这是我在解决这个问题的尝试,我希望其他人能找到更有效的方法:
select #t1.id1, #t1.adm1, #t1.dx1, #t2.id2, #t2.adm2, #t2.dx2, iif(#t1.dx1=#t2.dx2,1,0) as shared Into #t3 From #t1 cross join #t2
Select id1, adm1, dx1, id2, adm2, sum(shared) as In1In2, iif(sum(shared)=0,1,0) as In1Not2 into #t4 from #t3 group by id1, adm1, dx1, id2, adm2
Select id1, adm1, dx1, id2, adm2, sum(In1Not2) as nIn1Not2, into #t5 from #t4 group by id1, adm1, id2, adm2
Select id1, adm1, dx2, id2, adm2, iif(sum(shared)=0,1,0) as In2Not1 into #t6 from #t3 group by id1, adm1, dx2, id2, adm2
Select id1, adm1, id2, adm2, sum(In2Not1) as nIn2Not1 into #t7 from #t6 group by id1, adm1, id2, adm2
在接下来的步骤中计算出的值被合并成一个共同的表。这种尝试的问题在于,在100,000的t1上运行它,并且400,000条记录的t2耗时超过2个小时。
我认为这是一个需要交叉连接的例子 – user2001212 2013-02-27 21:50:38
以下是我如何使用交叉连接来处理它:从交叉连接找到彼此匹配的诊断。然后为每组案例计算匹配和不匹配的数量。然而,解决方案似乎非常耗时。 – user2001212 2013-02-27 22:07:29