如何计算自一天开始以来每小时的聚合不同 ID？答案

【问题标题】：how to count aggregate distinct IDs Per Hour Since Beginning of the day?如何计算自一天开始以来每小时的聚合不同 ID？
【发布时间】：2021-04-27 11:22:37
【问题描述】：

我需要编写一个查询，它计算不同的 ID 并随着时间的推移聚合它们。

例如

结果应该是

【问题讨论】：

标签： sql aggregate distinct vertica

【解决方案1】：

对于每个id，记录id第一次出现的时间，然后进行累加：

select hour, sum(count(*)) over (partition by day order by hour)
from (select day, id, min(hour) as hour
      from t
      group by day, id
     ) t
group by hour
order by hour;

注意：这假设您确实需要给定一天内的小时。

你也可以这样表达：

select day, hour, sum(cnt) over (partition by day order by hour)
from (select day, hour, count(*) as cnt
      from (select day, id, min(hour) as hour
            from t
            group by day, id
           ) t
      group by hour
     ) h
order by hour;

上面不包括一个小时，除非那个小时有一个新的 id。对于所有时间，您都可以使用窗口函数：

select hour,
       sum(sum( (seqnum = 1)::int ) over (partition by day order by hour)
from (select day, id,
             row_number() over (partition by day, id order by hour) as seqnum
      from t
     ) t
group by hour
order by hour;

你也可以这样表达：

select day, hour, sum(cnt) over (partition by day order by hour)
from (select day, hour, sum( (seqnum = 1)::int ) as cnt
      from (select day, id,
                   row_number() over (partition by day, id order by hour) as seqnum
            from t
           ) t
      group by hour, day
     ) dh
order by hour;

【讨论】：

1.它不在 vertica 2 中运行。您错过了内部查询中的小时列我收到此错误“错误：聚合函数调用不能包含分析函数调用”
@erezlale 。 . .哇！这是一个非常神秘的限制。我从未在其他数据库中听说过这样的问题——包括为 Vertica 提供代码库的 Postgres。

【解决方案2】：

我每小时的计数不同。

declare @tmp as table(RecordNumber int, Day date, Hour int, ID int)
insert into @tmp(RecordNumber,Day,Hour,ID) values(1,'4/27/2021',1,100),
(2,'4/27/2021',1,100)
,(3,'4/27/2021',1,101)
,(4,'4/27/2021',2,202)
,(5,'4/27/2021',2,203)
,(6,'4/27/2021',3,101)
,(7,'4/27/2021',8,305)
,(8,'4/27/2021',16,100)
,(9,'4/27/2021',22,30)
,(10,'4/27/2021',22,1000)
,(11,'4/27/2021',23,100)
,(12,'4/27/2021',23,50)


select distinct Day, Hour, count(distinct ID) ID_Count from @tmp
group by Day, Hour

输出

 Hour   ID_Count
   1    2
   2    2
   3    1
   8    1
  16    1
  22    2
  23    2

【讨论】：