How to calculate AVG value of column by number of DISTINCT user_ids and GROUP BY date in Postgresql

Question

I am looking to find the average duration per number of unique users, by date.

Data Table

| date       | user_id | duration |
|------------|---------|----------|
| 2018-10-02 | 1       | 5        |   
| 2018-10-02 | 1       | 10       |   
| 2018-10-02 | 2       | 15       |   
| 2018-10-03 | 1       | 20       |   
| 2018-10-03 | 2       | 30       |
| 2018-10-04 | 2       | 5        |

Expected Return

| date       | ave_duration_per_unique_user |
|------------|------------------------------|
| 2018-10-02 | 15                           |
| 2018-10-03 | 25                           |
| 2018-10-04 | 5                            |

The query below gives the average based on total duration for the date / total number of entries for the date. total number of entries for the date should be number of unique user_ids for the date.

SELECT
  DISTINCT date,
  AVG(session_duration)
FROM
  my_table
GROUP BY
  date, distinct_id
ORDER BY
  date

| date       | ave_duration_per_unique_user |
|------------|------------------------------|
| 2018-10-02 | 10                           | // should be 15
| 2018-10-03 | 25                           |
| 2018-10-04 | 5                            |

Answer 1

You can do the arithmetic as:

select date, sum(duration) / count(distinct user)
from my_table
group by date;

I will note that I would define "average per unique user" a bit differently. It would be (7.5 + 15) / 2 = 11.25 for the first date. However, that is not the calculation you have described.

Answer 2

gist showing table defs & data definition here , but here's the query you're looking for:

testdb=# select date, avg(user_total)::bigint from (
    select date, user_id, sum(duration) user_total from t group by date, user_id)_
         group by date order by date;
    date    | avg 
------------+-----
 2018-10-02 |  15
 2018-10-03 |  25
 2018-10-04 |   5
(3 rows)