Mysql 贝叶斯并按星级排序

Question

假设我有两个 table。 businesses 和 reviews 用于企业。

businesses table:

+----+-------+
| id | title |
+----+-------+

reviews table:

+----+-------------+---------+------+
| id | business_id | message | rate |
+----+-------------+---------+------+

每条评论都有 rate（1 到 5 颗星）
我想根据 Bayesian Ranking 的评论率对商家进行排序，条件是至少有 2 条评论。

这是我的查询：

SELECT b.id,
(SELECT COUNT(r.rate) as rr FROM reviews r WHERE r.business_id = b.id) as rr,
(SELECT
        ((COUNT(r.rate) / (COUNT(r.rate) + 2)) AVG(r.rate) + 
        (2 /(COUNT(r.rate) + 2)) 4)
    FROM reviews r where r.business_id = b.id AND rr > 2
) as score
FROM businesses b
order by score desc
LIMIT 4

这将输出我：

+------+----+------------+
| id   | rr | score      |
+------+----+------------+
| 992  | 14 | 4.31250000 |
+------+----+------------+
| 237  | 3  | 4.2000000  |
+------+----+------------+
| 19   | 5  | 4.0000000  |
+------+----+------------+
| 1009 | 12 | 3.9285142  |
+------+----+------------+

我有两个问题：

如您在 ((COUNT(r.rate) / (COUNT(r.rate) + 2)) AVG(r.rate) + (2 /(COUNT(r.rate) + 2)) 4) FROM reviews r where r.business_id = b.id AND rr > 2 ) 中所见，某些函数运行不止一次，例如 COUNT 或 AVG。它们是否运行在后台运行一次并可能缓存结果？或者运行每次调用？
是否有任何等效的查询，但更优化？

提前致谢。

Answer 1

我希望 MySQL 可以优化多个计数，但不确定。

但是您可以重新安排查询以加入子查询。这样您就不会为每一行执行 2 个子查询。

SELECT b.id,
        sub0.rr,
        sub0.score
FROM businesses b
INNER JOIN
(
    SELECT r.business_id,
            COUNT(r.rate) AS rr ,
            ((COUNT(r.rate) / (COUNT(r.rate) + 2)) AVG(r.rate) + (2 /(COUNT(r.rate) + 2)) 4) AS score
    FROM reviews r 
    GROUP BY r.business_id
    HAVING rr > 2 
) sub0
ON sub0.business_id = b.id 
ORDER BY score DESC
LIMIT 4

请注意，此处的结果略有不同，因为它将排除只有 2 条评论的记录，而您的查询仍将 return 它们，但分数为 NULL。我在您的原始查询中留下了明显缺失的运算符（即，在 AVG(r.rate) 之前和 4 之前）AS score .

使用上面的想法，您可以将其重新编码为 return 子查询中的计数和平均比率，并且只需使用那些 returned 列的值进行计算。

SELECT b.id,
        sub0.rr,
        ((rr / (rr + 2)) arr + (2 /(rr + 2)) 4) AS score
FROM businesses b
INNER JOIN
(
    SELECT r.business_id,
            COUNT(r.rate) AS rr ,
            AVG(r.rate) AS arr
    FROM reviews r 
    GROUP BY r.business_id
    HAVING rr > 2 
) sub0
ON sub0.business_id = b.id 
ORDER BY score DESC
LIMIT 4

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