SQL查询中的四分位数

Question

我有一个非常简单的表格，如下所示：

CREATE TABLE IF NOT EXISTS LuxLog (
  Sensor TINYINT,
  Lux INT,
  PRIMARY KEY(Sensor)
)

它包含来自不同传感器的数千条日志

我希望所有传感器都有Q1和Q3

我可以对每个数据进行一次查询，但最好对所有传感器进行一次查询（通过一次查询返回Q1和Q3）

我认为这将是一个相当简单的操作，因为四分位数被广泛使用，并且是频率计算中的主要统计变量之一。事实上，我发现了大量过于复杂的解决方案，而我希望找到一些简洁明了的解决方案

谁能给我一个提示

编辑：这是我在网上找到的一段代码，但它对我不起作用：

SELECT  SUBSTRING_INDEX(
        SUBSTRING_INDEX(
            GROUP_CONCAT(                 -- 1) make a sorted list of values
                Lux
                ORDER BY Lux
                SEPARATOR ','
            )
        ,   ','                           -- 2) cut at the comma
        ,   75/100 * COUNT(*)        --    at the position beyond the 90% portion
        )
    ,   ','                               -- 3) cut at the comma
    ,   -1                                --    right after the desired list entry
    )                 AS `75th Percentile`
    FROM    LuxLog
    WHERE   Sensor=12
    AND     Lux<>0

我得到1作为返回值，而它应该是一个可以除以10的数字（10,20,30…..1000）

Answer 1

请参见SQLFIDLE:http://sqlfiddle.com/#!9/accca6/2/6 注意：对于SQLFIDLE，我已经生成了100行，1到100之间的每个整数都有一行，但它是一个随机顺序（在excel中完成）

代码如下：

SET @number_of_rows := (SELECT COUNT(*) FROM LuxLog);
SET @quartile := (ROUND(@number_of_rows*0.25));
SET @sql_q1 := (CONCAT('(SELECT "Q1" AS quartile_name , Lux, Sensor FROM LuxLog ORDER BY Lux DESC LIMIT 1 OFFSET ', @quartile,')'));
SET @sql_q3 := (CONCAT('( SELECT "Q3" AS quartile_name , Lux, Sensor FROM LuxLog ORDER BY Lux ASC LIMIT 1 OFFSET ', @quartile,');'));
SET @sql := (CONCAT(@sql_q1,' UNION ',@sql_q3));
PREPARE stmt1 FROM @sql;
EXECUTE stmt1;

编辑：

SET @current_sensor := 101;
SET @quartile := (ROUND((SELECT COUNT(*) FROM LuxLog WHERE Sensor = @current_sensor)*0.25));
SET @sql_q1 := (CONCAT('(SELECT "Q1" AS quartile_name , Lux, Sensor FROM LuxLog WHERE Sensor=', @current_sensor,' ORDER BY Lux DESC LIMIT 1 OFFSET ', @quartile,')'));
SET @sql_q3 := (CONCAT('( SELECT "Q3" AS quartile_name , Lux, Sensor FROM LuxLog WHERE Sensor=', @current_sensor,' ORDER BY Lux ASC LIMIT 1 OFFSET ', @quartile,');'));
SET @sql := (CONCAT(@sql_q1,' UNION ',@sql_q3));
PREPARE stmt1 FROM @sql;
EXECUTE stmt1;

基本推理如下： 对于四分位1，我们希望从顶部得到25%，因此我们希望知道有多少行，即：

SET @number_of_rows := (SELECT COUNT(*) FROM LuxLog);

现在我们知道了行数，我们想知道其中25%是多少，这是这一行：

SET @quartile := (ROUND(@number_of_rows*0.25));

然后，为了找到一个四分位数，我们要按Lux对LuxLog表进行排序，然后获得行号“@quartile”，为了做到这一点，我们将偏移量设置为@quartile，表示我们要从行号@quartile开始选择，我们说limit 1表示我们只想检索一行。那就是：

SET @sql_q1 := (CONCAT('(SELECT "Q1" AS quartile_name , Lux, Sensor FROM LuxLog ORDER BY Lux DESC LIMIT 1 OFFSET ', @quartile,')'));

对于另一个四分位数，我们的做法（几乎）相同，但我们不是从顶部开始（从较高的值到较低的值），而是从底部开始（这解释了ASC）

但是现在我们只在变量@sql_q1和@sql_q3中存储了字符串，所以将它们连接起来，我们合并查询结果，准备查询并执行它

Answer 2

像这样的东西应该可以做到：

select
    ll.*,
    if (a.position is not null, 1,
        if (b.position is not null, 2, 
        if (c.position is not null, 3, 
        if (d.position is not null, 4, 0)))
    ) as quartile
from
    luxlog ll
    left outer join luxlog a on ll.position = a.position and a.lux > (select count(*)*0.00 from luxlog) and a.lux <= (select count(*)*0.25 from luxlog)
    left outer join luxlog b on ll.position = b.position and b.lux > (select count(*)*0.25 from luxlog) and b.lux <= (select count(*)*0.50 from luxlog)
    left outer join luxlog c on ll.position = c.position and c.lux > (select count(*)*0.50 from luxlog) and c.lux <= (select count(*)*0.75 from luxlog)
    left outer join luxlog d on ll.position = d.position and d.lux > (select count(*)*0.75 from luxlog)
;    

以下是完整的示例：

use example;

drop table if exists luxlog;

CREATE TABLE LuxLog (
  Sensor TINYINT,
  Lux INT,
  position int,
  PRIMARY KEY(Position)
);

insert into luxlog values (0, 1, 10);
insert into luxlog values (0, 2, 20);
insert into luxlog values (0, 3, 30);
insert into luxlog values (0, 4, 40);
insert into luxlog values (0, 5, 50);
insert into luxlog values (0, 6, 60);
insert into luxlog values (0, 7, 70);
insert into luxlog values (0, 8, 80);

select count(*)*.25 from luxlog;
select count(*)*.50 from luxlog;

select
    ll.*,
    a.position,
    b.position,
    if(
        a.position is not null, 1,
        if (b.position is not null, 2, 0)
    ) as quartile
from
    luxlog ll
    left outer join luxlog a on ll.position = a.position and a.lux >= (select count(*)*0.00 from luxlog) and a.lux < (select count(*)*0.25 from luxlog)
    left outer join luxlog b on ll.position = b.position and b.lux >= (select count(*)*0.25 from luxlog) and b.lux < (select count(*)*0.50 from luxlog)
    left outer join luxlog c on ll.position = c.position and c.lux >= (select count(*)*0.50 from luxlog) and c.lux < (select count(*)*0.75 from luxlog)
    left outer join luxlog d on ll.position = d.position and d.lux >= (select count(*)*0.75 from luxlog) and d.lux < (select count(*)*1.00 from luxlog)
;    


select
    ll.*,
    if (a.position is not null, 1,
        if (b.position is not null, 2, 
        if (c.position is not null, 3, 
        if (d.position is not null, 4, 0)))
    ) as quartile
from
    luxlog ll
    left outer join luxlog a on ll.position = a.position and a.lux > (select count(*)*0.00 from luxlog) and a.lux <= (select count(*)*0.25 from luxlog)
    left outer join luxlog b on ll.position = b.position and b.lux > (select count(*)*0.25 from luxlog) and b.lux <= (select count(*)*0.50 from luxlog)
    left outer join luxlog c on ll.position = c.position and c.lux > (select count(*)*0.50 from luxlog) and c.lux <= (select count(*)*0.75 from luxlog)
    left outer join luxlog d on ll.position = d.position and d.lux > (select count(*)*0.75 from luxlog)
;    

Answer 3

使用NTILE很简单，但它是一个Postgres函数。你基本上就是这样做的：

SELECT value_you_are_NTILING,
    NTILE(4) OVER (ORDER BY value_you_are_NTILING DESC) AS tiles
FROM
(SELECT math_that_gives_you_the_value_you_are_NTILING_here AS value_you_are_NTILING FROM tablename);

下面是我在SQLFIDLE上为您制作的一个简单示例：http://sqlfiddle.com/#!15/7f05a/1

在MySQL中，您将使用RANK。。。以下是SQLFIDLE：http://www.sqlfiddle.com/#!2/d5587/1（来自下面链接的问题）

MySQL RANK（）的使用来自这里回答的Stackoverflow:Rank function in MySQL

寻找萨尔曼A.的答案