BigQuery - 组合零散的事件答案

【问题标题】：BigQuery - combine fragmented eventsBigQuery - 组合零散的事件
【发布时间】：2020-02-26 10:24:50
【问题描述】：

这是一个示例数据：

create table activity as
select "2020-02-25T09:06:12" as datetime_start,  "2020-02-25T09:07:31" as datetime_end, 0 as flag uniuon all 
select "2020-02-25T09:16:08" as datetime_start,  "2020-02-25T09:17:31" as datetime_end, 0 as flag uniuon all 
select "2020-02-25T09:17:31" as datetime_start,  "2020-02-25T09:27:31" as datetime_end, 1 as flag uniuon all 
select "2020-02-25T09:27:31" as datetime_start,  "2020-02-25T09:32:41" as datetime_end, 1 as flag uniuon all 
select "2020-02-25T09:35:57" as datetime_start,  "2020-02-25T09:37:31" as datetime_end, 0 as flag uniuon all 
select "2020-02-25T09:49:23" as datetime_start,  "2020-02-25T09:51:16" as datetime_end, 0 as flag uniuon all 
select "2020-02-25T09:51:16" as datetime_start,  "2020-02-25T10:03:46" as datetime_end, 1 as flag uniuon all 
select "2020-02-25T10:03:46" as datetime_start,  "2020-02-25T10:05:57" as datetime_end, 1 as flag uniuon all 
select "2020-02-25T10:05:57" as datetime_start,  "2020-02-25T10:07:31" as datetime_end, 1 as flag uniuon all 
select "2020-02-25T10:07:31" as datetime_start,  "2020-02-25T10:10:22" as datetime_end, 1 as flag uniuon all 
select "2020-02-25T10:10:22" as datetime_start,  "2020-02-25T10:12:55" as datetime_end, 1 as flag uniuon all 
select "2020-02-25T10:12:55" as datetime_start,  "2020-02-25T10:20:17" as datetime_end, 1 as flag uniuon all 
select "2020-02-25T10:20:17" as datetime_start,  "2020-02-25T10:27:40" as datetime_end, 1 as flag uniuon all 
select "2020-02-25T10:27:40" as datetime_start,  "2020-02-25T10:39:51" as datetime_end, 1 as flag;

我正在寻找将根据标志列计算活动块的查询。
如果标志设置为 1，则直到标志更改为 0 之后的行需要合并到单个活动块中。

上面的例子产生了 6 个活动块。

2020-02-25T09:06:12 - 2020-02-25T09:07:31
2020-02-25T09:16:08 - 2020-02-25T09:17:31
2020-02-25T09:17:31 - 2020-02-25T09:32:41
2020-02-25T09:35:57 - 2020-02-25T09:37:31
2020-02-25T09:49:23 - 2020-02-25T09:51:16
2020-02-25T09:51:16 - 2020-02-25T10:39:51

【问题讨论】：

标签： sql database group-by google-bigquery gaps-and-islands

【解决方案1】：

这回答了问题的原始版本。

GMB 的答案可能有效，但它似乎是定制的，因为它硬编码了标志的值。我更喜欢更通用的方法：

with activity as (
    select "2020-02-25T09:06:12" as datetime_start,  "2020-02-25T09:07:31" as datetime_end, 0 as flag union all 
    select "2020-02-25T09:16:08" as datetime_start,  "2020-02-25T09:17:31" as datetime_end, 0 as flag union all 
    select "2020-02-25T09:17:31" as datetime_start,  "2020-02-25T09:27:31" as datetime_end, 1 as flag union all 
    select "2020-02-25T09:27:31" as datetime_start,  "2020-02-25T09:32:41" as datetime_end, 1 as flag union all 
    select "2020-02-25T09:35:57" as datetime_start,  "2020-02-25T09:37:31" as datetime_end, 0 as flag union all 
    select "2020-02-25T09:49:23" as datetime_start,  "2020-02-25T09:51:16" as datetime_end, 0 as flag union all 
    select "2020-02-25T09:51:16" as datetime_start,  "2020-02-25T10:03:46" as datetime_end, 1 as flag union all 
    select "2020-02-25T10:03:46" as datetime_start,  "2020-02-25T10:05:57" as datetime_end, 1 as flag union all 
    select "2020-02-25T10:05:57" as datetime_start,  "2020-02-25T10:07:31" as datetime_end, 1 as flag union all 
    select "2020-02-25T10:07:31" as datetime_start,  "2020-02-25T10:10:22" as datetime_end, 1 as flag union all 
    select "2020-02-25T10:10:22" as datetime_start,  "2020-02-25T10:12:55" as datetime_end, 1 as flag union all 
    select "2020-02-25T10:12:55" as datetime_start,  "2020-02-25T10:20:17" as datetime_end, 1 as flag union all 
    select "2020-02-25T10:20:17" as datetime_start,  "2020-02-25T10:27:40" as datetime_end, 1 as flag union all 
    select "2020-02-25T10:27:40" as datetime_start,  "2020-02-25T10:39:51" as datetime_end, 1 as flag
    )
select min(datetime_start) as datetime_stat,
       max(datetime_end) as datetime_end,
       flag
from (select a.*,
             countif( datetime_start <> prev_datetime_end OR
                      prev_flag <> flag
                    ) over (order by datetime_start) as grp
       from (select a.*,
                    lag(flag) over (order by datetime_start) as prev_flag,
                    lag(datetime_end) over (order by datetime_start) as prev_datetime_end
             from activity a
            ) a
) t
group by flag, grp

【讨论】：

@ronencozen 。 . .不。如果修改后的问题使答案无效，我不会查看它们。更好的方法是提出一个新问题。
我明白了，谢谢你对我最初的问题的一个很好的回答。

【解决方案2】：

这是一个gaps-and-island的变种。这是一种使用lag() 和窗口总和来定义连续1s 组的方法：

select
    min(datetime_start) datetime_stat,
    max(datetime_end) datetime_end,
    flag
from (
    select
        t.*,
        sum(case when flag = 1 and lag_flag = 1 then 0 else 1 end) 
            over(order by datetime_start) grp
    from (
        select 
            t.*,
            lag(flag) over(order by datetime_start) lag_flag
        from mytable t
    ) t
) t
group by flag, grp

【讨论】：

感谢您的及时回复。我将您的解决方案插入到我的查询中，但它没有输出我要查找的内容。
让我们以第 3 行和第 4 行组成的活动块编号 3 为例。我没有将它组合成一行，而是返回原始行，其中 datetime_end 等于前一行的 datetime_end。
@ronencozen：here is a db fiddle。这似乎为您的示例数据产生了正确的结果（这是一个 MySQL 小提琴，因为据我所知没有 BQ 小提琴，但逻辑是相同的，这是标准 SQL）。
这里是 BigQuery 沙盒的链接。 cloud.google.com/bigquery/docs/sandbox