Find groups in data by relative difference between records - sql

Question

I have some rows that are sorted by price:
| id | price |
|----|-------|
| 1 | 2.00 |
| 2 | 2.10 |
| 3 | 2.11 |
| 4 | 2.50 |
| 5 | 2.99 |
| 6 | 3.02 |
| 7 | 9.01 |
| 8 | 9.10 |
| 9 | 9.11 |
| 10 | 13.01 |
| 11 | 13.51 |
| 12 | 14.10 |
I need to group them in "price groups". An item belongs to a different group when difference in price between it and the previous item is greater than some fixed value, say 1.50.
So the expected result is something like this:
| MIN(price) | MAX(price) |
|------------|------------|
| 2.00 | 3.02 |
| 9.01 | 9.11 |
| 13.01 | 14.10 |
I'm not even sure how to call this type of grouping. Group by "rolling difference"? Not exactly...
Can this be done in SQL (or in Postgres in particular)?

Answer 1

Your results are consistent with looking at the previous value and saying a group starts when the difference is greater than 1.5. You can do this with lag(), a cumulative sum, and aggregation:
select min(price), max(price)
from (select t.*,
count(*) filter (where prev_price is null or prev_price < price - 1.5) over (order by price) as grp
from (select t.*,
lag(price) over (order by price) as prev_price
from t
) t
) t
group by grp

Answer 2

Thanks Gordon Linoff for his answer, it is exactly what I was after!
I ended up using this query here simply because I understand it better. I guess it is more noobish, but so am I.
Both queries sort a table of 1M rows into 34 groups in about a second. This query is a bit more performant on 11M rows, sorting them into 380 groups in 15 seconds, vs 23 seconds in Gordon's answer.
SELECT results.group_i, MIN(results.price), MAX(results.price), AVG(results.price)
FROM (
SELECT *,
SUM(new_group) OVER (ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS group_i
FROM (
SELECT annotated.*,
CASE
WHEN prev_price IS NULL OR price - prev_price > 1.5 THEN 1
ELSE 0
END AS new_group
FROM (
SELECT *,
LAG(price) OVER (ORDER BY price) AS prev_price
FROM prices
) AS annotated
) AS grouppable
) AS results
GROUP BY results.group_i
ORDER BY results.group_i;