我需要对连续的整数求和并按spoken_correctly
> 0进行分组。
我可以通过查看lag
和lead
找出哪些部分是连续的,但是然后不确定如何对连续组consecutive
的值求和。
即,我有两个组,其中连续的spoken_correctly
值>0。绿色的第一组有三行非零的spoken_correctly
,绿色的第二组有两个。 >
所需的输出:
此SQL在输出上方产生第一张图像:
select *, case when (q.times_spoken_correctly > 0 and (q.lag > 0 or q.lead > 0)) then 1 else 0 end as consecutive
from (
select *, lag(q.times_spoken_correctly) over (partition by q.profile_id order by q.profile_id) as lag, lead(q.times_spoken_correctly) over (partition by q.profile_id order by q.profile_id) as lead
from (
SELECT *
FROM ( VALUES (3, 0, '2019-01-15 19:15:06'),
(3, 0, '2019-01-15 19:15:07'),
(3, 1, '2019-01-15 19:16:06'),
(3, 2, '2019-01-15 19:16:10'),
(3, 2, '2019-01-15 19:17:06'),
(3, 0, '2019-01-15 19:17:11'),
(3, 0, '2019-01-15 19:39:06'),
(3, 3, '2019-01-15 19:40:10'),
(3, 4, '2019-01-15 19:40:45')
) AS baz ("profile_id", "times_spoken_correctly", "w_created_at")
) as q
) as q
答案 0 :(得分:1)
这是一个缺口和孤岛问题,可以通过使用row_number
形成序列组来解决。
select profile_id, count(*) as consec FROM
(
SELECT t.*, row_number() OVER ( PARTITION BY profile_id ORDER BY w_created_at ) -
row_number() OVER ( PARTITION BY profile_id, CASE times_spoken_correctly
WHEN 0 THEN 0 ELSE 1 END
ORDER BY w_created_at ) as seq --group zeros and non zeros
FROM t ORDER BY w_created_at
) s WHERE times_spoken_correctly > 0 --to count only "> zero" groups.
GROUP BY profile_id,seq;