Aggregate columns with additional (distinct) filters

Question

This code works as expected, but I it\'s long and creepy.

select p.name, p.played, w.won, l.lost from

(select users.name, count(games.name) as played
from u         


        

Answer 1

The aggregate FILTER clause in Postgres 9.4 or newer is shorter and faster:

SELECT u.name
     , count(*) FILTER (WHERE g.winner_id  > 0)    AS played
     , count(*) FILTER (WHERE g.winner_id  = u.id) AS won
     , count(*) FILTER (WHERE g.winner_id <> u.id) AS lost
FROM   games g
JOIN   users u ON u.id IN (g.player_1_id, g.player_2_id)
GROUP  BY u.name;

• The manual
• Postgres Wiki
• Depesz blog post

In Postgres 9.3 (or any version) this is still shorter and faster than nested sub-selects or CASE expressions:

SELECT u.name
     , count(g.winner_id  > 0 OR NULL)    AS played
     , count(g.winner_id  = u.id OR NULL) AS won
     , count(g.winner_id <> u.id OR NULL) AS lost
FROM   games g
JOIN   users u ON u.id IN (g.player_1_id, g.player_2_id)
GROUP  BY u.name;

Details:

• For absolute performance, is SUM faster or COUNT?

Answer 2

select users.name, 
       count(case when games.winner_id > 0 
                  then games.name 
                  else null end) as played,
       count(case when games.winner_id = users.id 
                  then games.name 
                  else null end) as won,
       count(case when games.winner_id != users.id 
                  then games.name 
                  else null end) as lost
from users inner join games 
     on games.player_1_id = users.id or games.player_2_id = users.id
group by users.name;

Answer 3

This is a case where correlated subqueries may simplify the logic:

select u.*, (played - won) as lost
from (select u.*,
             (select count(*)
              from games g
              where g.player_1_id = u.id or g.player_2_id = u.id
             ) as played,
             (select count(*)
              from games g
              where g.winner_id = u.id
             ) as won
      from users u
     ) u;

This assumes that there are no ties.