Finding the highest after grouping by month - sql

Question

In postgres, I want to output the persons who have the highest no. of "discussed" requests for each month, irrespective of the year i.e. there should be 12 outputs.
ID PERSON REQUEST DATE
4 datanoise opened 2010-09-02
5 marsuboss opened 2010-09-02
6 m3talsmith opened 2010-09-06
7 sferik opened 2010-09-08
8 sferik opened 2010-09-09
8 dtrasbo discussed 2010-09-09
8 brianmario discussed 2010-09-09
8 sferik discussed 2010-09-09
9 rsim opened 2011-09-09
.....more tuples to follow
*This is just a small part of the databse. also assume that the dataset is big enough that all months are represented in the date column.

Answer 1

Test data:
CREATE TEMPORARY TABLE foo( id SERIAL PRIMARY KEY, name INTEGER NOT NULL,
dt DATE NULL, request BOOL NOT NULL );
INSERT INTO foo (name,dt,request) SELECT random()*1000,
'2010-01-01'::DATE+('1 DAY'::INTERVAL)*(random()*3650), random()>0.5
FROM generate_series(1,100000) n;
SELECT * FROM foo LIMIT 10;
id | name | dt | request
----+------+------------+---------
1 | 110 | 2014-11-05 | f
2 | 747 | 2015-03-12 | t
3 | 604 | 2014-09-26 | f
4 | 211 | 2011-12-14 | t
5 | 588 | 2016-12-15 | f
6 | 96 | 2012-02-19 | f
7 | 17 | 2018-09-18 | t
8 | 591 | 2018-02-15 | t
9 | 370 | 2015-07-28 | t
10 | 844 | 2019-05-16 | f
Now you have to get the count per name and month, then get the max count, but that won't give you the name that has the maximum, which requires joining back with the previous result. In order to do the group by only once, it is done in a CTE:
WITH totals AS (
SELECT EXTRACT(month FROM dt) mon, name, count(*) cnt FROM foo
WHERE request=true GROUP BY name,mon
)
SELECT * FROM
(SELECT mon, max(cnt) cnt FROM totals GROUP BY mon) x
JOIN totals USING (mon,cnt);
If several names have the same maximum count, they will be returned both. To keep only one, you can use DISTICT ON:
WITH (same as above)
SELECT DISTINCT ON (mon) * FROM
(SELECT mon, max(cnt) cnt FROM totals GROUP BY mon) x
JOIN totals USING (mon,cnt) ORDER BY mon,name;
You can also use DISTINCT ON to keep only one row per month, specified by the ORDER clause, in this cas by count desc, so it keeps the highest count.
SELECT DISTINCT ON (mon) * FROM (
SELECT EXTRACT(month FROM dt) mon, name, count(*) cnt FROM foo
WHERE request=true GROUP BY name,mon
)x ORDER BY mon, cnt DESC;
...or you could hack an argmax() function by sticking the primary key into an array passed to max(), which means it will return the id of the row which has the maximum value:
SELECT mon, cntid[1] cnt, name FROM
(SELECT mon, max(ARRAY[cnt,id]) cntid FROM (
SELECT EXTRACT(month FROM dt) mon, name, count(*) cnt, min(id) id FROM foo
WHERE request=true GROUP BY name,mon
) x GROUP BY mon)y
JOIN foo ON (foo.id=cntid[2]);
Which one will be faster?...

Answer 2

given your table is named t01 and the colum date is date1 (and in string format):
create temp table t02 as
select extract(month from CAST(date1 as date)) as month, person, count(*) nb from t01 where request = 'discussed' group by 1, 2 ;
create temp table t03 as
select month, max(nb) max_nb from t02 group by 1 ;
the result is :
select month , person from t02 a natural join t03 b where a.nb = b.max_nb;
https://rextester.com/BYMM84335[ : run here]1

Answer 3

I would recommend distinct on. If you want to combine all the months into a single "uber-month":
select distinct on (extract(month from date)) person, extract(month from date), count(*) as num_discussed
from t
where request = 'discussed'
group by person, extract(month from date)
order by extract(month from date), num_discussed desc;
Distinct on is a very handy Postgres extension. It returns on row per "group", which is defined by the expressions in parentheses. The row is the "first" one determined by the order by clause.
If you want the highest month regardless of year:
select distinct on (extract(month from date)) person, date_trunc('month', date), count(*) as num_discussed
from t
where request = 'discussed'
group by person, date_trunc('month', date)
order by extract(month from date), num_discussed desc;