Sql 直方图数据的百分位数

下表显示了一些考试的学生成绩数据

CREATE TABLE grades
AS
  SELECT name, exams, grade_poor, grade_fair, grade_good, grade_vgood
  FROM ( VALUES
    ( 'arun'  , 8  , 1 , 4 , 2 , 1 ),
    ( 'neha'  , 10 , 3 , 2 , 1 , 4 ),
    ( 'ram'   ,  5 , 1 , 1 , 3 , 0 ),
    ( 'radha' ,  8 , 0 , 3 , 1 , 4 )
  ) AS t(name,exams,grade_poor,grade_fair,grade_good,grade_vgood);

等级的排列顺序为：vgood>good>fair>poor

有没有可能（或者说有意义）用这些数据找到每个学生的第50个百分位分数？例如，如果学生姓名

arun

我们将数据视为一系列年级类别，那么第50个百分位将是

grade\u fair

选择姓名、考试、，
SELECT name, exams,
       CASE WHEN 0.5 * exams <= grade_poor
                THEN 'grade_poor'
            WHEN 0.5 * exams <= grade_poor + grade_fair
                THEN 'grade_fair'
            WHEN 0.5 * exams <= grade_poor + grade_fair + grade_good
                THEN 'grade_good'
            ELSE 'grade_vgood' END AS median_grade;

当0.5*考试时，首先需要取消此项。我们可以这样做

SELECT name,
  ARRAY[grade_poor, grade_fair, grade_good, grade_vgood]
FROM grades

 name  |   array   
-------+-----------
 arun  | {1,4,2,1}
 neha  | {3,2,1,4}
 ram   | {1,1,3,0}
 radha | {0,3,1,4}

然后我们需要索引到等级。。。我们使用
横向交叉连接
。我们有4行，数组为4。我们要4*4排

SELECT name, grades, gs1.x, grades[gs1.x] AS gradeqty
FROM (
  SELECT name,
    ARRAY[grade_poor, grade_fair, grade_good, grade_vgood]
  FROM grades
) AS t(name, grades)
  CROSS JOIN LATERAL generate_series(1,4) AS gs1(x)
ORDER BY name, x;


 name  |  grades   | x |  gradeqty
-------+-----------+---+----------
 arun  | {1,4,2,1} | 1 |        1
 arun  | {1,4,2,1} | 2 |        4
 arun  | {1,4,2,1} | 3 |        2
 arun  | {1,4,2,1} | 4 |        1
 neha  | {3,2,1,4} | 1 |        3
 neha  | {3,2,1,4} | 2 |        2
 neha  | {3,2,1,4} | 3 |        1
 neha  | {3,2,1,4} | 4 |        4
 radha | {0,3,1,4} | 1 |        0
 radha | {0,3,1,4} | 2 |        3
 radha | {0,3,1,4} | 3 |        1
 radha | {0,3,1,4} | 4 |        4
 ram   | {1,1,3,0} | 1 |        1
 ram   | {1,1,3,0} | 2 |        1
 ram   | {1,1,3,0} | 3 |        3
 ram   | {1,1,3,0} | 4 |        0
(16 rows)

现在剩下的是，我们需要再次交叉连接横向
，以复制x（我们的等级），超过等级数量

SELECT name,
  gs1.x
FROM (
  SELECT name,
    ARRAY[grade_poor, grade_fair, grade_good, grade_vgood]
  FROM grades
) AS t(name, grades)
CROSS JOIN LATERAL generate_series(1,4) AS gs1(x)
CROSS JOIN LATERAL generate_series(1,grades[gs1.x]) AS gs2(x)
ORDER BY name, gs1.x;

 name  | x 
-------+---
 arun  | 1
 arun  | 2
 arun  | 2
 arun  | 2
 arun  | 2
 arun  | 3
 arun  | 3
 arun  | 4
 neha  | 1
 neha  | 1
 neha  | 1
 neha  | 2
 neha  | 2
 neha  | 3
 neha  | 4
 neha  | 4
 neha  | 4
 neha  | 4
 radha | 2
 radha | 2
 radha | 2
 radha | 3
 radha | 4
 radha | 4
 radha | 4
 radha | 4
 ram   | 1
 ram   | 2
 ram   | 3
 ram   | 3
 ram   | 3
(31 rows)

现在，我们按名称分组，然后使用

SELECT name, percentile_disc(0.5) WITHIN GROUP (ORDER BY gs1.x)
FROM (
  SELECT name,
    ARRAY[grade_poor, grade_fair, grade_good, grade_vgood]
  FROM grades
) AS t(name, grades)
CROSS JOIN LATERAL generate_series(1,4) AS gs1(x)
CROSS JOIN LATERAL generate_series(1,grades[gs1.x]) AS gs2(x)
GROUP BY name ORDER BY name;

 name  | percentile_disc 
-------+-----------------
 arun  |               2
 neha  |               2
 radha |               3
 ram   |               3
(4 rows)

想更进一步，让它更漂亮

SELECT name, (ARRAY['Poor', 'Fair', 'Good', 'Very Good'])[percentile_disc(0.5) WITHIN GROUP (ORDER BY gs1.x)]
FROM (
  SELECT name,
    ARRAY[grade_poor, grade_fair, grade_good, grade_vgood]
  FROM grades
) AS t(name, grades)
CROSS JOIN LATERAL generate_series(1,4) AS gs1(x)
CROSS JOIN LATERAL generate_series(1,grades[gs1.x]) AS gs2(x)
GROUP BY name
ORDER BY name;

 name  | array 
-------+-------
 arun  | Fair
 neha  | Fair
 radha | Good
 ram   | Good
(4 rows)

如果我们增加一个新用户，我们的输出会稍微多样化一些

INSERT INTO grades (name,grade_poor,grade_fair,grade_good,grade_vgood)
VALUES ('Bob', 0,0,0,100);

 name  |   array   
-------+-----------
 arun  | Fair
 Bob   | Very Good
 neha  | Fair
 radha | Good
 ram   | Good
(5 rows)

你想如何处理平局，比如neha和radha的情况？不太确定，掷硬币吧？对不起，你运气不好。PostgreSQL不能掷硬币。我想我们可以得出结论，没有解决方案！！当出现平局时-掷硬币=>概率可以是0.5。感谢详细解释-在使用有序集合聚合函数之前，我已经完成了所有步骤。这一个给出了错误-我甚至看不出来。但这可能是一个问题。SQL fiddle在所有方面都很差劲。永远不要使用它，ever.SQLfiddle使用PG9.3，有序集聚合函数不存在，这是一个很好的答案，很好地解释了这个非常复杂的查询如何处理如此简单的问题。当
查询时，它也会产生与低级的
情况完全相同的结果。