Question

我有一个包含三列的表：货币，使用该货币进行交易的交易所，日期

Currency  Exchange Date
USD       NewYork  01/12/20
USD       NewYork  01/11/20
USD       NewYork  01/10/20
USD       Montreal 01/10/20
CAD       Montreal 01/07/20
CAD       Montreal 01/06/20
CAD       Beijing  01/06/20

我需要回答以下问题：哪种交易所是该特定货币的领导者。

这意味着对于给定的货币，计算要交换的记录数量只返回最大值换句话说，查询结果应该类似于

Currency Exchange Frequency 
USD      NewYork  3
CAD      Montreal 2

Answer 1

如果要在聚合查询中使用最常见的行，请使用窗口函数：

select ce.*
from (select currency, exchange, count(*) as cnt,
             rank() over (partition by currency order by count(*) desc) as seqnum
      from t
      group by currency, exchange
     ) ce
where seqnum = 1;

注意：如果发生平局，则返回所有最大值。如果只需要一个，则使用row_number()代替rank()。

编辑：

在Postgres（我回答后添加）中，您可以使用distinct on：

select distinct on (currency) exchange, count(*) as cnt
from t
group by currency, exchange
order by currency, count(*) desc;

请注意，如果有关系，这不会返回重复项。

Answer 2

您可以使用窗口功能：

select *
from (
    select currency, exchange, count(*) frequency,
        rank() over(partition by currency order by count(*) desc) rn
    from mytable
    group by currency, exchange
) t
where rn = 1

Answer 3

您可以在场景中使用distinct on。只需通过分组计数 currency和exchange并按currency和count降序排列。因此查询将如下所示：

select
distinct on (currency)
currency,
exchange,
count(*)
from table1
group by 1,2
order by 1,3 desc

Answer 4

具有first_value()和max()窗口功能：

select distinct currency,
       first_value(exchange) over (partition by currency order by count(*) desc) exchange,
       max(count(*)) over (partition by currency) frequency
from tablename
group by currency, exchange

请参见demo。
结果：

> currency | exchange    | frequency
> :------- | :---------- | --------:
> CAD      | Montreal    |         2
> USD      | NewYork     |         3

如何计算SQL中的最高频繁值

4 个答案: