Question

假设我有一张如下表：

Mark - Green
Mark - Blue
Mark - Red
Adam - Yellow
Andrew - Red
Andrew - Green

我的目标是将用户“Mark”与数据库中的所有其他用户进行比较，以找出他最相似的其他用户。在这种情况下，他与安德鲁（2/3比赛）最相似，与亚当（0/3）比赛最不相似。在我发现哪个用户与Mark最相似之后，我想提取Andrew得到的但是Mark没有得到的。

这在MySQL中可行吗？我感谢所有的帮助，谢谢你们！

编辑：所有好帮助都超越了！十分感谢大家！我一定会查看你所有的贡献！

Answer 1

以下查询尝试列出与Mark匹配的所有用户。它基本上将表与Mark条目连接起来，并计算所有用户的公共条目。

SELECT ours.user, theirs.user, count(*) as `score`
FROM tableName as `theirs`, (SELECT *
                              FROM tableName
                              WHERE user = 'Mark') as `ours`
WHERE theirs.user != 'Mark' AND
      theirs.color = ours.color
GROUP BY theirs.user
ORDER BY score DESC

但是，如果有重复数据（即一个人选择两次相同的颜色），查询将无法运行。但是，如果您在评论中提到它不会发生，那不应该是一个问题。

可以修改查询以显示所有用户的分数：

SELECT ours.user as `myUser`, theirs.user as `theirUser`, count(*) as `score`
FROM tableName as `ours`, tableName as `theirs`
WHERE theirs.user != ours.user AND
      theirs.color = ours.color
GROUP BY ours.user, theirs.user
ORDER BY score DESC

让Q成为上述查询，为您提供最相似的用户。拥有该用户后，您可以使用它来显示它们之间的不同条目。这就是我们要做的事情：

SELECT * 
FROM tableName as theirs
WHERE user = 'Andrew'
AND NOT EXISTS (SELECT 1
                FROM tableName as ours
                WHERE ours.user = 'Mark'
                      AND ours.color = theirs.color)

从Andrew替换输入Mark和Q：

SELECT similar.myUser, theirs.user, theirs.color  
FROM tableName as theirs JOIN (Q) as similar
ON theirs.user = similar.theirUser
WHERE NOT EXISTS (SELECT 1
                  FROM tableName as ours
                  WHERE ours.user = similar.myUser
                        AND ours.color = theirs.color)

Here's the final query up and running。希望这是有道理的。

Answer 2

使用FULLTEXT INDEXES。您的查询将如下：

SELECT * FROM user WHERE MATCH (name,color) AGAINST ('Mark blue');

或者最简单的方法是使用LIKE搜索

 SELECT * FROM user WHERE name LIKE '%Mike%' OR color = 'blue'

您可以选择哪种方式更适合您

Answer 3

select 
    name,
    sum(case when t2.cnt > t1.cnt then t1.cnt else t2.cnt end) matches
from (
    select name, color, count(*) cnt
    from table
    where name <> 'Mark'
    group by name, color
) t1 left join (
    select color, count(*) cnt
    from table
    where name = 'Mark'
    group by color
) t2 on t2.color = t1.color
group by name
order by matches desc

派生表t1包含每个用户（Mark除外）的颜色数，t2包含与Mark相同的颜色。然后将这些表格连接在颜色上，并取2个计数中较小的一个，即如果Amy有2个红色且Mark有1个红色，则将1作为匹配数。最后按名称分组并返回最大的匹配数。

Answer 4

这应该让你接近。复杂性来自这样一个事实：您允许每个用户多次选择每种颜色，并要求在您要比较的其他用户中匹配每个相同的对。因此，我们真的很想知道用户每种颜色的总颜色选择数量，以及该数字与相同用户对同一颜色的数量的比较。

首先，我们创建一个派生关系，为我们做简单的数学运算（按颜色计算每个用户的选择数）：

 CREATE VIEW UserColorCounts (User, Color, TimesSeen) 
   AS SELECT User, Color, COUNT(*) FROM YourTable GROUP BY User, Color

其次，我们需要某种关系，将主要用户的每个颜色计数与每个次要用户的颜色计数进行比较：

 CREATE VIEW UserColorMatches (User, OtherUser, Color, TimesSeen, TimesMatched)
  AS SELECT P.User, S.User, P.Color, P.TimesSeen, LEAST(P.TimesSeen, S.TimesSeen)
  FROM UserColorCounts P LEFT OUTER JOIN UserColorCounts S
  ON P.Color = S.Color AND P.User <> S.User

最后，我们总计了每个主要用户的颜色计数，并与每个次要用户的匹配颜色计数进行比较：

 SELECT User, OtherUser, SUM(TimesMatched) AS Matched, SUM(TimesSeen) AS OutOf
    FROM UserColorMatches WHERE OtherUser IS NOT NULL
    GROUP BY User, OtherUser

Answer 5

   select match.name, count(*) as count 
     from table 
     join table as match 
           on match.name <> table.name 
          and table.name = 'mark' 
          and match.color = table.color 
    group by match.name 
    order by count(*) desc

Answer 6

以下查询会返回name和matching_name之间的匹配分数以及可获得的最高分数，以便您知道匹配所具有的％值。

此代码将color列中的重复值计为只有一个，因此如果您有两次记录Mark - Red，则只会计为1。

select
  foo.name, foo.matching_name, count(*) AS matching_score, goo.color_no AS max_score 
from 
(
  select 
    distinct a.name, a.color, b.name AS matching_name
  from 
    (
      select name, color from yourtable
    ) a
    left join yourtable b on a.color = b.color and a.name <> b.name
  where b.name is not null
) foo
left join ( select name, count(distinct color) AS color_no from yourtable group by name ) goo
  on foo.name = goo.name
group by foo.name, foo.matching_name

附加SQLFiddle以预览输出。

MySQL找到最相似的用户

6 个答案: