PostgreSQL MAX and GROUP BY

select *
from (
  select id, 
         year,
         thing,
         max(thing) over (partition by id) as max_thing
  from the_table
) t
where thing = max_thing

or:

select t1.id,
       t1.year,
       t1.thing
from the_table t1
where t1.thing = (select max(t2.thing) 
                  from the_table t2
                  where t2.id = t1.id);

or

select t1.id,
       t1.year,
       t1.thing
from the_table t1
  join ( 
    select id, max(t2.thing) as max_thing
    from the_table t2
    group by id
  ) t on t.id = t1.id and t.max_thing = t1.thing

or (same as the previous with a different notation)

with max_stuff as (
  select id, max(t2.thing) as max_thing
  from the_table t2
  group by id
) 
select t1.id, 
       t1.year,
       t1.thing
from the_table t1
  join max_stuff t2 
    on t1.id = t2.id 
   and t1.thing = t2.max_thing

The shortest (and possibly fastest) query would be with DISTINCT ON, a PostgreSQL extension of the SQL standard DISTINCT clause:

SELECT DISTINCT ON (1)
       id, count, year
FROM   tbl
ORDER  BY 1, 2 DESC, 3;

The numbers refer to ordinal positions in the SELECT list. You can spell out column names for clarity:

SELECT DISTINCT ON (id)
       id, count, year
FROM   tbl
ORDER  BY id, count DESC, year;

The result is ordered by id etc. which may or may not be welcome. It's better than "undefined" in any case.

It also breaks ties (when multiple years share the same maximum count) in a well defined way: pick the earliest year. If you don't care, drop year from the ORDER BY. Or pick the latest year with year DESC.

For many rows per id, other query techniques are (much) faster. See:

• Select first row in each GROUP BY group?
• Optimize GROUP BY query to retrieve latest row per user