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我试图找出哪些类之间的重叠最多。数据存储在 MySQL 中，每个学生在他/她参加的每个课程的数据库中都有一个完全独立的行（我没有配置它，我无法更改它）。我粘贴了下表的简化版本。实际上有大约 20 种不同的课程。

CREATE TABLE classes
(`student_id` int, `class` varchar(13));
INSERT INTO classes
(`student_id`, `class`)
VALUES
(55421, 'algebra'),
(27494, 'algebra'),
(64934, 'algebra'),
(65364, 'algebra'),
(21102, 'algebra'),
(90734, 'algebra'),
(20103, 'algebra'),
(57450, 'gym'),
(76411, 'gym'),
(24918, 'gym'),
(65364, 'gym'),
(55421, 'gym'),
(89607, 'world_history'),
(54522, 'world_history'),
(49581, 'world_history'),
(84155, 'world_history'),
(55421, 'world_history'),
(57450, 'world_history');

我最终想使用 Circos（此处为背景），但我对任何能让我理解并向人们展示重叠最多和最少重叠的方法感到满意。这不是我的想法，但我在想我可以为每门课程使用一个包含一行和一列的输出表，并列出不同类相交的重叠数量。每门课程与其自身相交的位置可以显示与任何其他类别没有重叠的人数。

Excel 中 3x3 矩阵的屏幕截图

4

2 回答 2

1

您可以通过生成表示链接的结果来做到这一点： src -> dst = nb

1) 获取矩阵

select c1.class src_class, c2.class dst_class
from (select distinct class from classes) c1
join (select distinct class from classes) c2
order by src_class, dst_class

“选择不同的类”不是生成矩阵所必需的，您可以直接选择类和GROUP BY。但是，在第 2 步，我们需要独特的结果。

结果 ：

src_class      dst_class
-----------------------------
algebra        algebra
algebra        gym
algebra        world_history
gym            algebra
gym            gym
gym            world_history
world_history  algebra
world_history  gym
world_history  world_history

2）加入匹配源和目标的学生列表

select c1.class src_class, c2.class dst_class, count(v.student_id) overlap
from (select distinct class from classes) c1
join (select distinct class from classes) c2
left join classes v on
(
    v.class = c1.class
    and v.student_id in (select student_id from classes
                         where class = c2.class)
)
group by src_class, dst_class
order by src_class, dst_class

不同的值（第 1 步）允许我们获取所有类，即使它们不是链接（并改为 0）。

结果 ：

src_class      dst_class      overlap
-------------------------------------
algebra        algebra           7
algebra        gym               2
algebra        world_history     1
gym            algebra           2
gym            gym               5
gym            world_history     2
world_history  algebra           1
world_history  gym               2
world_history  world_history     6

3 - 如果类相等，则进行不同的计算

select c1.class src_class, c2.class dst_class, count(v.student_id) overlap
from (select distinct class from classes) c1
join (select distinct class from classes) c2
left join classes v on
(
    v.class = c1.class and
    (
        -- When classes are equals
        -- Students presents only in that class
        (c1.class = c2.class
         and 1 = (select count(*) from classes
                  where student_id = v.student_id))
    or
        -- When classes are differents
        -- Students present in both classes
        (c1.class != c2.class
         and v.student_id in (select student_id from classes
                              where class = c2.class))
    )
)
group by src_class, dst_class
order by src_class, dst_class

结果 ：

src_class      dst_class      overlap
-------------------------------------
algebra        algebra           5
algebra        gym               2
algebra        world_history     1
gym            algebra           2
gym            gym               2
gym            world_history     2
world_history  algebra           1
world_history  gym               2
world_history  world_history     4
于 2016-05-10T22:40:10.303 回答
1

只需使用自联接和聚合：

select c1.class, c2.class, count(*)
from classes c1 join
     classes c2
     on c1.student_id = c2.student_id
group by c1.class, c2.class;

这不会以完全相同的格式生成它。

于 2016-05-10T21:58:12.793 回答