如何计算SQL中的最频繁值

我有一个表，它有三列：货币，交易这种货币的交易所，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

我需要回答一个问题，哪家交易所是这种特殊货币的领导者

这意味着对于给定的货币，计算要交换的记录数 并且只返回最大值 换句话说，查询的结果应该是

Currency Exchange Frequency 
USD      NewYork  3
CAD      Montreal 2 

---

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;

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

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

注意：如果发生连接，则返回所有最大值。如果您只需要一个，请使用

行号（）

而不是

rank（）

编辑：

在Postgres中（在我回答后添加），您可以在上使用

distinct：

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

请注意，如果存在关联，则不会返回重复项。
如果要在聚合查询中使用最常用的行，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;

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

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

注意：如果发生连接，则返回所有最大值。如果您只需要一个，请使用
行号（）
而不是
rank（）

编辑：

在Postgres中（在我回答后添加），您可以在

上使用

distinct：

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

请注意，如果存在关联，则不会返回重复项。
您可以使用窗口功能：

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;

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

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

您可以使用窗口功能：

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;

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

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

使用
first\u value（）
和
max（）
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;

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

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

请参阅。

结果:

使用
first\u value（）
和
max（）
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;

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

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

请参阅。

结果:

在场景中，您可以在

上使用

distinct。只需分组计算即可
货币
和
交换
并按
货币
和
计数
降序排序。
因此，查询如下所示：

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

在场景中，您可以在

上使用

distinct。只需分组计算即可
货币
和
交换
并按
货币
和
计数
降序排序。
因此，查询如下所示：

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

而
3
和
7
从何而来？Gordon Linoff它们来自于它们出现在table@OlliePugh . . . 它们与您的示例数据不匹配。请用您正在运行的数据库标记您的问题：mysql、oracle、postgresql…？@GordonLinoff这不是我的示例数据lolAnd
3
和
7
来自何处？Gordon Linoff它们来自它们在数据库中出现的次数table@OlliePugh . . . 它们与您的示例数据不匹配。请用您正在运行的数据库标记您的问题：mysql、oracle、postgresql…？@GordonLinoff这不是我的示例数据lol