总和分组连续整数

时间:2019-01-21 04:45:56

标签: sql postgresql group-by gaps-and-islands

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laglead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。 >

enter image description here

所需的输出:

enter image description here

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

1 个答案:

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;

Demo