SQL自连接和聚合

Question

我在 postgres 中有一个具有以下结构的表

表路径：乘客、出发地、目的地、日期、月份、年份

我想根据一年中在一条路线上行驶的乘客数量找到前 3 条路线。路线上的乘客总数 (A <-> B) = 乘客总数 (A -> B) + 乘客总数 ( B->A )

聚合路线上乘客数量的最佳/最优方法是什么，表格行数约为 1.5 亿行。

谢谢

Answer 1

对此有两种方法。一个是聚合，另一个是连接。

select least(origin, dest) as od1, greatest(origin, dest) as od2, sum(passengers) as numpassengers
from path t
group by least(origin, dest), greatest(origin, dest)
order by numpassengers
limit 3;

另一种是自连接。如果每个方向只有一行，则无需聚合即可执行此操作：

select p1.origin, p1.dest, p1.passengers + p2.passengers as numpassengers
from path p1 join
     path pt2
     on p1.origin = p2.dest and p1.dest = p2.origin
where p1.origin < p1.dest
order by numpassengers desc
limit 3;

否则，您需要自联接和聚合，因此第一种方法可能更快：

select p1.origin, p1.dest, sum(p1.passengers + p2.passengers) as numpassengers
from path p1 join
     path pt2
     on p1.origin = p2.dest and p1.dest = p2.origin
where p1.origin < p1.dest
group by p1.origin, p1.dest
order by numpassengers desc
limit 3;

我不知道哪个会更有效率。但是，我怀疑按总和计算的前 3 条路线将在每个方向的前 100 条中。如果是这样，请在 numpassengers 上建立一个索引，然后尝试：

select least(origin, dest) as od1, greatest(origin, dest) as od2, sum(passengers) as numpassengers
from path t cross join
     (select min(passengers) as cutoff
      from (select distinct passengers
            from path
            order by passengers desc
            limit 100
           ) t
     ) minp
where numpassengers >= minp.cutoff
group by least(origin, dest), greatest(origin, dest)
order by numpassengers
limit 3;

截止的计算应该只使用索引，并大大减少其余查询的负载。

编辑：

如果您没有least()and greatest()，只需使用case语句：

select (case when origin < dest then origin else dest end) as od1,
       (case when origin < dest then dest else origin end)  as od2,
       sum(passengers) as numpassengers
from path t
group by 1, 2
order by numpassengers
limit 3;

您可以重复case的语句group by。但是 Amazon Redshift 允许您在group by子句中引用列别名或位置。