Question

我的数据如下：

Person  |  Group
-----------------
Andy    |  1
Andy    |  2
Doug    |  2
Jack    |  1
Carl    |  2
Joe     |  1
Joe     |  2

我需要这些数据的输出，如下所示：

num_persons  |  num_persons_in_both_groups  |  overlap
--------------------------------------------------------
5            |  2                           |  40%

我可以拿到num_persons，但对num_persons_in_both_groups感到困惑。如何使用SQL获得此信息？假设它是MySQL。 overlap很简单。

谢谢！

Answer 1

好问题！请尝试以下操作：

SELECT SUM(GroupCount) AS num_persons
    , SUM(CASE WHEN PersonsInGroups > 1 THEN 1 ELSE 0 END) AS num_persons_in_both_groups
    , (SUM(CASE WHEN PersonsInGroups > 1 THEN 1 ELSE 0 END) / SUM(GroupCount))*100 AS overlap
FROM
(
SELECT COUNT(1) AS PersonsInGroups
    , 1 AS GroupCount
FROM t
GROUP BY Person
) x

根据评论，实际上您可以这样写：

SELECT COUNT(1) AS num_persons
    , SUM(CASE WHEN PersonsInGroups > 1 THEN 1 ELSE 0 END) AS num_persons_in_both_groups
    , (SUM(CASE WHEN PersonsInGroups > 1 THEN 1 ELSE 0 END) / COUNT(1))*100 AS overlap
FROM
(
SELECT COUNT(1) AS PersonsInGroups
FROM t
GROUP BY Person
) x

Answer 2

我建议两种聚合级别：

select count(*) as num_persons,
       sum(group_1 > 0 and group_2 > 0) as in_both,
       avg(group_1 > 0 and group_2 > 0) as ratio
from (select person,
             max( group = 1 ) as group_1,
             max( group = 2 ) as group_2
      from t
      group by person
     ) p

我喜欢它，因为它很灵活。

但是，您也可以在没有子查询的情况下执行此操作，假设恰好有两个组：

select count(distinct person) as num_persons,
       count(*) - count(distinct person) as num_in_both,
       (count(*) - count(distinct person)) / count(distinct person) as ratio
from t;

计算两组的人数

2 个答案: