在Postrgres聚合中过滤

Question

我在Postgres有一张桌子叫tasks.它记录了机械土耳其式的任务.它包含以下列:

entity_name, text (the thing being reviewed)
reviewer_email, text (the email address of the person doing the reviewing)
result, boolean (the entry provided by the reviewer)

需要检查的每个实体都会生成两个任务行,每个行分配给不同的审阅者.当两个评论者都不同意时(例如他们的价值result不相等),应用程序将启动第三个任务,分配给主持人.主持人始终拥有相同的电子邮件域.

我试图获得每次评论者的评论,审稿人被主持人否决,或者由主持人确认.我认为我相当接近,但最后一点证明是棘手的:

SELECT
  reviewer_email,
  COUNT(*) FILTER(
    WHERE entity_name IN (
      SELECT entity_name
      FROM tasks
      GROUP BY entity_name
      HAVING
        COUNT(*) FILTER (WHERE result IS NOT NULL) = 3 -- find the entities that have exactly three reviews
      AND
        -- this is the tricky part: 
        -- need something like:
        -- WHERE current_review.result = moderator_review.result
    )
  ) AS overruled_count
FROM
  tasks
WHERE
  result IS NOT NULL
GROUP BY
  reviewer_email
HAVING
  reviewer_email NOT LIKE '%@moderators-domain.net'

样本数据:

id | entity_name | reviewer_email             | result
 1 | apple       | bob@email.net              | true
 2 | apple       | alice@email.net            | false
 3 | apple       | mod@@moderators-domain.net | true
 4 | pair        | bob@email.net              | true
 5 | pair        | alice@email.net            | false
 6 | pair        | mod@@moderators-domain.net | false
 7 | kiwi        | bob@email.net              | true
 8 | kiwi        | alice@email.net            | true

期望的结果:

reviewer_email  | overruled_count | affirmed_count
bob@email.net   |               1 |              1
alice@email.net |               1 |              1

Bob和Alice各自做了三次评论.在一次审查中,他们同意,因此没有适度.他们对其他两次评论持不同意见,并被推翻一次,并由主持人确认一次.

我相信上面的代码让我走在正确的轨道上,但我肯定对其他方法感兴趣.

Answer 1

我认为这比你可能意识到的更难.以下内容将主持人审核附加到每个非主持人审核:

select t.*, tm.result as moderator_result
from tasks t join
    tasks tm
    on t.entity_name = tm.entity_name
where t.reviewer_email NOT LIKE '%@moderators-domain.net' and
      tm.reviewer_email LIKE '%@moderators-domain.net';

从这里,我们可以汇总您想要的结果:

select reviewer_email,
       sum( (result = moderator_result)::int ) as moderator_agrees,
       sum( (result <> moderator_result)::int ) as moderator_disagrees
from (select t.*, tm.result as moderator_result
      from tasks t join
           tasks tm
           on t.entity_name = tm.entity_name
      where t.reviewer_email NOT LIKE '%@moderators-domain.net' and
            tm.reviewer_email LIKE '%@moderators-domain.net'
     ) t
group by reviewer_email;

可能有一种方法可以使用filter甚至是窗口功能.这种方法对我来说似乎最自然.

我应该注意,子查询当然没有必要:

select t.reviewer_email,
       sum( (t.result = tm.result)::int ) as moderator_agrees,
       sum( (t.result <> tm.result)::int ) as moderator_disagrees
from tasks t join
     tasks tm
     on t.entity_name = tm.entity_name
where t.reviewer_email NOT LIKE '%@moderators-domain.net' and
      tm.reviewer_email LIKE '%@moderators-domain.net'
group by t.reviewer_email;