Subtracting values by id and a previous counter - postgresql

Question

Here is a Snippet of my data :-
customers order_id order_date order_counter
1 a 1/1/2018 1
1 b 1/4/2018 2
1 c 3/8/2018 3
1 d 4/9/2019 4
I'm trying to get the average number of days between the order time for each customer. So for the following Snippet the average number of days should be 32.66 days as there were 3,62,32 number of days between each order, sum it, and then divide by 3.
My data has Customers that may have more than 100+ orders .

Answer 1

You could use LAG function:
WITH cte AS (
SELECT customers,order_date-LAG(order_date) OVER(PARTITION BY customers ORDER BY order_counter) AS d
FROM t
)
SELECT customers, AVG(d)
FROM cte
WHERE d IS NOT NULL
GROUP BY customers;
db<>fiddle demo

Answer 2

With a self join, group by customer and get the average difference:
select
t.customers,
round(avg(tt.order_date - t.order_date), 2) averagedays
from tablename t inner join tablename tt
on tt.customers = t.customers and tt.order_counter = t.order_counter + 1
group by t.customers
See the demo.
Results:
| customers | averagedays |
| --------- | ----------- |
| 1 | 32.67 |

Answer 3

Please check below query.
I tried to insert data of two customers so that we can check that average for every customer is coming correct.
DB Fiddle Example: https://www.db-fiddle.com/
CREATE TABLE test (
customers INTEGER,
order_id VARCHAR(1),
order_date DATE,
order_counter INTEGER
);
INSERT INTO test
(customers, order_id, order_date, order_counter)
VALUES
('1', 'a', '2018-01-01', '1'),
('1', 'b', '2018-01-04', '2'),
('1', 'c', '2018-03-08', '3'),
('1', 'd', '2018-04-09', '4'),
('2', 'a', '2018-01-01', '1'),
('2', 'b', '2018-01-06', '2'),
('2', 'c', '2018-03-12', '3'),
('2', 'd', '2018-04-15', '4');
commit;
select customers , round(avg(next_order_diff),2) as average
from
(
select customers , order_date , next_order_date - order_date as next_order_diff
from
(
select customers ,
lead(order_date) over (partition by customers order by order_date) as next_order_date , order_date
from test
) a
where next_order_date is not null
) a
group by customers
order by customers
;

Answer 4

Another option. I would myself like the answer from #forpas except that it depends on the monotonically increasing value for order_counter (what happens when an order is deleted). The following accounts for that by actually counting the number of order pairs. It also accounts for customers have places only 1 order, returning NULL as the average.
select customers, round(sum(nd)::numeric/n, 2) avg_days_to_order
from (
select customers
, order_date - lag(order_date) over(partition by customers order by order_counter) nd
, count(*) over (partition by customers) - 1 n
from test
)d
group by customers, n
order by customers;