基于不同列的列重叠

Question

我在 Postgres 中有一个简单的 table，其中包含星期几和人员。

INSERT INTO mytable (day, person)
values
('Monday', 'A'),
('Monday', 'A'),
('Monday', 'B'),
('Tuesday', 'A'),
('Wednesday', 'A'),
('Wednesday', 'B'),
('Thursday', 'B'),
('Thursday', 'B');

我想找出每一天重叠的人数。因此，例如，星期一有 A 和 B。星期二只有 B，因此会产生如下一行：

('Monday', 'Tuesday', 1)

最终输出应如下所示：

('Monday', 'Monday', 2),
('Monday', 'Tuesday', 1),
('Monday', 'Wednesday', 2),
('Monday', 'Thursday', 1),
('Tuesday', 'Tuesday', 1),
('Tuesday', 'Monday', 1),
('Tuesday', 'Wednesday', 1),
('Tuesday', 'Thursday', 0),
('Wednesday', 'Wednesday', 2),
('Wednesday', 'Monday', 2),
('Wednesday', 'Tuesday', 1),
('Wednesday', 'Thursday', 1),
('Thursday', 'Thursday', 1),
('Thursday', 'Monday', 1),
('Thursday', 'Tuesday', 0),
('Thursday', 'Wednesday', 1)

我一直在尝试交叉连接和相交，但似乎没有产生正确的结果。

Answer 1

使用自连接：

select t1.day, t2.day, count(distinct t2.person) as num_overlaps
from mytable t1 join
     mytable t2
     on t1.person = t2.person
group by t1.day, t2.day
order by t1.day, t2.day;

编辑：

我看到你也想包括零重叠。这有点棘手。为此，您需要生成所有天的组合，然后加入数据：

select d1.day, d2.day, count(distinct t2.person)
from (select distinct day from mytable) d1 cross join
     (select distinct day from mytable) d2 left join
     mytable t1
     on t1.day = d1.day left join
     mytable t2
     on t2.day = d2.day and t2.person = t1.person
group by d1.day, d2.day
order by d1.day, d2.day;

Here 是一个 db<>fiddle.

基于不同列的列重叠

Overlap of a column based on a different column

sql

postgresql

postgresql-9.5