Question

对于第1列中的每个唯一字符串，第2列中最常见的字符串是什么？

例如表格：

1 | 2
-----
A   a
A   a
A   a
A   b
B   b
B   b
B   b
B   a
B   c
C   c
C   d
C   a

结果应该如下：

X | Most common | Weighting
A        a         0.75
B        b         0.60
C        a         0.33

我想使用GROUP BY子句，但我不知道任何适用于字符串的聚合函数。另外，我知道在关系方面如何处理已经有些含糊不清（比如C）。在我的申请中，虽然我只关心加权> 0.50的情况，但模糊性并不重要。

我正在使用SSMS 2014。

Answer 1

下面的CTE计算表格中每条记录的权重，作为计数的商。然后，我们可以使用行号保留每个col1分区的第一条记录。请注意，我不会处理关系的情况，尽管我们可以轻松地添加另一个排序来打破平局。

WITH cte AS (
    SELECT col1, col2,
        1.0 * COUNT(*) OVER (PARTITION BY col1, col2) /
              COUNT(*) OVER (PARTITION BY col1) weighting
    FROM yourTable
)

SELECT col1, col2, weighting
FROM
(
    SELECT *,
        ROW_NUMBER() OVER (PARTITION BY col1 ORDER BY weighting DESC) rn
    FROM cte
) t
WHERE rn = 1
ORDER BY col1;

Demo

Answer 2

所有这些答案看起来都很复杂：

select col1, col2, col2_cnt * 1.0 / col1_cnt
from (select col1, col2,
             count(*) as col2_cnt,
             sum(count(*)) over (partition by col1) as col1_cnt,
             row_number() over (partition by col1 order by count(*) desc) as seqnum
      from t
      group by col1, col2
     ) t
where seqnum = 1

Answer 3

count应该相当简单;这是一个快速选择：

;WITH CTE
AS (
    SELECT nm1
        ,nm2
        ,count(*) AS ct
    FROM #a
    GROUP BY nm1
        ,nm2
    )
    ,CTE2
AS (
    SELECT *
        ,ROW_NUMBER() OVER (
            PARTITION BY nm1 ORDER BY ct DESC
            ) rn
        ,ct * 1.0 / (sum(ct) OVER (PARTITION BY nm1)) AS wt
    FROM CTE
    )
SELECT nm1
    ,nm2
    ,wt
FROM CTE2
WHERE rn = 1

请注意，如果您有联系，row_number可能无法预测 - 如果您希望在有联系时返回两个值，请改用rank。

Answer 4

使用count和row_number窗口函数执行此操作的一种方法。

select top 1 with ties col1,col2,weighting
from (select col1,col2,1.0*count(*) over(partition by col1,col2)/count(*) over(partition by col1) as weighting
      from t
     ) t
order by row_number() over(partition by col1 order by weighting desc,col2) --in case of ties the row with least col2 value will be picked up

如何按SQL Server中最常见的字符串进行分组？

4 个答案:

Demo