如何计算 SQL 中出现频率最高的值

Question

我有一个 table，它包含三列：货币、交易该货币的交易所、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

我需要回答哪个交易所是该特定货币的领导者的问题。

这意味着对于给定的货币，计算兑换多少条记录并且 return 只有最大值换句话说，查询的结果应该是这样的

Currency Exchange Frequency 
USD      NewYork  3
CAD      Montreal 2

Answer 1

如果你想要聚合查询中最常见的行，那么使用window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：如果出现平局，这 return 都是最大值。如果你只想要一个，那么使用 row_number() 而不是 rank().

编辑：

在Postgres中（我回答后添加的），你可以使用distinct on:

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

请注意，如果有联系，这不会 return 重复。

Answer 2

您可以使用 window 函数：

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

使用 first_value() 和 max() window 函数：

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

参见demo。
结果：

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

Answer 4

场景中可以使用distinct on。只需分组计数 currency 和 exchange 并按 currency 和 count 降序排列。所以查询将如下所示：

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

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