sql - 取某列的平均值，然后过滤掉所有小于平均值的行，然后统计行数

Question

这是我已经研究了一段时间但没有结果的 sql 问题。

架构是

CREATE TABLE states (
    statecode text primary key,
    population_2010 integer,
    population_2000 integer,
    population_1950 integer,
    population_1900 integer,
    landarea real,
    name text,
    admitted_to_union text
);
CREATE TABLE counties(
    name text,
    statecode text references states(statecode),
    population_1950 integer,
    population_2010 integer
);
CREATE TABLE senators(
    statecode text references states(statecode),
    name text primary key,
    affiliation text,
    took_office integer,
    born integer);
CREATE TABLE committees(
    id text primary key,
    parent_committee text references committees(id),
    name text,
    chairman text references senators(name),
    ranking_member text references senators(name)
);

问题是: 返回有多少个州的数量超过县的平均数量

我知道如何计算平均值:

select avg(state_count) 
from
    (select count(*) as state_count
     from counties C
     group by C.statecode)

但我想知道我是否可以重用子查询。

像这样(下面的不行):

select count(*) as state_count
from counties C
group by C.statecode
having count(*) > avg(state_count)

这个查询还是没有统计行数，有没有优雅的办法？

编辑: 刚找到示例答案

SELECT COUNT(*)
FROM (SELECT statecode, COUNT(*) AS counts
      FROM counties
      GROUP BY statecode) s
WHERE s.counts > (SELECT AVG(t.counts) 
                  FROM  (SELECT COUNT(*) AS counts
                         FROM counties 
                         GROUP BY statecode) t);

但是我还是很好奇有没有更好的方法呢？

Answer 1

您提供的示例答案中不经济的部分是这个

SELECT COUNT(*) AS counts
FROM counties 
GROUP BY statecode

这部分出现了两次，一次是为了计算平均值，一次是为了找到大于平均值的计数。所以这是我使用 CTE 的尝试，它为这两个目的重用了上面的代码:

WITH c AS
(
   SELECT COUNT(*) AS counts
   FROM counties 
   GROUP BY statecode
)
SELECT COUNT(*) FROM c WHERE counts > (SELECT  Avg(CAST(counts AS decimal)) 
                                       FROM c)