计算从开始和结束时间导出的每分钟会话

Question

我有一个表格，其中包含用户活动的记录，其中涵盖了由开始和结束时间指示的跨度。我正在寻找前一天每单位时间内系统中活动用户的数量。

最大会话长度是一个小时，并且它们不跨越小时边界。会话可以结束，并在同一分钟内开始新会话。

下面是查询的一个精简版：

with minutes AS (
    -- ignore this...it generates a day's worth of timestamps for each minute 
    -- it's hairy but is what I'm stuck with on redshift 
    select (dateadd(minute, -row_number() over (order by true), sysdate::date)) as minute 
     from seed_table limit 1440 
), 
sessions as (
    select sid, ts_start, ts_end 
    from user_sessions s 
    where ts_end >= sysdate::date-'1 day'::interval 
     and ts_start < sysdate::date 
) 
select m.minute, count(distinct(s.sid)) 
from minutes m 
left join sessions s on s.ts_end >= m.minute and s.ts_start < m.minute+'1 min'::interval 
group by 1 

我试图避开那个讨厌左连接：

-> XN Nested Loop Left Join DS_BCAST_INNER (cost=6913826151.95..4727012848741.55 rows=410434560 width=166) 
    Join Filter: (("inner".ts_start < ("outer"."minute" + '00:01:00'::interval)) AND ("inner".ts_end >= "outer"."minute")) 

下面是基于戈登·利诺夫的答案是什么，几乎为我工作。当用户的会话在一分钟之内转换时，它就会被计算在内。看起来像正确的方向。原来的查询可能会因为同样的原因而计算在内，但有机会在一分钟内获得不同会话ID的计数来解决这个问题。

select minute, sum(count) over (order by minute rows unbounded preceding) as users 
from (
    select minute, sum(count) as count 
    from (
     (
      select date_trunc('minute', ts_start) as minute, count(*) as count 
      from sessions 
      group by 1 
     ) union all (
      select date_trunc('minute', ts_end) as minute, - count(*) as count 
      from sessions 
      group by 1 
     ) 
    ) s1 
    group by minute 
) s2 
order by minute; 

为了便于比较，这里有一个小时的数据的时序结果：

1. 原始查询时间：81301.345毫秒
2. 总和，在查询时间：36242.342毫秒

Answer 1

通过计算每分钟的启动和停止次数，然后计算累计和，您可以更快地做到这一点。结果是这样的：

select minute, sum(cnt) over (order by minute) 
from ((select date_trunc('minute', ts_start) as minute, count(*) as cnt 
     from sessions 
     group by 1 
    ) union all 
     (select date_trunc('minute', ts_end), - count(*) 
     from sessions 
     group by 1 
    ) 
    ) s 
group by minute 
order by minute;