SQL / Presto SQL：在同一列中按组求和

Question

我正在尝试解决如下问题：

有一张这样的表：

logtime	name	seconds	flag
1629302433	a	30	1-1
1629302463	a	30	1-1
1629302483	a	20	0-1
1629302513	a	30	1-1
1629302533	a	20	0-1
1629302553	a	30	1-1

当 flag = 0-1 时，数据将被分成 3 部分，并按每个部分求和秒列值，如下所示： （logtime 是时间戳）

name	seconds
a	60
a	30
a	30

Answer 1

将每行所属的组数计算为标志“0-1”出现次数的运行总和。然后按名称和组号聚合组。

演示：

with mytable as (
SELECT * FROM (
    VALUES
(1629302433, 'a', 30, '1-1'),
(1629302463, 'a', 30, '1-1'),
(1629302483, 'a', 20, '0-1'),
(1629302513, 'a', 30, '1-1'),
(1629302533, 'a', 20, '0-1'),
(1629302553, 'a', 30, '1-1')
) AS t (logtime, name, seconds, flag)
)

select name, 
       sum(seconds) seconds
from
(--calculate group number as running sum of 0-1 occurances
select logtime, name, seconds, flag,
       sum(case when flag='0-1' then 1 else 0 end) over(partition by name order by logtime) as group_nbr
  from mytable
)s
where flag='1-1' --do not sum '0-1' records
group by name, group_nbr 
order by name, group_nbr --remove ordering if not necessary

结果：

name    seconds 
a       60
a       30
a       30

Answer 2

您可以使用lag() 函数查找值变化的位置，然后进行累积求和以分配组，然后对组求和：

WITH dataset AS (
  SELECT * 
  FROM 
    (
      VALUES 
        (1629302433,    'a',    30, '1-1'),
        (1629302463,    'a',    30, '1-1'),
        (1629302483,    'a',    20, '0-1'),
        (1629302513,    'a',    30, '1-1'),
        (1629302533,    'a',    20, '0-1'),
        (1629302553,    'a',    30, '1-1')
    ) AS t (logtime,    name,   seconds,    flag)
) 

select name, sum(seconds) seconds
from (
         select *,
                sum(case when flag = prev_flag then 0 else 1 end) over (partition by name order by logtime) as grp
         from (
                  select logtime,
                         name,
                         seconds,
                         flag,
                         lag(flag) over (partition by name order by logtime) as prev_flag
                  from dataset
              )
     )
where flag = '1-1'
group by name, grp

输出：

name	seconds
a	60
a	30
a	30