Question

我希望联接两个表，但是仅通过左表出现次数来获得联接列的平均值

文档：

+-----+-----+-------+
| dId | name| score |
+-----+-----+-------+
| A   | n1  | 100   |
| B   | n1  | 70    |
+-----+-----+-------+

实体：

+------+------------+-----+
| ename| details    | dId |
+------+------------+-----+
| e1   | a          |   A |
| e2   | a          |   A |
| e3   | b          |   A |
| e4   | c          |   B |
+------+------------+-----+

预期输出：

+------+--------+---------------+
| name | average| entities      |
+------+--------+---------------+
| n1    | 85    |e1, e2, e3, e4 |
+------+--------+---------------+

因为（100 + 70）/ 2 = 85

当前输出：

+------+--------+---------------+
| name | average| entities      |
+------+--------+---------------+
| n1    | 92.5  |e1, e2, e3, e4 |
+------+--------+---------------+

因为（100 + 100 + 100 + 70）/ 4 = 92.5

当前查询：

SELECT
  docT.name,
  AVG(docT.score),
  STRING_AGG(entityT.ename)
FROM
  document_sentiment docT
JOIN
  entity_sentiment entityT
ON
  docT.dId = entityT.dId
GROUP BY
  docT.cname

如何获得预期输出中的分数？

Answer 1

尝试以下代码

select name, ename, avg(score) as score
from (SELECT
  docT.name,
  doct.Did,
  MAX(docT.score) as score,
  STRING_AGG(entityT.ename) as ename
FROM
  document_sentiment docT
JOIN
  entity_sentiment entityT
ON
  docT.dId = entityT.dId
GROUP BY
  docT.cname, doct.Did
) sub
group by name, ename

Answer 2

尝试一下

select  t.name, av,  
    GROUP_CONCAT(DISTINCT entityT.name ORDER BY entityT.name SEPARATOR ', ') AS entities
from (
    SELECT docT.dId, docT.name,
          AVG(docT.score) av
    FROM document_sentiment docT
    GROUP BY docT.name) T
JOIN entity_sentiment entityT ON T.dId = entityT.dId
GROUP BY T.name

SQL Fiddle

Answer 3

以下是用于BigQuery标准SQL

#standardSQL
SELECT
  docT.name,
  AVG(docT.score) average,
  STRING_AGG(entityT.ename) entities
FROM `project.dataset.document_sentiment` docT
JOIN (
  SELECT dId, STRING_AGG(ename) ename
  FROM `project.dataset.entity_sentiment`
  GROUP BY dId
) entityT
ON docT.dId = entityT.dId
GROUP BY docT.name

您可以使用问题中的示例数据来测试，玩游戏，如下例所示

#standardSQL
WITH `project.dataset.document_sentiment` AS (
  SELECT 'A' dId, 'n1' name, 100 score UNION ALL
  SELECT 'B', 'n1', 70 
), `project.dataset.entity_sentiment` AS (
  SELECT 'e1' ename, 'a' details, 'A' dId UNION ALL
  SELECT 'e2', 'a', 'A' UNION ALL
  SELECT 'e3', 'b', 'A' UNION ALL
  SELECT 'e4', 'c', 'B' 
)
SELECT
  docT.name,
  AVG(docT.score) average,
  STRING_AGG(entityT.ename) entities
FROM `project.dataset.document_sentiment` docT
JOIN (
  SELECT dId, STRING_AGG(ename) ename
  FROM `project.dataset.entity_sentiment`
  GROUP BY dId
) entityT
ON docT.dId = entityT.dId
GROUP BY docT.name  

Row name    average     entities     
1   n1      85.0        e1,e2,e3,e4

Answer 4

这很棘手。我认为窗口函数可能是最简单的解决方案：

SELECT docT.name, docT.avg_score,
       STRING_AGG(entityT.ename)
FROM (SELECT docT.*,
             AVG(docT.score) OVER (PARTITION BY docT.cname) as avg_score
      FROM document_sentiment docT
     ) docT JOIN
     entity_sentiment entityT
    ON docT.dId = entityT.dId
GROUP BY docT.cname, docT.avg_score;

为什么这很棘手？好吧，如果您按cname进行汇总，则您将损失dId而无法进行JOIN。

预聚合无法解决问题。幸运的是，这可以通过使用窗口函数来解决。

SQL-连接两个表，但仅通过左表出现次数来获得连接列的平均值

4 个答案: