如何在表达式中多次使用 COUNT() 实现 OVER？

Question

我对为解决 LeetCode 中的问题而编写的查询有疑问。问题是：

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| ad_id         | int     |
| user_id       | int     |
| action        | enum    |
+---------------+---------+

(ad_id, user_id) is the primary key for this table.

Each row of this table contains the ID of an Ad, the ID of a user and the action taken by this user regarding this Ad. The action column is an ENUM type of ('Clicked', 'Viewed', 'Ignored').

A company is running Ads and wants to calculate the performance of each Ad.

Performance of the Ad is measured using Click-Through Rate (CTR) where:

CTR = { 0 if no ad clicks, Ad clicks / (Ad clicks + Ad views) otherwise

Write an SQL query to find the ctr of each Ad.

Round ctr to 2 decimal points. Order the result table by ctr in descending order and by ad_id in ascending order in case of a tie.

The query result format is in the following example:

Ads table:

+-------+---------+---------+
| ad_id | user_id | action  |
+-------+---------+---------+
| 1     | 1       | Clicked |
| 2     | 2       | Clicked |
| 3     | 3       | Viewed  |
| 5     | 5       | Ignored |
| 1     | 7       | Ignored |
| 2     | 7       | Viewed  |
| 3     | 5       | Clicked |
| 1     | 4       | Viewed  |
| 2     | 11      | Viewed  |
| 1     | 2       | Clicked |
+-------+---------+---------+

这是一个 fiddle 示例数据和我尝试的解决方案。尝试的解决方案转载如下：

SELECT DISTINCT t.ad_id, ROUND(
    IF(
        COUNT(c.ad_id) OVER (PARTITION BY t.ad_id) = 0, 
        0,
      COUNT(c.ad_id) OVER (PARTITION BY t.ad_id) * 100 / ( COUNT(c.ad_id) OVER (PARTITION BY t.ad_id) + COUNT(v.ad_id) OVER (PARTITION BY t.ad_id) )
    ), 2) as ctr
FROM Ads as t
LEFT JOIN Ads as c ON c.ad_id=t.ad_id AND c.user_id=t.user_id AND c.action='Clicked'
LEFT JOIN Ads as v ON v.ad_id=t.ad_id AND v.user_id=t.user_id AND v.action='Viewed'
GROUP BY t.ad_id, c.ad_id, v.ad_id
ORDER BY ctr DESC, t.ad_id

此查询的结果：

ad_id   ctr
1   50.00
2   50.00
3   50.00
5   0.00

正确的结果应该是：

ad_id ctr
1, 66.67
3, 50.00
2, 33.33
5, 0.00

通过查看示例数据，我的猜测是 COUNT() 实际上并未像我预期的那样按 t.ad_id 进行分区。 50% 的点击率结果可以解释为我的点击率计算计算了所有 'Clicked' 和所有 'Viewed' 实例。（另一方面，删除 CTR 计算中的 OVER 语句 - 只是计算，而不是条件 - 不会产生上面的结果，正如我的假设所暗示的那样。所以我不确定这一点。）

我使用OVER的方式有问题吗？我的逻辑在这里有缺陷吗？

另外，我还有一个额外的问题：我选择在这里使用 JOIN 是因为我假设 JOIN 比使用子查询更快。这是一个公平的假设吗？我正在为数据分析师 1 面试而学习 - 你认为面试官会关心我是否使用 JOIN 与子查询吗？

编辑：感谢 forpas 的解释，我能够想出一个比我原来的解决方案简单得多的解决方案。我认为 forpas 在他下面的回答中的解决方案可能仍然更可取，因为它明确处理 table.

中的 NULL。

SELECT ad_id, ROUND(IF(
    SUM(action='Clicked') = 0,
    0,
    SUM(action='Clicked') * 100 / ( SUM(action='Clicked') + SUM(action='Viewed'))
), 2) as ctr
FROM Ads
GROUP BY ad_id
ORDER BY ctr DESC, ad_id

Answer 1

您可以使用条件聚合来完成：

SELECT ad_id,
  ROUND(100 * COALESCE(SUM(action = 'Clicked') / SUM(action IN ('Clicked', 'Viewed')), 0), 2) ctr
FROM Ads
GROUP BY ad_id
ORDER BY ctr DESC, ad_id;

您可以使用 SUM() window 函数获得相同的结果，但我不认为这对性能或可读性更好：

SELECT DISTINCT ad_id,
  ROUND(
    100 * 
    COALESCE(
      SUM(action = 'Clicked') OVER (PARTITION BY ad_id) / 
      SUM(action IN ('Clicked', 'Viewed')) OVER (PARTITION BY ad_id)
      , 0
    )
    , 2
  ) ctr
FROM Ads
ORDER BY ctr DESC, ad_id;

参见demo。
结果：

> ad_id |   ctr
> ----: | ----:
>     1 | 66.67
>     3 | 50.00
>     2 | 33.33
>     5 |  0.00

Answer 2

SELECT t1.ad_id,
(CASE 
 WHEN t2.clicked/t1.total IS NULL THEN 0 
 ELSE round((t2.clicked/t1.total)*100,2) END) as ctr 
FROM
(SELECT ad_id,SUM(CASE WHEN action IN ('Viewed','Clicked') THEN 1 ELSE 0 END) as total
FROM Ads
GROUP BY 1
)t1
LEFT JOIN 
(SELECT ad_id,SUM(CASE WHEN action IN('Clicked') THEN 1 ELSE 0 END) as clicked
FROM Ads
GROUP BY 1)t2
ON t1.ad_id = t2.ad_id
ORDER BY 2 DESC, ad_id;

如何在表达式中多次使用 COUNT() 实现 OVER？

How do I implement OVER using COUNT() multiple times in an expression?

mysql

sql

window-functions