Use two aggregate functions in the same query

12,313

Solution 1

Firstly you might want to read this article about reasons to use ANSI 92 Joins instead of the aged ANSI 89 as used above. Secondly, SQLLite does support the GROUP_CONCAT function so you can use this.

You just neeed to add your second query as subquery into the first to get the desired result:

SELECT  Talks.TalkID, 
        Talks.Title, 
        ts.Speakers, 
        COUNT(*) AS SimilarTalks
FROM    Talks
        INNER JOIN SimilarTalks 
            ON Talks.TalkID = SimilarTalks.SimilarTo
        INNER JOIN Talks t2
            ON SimilarTalks.TalkID = t2.TalkID
            AND t2.Starred = 1
        INNER JOIN
        (   SELECT  TalkID, GROUP_CONCAT(Speaker, ',') AS Speakers
            FROM    TalkSpeaker
            GROUP BY TalkID
        ) ts
            ON ts.TalkID = Talks.TalkID
WHERE   Talks.Starred = 0
GROUP BY Talks.TalkID, Talks.Title, ts.Speakers
ORDER BY COUNT(*) DESC
LIMIT 2;

Example on SQL Fiddle

EDIT

You could also do this without a subquery using DISTINCT:

SELECT  Talks.TalkID, 
        Talks.Title, 
        GROUP_CONCAT(DISTINCT ts.Speaker) AS Speakers,
        COUNT(DISTINCT t2.TalkID) AS SimilarTalks
FROM    Talks
        INNER JOIN SimilarTalks 
            ON Talks.TalkID = SimilarTalks.SimilarTo
        INNER JOIN Talks t2
            ON SimilarTalks.TalkID = t2.TalkID
            AND t2.Starred = 1
        INNER JOIN TalkSpeaker ts
            ON ts.TalkID = Talks.TalkID
WHERE   Talks.Starred = 0
GROUP BY Talks.TalkID, Talks.Title
ORDER BY COUNT(DISTINCT t2.TalkID) DESC
LIMIT 2;

However I see no benefit at all in this method, and it is likely to be less efficient (I have not tested so can't be certain)

Solution 2

First, to get just the IDs of the desired talks, remove the other fields from your first query:

SELECT unstarred.talkID
FROM talks AS starred
  JOIN similarTalks AS s ON starred.talkID = s.talkID
  JOIN talks AS unstarred ON s.similarTo = unstarred.talkID
WHERE starred.starred
  AND NOT unstarred.starred
GROUP BY unstarred.talkID
ORDER BY COUNT(*) DESC
LIMIT 2

Then, use this as a subquery to get the information about the desired talks:

SELECT t.title AS Title,
       group_concat(s.speaker, ', ') AS Speakers
FROM talks AS t JOIN talkspeaker AS s ON t.talkID = s.talkID
WHERE t.talkID IN (SELECT unstarred.talkID
                   FROM talks AS starred
                     JOIN similarTalks AS s ON starred.talkID = s.talkID
                     JOIN talks AS unstarred ON s.similarTo = unstarred.talkID
                   WHERE starred.starred
                     AND NOT unstarred.starred
                   GROUP BY unstarred.talkID
                   ORDER BY COUNT(*) DESC
                   LIMIT 2)
GROUP BY t.talkID
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12,313
Samik R
Author by

Samik R

Updated on June 11, 2022

Comments

  • Samik R
    Samik R almost 2 years

    Consider the following tables:

    [Table: talks]
    talkID | title        | starred
    -------+--------------+--------
    1      | talk1-title  | 1
    2      | talk2-title  | 1
    3      | talk3-title  | 0
    4      | talk4-title  | 0
    5      | talk5-title  | 0
    
    [Table: talkspeaker]
    talkID | speaker
    -------+---------
    1      | Speaker1
    1      | Speaker2
    2      | Speaker3
    3      | Speaker4
    3      | Speaker5
    4      | Speaker6
    5      | Speaker7
    5      | Speaker8
    
    [Table: similartalks]
    talkID | similarTo
    -------+----------
    1      | 3
    1      | 4
    2      | 3
    2      | 4
    2      | 5
    3      | 2
    4      | 5
    5      | 3
    5      | 4
    

    What I want to do is: Given the set of starred talks, I would like to select the top 2 of the unstarred talks (starred = 0) and their titles and speakers that are most similar to the set of starred talks. The problem is that getting the speakers requires using an aggregate function, and so does getting the most similar talks.

    Without the speakers in the fray, I have been able to get the most similar talks using the following query:

    select t2.talkID, t2.title, count(*) as count 
    from similarTalks s, talks t1, talks t2
    where s.talkID = t1.talkID
    and t1.Starred = 1
    and s.similarTo = t2.TalkID
    and t2.Starred = 0
    group by t2.title, t2.talkID
    order by count desc
    limit 2
    

    Generally, I use the following aggregate function for getting the speakers, with appropriate group by columns (assume t = talkspeaker):

    group_concat(t.speaker, ', ') as Speakers
    

    as in

    select t1.title, group_concat(t2.speaker, ', ') as Speakers 
    from talks t1, talkspeaker t2
    where t1.talkID = t2.talkID
    group by t1.title
    

    But I am not able to combine the two things together. It might matter that I am planning to run this query in a sqlite database (that is where the group_concat function comes from). The answer to the top 2 unstarred talks most similar to starred talks seem to be with talkIDs 3 and 4.

  • Samik R
    Samik R over 11 years
    Thanks for pointing me out to the SQL92 syntax. I use SQL on and off, so wasn't really aware of this new syntax. Also thanks for introducing me to SQL fiddle - seems like a useful service. In my tests, the second query seem to consistently run faster (presumably because of no subquery, but not sure), but is returning different (and incorrect) result than the first one.
  • Samik R
    Samik R over 11 years
    Thanks for the response. This seem to take around the same time as the first query in @GarethD's response. I like having the count (of similar talks) information outside though.
  • Samik R
    Samik R over 11 years
    Since I did not hear back, I ended up using the first query here. Thanks again - and if you do have some time, I will appreciate if you can comment on the second query.
  • GarethD
    GarethD over 11 years
    Haha, I spent about 20 minutes looking at this yesterday, and just spent another 20, breaking down the second query because I could not for the life of me work out why it wasn't working properly. Both queries were missing DESC after the ORDER BY, but the SQL Fiddle wasn't. I could not see the wood for the trees... I've updated the answer.
  • Samik R
    Samik R over 11 years
    I so much appreciate taking time to investigate this. I will take a look.