Aggregate highest prices per client of salesmen - sql

Question

I have a table like this:
SELECT * FROM orders;
client_id | order_id | salesman_id | price
-----------+----------+-------------+-------
1 | 167 | 1 | 65
1 | 367 | 1 | 27
2 | 401 | 1 | 29
2 | 490 | 2 | 48
3 | 199 | 1 | 68
3 | 336 | 2 | 22
3 | 443 | 1 | 84
3 | 460 | 2 | 92
I want to find the an array of order_ids for each of the highest priced sales for each unique salesman and client pair. In this case I want the resulting table:
salesman_id | order_id
-------------+----------------
1 | {167, 401, 443}
2 | {490, 460}
So far I have an outline for a query:
SELECT salesman_id, max_client_salesman(order_id)
FROM orders
GROUP BY salesman_id;
However I'm having trouble writing the aggregate function max_client_salesman.
The documentation online for aggregate functions and arrays in postgres is very minimal. Any help is appreciated.

Answer 1

Standard SQL
I would combine the window function last_value() or firstvalue() with DISTINCT to the get the orders with the highest price per (salesman_id, client_id) efficiently and then aggregate this into the array you are looking for with the simple aggregate function array_agg().
SELECT salesman_id
,array_agg(max_order_id) AS most_expensive_orders_per_client
FROM (
SELECT DISTINCT
salesman_id, client_id
,last_value(order_id) OVER (PARTITION BY salesman_id, client_id
ORDER BY price
ROWS BETWEEN UNBOUNDED PRECEDING
AND UNBOUNDED FOLLOWING) AS max_order_id
FROM orders
) x
GROUP BY salesman_id
ORDER BY salesman_id;
Returns:
salesman_id | most_expensive_orders_per_client
-------------+------------------------------------
1 | {167, 401, 443}
2 | {490, 460}
SQL Fiddle.
If there are multiple highest prices per (salesman_id, client_id), this query pick only one order_id arbitrarily - for lack of definition.
For this solution it is essential to understand that window functions are applied before DISTINCT. How you to combine DISTINCT with a window function:
PostgreSQL: running count of rows for a query 'by minute'
For an explanation on ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING refer to this closely related answer on dba.SE.
Simper with non-standard DISTINCT ON
PostgreSQL implements, as extension to the SQL standard, DISTINCT ON. With it you can very effectively select rows unique according to a defined set of columns.
It won't get simpler or faster than this:
SELECT salesman_id
,array_agg(order_id) AS most_expensive_orders_per_client
FROM (
SELECT DISTINCT ON (1, client_id)
salesman_id, order_id
FROM orders
ORDER BY salesman_id, client_id, price DESC
) x
GROUP BY 1
ORDER BY 1;
SQL Fiddle.
I also use positional parameters for shorter syntax. Details:
Select first row in each GROUP BY group?

Answer 2

I think you want the Postgres function array_agg in combination with row_number() However, your description of the query does not make sense to me.
The following gets clients and salesmen and the list of orders for the highest priced order by salesman:
select client_id, salesman_id, array_agg(order_id)
from (select o.*,
row_number() over (partition by salesman_id order by price desc) as sseqnum,
row_number() over (partition by client_id order by price desc) as cseqnum
from orders o
) o
where sseqnum = 1
group by salesman_id, client_id
I don't know what you mean by "highest priced sales for each salesman and client". Perhaps you want:
where sseqnum = 1 or cseqnum = 1