如何计算Postgres中的中位数？

Question

I have created a basic database (picture attached) Database , I am trying to find the following:

"Median total amount spent per user in each calendar month"

I tried the following, but getting errors:

SELECT 
user_id,
AVG(total_per_user)
FROM (SELECT user_id,
        ROW_NUMBER() over (ORDER BY total_per_user DESC) AS desc_total,
        ROW_NUMBER() over (ORDER BY total_per_user ASC) AS asc_total
      FROM (SELECT EXTRACT(MONTH FROM created_at) AS calendar_month,
            user_id,    
            SUM(amount) AS total_per_user
            FROM transactions
            GROUP BY calendar_month, user_id) AS total_amount   
      ORDER BY user_id) AS a
WHERE asc_total IN (desc_total, desc_total+1, desc_total-1)
GROUP BY user_id
;

Answer 1

In Postgres, you could just use aggregate function percentile_cont() :

select 
    user_id,
    percentile_cont(0.5) within group(order by total_per_user) median_total_per_user
from (
    select user_id, sum(amount) total_per_user
    from transactions
    group by date_trunc('month', created_at), user_id
) t
group by user_id

Note that date_trunc() is probably closer to what you want than extract(month from...) - unless you do want to sum amounts of the same month for different years together, which is not how I understood your requirement.

Answer 2

Just use percentile_cont() . I don't fully understand the question. If you want the median of the monthly spending, then:

SELECT user_id,
       PERCENTILE_CONT(0.5) WITHIN GROUP (ORDER BY total_per_user
        ROW_NUMBER() over (ORDER BY total_per_user DESC) AS desc_total,
        ROW_NUMBER() over (ORDER BY total_per_user ASC) AS asc_total
FROM (SELECT DATE_TRUNC('month', created_at) AS calendar_month,
             user_id, SUM(amount) AS total_per_user
      FROM transactions t
      GROUP BY calendar_month, user_id
     ) um   
GROUP BY user_id;

There is a built-in function for median. No need for fancier processing.