Question

我是Postgres和数据库的新手，所以很抱歉。我进行了查询，以从一张表中获得每所学校的学生人数。现在我有了下表：

school_probs:

school_code(PK bigint)  schoolName(text)    probs(numeric)
1                       CAA             {0.05,0.08,0.18,0.3,0.11,0.28}
2                       CAS             {0.06,0.1,0.295,0.36,0.12,0.065}
3                       CBA             {0.05,0.11,0.35,0.32,0.12,0.05}
4                       CL              {0.07,0.09,0.24,0.4,0.06,0.09}

我该如何将每个学校的计数乘以“概率”列中的每个数字。例如：我们在学校“ CAA”的学生总数为198，则概率分布为 (0.05*198, 0.08*198, 0.18*198, 0.3*198, 0.11*198, 0.28*198)。有了结果，我就可以给学生评分。

我要获取计数的查询如下：

SELECT simulated_records.school, COUNT(simulated_records.school) as studentCount INTO CountSchool
FROM simulated_records, school_probs
WHERE simulated_records.school = school_probs.school
GROUP BY simulated_records.school;

Answer 1

要将一个数组中的元素与一个常数相乘，您需要取消嵌套，相乘和聚合。有一些警告需要等待。考虑：

PostgreSQL unnest() with element number

最好使用ARRAY constructor。

话虽如此，并就您未公开的表设计做出一些假设，但我还将简化计数：

Aggregate a single column in query with many columns

到达：

SELECT *, ARRAY(SELECT unnest(p.probs) * r.student_ct) AS scaled_probs
FROM   school_probs p
LEFT   JOIN  (
   SELECT school, COUNT(*)::int AS student_ct
   FROM   simulated_records
   GROUP  BY 1
   ) r USING (school);

或者，将NULL数组表示为NULL数组：

SELECT *
FROM   school_probs p
LEFT   JOIN  (
   SELECT school, COUNT(*)::int AS student_ct
   FROM   simulated_records
   GROUP  BY 1
   ) r USING (school)
LEFT   JOIN LATERAL (
   SELECT ARRAY(SELECT unnest(p.probs) * r.student_ct) AS scaled_probs
   ) p1 ON p.probs IS NOT NULL;

db <>提琴here

我建议这种简单形式仅在Postgres 10或更高版本的SELECT列表中带有一个set-returning函数，因为：

What is the expected behaviour for multiple set-returning functions in select clause?

将列中的数组与另一个查询的计数结果相乘

1 个答案: