SQL查找对等关系

时间：2018-09-28 19:33:39

标签： sql sql-server select sql-server-2017 dataexplorer

我正在尝试使用Stack Exchange Data Explorer（SEDE）查找一种情况，其中Stack Overflow上的两个不同的用户已经接受了彼此的答案。例如：

Post A { Id: 1, OwnerUserId: "user1", AcceptedAnswerId: "user2" }

和

Post B { Id: 2, OwnerUserId: "user2", AcceptedAnswerId: "user1" }

我目前有一个查询，可以找到两个在问题上进行过合作的用户作为发问者-回答者，但它不能确定这种关系是否是对等的：

SELECT user1.Id AS User_1, user2.Id AS User_2
FROM Posts p
INNER JOIN Users user1 ON p.OwnerUserId = user1.Id
INNER JOIN Posts p2 ON p.AcceptedAnswerId = p2.Id
INNER JOIN Users user2 ON p2.OwnerUserId = user2.Id
WHERE p.OwnerUserId <> p2.OwnerUserId
AND p.OwnerUserId IS NOT NULL
AND p2.OwnerUserId IS NOT NULL
AND user1.Id <> user2.Id
GROUP BY user1.Id, user2.Id HAVING COUNT(*) > 1;

对于不熟悉该架构的任何人，都有两个这样的表：

Posts
--------------------------------------
Id                      int
PostTypeId              tinyint
AcceptedAnswerId        int
ParentId                int
CreationDate            datetime
DeletionDate            datetime
Score                   int
ViewCount               int
Body                    nvarchar (max)
OwnerUserId             int
OwnerDisplayName        nvarchar (40)
LastEditorUserId        int
LastEditorDisplayName   nvarchar (40)
LastEditDate            datetime
LastActivityDate        datetime
Title                   nvarchar (250)
Tags                    nvarchar (250)
AnswerCount             int
CommentCount            int
FavoriteCount           int
ClosedDate              datetime
CommunityOwnedDate      datetime

和

Users
--------------------------------------
Id                      int
Reputation              int
CreationDate            datetime
DisplayName             nvarchar (40)
LastAccessDate          datetime
WebsiteUrl              nvarchar (200)
Location                nvarchar (100)
AboutMe                 nvarchar (max)
Views                   int
UpVotes                 int
DownVotes               int
ProfileImageUrl         nvarchar (200)
EmailHash               varchar (32)
AccountId               int

5 个答案:

答案 0 :(得分：2)

最简单形式的查询（这样它就不会超时查询16M个问题）将是：

WITH accepter_acceptee(a, b) AS (
    SELECT q.OwnerUserId, a.OwnerUserId
    FROM Posts AS q
    INNER JOIN Posts AS a ON q.AcceptedAnswerId = a.Id
    WHERE q.PostTypeId = 1 AND q.OwnerUserId <> a.OwnerUserId
), collaborations(a, b, type) AS (
    SELECT a, b, 'a accepter b' FROM accepter_acceptee
    UNION ALL
    SELECT b, a, 'a acceptee b' FROM accepter_acceptee
)
SELECT a, b, COUNT(*) AS [collaboration count]
FROM collaborations
GROUP BY a, b
HAVING COUNT(DISTINCT type) = 2
ORDER BY a, b

结果：

答案 1 :(得分：1)

使用Salman A's answer中的技术，改进了排序并添加了一些更有用的列。

结合my other answer中的查询，它显示了一些有趣的关系。

See it in SEDE. 

WITH QandA_users AS (
    SELECT      q.OwnerUserId   AS userQ
                , a.OwnerUserId AS userA
    FROM        Posts q
    INNER JOIN  Posts a         ON q.AcceptedAnswerId = a.Id
    WHERE       q.PostTypeId    = 1
),
pairsUnion (user1, user2, whoAnswered) AS (
    SELECT  userQ, userA, 'usr 2 answered'
    FROM    QandA_users
    WHERE   userQ <> userA
    UNION ALL
    SELECT  userA, userQ, 'usr 1 answered'
    FROM    QandA_users
    WHERE   userQ <> userA
),
collaborators AS (
    SELECT      user1, user2, COUNT(*) AS [Reciprocations]
    FROM        pairsUnion
    GROUP BY    user1, user2
    HAVING COUNT (DISTINCT whoAnswered) > 1
)
SELECT
            'site://u/' + CAST(c.user1 AS NVARCHAR) + '|Usr ' + u1.DisplayName      AS [User 1]
            , 'site://u/' + CAST(c.user2 AS NVARCHAR) + '|Usr ' + u2.DisplayName    AS [User 2]
            , c.Reciprocations                                                      AS [Reciprocal Accptd posts]
            , (SELECT COUNT(*)  FROM QandA_users qau  WHERE qau.userQ = c.user1)    AS [Usr 1 Qstns wt Accptd]
            , (SELECT COUNT(*)  FROM QandA_users qau  WHERE qau.userQ = c.user1  AND qau.userA = c.user2) AS [Accptd Ansr by Usr 2]
            , (SELECT COUNT(*)  FROM QandA_users qau  WHERE qau.userA = c.user2)    AS [Usr 2 Ttl Accptd Answrs]
FROM        collaborators c
INNER JOIN  Users u1        ON u1.Id = c.user1
INNER JOIN  Users u2        ON u2.Id = c.user2
ORDER BY    c.Reciprocations DESC
            , u1.DisplayName
            , u2.DisplayName

结果如下：

results

答案 2 :(得分：0)

这就是我要做的。以下是一些简化的数据：

if object_id('tempdb.dbo.#Posts') is not null drop table #Posts
create table #Posts
(
    PostId char(1),
    OwnerUserId int,
    AcceptedAnswerUserId int
)

insert into #Posts
values
('A', 1, 2),
('B', 2, 1),
('C', 2, 3),
('D', 2, 4),
('E', 3, 1),
('F', 4, 1)

出于我们的目的，我们并不真正在意PostId，而我们的出发点是一组成对的帖子拥有者（OwnerUserId）和接受的答卷者（{ {1}}。

（尽管不是必需的，您可以像这样可视化设置）

AcceptedAnswerUserId

现在，我们要查找该集合中所有具有相反两个字段的条目。即所有者（如果一个帖子是另一个帖子的已接受答案）。因此，在一对是（1，2）的情况下，我们想找到（2，1）。

我使用左联接进行了此操作，因此您可以看到它省略的行，但是将其更改为内部联接会将其限制为您所描述的集合。您可以随意获取信息（通过从帽子中挑选任何一列，或者如果您希望将它们放在一行中，则从其中一个表中返回两列）。

select distinct OwnerUserId, AcceptedAnswerUserId
from #Posts

编辑。如果要排除自己的答案，只需在select u1.OwnerUserId, u1.AcceptedAnswerUserId, u2.OwnerUserId, u2.AcceptedAnswerUserId from #Posts u1 left outer join #Posts u2 on u1.AcceptedAnswerUserId = u2.OwnerUserId and u1.OwnerUserId = u2.AcceptedAnswerUserId 子句中添加and u1.AcceptedAnswerUserId != u1.OwnerUserId。

就个人而言，我总是觉得很有趣的是，SQL和关系代数在集合论中扎根多么深，但是在SQL中进行基于集合的操作却感觉很笨拙。通常是因为为了保持顺序的缺乏，您必须在单个列中表示集合成员。但是然后要比较SQL中的集合成员，您需要将集合成员表示为单独的列。

现在考虑一下，您如何将其扩展到对同一则帖子发表评论的三合会用户？

答案 3 :(得分：0)

一个CTE和简单的inner joins就可以了。正如我在其他答案中观察到的那样，不需要那么多的代码。请注意我的很多评论。

链接到 StackExchange Data Explorer 并保存示例结果

with questions as ( -- this is needed so that we have ids of users asking and answering
select
   p1.owneruserid as question_userid
 , p2.owneruserid as answer_userid
 --, p1.id -- to view sample ids
from posts p1
inner join posts p2 on -- to fetch answer post
  p1.acceptedanswerid = p2.id
)
select distinct -- unique pairs
    q1.question_userid as userid1
  , q1.answer_userid as userid2
  --, q1.id, q2.id -- to view sample ids
from questions q1
inner join questions q2 on
      q1.question_userid = q2.answer_userid -- accepted answer from someone
  and q1.answer_userid = q2.question_userid -- who also accepted our answer
  and q1.question_userid <> q1.answer_userid -- and we aren't self-accepting

以帖子为例：

尽管如此，由于大型数据集和distinct部分，StackExchange可能会使您超时。如果您想查看一些数据，请删除distinct并在开始时添加top N：

with questions as (
...
)
select top 3 ...

答案 4 :(得分：0)

ETA：糟糕。误解了问题； Op想要 Accepted 答案，而以下是任何互惠答案。 （修改起来很容易，但我还是对后者更感兴趣。）

由于数据集非常大（并且需要不使SEDE超时），我选择限制AMAP集并从那里构建。

所以这个查询：

  1. 如果存在对等关系，则仅返回任何行。
  2. 返回所有此类问答对。
  3. 排除自我答案。
  4. 利用SEDE's query parameters and magic columns的可用性。

See it live in SEDE. 

-- UserA: Enter ID of user A
-- UserB: Enter ID of user B
WITH possibleAnswers AS (
    SELECT
                a.Id                AS aId
                , a.ParentId        AS qId
                , a.OwnerUserId   
                , a.CreationDate
    FROM        Posts a
    WHERE       a.PostTypeId        = 2  --  answers
    AND         a.OwnerUserId       IN (##UserA:INT##, ##UserB:INT##)
),
possibleQuestions AS (
    SELECT
                q.Id                AS qId
                , q.OwnerUserId   
                , q.Tags
    FROM        Posts q
    INNER JOIN  possibleAnswers pa  ON q.Id = pa.qId
    WHERE       q.PostTypeId        = 1  --  questions
    AND         q.OwnerUserId       IN (##UserA:INT##, ##UserB:INT##)
    AND         q.OwnerUserId       != pa.OwnerUserId  --  No self answers
)
SELECT 
            pa.OwnerUserId          AS [User Link]
            , 'answers'             AS [Action]
            , pq.OwnerUserId        AS [User Link]
            , pa.CreationDate       AS [at]
            , pq.qId                AS [Post Link]
            , pq.Tags
FROM        possibleQuestions pq
INNER JOIN  possibleAnswers pa      ON pq.qId = pa.qId
WHERE       pq.OwnerUserId          =  ##UserB:INT##
AND         EXISTS (SELECT * FROM possibleQuestions pq2  WHERE pq2.OwnerUserId =  ##UserA:INT##)

UNION ALL SELECT 
            pa.OwnerUserId          AS [User Link]
            , 'answers'             AS [Action]
            , pq.OwnerUserId        AS [User Link]
            , pa.CreationDate       AS [at]
            , pq.qId                AS [Post Link]
            , pq.Tags
FROM        possibleQuestions pq
INNER JOIN  possibleAnswers pa      ON pq.qId = pa.qId
WHERE       pq.OwnerUserId          =  ##UserA:INT##
AND         EXISTS (SELECT * FROM possibleQuestions pq2  WHERE pq2.OwnerUserId =  ##UserB:INT##)

ORDER BY    pa.CreationDate

它产生的结果如下（单击以查看大图）：

results

有关所有此类用户对的列表，请参见this SEDE query。

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