数据库

数据库由两个表组成。第一个表是用户，我正在尝试比较。

id | self_ranking
----------------------------------
1       | 9
2       | 3
3       | 2

第二个表格是用户对特定项目的一系列分数。

id | user_id | item_id | score
----------------------------------
1  | 1       | 1       | 4
2  | 1       | 2       | 5
3  | 1       | 3       | 8
4  | 1       | 4       | 3

任务

我想找到给定的“最相似”的用户，同等地评估所有排名的项目（以及自我评分）。因此，完全匹配将是以完全相同的方式对所有相同项目进行排序的用户。对自己的评价是相同的，而下一个最佳选择是一个项目的排名略有差异。

我遇到了困难：

以有效的方式加入两个表格
处理不是所有用户都排名相同的事实。我们只想比较相同项目的排名。

有人可以帮我构建合理的查询吗？我对MySQL并不十分强大，如果这个答案显而易见，那就很抱歉。

输出

如果用户4已将自己排名为8且项目1 => 4且2 => 5，则我希望查询用户4的最近用户返回1，即最近用户的user_id。 / p>

Answer 1

在@ eggyal方法的细微改进中，我合并了我们能够匹配的项目数量。

SELECT   u2.user_id

-- join our user to their scores
FROM     (users u1 JOIN scores s1 USING (user_id))

-- and then join other users and their scores
    JOIN (users u2 JOIN scores s2 USING (user_id))
      ON s1.item_id  = s2.item_id
     AND u1.user_id != u2.user_id

-- filter for our user of interest
WHERE    u1.user_id = ?

-- group other users' scores together
GROUP BY u2.user_id

-- subtract the degree of difference in correlating scores from the number of correlating scores
ORDER BY (SUM(s1.item_id = s2.item_id) - 
  ( SUM(ABS(s2.score - s1.score) + ABS(u2.self - u1.self) ) ) ) DESC

Answer 2

SELECT   u2.user_id

-- join our user to their scores
FROM     (users u1 JOIN scores s1 USING (user_id))

-- and then join other users and their scores
    JOIN (users u2 JOIN scores s2 USING (user_id))
      ON s1.item_id  = s2.item_id
     AND u1.user_id != u2.user_id

-- filter for our user of interest
WHERE    u1.user_id = ?

-- group other users' scores together
GROUP BY u2.user_id

-- and here's the magic: order in descending order of "distance" between
-- our selected user and all of the others: you may wish to weight
-- self_ranking differently to item scores, in which case just multiply
-- appropriately
ORDER BY SUM(ABS(s2.score - s1.score))
       + ABS(u2.self_ranking - u1.self_ranking) DESC

MySQL Query查找最相似的数字行

数据库

任务

输出

2 个答案: