在 MySQL (BigQuery) 中查找所有平均结果的最大数量

Question

我按天计算了物业的平均价格。我想找到平均每日房价最高的房产：

select Prop_Name as property, arrival_date as arrival, round(avg(Rate)) as rate
from [SomeDataBase] 
where [Timestamp] >= '2015-06-15 00:00:00.000' 
and [Timestamp] <= '2015-06-20 23:59:59.000'  
group by property, arrival 
order by arrival asc limit 10;

所以我返回以下内容：

Property             Date              Rate
Prop_One    2015-06-15 00:00:00 UTC    281.0     
Prop_Two    2015-06-15 00:00:00 UTC    343.0     
Prop_Three  2015-06-15 00:00:00 UTC    266.0     
Prop_One    2015-06-15 00:00:00 UTC    87.0  
Prop_Three  2015-06-15 00:00:00 UTC    132.0     
Prop_Two    2015-06-15 00:00:00 UTC    80.0 

我想返回每天的最高费率而不重复：

Prop_Three  2015-06-15 00:00:00 UTC    400.0     
Prop_One    2015-06-16 00:00:00 UTC    586.0     
Prop_Three  2015-06-17 00:00:00 UTC    190.0     
Prop_Two    2015-06-18 00:00:00 UTC    180.0    

Answer 1

我认为 bigquery 支持row_number()，所以我认为这会起作用：

select t.*
from (select Prop_Name as property, arrival_date as arrival,
             round(avg(Rate)) as rate,
             row_number() over (partition by arrival_date order by avg(rate) desc) as seqnum
      from [SomeDataBase] 
      where [Timestamp] >= '2015-06-15' and
            [Timestamp] < '2015-06-21'  
      group by property, arrival 
     ) t
where seqnum = 1;

编辑：

啊。以上是标准 SQL，但以下可能适用于 Bigquery：

select t.*
from (select t.*,
             row_number() over (partition by arrival_date order by rate desc) as seqnum
      from (select Prop_Name as property, arrival_date as arrival,
                   round(avg(Rate)) as rate
            from [SomeDataBase] 
            where [Timestamp] >= '2015-06-15' and
                  [Timestamp] < '2015-06-21'  
            group by property, arrival 
           ) t
     ) t
where seqnum = 1;