ORACLE的简单解决方案:

SELECT ROUND(MEDIAN(Lat_N), 4) FROM Station;

简单的解决方案，理解MySQL:

select case MOD(count(lat_n),2) 
when 1 then (select round(S.LAT_N,4) from station S where (select count(Lat_N) from station where Lat_N < S.LAT_N ) = (select count(Lat_N) from station where Lat_N > S.LAT_N))
else (select round(AVG(S.LAT_N),4) from station S where 1 = (select count(Lat_N) from station where Lat_N < S.LAT_N ) - (select count(Lat_N) from station where Lat_N > S.LAT_N))
end from station;

解释

STATION是表名。LAT_N是具有数值的列名

假设站表中有101条记录(奇数)。这意味着如果表以asc或desc排序，则中位数是第51条记录。

In above query for every S.LAT_N of S table I am creating two tables. One for number of LAT_N values less than S.LAT_N and another for number of LAT_N values greater than S.LAT_N. Later I am comparing these two tables and if they are matched then I am selecting that S.LAT_N value. When I check for 51st records there are 50 values less than 51st record and there 50 records greater than 51st record. As you see, there are 50 records in both tables. So this is our answer. For every other record there are different number of records in two tables created for comparison. So, only 51st record meets the condition.

现在假设站表中有100条记录(偶数)。这意味着如果表以asc或desc排序，则中位数是第50条和第51条记录的平均值。

Same as odd logic I am creating two tables. One for number of LAT_N values less than S.LAT_N and another for number of LAT_N values greater than S.LAT_N. Later I am comparing these two tables and if their difference is equal to 1 then I am selecting that S.LAT_N value and find the average. When I check for 50th records there are 49 values less than 50th record and there 51 records greater than 50th record. As you see, there is difference of 1 record in both tables. So this(50th record) is our 1st record for average. Similarly, When I check for 51st records there are 50 values less than 51st record and there 49 records greater than 51st record. As you see, there is difference of 1 record in both tables. So this(51st record) is our 2nd record for average. For every other record there are different number of records in two tables created for comparison. So, only 50th and 51st records meet the condition.

关心奇数的计数-给出中间两个值的平均值。

SELECT AVG(val) FROM
  ( SELECT x.id, x.val from data x, data y
      GROUP BY x.id, x.val
      HAVING SUM(SIGN(1-SIGN(IF(y.val-x.val=0 AND x.id != y.id, SIGN(x.id-y.id), y.val-x.val)))) IN (ROUND((COUNT(*))/2), ROUND((COUNT(*)+1)/2))
  ) sq

按维度分组的中位数:

SELECT your_dimension, avg(t1.val) as median_val FROM (
SELECT @rownum:=@rownum+1 AS `row_number`,
   IF(@dim <> d.your_dimension, @rownum := 0, NULL),
   @dim := d.your_dimension AS your_dimension,
   d.val
   FROM data d,  (SELECT @rownum:=0) r, (SELECT @dim := 'something_unreal') d
  WHERE 1
  -- put some where clause here
  ORDER BY d.your_dimension, d.val
) as t1
INNER JOIN  
(
  SELECT d.your_dimension,
    count(*) as total_rows
  FROM data d
  WHERE 1
  -- put same where clause here
  GROUP BY d.your_dimension
) as t2 USING(your_dimension)
WHERE 1
AND t1.row_number in ( floor((total_rows+1)/2), floor((total_rows+2)/2) )

GROUP BY your_dimension;

如果MySQL有ROW_NUMBER，那么MEDIAN是(受SQL Server查询的启发):

WITH Numbered AS 
(
SELECT *, COUNT(*) OVER () AS Cnt,
    ROW_NUMBER() OVER (ORDER BY val) AS RowNum
FROM yourtable
)
SELECT id, val
FROM Numbered
WHERE RowNum IN ((Cnt+1)/2, (Cnt+2)/2)
;

如果您有偶数个条目，则使用IN。

如果你想找到每个组的中位数，那么只需要在你的OVER子句中PARTITION BY组。

Rob

我发现接受的解决方案在我的MySQL安装上不起作用，返回一个空集，但这个查询在我测试的所有情况下都适用:

SELECT x.val from data x, data y
GROUP BY x.val
HAVING SUM(SIGN(1-SIGN(y.val-x.val)))/COUNT(*) > .5
LIMIT 1

我发现这个答案非常有用——https://www.eversql.com/how-to-calculate-median-value-in-mysql-using-a-simple-sql-query/

SET @rowindex := -1;

SELECT
   AVG(g.grade)
FROM
   (SELECT @rowindex:=@rowindex + 1 AS rowindex,
       grades.grade AS grade
    FROM grades
    ORDER BY grades.grade) AS g
WHERE
g.rowindex IN (FLOOR(@rowindex / 2) , CEIL(@rowindex / 2));