Question

我有一个表（DeviceOS2），并希望每个ID的每一列（操作系统和设备）获得最频繁的值。

ID      OS      Device

123     OSX     Mac 
123     OSX     PC  
123     OSX     PC  
123     Android Tablet

所需结果：

ID      OS      Device

123     OSX     PC

但是，我的代码现在为我提供了以下信息：

ID       OS            Device

123      Android       Tablet
123      OSX           Mac
123      OSX           PC

看起来它拾取了每个组合。

当前代码（T-SQL）：

Select 
ID,
OS,
Device

FROM(
Select 
ID,
OS,
Device
FROM DeviceOS2
Group By ID,OS,Device) a 
Group By ID,OS,Device

Answer 1

您可以使用：

SELECT TOP 1 WITH TIES *
FROM tab
ORDER BY COUNT(*) OVER(PARITIION BY ID,OS) DESC

Answer 2

这称为模式。您可以使用窗口功能：

select o.*
from (select os, device, count(*) as cnt,
             row_number() over (partition by os order by count(*) desc) as seqnum
      from DeviceOS2
      group by os, device
     ) o
where seqnum = 1;

如果您想要最频繁的组合，请使用：

select os, device, count(*) as cnt
from DeviceOS2
group by os, device
order by count(*) desc
fetch first 1 row only;

（或根据需要使用select top (1)）。

编辑：

针对您编辑的问题：

select o.*
from (select os, device, count(*) as cnt,
             row_number() over (partition by os order by count(*) desc) as seqnum
      from DeviceOS2
      group by os, device
     ) o
where seqnum = 1;

如果您想要最频繁的组合，那么查询会稍微复杂一些。一种方法是两种聚合：

select o.id,
       max(case case when o.seqnum = 1 then os end) as most_frequent_os,
      max(case case when d.seqnum = 1 then device end) as most_frequent_device
from (select id, os, count(*) as cnt,
             row_number() over (partition by id order by count(*) desc) as seqnum
      from DeviceOS2
      group by id, os
     ) o join
     (select id, device, count(*) as cnt,
             row_number() over (partition by id order by count(*) desc) as seqnum
      from DeviceOS2
      group by id, device
     ) d
     on d.id = o.id

Answer 3

尝试一下：

select top 1 with ties a.ID, a.OS,a.Device
from (
select d.ID, d.OS, d.Device, ROW_NUMBER () over (partition by d.OS, d.Device order by id) rnk
from DeviceOS2 d)a
order by a.rnk desc

更新

如果每个ID需要最频繁的ID：

select c.ID,c.OS,c.Device from (
select d.ID, d.OS, d.Device, ROW_NUMBER () over (partition by d.id, d.OS, d.Device order by id) rnk
from DeviceOS2 d)c
join 
(
select  a.ID,max(a.rnk) AS rnk
from (
select d.ID, d.OS, d.Device, ROW_NUMBER () over (partition by d.id, d.OS, d.Device order by id) rnk
from DeviceOS2 d)a
group by a.ID) a
on c.ID = a.ID and a.rnk = c.rnk

每组获得最频繁的价值

3 个答案: