mysql - SQL分类

Question

我有一个系统可以跟踪用户查看的文档。每个文档都有它的 ID 和它所属的集群。我的系统跟踪 session ID 和查看次数。我现在想构建一个 SQL 查询，它会给我两列 - session ID 和分类集群。分类算法很简单:

1. select all sessions
2. for each session S
   I. prepare an accumulator ACC for clusters
   II. select the clusters of viewed documents for this session
   III. for each cluster C accumulate the cluster count ( ACC[C]++ )
   IV. find the maximum in the ACC. That is the cluster that the session was classified to

表结构如下，我用的是MySQL 5.5.16:

session

+-------+-----------+--------------------+
| ID    | sessionID | classified_cluster |
+-------+-----------+--------------------+

session 文档

+-------+-----------+------------+
| ID    | sessionID | documentID |
+-------+-----------+------------+

集群

+-------+-------+
| ID    | label |
+-------+-------+

集群文档

+-------+-----------+------------+
| ID    | clusterID | documentID |
+-------+-----------+------------+

所以基本上，我想为每个 session 选择集群，计算每个集群在查看文档中的出现次数，并找到最大出现次数。那么出现次数最多的集群的 ID 就是 session 的结果，因此最终结果集包含 session ID 和出现次数最多的集群:

结果

+-----------+-----------------------+
| sessionID | classifiedIntoCluster |
+-----------+-----------------------+

我设法通过以下查询获取每个 session (步骤 2/II.)的已查看文档的集群:

SELECT SD.session_id, CD.cluster_id 
FROM cluster_document AS CD 
INNER JOIN session_document AS SD 
ON CD.document_id = SD.document_id
WHERE session_id IN (SELECT session_id FROM session) 

我无法弄清楚其余部分。这甚至可以用于嵌套的 SELECT 查询吗？我应该使用光标吗？如果是的话，有人可以用光标展示一个例子吗？任何帮助都感激不尽。

编辑 #1:添加了 C# 实现、MySQL 转储和预期结果

C# 实现

    private void ClassifyUsers() {
        int nClusters = Database.SelectClusterCount(); //get number of clusters
        DataSet sessions = Database.SelectSessions(); //get all sessions
        foreach (DataRow session in sessions.Tables[0].Rows) { //foreach session
           int[] acc = new int[nClusters]; //prepare an accumulator for each known cluster
           string s_id = session["session_id"].ToString();
           DataSet sessionClusters = Database.SelectSessionClusters(s_id); //get clusters for this session

           foreach (DataRow cluster in sessionClusters.Tables[0].Rows) { //for each cluster
               int c = Convert.ToInt32(cluster["cluster_id"].ToString()) - 1;
               acc[c]++; //accumulate the cluster count
           }

           //find the maximum in the accumulator -> that is the most relevant cluster
           int max = 0;
           for (int j = 0; j < acc.Length; j++) {
               if (acc[j] >= acc[max]) max = j;
           }
           max++;
           Database.UpdateSessionCluster(s_id, max); //update the session with its new assigned cluster
       }
    }

表结构、测试数据和预期结果

Table structure and test data

Expected result

编辑 #2:添加了一个较小的数据集和进一步的算法演练

这是一个较小的数据集:

session

session id    |  cluster
abc                 0
def                 0
ghi                 0
jkl                 0       
mno                 0

集群

cluster_id  | label
1               A
2               B
3               C
4               D
5               E

SESSION_DOCUMENT

id      | session_id    |   document_id
1           abc             1
2           def             5
3           jkl             3
4           ghi             4
5           mno             2
6           def             2
7           abc             5
8           ghi             3

CLUSTER_DOCUMENT

id      | cluster_id    |   document_id
1           1                  2
2           1                  3
3           2                  5
4           3                  5
5           3                  1
6           4                  3
7           5                  2
8           5                  4

算法详解

第 1 步:获取 session 查看的文档的集群

session_id  |  cluster_id   | label     | document_id   
abc             3               C           1
abc             2               B           5
abc             3               C           5
-----
def             2               B           5
def             3               C           5   
def             1               A           2
def             5               E           2   
----
ghi             5               E           4   
ghi             1               A           3   
ghi             4               D           3   
----
jkl             1               A           3   
jkl             4               D           3   
----
mno             1               A           2
mno             5               E           2

第 2 步:计算簇的出现次数

session_id |    cluster_id  | label |   occurrence
abc             3               C           2   <--- MAX
abc             2               B           1
----
def             2               B           1
def             3               C           1   
def             1               A           1
def             5               E           1   <--- MAX
----
ghi             5               E           1   
ghi             1               A           1   
ghi             4               D           1   <--- MAX
----
jkl             1               A           1   
jkl             4               D           1   <--- MAX
----
mno             1               A           1   
mno             5               E           1   <--- MAX

第 3 步(最终结果):为每个 session 找到最大发生的簇(见上文)并构建最终结果集(session_id、cluster_id):

session_id |    cluster_id  
abc                 3           
def                 5
ghi                 4
jkl                 4
mno                 5

编辑 #3:已接受的答案说明

给出的两个答案都是正确的。他们都提供了解决问题的方法。我给了 Mosty Mostacho 可接受的答案，因为他首先交付了解决方案，并提供了带有 VIEW 的解决方案的另一个版本。来自 mankuTimma 的解决方案与 Mosty Mostacho 的解决方案质量相同。因此，我们有两个同样好的解决方案，我只是选择了 Mosty Mostacho，因为他是第一个。

感谢他们的贡献。.

Answer 1

好吧，我对如何在有很多相等的情况下选择一个事件表示怀疑，但查看 C# 代码似乎这个选择是不确定的。

现在，给定样本数据，第 2 步实际导致:

+------------+------------+-------+------------+
| SESSION_ID | CLUSTER_ID | LABEL | OCCURRENCE |
+------------+------------+-------+------------+
| abc        |          3 | C     |          2 |
| def        |          1 | A     |          1 |
| def        |          2 | B     |          1 |
| def        |          3 | C     |          1 |
| def        |          5 | E     |          1 |
| ghi        |          1 | A     |          1 |
| ghi        |          4 | D     |          1 |
| ghi        |          5 | E     |          1 |
| jkl        |          1 | A     |          1 |
| jkl        |          4 | D     |          1 |
| mno        |          1 | A     |          1 |
| mno        |          5 | E     |          1 |
+------------+------------+-------+------------+

因此，继续处理这些数据，我得到了该 session ID 的 session_id 和 max(cluster_id)，结果是:

+------------+------------+
| SESSION_ID | CLUSTER_ID |
+------------+------------+
| abc        |          3 |
| def        |          5 |
| ghi        |          5 |
| jkl        |          4 |
| mno        |          5 |
+------------+------------+

max(cluster_id) 就是用来执行非确定性选择的。这是查询:

select s1.session_id, max(s1.cluster_id) as cluster_id from (
  select sd.session_id, cd.cluster_id, count(*) as Occurrence
  from session_document sd
  join cluster_document cd
  on sd.document_id = cd.document_id
  join cluster c
  on c.cluster_id = cd.cluster_id
  group by sd.session_id, cd.cluster_id, c.label
) as s1
left join (
  select sd.session_id, count(*) as Occurrence
  from session_document sd
  join cluster_document cd
  on sd.document_id = cd.document_id
  join cluster c
  on c.cluster_id = cd.cluster_id
  group by sd.session_id, cd.cluster_id, c.label
) as s2
on s1.session_id = s2.session_id and s1.occurrence < s2.occurrence
where s2.occurrence is null
group by s1.session_id

也许添加一个 View 会提高性能(替换上面的查询):

create view MaxOccurrences as (
  select sd.session_id, cd.cluster_id, count(*) as Occurrence
  from session_document sd
  join cluster_document cd
  on sd.document_id = cd.document_id
  join cluster c
  on c.cluster_id = cd.cluster_id
  group by sd.session_id, cd.cluster_id, c.label
);

select s1.session_id, max(s1.cluster_id) as cluster_id
from MaxOccurrences as s1
left join MaxOccurrences as s2
on s1.session_id = s2.session_id and s1.occurrence < s2.occurrence
where s2.occurrence is null
group by s1.session_id

让我知道它是否有效。