Mysql貝葉斯和按星級評分

Question

說我有兩張桌子。 businesses和reviews爲企業。

businesses表：

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

reviews表：

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

每個評論有一個rate（1到5星）
我想他們的評論率對企業進行排序，根據Bayesian Ranking條件至少有2條評論。

這裏是我的查詢：

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 

這將輸出我：

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

我有兩個問題：

1. 當你看到在((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)一些功能正在運行更多比一次，如COUNT或AVG。他們是否在後臺運行一次，也許緩存resuslt？或運行每一個電話？

2. 是否有任何等效查詢，但更優化？

在此先感謝。

Answer 1

我希望MySQL能夠優化多重計數，但不能確定。

但是，您可以重新安排您查詢加入反對子查詢。這樣你不會爲每一行執行2個子查詢。

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 

注意，這裏的結果是非常略有不同，因爲它會排除只有2條評論記錄，而您的查詢仍然會返回，但他們的得分爲NULL。我已經離開了明顯缺少的運營商（即AVG（r.rate）之前和之前4）您的原始查詢AS評分。

使用上面的想法，您可以重新編碼它以返回子查詢中的計數和平均速率，並僅使用那些返回的列的值來計算。

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