直方图数据中的百分位数 [英] percentiles from histogram data

问题描述

下表捕获了许多考试中的学生成绩数据。

The following table captures student grade data over a number of exams.

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);

等级排序的意义是vgood> good> fair>贫穷

the grades are ordered in the sense that vgood > good > fair > poor

是否有可能（或有道理）使用此数据为每个学生找到第50个百分位？例如，如果学生姓名为 arun ，如果我们将数据视为一系列成绩类别，则第50个百分位将为 grade_fair 。

would it be possible ( or would it make sense) to find the 50th percentile grade for each student with this data ? For example - in case of student name arun if we think of the data as a series of grade categories - the 50th percentile would be grade_fair.

推荐答案

首先，您需要取消此项。我们可以这样做...

First you need to unpivot this. We can do that like this...

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}

然后我们需要对成绩进行索引...与 CROSS JOIN LATERAL 。我们有4行，每行有4个数组。我们要4 * 4行。

Then we need to index into grades... We do that with a CROSS JOIN LATERAL. We have 4 rows with an array of 4. We want 4*4 rows.

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，我们需要再次交叉加入横向以重现x（我们的成绩），超过等级

Now what remains, is we need to CROSS JOIN LATERAL again to reproduce x (our grade), over gradeqty

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，然后我们使用有序集合聚合函数 percent_disc 完成工作。

Now we GROUP BY name and then we use an Ordered-Set Aggregate Functions percent_disc to finish the job..

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)

想进一步介绍它并使它漂亮。 。

Want to go into it further and make it pretty...

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)

如果我们千斤顶，我们可以得到稍微更多些的输出

We can get a slightly more varied out put if we jack up a new user.

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)

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