SQL 与分组相交

Question

鉴于这两个 tables/sets 具有不同的项目组，~~我如何找到 set1 中的哪些组跨越 set2 中的多个组？~~ 如何找到 set1 中无法被 set2 中的单个组覆盖的组？

例如对于下表，A (1,2,5) 是唯一跨越 s1(1,2,3) 和 s2(2,3,4,5)。 B 和 C 不是答案，因为它们都包含在一个组中 s2。

我更愿意使用 SQL（Sql Server 2008 R2 可用）。

谢谢。

set1                            set2
 +---------+----------+          +---------+----------+
 | group   |  item    |          | group   |  item    |
 `````````````````````+          `````````````````````+
 |   A     |    1     |          |   s1    |    1     |
 |   A     |    2     |          |   s1    |    2     |
 |   A     |    5     |          |   s1    |    3     |
 |   B     |    4     |          |   s2    |    2     |
 |   B     |    5     |          |   s2    |    3     |
 |   C     |    3     |          |   s2    |    4     |
 |   C     |    5     |          |   s2    |    5     |
 +---------+----------+          +---------+----------+

使用这个 sqlfiddle 试试：http://sqlfiddle.com/#!6/fac8a/3

或者使用下面的脚本生成临时表来尝试答案：

create table #set1 (grp varchar(5),item int)
create table #set2 (grp varchar(5),item int)

insert into #set1 select 'a',1 union select 'a',2 union select 'a',5 union select 'b',4 union select 'b',5 union select 'c',3 union select 'c',5
insert into #set2 select 's1',1 union select 's1',2 union select 's1',3 union select 's2',2 union select 's2',3 union select 's2',4 union select 's2',5

select * from #set1
select * from #set2

--drop table #set1
--drop table #set2

Answer 1

Select 来自 set1 的组，在 set2 中没有组，而 set1 中的所有项目都存在于 set2 中：

select s1.grp from set1 s1
where not exists(
  select * from set2 s2 where not exists(
    select item from set1 s11 
    where s11.grp = s1.grp 
    except 
    select item from set2 s22
    where s22.grp = s2.grp))
group by s1.grp

Answer 2

您可以通过以下查询找到解决方案：

SELECT A.GROUP AS G1, A.ITEM AS T1, B.GROUP, B.ITEM
FROM SET1 A RIGHT JOIN SET2 B ON A.ITEM=B.ITEM
WHERE A.GROUP IS NULL

Answer 3

你能检查一下吗

SELECT DISTINCT a.Group1, a.Item, b.CNT
FROM SET1 a
INNER JOIN
(SELECT GroupA, COUNT(*) CNT
 FROM
 (
    SELECT DISTINCT a.Group1 GroupA, b.Group1 GroupB
    FROM SET1 a
        INNER JOIN SET2 b ON a.Item = b.Item
 ) a GROUP BY GroupA
) b ON a.Group1 = b.GroupA
 WHERE b.CNT > 1

Answer 4

感谢您的评论。我相信以下编辑后的查询会起作用：

Select distinct grp1, initialRows, max(MatchedRows) from 
  (
  select a.grp as grp1, b.grp as grp2
  , count(distinct case when b.item is not null then a.item end) as MatchedRows
  , d.InitialRows
  from set1 a
  left join set2 b
  on a.item = b.item
  left join 
    (select grp, count(distinct Item) as InitialRows from set1
     group by grp) d
  on a.grp = d.grp
  group by a.grp, b.grp, InitialRows
) c
group by grp1, InitialRows
having max(MatchedRows) < InitialRows

Answer 5

好的。这很丑陋，但它应该有效。我在 fiddle 试过了。我觉得可以通过windowing来完成，但是我得想一想。

这是丑陋的现在。

WITH d1 AS (
SELECT set1.grp
     , COUNT(*) cnt
  FROM set1
 GROUP BY set1.grp
), d2 AS (  
SELECT set1.grp grp1
     , set2.grp grp2
     , COUNT(set1.item) cnt
  FROM set1
 INNER JOIN set2
    ON set1.item = set2.item
 GROUP BY set1.grp
     , set2.grp
 )
SELECT grp
  FROM d1
EXCEPT  
SELECT d1.grp 
  FROM d1
 INNER JOIN d2
    ON d2.grp1 = d1.grp
   AND d2.cnt = d1.cnt

Answer 6

我认为这会成功。子查询 returns 每个 set1 组设置 2 个组，通过计算匹配项并将匹配项计数与 set1 组计数进行比较，匹配 set1 中的所有项目。

select s.grp from #set1 s
group by s.grp
having not exists ( 
    select s2.grp from #set2 s2 inner join #set1 s1 on s2.item = s1.item
    where s1.grp = s.grp
    group by s2.grp
    having count(s.item) = count(s2.item)   
    )

Answer 7

与罗伯特公司基本相同
我没有从他的回答中得到这个 - 独立提出这个

    select set1.group  
      from set1
except
    select set1count.group 
      from ( select set1.group            , count(*) as [count]  
               from set1 
           ) as set1count 
      join ( select set1.group as [group1], count(*) as [count]  
               from set1 
               join set2 
                 on set2.item = set1.item 
              group by set1.group, set2.group -- this is the magic
           ) as set1count 
        on set1count.group = set2count.[group1] -- note no set2.group match
       and set1count.count = set12count.count -- the items in set1 are in at least on set2 group

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