Select 在两列中唯一，分组为

Question

I have a table:

CREATE TABLE stats_test
(
  id1 bigint,
  id2 bigint,
  date timestamp with time zone
);    

And data inside:

 id1 | id2 |          date
-----+-----+------------------------
   1 |   2 | 2020-12-01 00:00:00+00
   2 |   1 | 2020-12-01 00:00:00+00
   3 |   4 | 2020-11-01 00:00:00+00
   4 |   3 | 2020-11-01 00:00:00+00
   1 |   3 | 2020-12-01 00:00:00+00
   1 |   3 | 2020-11-01 00:00:00+00

With this query i get results:

SELECT EXTRACT(YEAR FROM date), EXTRACT(MONTH FROM date), 
COUNT(DISTINCT id1) AS unique_id1, COUNT(DISTINCT id2) AS unique_id2 
FROM stats_test GROUP BY EXTRACT(YEAR FROM date), EXTRACT(MONTH FROM date);

 date_part | date_part | unique_id1 | unique_id2
-----------+-----------+------------+------------
      2020 |        11 |          3 |          2
      2020 |        12 |          2 |          3

How to get another column with count unique ids from the set of both column (id1, id2) grouped by year and month?

 date_part | date_part | unique_id1 | unique_id2 | unique_both_ids
-----------+-----------+------------+------------+----------------
      2020 |        11 |          3 |          2 |
      2020 |        12 |          2 |          3 |    

Answer 1

count(distinct..) only allows a single expression (so count(distinct id1,id2) is rejected), but you can use an anonymous row expression to overcome that limitation:

select extract(year from date) as year, 
       extract(month from date) as month,
       count(distinct id1) as unique_id1, 
       count(distinct id2) as unique_id2,
       count(distinct (id1,id2)) as unique_both_ids
from stats_test 
group by extract(year from date), extract(month from date);

Note that 1,2 and 2,1 would be considered two different things. If you want them to be treated the same use: count(distinct (least(id1,id2), greatest(id1,id2)))