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

Question

I want to calculate the median and average for given coordinates in a specific radius.

The important attributes are: - latitude - longitude - price

The sql command to calculate the average is:

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;

My question is how can I calculate the median for the price in the given coordinates and radius. MySQL has no median() function.

EDIT: Now I have tried the code from Simple way to calculate median with MySQL

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;

It works for all the 200'000 records, but when I will add the WHERE distance <= 20 is the mysql - request overloaded.

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;

Is there somewhere a misstep?

Answer 1

The problem is with the table scan to compute the distances, not with the median.

• Put the data in a TEMPORARY TABLE so you don't have to evaluate it 3 times (avg, count, and median).
• Add a "bounding box" to the innermost WHERE to limit the checks to a 20x20 "square".
• INDEX(latitude)
• Use HAVING distance < 20 instead of needing yet-another subquery.