MySQL: 如何计算特定半径（经度/纬度）内的中位数

Question

我想计算给定坐标在特定半径内的中位数和平均值。

重要的属性是： - 纬度 - 经度 - 价格

计算平均值的sql命令是：

SELECT avg(price) as average
FROM (SELECT r.*,
            ( 6371 * acos( cos( radians(37.3541079) ) * cos( radians( ANY_VALUE(`latitude` )) ) * cos( radians( ANY_VALUE(`longitude`) ) - radians(-121.9552356) ) + sin( radians(37.3541079) ) * sin( radians( ANY_VALUE(`latitude`) ) ) ) ) AS distance 
      FROM `Rental` r
     ) r
WHERE distance <= 20;

我的问题是如何计算给定坐标和半径内价格的中位数。 MySQL 没有 median() 函数。

编辑： 现在我已经尝试了 Simple way to calculate median with MySQL

中的代码

SELECT AVG(middle_values) AS 'median' FROM (
  SELECT t1.price AS 'middle_values' FROM
    (
      SELECT @row:=@row+1 as `row`, x.price
      FROM rental AS x, (SELECT @row:=0) AS r
      WHERE 1
      -- put some where clause here
      ORDER BY x.price
    ) AS t1,
    (
      SELECT COUNT(*) as 'count'
      FROM rental x
      WHERE 1
      -- put same where clause here
    ) AS t2
    -- the following condition will return 1 record for odd number sets, or 2 records for even number sets.
    WHERE t1.row >= t2.count/2 and t1.row <= ((t2.count/2) +1)) AS t3;

它适用于所有 200'000 条记录，但是当我添加 WHERE distance <= 20 时 mysql - 请求过载。

SELECT AVG(middle_values) AS 'median' FROM (
  SELECT t1.price AS 'middle_values' FROM
    (
      SELECT @row:=@row+1 as `row`, x.price
      FROM rental AS x, (SELECT @row:=0) AS r, (SELECT a.*,
            ( 6371 * acos( cos( radians(37.3541079) ) * cos( radians( ANY_VALUE(`latitude` )) ) * cos( radians( ANY_VALUE(`longitude`) ) - radians(-121.9552356) ) + sin( radians(37.3541079) ) * sin( radians( ANY_VALUE(`latitude`) ) ) ) ) AS distance 
      FROM `Rental` a
     ) a
      WHERE distance <= 20
      -- put some where clause here
      ORDER BY x.price
    ) AS t1,
    (
      SELECT COUNT(*) as 'count'
      FROM rental x, (SELECT a.*,
            ( 6371 * acos( cos( radians(37.3541079) ) * cos( radians( ANY_VALUE(`latitude` )) ) * cos( radians( ANY_VALUE(`longitude`) ) - radians(-121.9552356) ) + sin( radians(37.3541079) ) * sin( radians( ANY_VALUE(`latitude`) ) ) ) ) AS distance 
      FROM `Rental` a
     ) a
      WHERE distance <= 20
      -- put same where clause here
    ) AS t2
    -- the following condition will return 1 record for odd number sets, or 2 records for even number sets.
    WHERE t1.row >= t2.count/2 and t1.row <= ((t2.count/2) +1)) AS t3;

有什么地方失误了吗？

Answer 1

问题在于计算距离的 table 扫描，而不是中位数。

将数据放在 TEMPORARY TABLE 中，这样您就不必对其进行 3 次评估（平均值、计数和中位数）。
将 "bounding box" 添加到最里面的 WHERE 以将检查限制为 20x20 "square"。
INDEX(latitude)
使用 HAVING distance < 20 而不是需要另一个子查询。

MySQL: 如何计算特定半径（经度/纬度）内的中位数

MySQL: How can I calculate the median in a specific radius (longitude / latitude)

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