Question

Numbers表保留数字的值及其频率。

+----------+-------------+
|  Number  |  Frequency  |
+----------+-------------|
|  0       |  7          |
|  1       |  1          |
|  2       |  3          |
|  3       |  1          |
+----------+-------------+

在此表中，数字为0、0、0、0、0、0、0、1、2、2、2、3，因此中位数为（0 + 0）/ 2 = 0。找到给定数字频率的中位数（显示输出）？

+--------+
| median |
+--------|
| 0.0000 |
+--------+

我找到了以下解决方案here。但是，我无法理解。有人可以解释一下解决方案和/或发布带有解释的其他解决方案吗？

SELECT AVG(n.Number) AS median 
FROM Numbers n LEFT JOIN
(
SELECT Number, @prev := @count AS prevNumber, (@count := @count + Frequency) AS countNumber
FROM Numbers, 
(SELECT @count := 0, @prev := 0, @total := (SELECT SUM(Frequency) FROM Numbers)) temp ORDER BY Number
) n2
ON n.Number = n2.Number
WHERE 
(prevNumber < floor((@total+1)/2) AND countNumber >= floor((@total+1)/2))
OR
(prevNumber < floor((@total+2)/2) AND countNumber >= floor((@total+2)/2))

以下是可重复性的SQL脚本：

CREATE TABLE `Numbers` (
  `Number` INT NULL,
  `Frequency` INT NULL);

INSERT INTO `Numbers` (`Number`, `Frequency`) VALUES ('0', '7');
INSERT INTO `Numbers` (`Number`, `Frequency`) VALUES ('1', '1');
INSERT INTO `Numbers` (`Number`, `Frequency`) VALUES ('2', '3');
INSERT INTO `Numbers` (`Number`, `Frequency`) VALUES ('3', '1');

谢谢！

Answer 1

您可以使用累加和，然后取中点。我认为逻辑看起来像这样：

select avg(number)
from (select t.*, (@rf := @rf + frequency) as running_frequency
      from (select t.* from t order by number) t cross join
           (select @rf := 0) params
     ) t
where running_frequency - frequency >= ceil(@rf / 2) and
      running_frequency <= ceil((@rf + 1) / 2);

如何找到给定数字的中位数频率？

1 个答案: