsCTE S
GROUP BY S.Id
)
SELECT COUNT(*) As Matches, (SELECT MAX(NumberOfWords) FROM WordsPerString) as Total
FROM StringsCTE S1
INNER JOIN StringsCTE S2
ON S1.Word = S2.Word AND S1.Id <> S2.Id
WHERE S1.Id = 1 AND
NOT EXISTS -- Not already matched
(SELECT * FROM StringsCTE S3 WHERE S3.Word = S2.Word AND S3.Id <> S1.ID AND S3.Position < S2.Position)
| MATCHES | TOTAL |
|---------|-------|
| 2 | 5 |
使用此解决方案，我假设您希望删除重复项。切换第一个和第二个参数对结果没有影响 它返回一个值，而不是百分比，因为函数只能返回1个值或一个表。我假设你想要0到1之间的值，如果你乘以100，2/3=0.67或67%
CREATE function f_functionx
(
@str1 varchar(2000),
@str2 varchar(2000)
)
returns decimal(5,2)
as
BEGIN
DECLARE @returnvalue decimal(5,2)
DECLARE @list1 table(value varchar(50))
INSERT @list1
SELECT t.c.value('.', 'VARCHAR(2000)')
FROM (
SELECT x = CAST('<t>' +
REPLACE(@str1, ' ', '</t><t>') + '</t>' AS XML)
) a
CROSS APPLY x.nodes('/t') t(c)
DECLARE @list2 table(value varchar(50))
INSERT @list2
SELECT t.c.value('.', 'VARCHAR(2000)')
FROM (
SELECT x = CAST('<t>' +
REPLACE(@str2, ' ', '</t><t>') + '</t>' AS XML)
) a
CROSS APPLY x.nodes('/t') t(c)
;WITH isect as
(
SELECT count(*) match FROM
(
SELECT value FROM @list1
INTERSECT
SELECT value FROM @list2
) x
), total as
(
SELECT max(cnt) cnt
FROM
(
SELECT count(distinct value) cnt FROM @list1
UNION ALL
SELECT count(distinct value) FROM @list2
) x
)
SELECT
@returnvalue = cast(isect.match as decimal(9,2)) / total.cnt
FROM total
CROSS JOIN isect
RETURN @returnvalue
END
GO
1
0.67
使用此解决方案，我假设您希望删除重复项。切换第一个和第二个参数对结果没有影响 它返回一个值，而不是百分比，因为函数只能返回1个值或一个表。我假设你想要0到1之间的值，如果你乘以100，2/3=0.67或67%
CREATE function f_functionx
(
@str1 varchar(2000),
@str2 varchar(2000)
)
returns decimal(5,2)
as
BEGIN
DECLARE @returnvalue decimal(5,2)
DECLARE @list1 table(value varchar(50))
INSERT @list1
SELECT t.c.value('.', 'VARCHAR(2000)')
FROM (
SELECT x = CAST('<t>' +
REPLACE(@str1, ' ', '</t><t>') + '</t>' AS XML)
) a
CROSS APPLY x.nodes('/t') t(c)
DECLARE @list2 table(value varchar(50))
INSERT @list2
SELECT t.c.value('.', 'VARCHAR(2000)')
FROM (
SELECT x = CAST('<t>' +
REPLACE(@str2, ' ', '</t><t>') + '</t>' AS XML)
) a
CROSS APPLY x.nodes('/t') t(c)
;WITH isect as
(
SELECT count(*) match FROM
(
SELECT value FROM @list1
INTERSECT
SELECT value FROM @list2
) x
), total as
(
SELECT max(cnt) cnt
FROM
(
SELECT count(distinct value) cnt FROM @list1
UNION ALL
SELECT count(distinct value) FROM @list2
) x
)
SELECT
@returnvalue = cast(isect.match as decimal(9,2)) / total.cnt
FROM total
CROSS JOIN isect
RETURN @returnvalue
END
GO
1
0.67
SQL在字符串操作方面不是很强，如果您允许在服务器上安装一个CLR例程，您可能需要考虑CLR例程。@ Spky，是的，我可以安装它，任何想法或有用的链接来实现一个解决方案。这里有一些关于堆栈溢出的链接，在C语言中它非常容易，不幸的是，这里有一个这样的链接。我仅限于SQL，出于某些原因，我不想在C#中这样做。如果您仅限于SQL，请参阅下面我的答案。使用split函数创建两个临时表，然后连接它们。最慢的部分将是拆分函数，但除非你有一个非常大的单词List.Sql在字符串操作中不是很强，否则你可能不会注意到，如果你可以在你的服务器上安装一个CLR例程，你可能想考虑一个CLR例程。有什么想法或有用的链接来实现一个解决方案吗？这里有一些关于堆栈溢出的链接，在C#中这其实很容易，这里有一个这样的链接不幸的是，我限于SQL，出于某些原因，我不想在C#中这样做。如果你限于SQL，请参阅下面我的答案。使用split函数创建两个临时表，然后连接它们。最慢的部分是split函数，但除非您有一个非常大的单词列表，否则可能不会引起注意。感谢您的回答，这将只返回相似单词的计数，有没有办法在不重新调用split函数的情况下也返回单词1和单词2的计数？感谢更新答案，但是它返回的结果不正确，匹配项和FromW1始终相等。匹配项将始终相等，但代码假定#words1大于words 2。我将很快调整代码以返回两个表中较大的一个表counts谢谢您的回答，这将只返回相似单词的计数，是否有任何方法也可以返回单词1和单词2的计数而不重新调用拆分函数？感谢更新答案，但返回的结果不正确，匹配项和FromW1始终相等。匹配项将始终相等，但代码假定#字1大于字2。我将很快调整代码以返回两个表中较大的一个，这两个值位于stringtable中：“xx xx xx Golden horses Malaysia”，“xx xx xx xx xx xx”，它将返回5列，其中包含10个匹配项。所以它有200%match@t-clausen.dk更新了答案以解决您的问题。请在stringtable中尝试这两个值：“xx xx马来西亚金马”，“xx xx xx xx xx”，它将返回5列，其中10个匹配项。所以它有200%match@t-clausen.dk更新了答案以解决您的问题。这似乎是一个天才的答案，不幸的是，我无法比较性能，因为大多数记录都包含
&XML解析：。。。非法名称字符&&XML解析：。。。非法名称字符&