问题
Suppose I have this table:
select * from window_test;
k | v
---+---
a | 1
a | 2
b | 3
a | 4
Ultimately I want to get:
k | min_v | max_v
---+-------+-------
a | 1 | 2
b | 3 | 3
a | 4 | 4
But I would be just as happy to get this (since I can easily filter it with distinct
):
k | min_v | max_v
---+-------+-------
a | 1 | 2
a | 1 | 2
b | 3 | 3
a | 4 | 4
Is it possible to achieve this with PostgreSQL 9.1+ window functions? I'm trying to understand if I can get it to use separate partition for the first and last occurrence of k=a
in this sample (ordered by v
).
回答1:
This returns your desired result with the sample data. Not sure if it will work for real world data:
select k,
min(v) over (partition by group_nr) as min_v,
max(v) over (partition by group_nr) as max_v
from (
select *,
sum(group_flag) over (order by v,k) as group_nr
from (
select *,
case
when lag(k) over (order by v) = k then null
else 1
end as group_flag
from window_test
) t1
) t2
order by min_v;
I left out the DISTINCT
though.
回答2:
EDIT: I've came up with the following query — without window functions at all:
WITH RECURSIVE tree AS (
SELECT k, v, ''::text as next_k, 0 as next_v, 0 AS level FROM window_test
UNION ALL
SELECT c.k, c.v, t.k, t.v + level, t.level + 1
FROM tree t JOIN window_test c ON c.k = t.k AND c.v + 1 = t.v),
partitions AS (
SELECT t.k, t.v, t.next_k,
coalesce(nullif(t.next_v, 0), t.v) AS next_v, t.level
FROM tree t
WHERE NOT EXISTS (SELECT 1 FROM tree WHERE next_k = t.k AND next_v = t.v))
SELECT min(k) AS k, v AS min_v, max(next_v) AS max_v
FROM partitions p
GROUP BY v
ORDER BY 2;
I've provided 2 working queries now, I hope one of them will suite you.
SQL Fiddle for this variant.
Another way how to achieve this is to use a support sequence.
Create a support sequence:
CREATE SEQUENCE wt_rank START WITH 1;
The query:
WITH source AS ( SELECT k, v, coalesce(lag(k) OVER (ORDER BY v), k) AS prev_k FROM window_test CROSS JOIN (SELECT setval('wt_rank', 1)) AS ri), ranking AS ( SELECT k, v, prev_k, CASE WHEN k = prev_k THEN currval('wt_rank') ELSE nextval('wt_rank') END AS rank FROM source) SELECT r.k, min(s.v) AS min_v, max(s.v) AS max_v FROM ranking r JOIN source s ON r.v = s.v GROUP BY r.rank, r.k ORDER BY 2;
回答3:
Would this not do the job for you, without the need for windows, partitions or coalescing. It just uses a traditional SQL trick for finding nearest tuples via a self join, and a min on the difference:
SELECT k, min(v), max(v) FROM (
SELECT k, v, v + min(d) lim FROM (
SELECT x.*, y.k n, y.v - x.v d FROM window_test x
LEFT JOIN window_test y ON x.k <> y.k AND y.v - x.v > 0)
z GROUP BY k, v, n)
w GROUP BY k, lim ORDER BY 2;
I think this is probably a more 'relational' solution, but I'm not sure about its efficiency.
来源:https://stackoverflow.com/questions/10614505/window-functions-and-more-local-aggregation