将 SELECT 结果与另一个 table 的值相乘

Question

我有 3 个 table 来自不同学校的学生数据。我应该在 table(done) 中计算每所学校的学生人数，然后使用给定的概率计算出每组概率有多少学生。我从另一个查询中获得了每所学校学生的结果。

示例：我使用查询来查找学校的学生总数 "CAA"。如果是198，那么概率分布就是（0.05*198、0.08*198、0.18*198、0.3*198、0.11*198、0.28*198）。我如何从 table 中获取这些值并与我的结果相乘？

我希望这是有道理的，请原谅我是 Postgres 的新手。我已附上我的查询以获取两个相关 table 的计数和照片。

simulated_records:

record_id(PK bigint)   Status(text)    grade(text
 1                     CL              -
 2                     CEC             -
 3                     CEC             -
 4                     CEC             -
 5                     CAA             -
 6                     CAS             -
 7                     CAA             -
 8                     CAA             -
 9                     CAA             -
 10                    CL              -

school_probs:

school_code(PK bigint)  school(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}

grade_values:

id(PK integer)   score(text)    grade(text
1                95-100         A
2                90-94          A-
3                80-89          B+
4                70-79          B
5                60-69          C
6                0-59           D

我的查询：

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

我应该能够做的是根据每所学校的学生数量并使用概率（使用函数分解多少孩子应该有什么成绩）在模拟记录中填充成绩table .

另外实际模拟记录table有2000多条记录

预期结果：simulated_records 中的空成绩列将根据学生人数和 school_probs 的概率填充成绩。

record_id(PK bigint)   Status(text)    grade(text
 1                     CL              A
 2                     CEC             B
 3                     CEC             B
 4                     CEC             A
 5                     CAA             C
 6                     CAS             B
 7                     CAA             D
 8                     CAA             A
 9                     CAA             A
 10                    CL              C

Answer 1

不确定您期望的结果是什么，但这里是：

select school, factor, CountSchool
from (
      SELECT simulated_records.school, COUNT(simulated_records.school) as CountSchool
      FROM simulated_records, school_probs
      WHERE simulated_records.school = school_probs.school
      GROUP BY simulated_records.school
     ) eq
cross join (
    values 
         (0.05), (0.08), (0.18), (0.3), (0.11), (0.28)
    ) t (factor)

这会产生这样的结果：

    school  factor  count_of
1   CAA 0.0500  198
2   CAA 0.0800  198
3   CAA 0.1800  198
4   CAA 0.3000  198
5   CAA 0.1100  198
6   CAA 0.2800  198

nb：您尚未说明将如何存储 "factor" 信息。

Answer 2

demo:db<>fiddle

SELECT 
    school,
    json_object_agg(gv.grade, s.students)                           -- 4
FROM (
    SELECT 
        s.school, 
        ceil(unnest(sp.probs) * s.students_per_school) students,    -- 2   
        generate_series(1,6) gs                                     -- 3
    FROM (
        SELECT 'CAA'::text as school, 198 as students_per_school    -- 1
    ) s
    JOIN school_probs sp ON s.school = sp.school                      
) s
JOIN grade_values gv ON gv.id = s.gs
GROUP BY school

结果：

school   json_object_agg
CAA      { "A" : 10, "A-" : 16, "B+" : 36, "B" : 60, "C" : 22, "D" : 56 }

希望这是您期望的结果：

从某处获取数据（模拟您的 table 或子查询或其他）
1. unnest 将您的 numeric probs 数组扩展为每个元素一行。这是主要技巧
2. 与您的学生相乘
3. ceil 向上舍入（否则你可以做 floor 向下舍入或 round 进行正常舍入）
generate_series 添加一个 integer 列以便能够加入 grade_values table
现在每个学校的每个年级都有一行（请参阅下面的编辑）。如果这是您的预期结果，就是这样。如果您想将每个学校的成绩值汇总到一行中，您可以这样做 json_object_agg

编辑：没有聚合的版本：

SELECT 
    school,
    gv.grade, 
    s.students
FROM (
    SELECT 
        s.school, 
        ceil(unnest(sp.probs) * s.students_per_school) students, 
        generate_series(1,6) gs
    FROM (
        SELECT 'CAA'::text as school, 198 as students_per_school
    ) s
    JOIN school_probs sp ON s.school = sp.school
) s
JOIN grade_values gv ON gv.id = s.gs

结果：

school   grade   students
CAA      A       10
CAA      A-      16
CAA      B+      36
CAA      B       60
CAA      C       22
CAA      D       56

将 SELECT 结果与另一个 table 的值相乘

multiply SELECT results with values from another table

sql

postgresql

postgresql-9.1