Question

我正在尝试为给定的列ID合并BigQuery中的重叠间隔，由于某些事件正在进行中，因此数据还包含null值（如stop_time），我想合并这些间隔，以便它也可以处理null值。

输入：

输出：

Answer 1

这是一种空白和岛屿问题。您可以通过累积最大值end_time并与start_time进行比较来找出岛屿的起点。

还有一些其他逻辑可以处理NULL end_time。由于数据全部（大概）是过去的，因此我们将在将来的某个时间替换它，然后将其转换回NULL。

所以：

select id,
       min(start_time) as start_time,
       nullif(max(end_time), datetime('2100-01-01')) as end_time,
       count(*) as cnt
from (select t.*,
             countif( prev_max_end_time is null or prev_max_end_time < start_time) over (partition by id order by start_time) as grouping
      from (select t.* except (end_time),
                   coalesce(end_time, datetime('2100-01-01')) as end_time,
                   max(coalesce(end_time, datetime('2100-01-01'))) over
                       (partition by id
                        order by start_time
                        rows between unbounded preceding and 1 preceding
                       ) as prev_max_end_time
            from t
           ) t
    ) t
group by id, grouping
order by id, min(start_time);

Answer 2

我喜欢使用（我自己使用这种方法）带有窗口函数的解决方案，但这是另一个，只是为了好玩：）

with

sample_data as (
    select
        *
    from
        unnest(
            array[
                struct(1 as id, 'a' as unique_id, datetime'2019-07-21 20:30:05' as start_time, datetime'2019-07-21 20:30:10' as end_time),
                struct(1, 'b', datetime'2019-07-21 20:30:07', datetime'2019-07-21 20:30:17'),
                struct(1, 'c', datetime'2019-07-21 20:30:21', datetime'2019-07-21 20:30:25'),
                struct(1, 'd', datetime'2019-07-21 20:30:28', datetime'2019-07-21 20:30:33'),
                struct(1, 'e', datetime'2019-07-21 20:30:30', datetime'2019-07-21 20:30:36'),
                struct(1, 'f', datetime'2019-07-21 20:30:29', datetime'2019-07-21 20:30:45'),
                struct(1, 'g', datetime'2019-07-21 20:30:50', datetime'2019-07-21 20:30:53'),
                struct(1, 'h', datetime'2019-07-21 20:30:52', datetime'2019-07-21 20:30:59'),
                struct(1, 'i', datetime'2019-07-21 20:30:56', null),
                struct(2, 'j', datetime'2019-07-21 20:30:07', datetime'2019-07-21 20:30:14'),
                struct(2, 'k', datetime'2019-07-21 20:30:18', datetime'2019-07-21 20:30:30'),
                struct(2, 'l', datetime'2019-07-21 20:30:21', datetime'2019-07-21 20:30:30'),
                struct(2, 'm', datetime'2019-07-21 20:30:41', null),
                struct(2, 'n', datetime'2019-07-21 20:30:48', null),
                struct(3, 'o', datetime'2019-07-21 20:30:20', null),
                struct(3, 'p', datetime'2019-07-21 20:30:30', null)
            ]
        )
),

joins as (
    select
        sd1.*,
        sd2.id as id2,
        sd2.unique_id as unique_id2
    from
        sample_data as sd1
        left join sample_data as sd2 on sd1.id = sd2.id
                                        and sd1.start_time <= coalesce(sd2.end_time, datetime'2099-01-01')
                                        and coalesce(sd1.end_time, datetime'2099-01-01') >= sd2.start_time
    order by
        1, 2, 3, 6
),

by_groups as (
    select
        j1.id,
        j1.unique_id,
        j1.start_time,
        j1.end_time,
        array_agg(distinct j2.unique_id order by j2.unique_id) as group_id
    from
        joins as j1
        left join joins as j2 on    j1.id = j2.id
                                    and j1.unique_id2 = j2.unique_id2
                                    and j2.unique_id != j2.unique_id2
    group by
        1, 2, 3, 4
)

select
    id,
    array_length(group_id) as number_of_unique_id,
    min(start_time) as start_time,
    nullif(
        max(coalesce(
                end_time,
                datetime'2099-01-01')),
        datetime'2099-01-01') as end_time
from
    by_groups
group by
    1, 2, array_to_string(group_id, ',')

合并重叠间隔并包含空值

2 个答案: