MySQL get weighted result - mysql

Question

I have a Table with the following structure.
The Table has mostly records where gender = 1.
I'm looking for a solution to get a result set where on top around 60% of records have gender = 1 and around 40% with gender = 2 mixed, ordered by popularity desc.
The amount of member with gender = 2 is much less, which means after the result set should only have gender = 1 records.
Member table
id | nickname | gender | popularity
1 | jake | 1 | 80
2 | mike | 1 | 88
3 | dave | 1 | 75
4 | jenny | 2 | 85
5 | peter | 1 | 83
6 | nina | 2 | 88
7 | mister | 1 | 77
8 | drake | 1 | 80
Result should be something like, it must not meet exactly weighted list. the goal is to see mixed results of both genders.
id | nickname | gender | popularity
2 | mike | 1 | 88
5 | peter | 1 | 83
6 | nina | 2 | 88
1 | jake | 1 | 80
8 | drake | 1 | 80
4 | jenny | 2 | 85
7 | mister | 1 | 77
3 | dave | 1 | 75
My so far best result was (it don't take care about the 40:60 split):
SET #rank=0;
SET #rank2=0;
SELECT * FROM (
SELECT #rank:=#rank+1 AS rank, q.* FROM (SELECT * FROM test WHERE gender = 1 ORDER BY popularity DESC) AS q
UNION
SELECT #rank2:=#rank2+1 AS rank, q.* FROM (SELECT * FROM test WHERE gender = 2 ORDER BY popularity DESC) AS q
) AS r ORDER BY rank;

Answer 1

Please try...
SET #gender1Count = SELECT COUNT( * )
FROM tblMember
WHERE gender = 1;
SET #gender2Count = SELECT COUNT( * )
FROM tblMember
WHERE gender = 2;
SET #totalCount = SELECT COUNT( * )
FROM tblMember;
SELECT id AS id,
nickname AS nickname,
gender AS gender,
popularity AS popularity
FROM tblMember
JOIN ( SELECT id AS id
FROM tblMember
WHERE gender = 1
ORDER BY popularity DESC
LIMIT CASE
WHEN #gender1Count > #totalCount * 3 / 5
ROUND( #gender2Count * 3 / 2 )
ELSE
#gender1Count
END
UNION
SELECT id AS id
FROM tblMember
WHERE gender = 2
ORDER BY popularity DESC
LIMIT CASE
WHEN #gender1Count > #totalCount * 3 / 5
#gender2Count
ELSE
ROUND( #gender1Count * 2 / 3 )
END
) nominees ON tblMember.id = nominees.id
ORDER BY popularity DESC;
The above will give you a list where 60% of entries are gender = 1 and 40% are gender = 2. Please note that this is not the same as 60% or more of the total list as gender = 1 with the balance gender = 2 (or 40% or more of the total list as gender = 2 and the balance gender = 1).
It does this by forming a list of those whose gender equals 1 and sorting it into descending order of popularity. It then determines how many of the top entries it will grab from this list using LIMIT by checking if the count of gender = 1 members exceeds 60% (3/5ths) of the list. If it does then we will need to reduce the number of gender = 1 records to be retrieved to 3/2 times the count of gender = 2 members. The id's of the chosen records are then returned.
(A quickish explanation for those who aren't great at fractions, 40% is the same as 2/5 (two fifths). If gender = 2 has two fifths of the final list then gender = 1 must have the other three fifths (3/5). To find the size of 3/5ths of the list we start with the known 2/5ths (the count of gender = 2) and divide that into 2 so that we know the size of 1/5th of the list. We can then multiply this 1/5 by 3 to determine how many record will make up 3/5ths (60%) of our list.)
Similar logic is used to form the list of gender = 2 members to be included in the final list.
(Please note that the records at the end of each list will likely have popularity values equal to those of the most popular excluded members whose gender corresponds to each list. In the absence of any subsorting in the formation of the two lists the selection of those that are or are not chosen will be arbitrary (and essentially semirandom).)
The two lists are then joined using the UNION operator in what is a simple type of vertical join. (Note : The more familiar INNER JOIN, LEFT JOIN, etc., are all types of horizontal joins).
An inner JOIN is then performed upon our list of amalgamated id's with our original table, giving us our 60% / 40% list. Finally, this list is sorted into descending order of popularity.
If you have any questions or comments, then please feel free to post a Comment accordingly.