Question

这里我们比较基因表记录如下：首先将基因a和所有基因a，b，c进行比较，例如aa，ab，ac同样需要b然后ba，bb，bc等...... ....

所以这里的匹配结果是n和b是2因为匹配记录是589,822共同基因项，因为b c count是1，因为匹配记录586和所有其他组合它应该是零。

goterm  gene auto
--------------------
589     a    1
822     a    2
478     a    3
586     b    4
589     b    5
600     c    6
586     c    7
822     b    8

查询：

select count(*), 
       x.gene,
       x.ng 
 from (select t.gene,
              v.gene as ng 
        from (select distinct gene 
                from gene) as t 
  cross join (select distinct gene from gene) as v) as x 
   left join (select (g.gene),(n.gene) as ng from gene g 
        join gene n on n.goterm=g.goterm where g.auto<n.auto ) as y on y.gene = x.ng 
                                                                   and y.ng = x.gene
group by x.gene,x.ng

最后，上述查询的输出是：

count  gene    gene
1      a       a
2      b       a
1      c       a
1      a       b
1      b       b
1      c       b
1      a       c
1      b       c
1      c       c

但输出必须是：

count  gene    gene
0      a       a
2      b       a  
0      c       a
0      a       b
0      b       b
1      c       b
0      a       c
0      b       c
0      c       c

Answer 1

不确定你的逻辑究竟是如何工作的（尤其是auto上的比较），但是这个查询产生了你正在寻找的结果：

declare @t table (goterm  int, gene char(1), auto int identity)
insert @t
          select 589,     'a'
union select all 822,     'a'
union select all 478,     'a'
union select all 586,     'b'
union select all 589,     'b'
union select all 600,     'c'
union select all 586,     'c'
union select all 822,     'b'

select  t1.gene
,       t2.gene
,       (
        select  COUNT(*)
        from    @t t3
        join    @t t4
        on      t3.goterm = t4.goterm
                and t3.auto > t4.auto
        where   t3.gene = t1.gene
                and t4.gene = t2.gene
        ) as Total
from    (
        select  distinct gene 
        from    @t
        ) t1
cross join
        (
        select  distinct gene 
        from    @t
        ) t2
order by
        t2.gene
,       t1.gene

打印：

gene    gene    Total
a       a       0
b       a       2
c       a       0
a       b       0
b       b       0
c       b       1
a       c       0
b       c       0
c       c       0

t1和t2用于创建基因组合矩阵。 Total查询会查找该组合的匹配数。 t3.auto > t4.auto条件让我感到困惑，因此您可能需要仔细检查。

Answer 2

select combo1.gene, combo2.gene, count (gene2.goterm)
from (select distinct gene from gene) combo1
cross join (select distinct gene from gene) combo2
join gene gene1 on combo1.gene = gene1.gene
left join gene gene2 on combo2.gene = gene2.gene
    and gene1.goterm = gene2.goterm and gene1.auto < gene2.auto
group by combo1.gene, combo2.gene
order by combo1.gene, combo2.gene

编辑：意识到我毕竟不需要DISTINCT。

Answer 3

尝试在查询中将count(*)更改为count(y.gene)。

SQL Query比较单个表中的行

3 个答案: