計算從開始和結束時間導出的每分鐘會話

Question

我有一個表格，其中包含用戶活動的記錄，其中涵蓋了由開始和結束時間指示的跨度。我正在尋找前一天每單位時間內系統中活動用戶的數量。

最大會話長度是一個小時，並且它們不跨越小時邊界。會話可以結束，並在同一分鐘內開始新會話。

下面是查詢的一個精簡版：

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 

我試圖避開那個討厭左連接：

-> 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")) 

下面是基於戈登·利諾夫的答案是什麼，幾乎爲我工作。當用戶的會話在一分鐘之內轉換時，它就會被計算在內。看起來像正確的方向。原來的查詢可能會因爲同樣的原因而計算在內，但有機會在一分鐘內獲得不同會話ID的計數來解決這個問題。

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; 

爲了便於比較，這裏有一個小時的數據的時序結果：

1. 原始查詢時間：81301.345毫秒
2. 總和，在查詢時間：36242.342毫秒

Answer 1

通過計算每分鐘的啓動和停止次數，然後計算累計和，您可以更快地做到這一點。結果是這樣的：

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;