SQL函數返回一個組中多個列的「最常見值」按

Question

我正在尋找最簡單的方法來返回分組的select語句的多個列結果中最常見的值。我在網上找到的所有東西都指向RANK中選擇的單個項目，或者在GROUP BY之外單獨處理每個列。

樣本數據：

SELECT 100 as "auser", 
'A' as "instance1", 'M' as "instance2" 
union all select 100, 'B', 'M' 
union all select 100,'C', 'N' 
union all select 100, 'B', 'O' 
union all select 200,'D', 'P' 
union all select 200, 'E', 'P' 
union all select 200,'F', 'P' 
union all select 200, 'F', 'Q' 

樣本數據結果：

auser instance1 instance2 
100  A   M 
100  B   M 
100  C   N 
100  B   O 
200  D   P 
200  E   P 
200  F   P 
200  F   Q 

查詢邏輯（我看到它在我的頭上）：

SELECT auser, most_common(instance1), most_common(instance2) 
FROM datasample 
GROUP BY auser; 

期望的結果：

100  B   M 
200  F   P 

Answer 1

解決此問題的方法使用了嵌套窗口函數。最裏面的子查詢計算每列的計數。下一個子查詢將這些排序（使用row_number()）。外部查詢然後使用條件聚合得到你想要的結果：

select auser, MAX(case when seqnum1 = 1 then instance1 end), 
     MAX(case when seqnum2 = 1 then instance2 end) 
from (select t.*, 
      ROW_NUMBER() over (partition by auser order by cnt1 desc) as seqnum1, 
      ROW_NUMBER() over (partition by auser order by cnt2 desc) as seqnum2 
     from (select t.*, 
        count(*) over (partition by auser, instance1) as cnt1, 
        COUNT(*) over (partition by auser, instance2) as cnt2 
      from t 
      ) t 
    ) t 
group by auser 

Answer 2

我不知道如果我能找到的東西更優雅，但如果你在SQL這可能做2005+（因爲我使用ranking function和CTEs）：

with instance1 as (
    select auser, instance1 
     , row_number() over (partition by auser order by count(*) desc, instance1) as row_num 
    from datasample 
    group by auser, instance1 
), instance2 as (
    select auser, instance2 
     , row_number() over (partition by auser order by count(*) desc, instance2) as row_num 
    from datasample 
    group by auser, instance2 
) 
select a.auser, a.instance1, b.instance2 
from instance1 as a 
    join instance2 as b on a.auser = b.auser 
where a.row_num = 1 
    and b.row_num = 1 
order by a.auser; 

我m不知道你希望如何處理空值，並且將row_num等同性移動到連接條件不會更改我的框中的執行計劃。

如果您使用的是SQL Server 2000，那麼可以使用count和"triangular join"來替換這些帶有派生表的CTE，並僞造row_number（）。