获得每组最常见的价值

Question

I have a table (DeviceOS2) and would like to get the most frequent value for each column (OS and Device) per ID.

ID      OS      Device

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

Desired result:

ID      OS      Device

123     OSX     PC  

However, my code now gets me the following:

ID       OS            Device

123      Android       Tablet
123      OSX           Mac
123      OSX           PC

Looks like it picks up every combination.

Current code (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

You could use:

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

Answer 2

This is called the mode . You can use window functions:

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;

If you want the most frequent combination, then use:

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

(or use select top (1) if you prefer).

EDIT:

For your edited question:

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;

If you want the most frequent combination, then the query is a bit more complicated. One method is two aggregations:

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

Try this:

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

Update

If you need the most frequent one for each 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