Question

I have a postgres table with customer ID’s, dates, and integers. I need to find the average of the top 3 records for each customer ID that have dates within the last year. I can do it with a single ID using the SQL below (id is the customer ID, weekending is the date, and maxattached is the integer).

One caveat: the maximum values are per month, meaning we’re only looking at the highest value in a given month to create our dataset, thus why we’re extracting month from the date.

SELECT 
  id,
  round(avg(max),0) 
FROM 
  (
   select 
     id,
     extract(month from weekending) as month,
     extract(year from weekending) as year,
     max(maxattached) as max 
   FROM 
     myTable 
   WHERE
     weekending >= now() - interval '1 year' AND 
     id=110070 group by id,month,year 
   ORDER BY
     max desc limit 3
   ) AS t 
GROUP BY id;

How can I expand this query to include all ID’s and a single averaged number for each one?

Here is some sample data:

ID     | MaxAttached | Weekending
110070 | 5           | 2011-11-10
110070 | 6           | 2011-11-17
110071 | 4           | 2011-11-10
110071 | 7           | 2011-11-17
110070 | 3           | 2011-12-01
110071 | 8           | 2011-12-01
110070 | 5           | 2012-01-01
110071 | 9           | 2012-01-01

So, for this sample table, I would expect to receive the following results:

ID     | MaxAttached

110070 | 5           
110071 | 8

This averages the highest value in a given month for each ID (6,3,5 for 110070 and 7,8,9 for 110071)

Note: postgres version 8.1.15

Answer 1

First – get the max(maxattached) for every customer and month:

SELECT id,
       max(maxattached) as max_att         
FROM myTable 
WHERE weekending >= now() - interval '1 year' 
GROUP BY id, date_trunc('month',weekending);

Next – for every customer rank all his values:

SELECT id,
       max_att,
       row_number() OVER (PARTITION BY id ORDER BY max_att DESC) as max_att_rank
FROM <previous select here>;

Next – get the top 3 for every customer:

SELECT id,
       max_att
FROM <previous select here>
WHERE max_att_rank <= 3;

Next – get the avg of the values for every customer:

SELECT id,
       avg(max_att) as avg_att
FROM <previous select here>
GROUP BY id;

Next – just put all the queries together and rewrite/simplify them for your case.

UPDATE: Here is an SQLFiddle with your test data and the queries: SQLFiddle.

UPDATE2: Here is the query, that will work on 8.1 :

SELECT customer_id,
       (SELECT round(avg(max_att),0)
        FROM (SELECT max(maxattached) as max_att         
              FROM table1
              WHERE weekending >= now() - interval '2 year' 
                AND id = ct.customer_id
              GROUP BY date_trunc('month',weekending)
              ORDER BY max_att DESC
              LIMIT 3) sub 
        ) as avg_att
FROM customer_table ct;

The idea – to take your initial query and run it for every customer (customer_table – table with all unique id for customers).

Here is SQLFiddle with this query: SQLFiddle.

Only tested on version 8.3 (8.1 is too old to be on SQLFiddle).