无法让Percentile_Cont（）在Postgresql中工作

Question

我试图使用公共表表达式使用PostgreSQL中的percentile_cont（）函数计算百分位数。目标是找到前1％的帐户关于他们的余额（这里称为金额）。我的逻辑是找到99％的百分位数，这将返回那些帐户余额超过其同行99％的人（从而找到1个百分点）

这是我的查询

--ranking subquery works fine
with ranking as(
       select a.lname,sum(c.amount) as networth  from customer a
       inner join 
       account b on a.customerid=b.customerid
       inner join 
       transaction c on b.accountid=c.accountid 
       group by a.lname order by sum(c.amount)
 )
select lname, networth, percentile_cont(0.99) within group 
order by networth over (partition by lname) from ranking ;

我一直收到以下错误。

ERROR:  syntax error at or near "order"
LINE 2: ...ame, networth, percentile_cont(0.99) within group order by n..

我想也许我忘记了一个闭合支架等，但我似乎无法弄清楚在哪里。我知道它可能是order关键字的东西，但我不知道该怎么做。你能帮我修一下这个错误吗？

Answer 1

这也使我绊倒了。

事实证明，postgres 9.3不支持percentile_cont，只有9.4+。

https://www.postgresql.org/docs/9.4/static/release-9-4.html

所以你必须使用这样的东西：

with ordered_purchases as (
  select
      price,
      row_number() over (order by price) as row_id,
      (select count(1) from purchases) as ct
  from purchases
)

select avg(price) as median
from ordered_purchases
where row_id between ct/2.0 and ct/2.0 + 1

关于https://www.periscopedata.com/blog/medians-in-sql的查询（部分：“Postgres中位数”）

Answer 2

您缺少within group (order by x)部分中的括号。

试试这个：

with ranking
as (
    select a.lname,
        sum(c.amount) as networth
    from customer a
    inner join account b on a.customerid = b.customerid
    inner join transaction c on b.accountid = c.accountid
    group by a.lname
    order by networth
    )
select lname,
    networth,
    percentile_cont(0.99) within group (
        order by networth
        ) over (partition by lname)
from ranking;

Answer 3

我想指出你不需要子查询：

select c.lname, sum(t.amount) as networth,
       percentile_cont(0.99) within group (order by sum(t.amount)) over (partition by lname)
from customer c inner join
     account a
     on c.customerid = a.customerid inner join
     transaction t
     on a.accountid = t.accountid
group by c.lname
order by networth;

此外，当使用表别名（应该总是）时，表缩写比任意字母更容易遵循。