艾伦在SQL中的Interval Algebra操作

Question

我一直在努力解决SQL中一些棘手的问题,我需要从事件间隔推断出资产利用率,并且刚刚了解了Allen的Interval Algebra,这似乎是解决这些问题的关键.

代数描述了区间之间的13种关系,下面的图像显示了前七个,其余的是逆的(即x之前的y,y遇到x等)

但我很难找到如何实施相关的操作.

根据我的示例数据,如何从SQL或PLSQL中的以下三种类型的操作中获取结果？

1. 防脱离
2. 降低
3. 找到差距

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请参阅我的SQLFiddle链接:http://sqlfiddle.com/#!4/cf0cc

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原始数据

   start end width
[1]     1  12    12
[2]     8  13     6
[3]    14  19     6
[4]    15  29    15
[5]    19  24     6
[6]    34  35     2
[7]    40  46     7

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操作1 - 分离结果

我想要一个查询来disjoint set从上面的数据中返回,其中所有重叠的间隔都被分成行,这样就不存在重叠.

我该如何处理这个SQL？

     start end width
[1]      1   7     7
[2]      8  12     5
[3]     13  13     1
[4]     14  14     1
[5]     15  18     4
[6]     19  19     1
[7]     20  24     5
[8]     25  29     5
[9]     34  35     2
[10]    40  46     7

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操作2 - 减少结果

如何减少/展平间隔,以便它们是:

• 不为空(即它们具有非空宽度);
• 不重叠;
• 从左到右排序;
• 甚至不相邻(即两个连续范围之间必须有非空的间隙)

对于我的例子,这看起来像:

    start end width
[1]     1  29    29
[2]    34  35     2
[3]    40  46     7

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操作3 - 差距结果

另外,我如何找到差距？

   start end width
[1]    30  33     4
[2]    36  39     4

Answer 1

这是一个SQLFiddle演示 首先创建临时表以简化查询,尽管您可以将这些创建查询放入最终查询并在没有临时表的情况下执行:

create table t as select * from
(
select null s ,"start"-1 as e  from data
union all
select "start" s,null e  from data
union all
select "end"+1 s ,null e  from data
union all
select null s ,"end" e  from data
) d where exists (select "start" 
                  from data where d.s between data."start" and data."end"
                               or d.e between data."start" and data."end"
                                );
--Operation 1 - Disjoined Result   
create table t1 as select s,e,e-s+1 width from
(
select distinct s,(select min(e) from t where t.e>=t1.s) e from t t1
) t2 where t2.s is not null and t2.e is not null;

--Operation 2 - Reduced Result
create table t2 as 
select s,e,e-s+1 width from
(
select s,(select min(d2.e) from t1 d2 where d2.s>=d.s and not exists
          (select s from t1 where t1.s=d2.e+1) ) e
from
t1 d where not exists(select s from t1 where t1.e=d.s-1) 
) t2;

--Temp table for Operation 3 - Gaps
create table t3 as 
select null s, s-1 e from t2
union all
select e+1 s, null e from t2;

现在这里是查询:

--Operation 1 - Disjoined Result
select * from t1 order by s;

--Operation 2 - Reduced Result


select * from t2 order by s;

--Operation 3 - Gaps

select s,e,e-s+1 width 
from
(
select s,(select min(e) from t3 where t3.e>=d.s) e from t3 d
) t4 where s is not null and e is not null
order by s;