Mysql Bayesian and sort by star ratings - mysql

Question

Say I have two tables. businesses and reviews for businesses.
businesses table:
+----+-------+
| id | title |
+----+-------+
reviews table:
+----+-------------+---------+------+
| id | business_id | message | rate |
+----+-------------+---------+------+
each review has a rate ( 1 to 5 stars )
I want to sort businesses by their reviews rates, based on Bayesian Ranking with condition of having at least 2 reviews.
Here is my query:
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
this will output me:
+------+----+------------+
| id | rr | score |
+------+----+------------+
| 992 | 14 | 4.31250000 |
+------+----+------------+
| 237 | 3 | 4.2000000 |
+------+----+------------+
| 19 | 5 | 4.0000000 |
+------+----+------------+
| 1009 | 12 | 3.9285142 |
+------+----+------------+
I have two questions:
as you see in ((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 ) some functions are running more than once, like COUNT or AVG. are they running once in background and maybe caches the resuslt? OR run for every single call?
is there any equivalent query for this but more optimize?
thanks in advance.

Answer 1

I would hope that MySQL would optimise the multiple counts away, but not certain.
However you could rearrange you query to join against a sub query. This way you are not performing 2 sub queries for every row.
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
Note that the result here are very slightly different as it will exclude records with only 2 reviews, while your query will still return them but with a score of NULL. I have left in the apparent missing operators (ie, before AVG(r.rate) and before 4) AS score from your original query.
Using the above idea you could recode it to return both the count and the average rate in the sub query, and just use the values of those returned columns for the calculation.
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