用MySQL计算中位数最简单(希望不会太慢)的方法是什么?我已经使用AVG(x)来寻找平均值，但我很难找到一个简单的方法来计算中位数。现在，我将所有的行返回到PHP，进行排序，然后选择中间的行，但是肯定有一些简单的方法可以在一个MySQL查询中完成它。
示例数据:
id | val
--------
1 4
2 7
3 2
4 2
5 9
6 8
7 3
对val排序得到2 2 3 4 7 8 9，因此中位数应该是4，而SELECT AVG(val) == 5。
当前回答
来自: http://mdb-blog.blogspot.com/2015/06/mysql-find-median-nth-element-without.html
我建议另一种方法，不用连接， 但是处理字符串
我没有用大数据表来检验， 但是中小型的桌子就可以了。
这里的好处是，它也可以通过GROUPING工作，所以它可以返回几个项目的中值。
下面是测试表的测试代码:
DROP TABLE test.test_median
CREATE TABLE test.test_median AS
SELECT 'book' AS grp, 4 AS val UNION ALL
SELECT 'book', 7 UNION ALL
SELECT 'book', 2 UNION ALL
SELECT 'book', 2 UNION ALL
SELECT 'book', 9 UNION ALL
SELECT 'book', 8 UNION ALL
SELECT 'book', 3 UNION ALL
SELECT 'note', 11 UNION ALL
SELECT 'bike', 22 UNION ALL
SELECT 'bike', 26
求每组中位数的代码:
SELECT grp,
SUBSTRING_INDEX( SUBSTRING_INDEX( GROUP_CONCAT(val ORDER BY val), ',', COUNT(*)/2 ), ',', -1) as the_median,
GROUP_CONCAT(val ORDER BY val) as all_vals_for_debug
FROM test.test_median
GROUP BY grp
输出:
grp | the_median| all_vals_for_debug
bike| 22 | 22,26
book| 4 | 2,2,3,4,7,8,9
note| 11 | 11
其他回答
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
根据魔术贴的答案，对于那些必须根据另一个参数分组的东西做中位数的人来说
SELECT grp_field, t1。val FROM ( SELECT grp_field， @rownum:=IF(@s = grp_field， @rownum + 1,0) AS row_number， @s:=IF(@s = grp_field， @s, grp_field) AS sec, d.val FROM data d， (SELECT @rownum:=0， @s:=0 ORDER BY grp_field, d.val )作为t1 JOIN ( SELECT grp_field, count(*)为total_rows 数据d GROUP BY grp_field )为t2 在t1。Grp_field = t2.grp_field 在t1.row_number =地板(total_rows / 2) + 1;
我刚刚在网上的评论中找到了另一个答案:
对于几乎所有SQL中的中位数: SELECT x.val from data x, data y GROUP BY x.val 总和(符号(1-SIGN (y.val-x.val))) = (COUNT (*) + 1) / 2
确保列有良好的索引，并且索引用于筛选和排序。与解释计划核对。
select count(*) from table --find the number of rows
计算“中值”行号。可能使用:median_row = floor(count / 2)。
然后把它从列表中挑出来:
select val from table order by val asc limit median_row,1
这将返回您想要的值的一行。
对于一个表站和列lat_n，下面是MySQL代码来获得中位数:
set @rows := (select count(1) from station);
set @v1 := 0;
set @sql1 := concat('select lat_n into @v1 from station order by lat_n asc limit 1 offset ', ceil(@rows/2) - 1);
prepare statement1 from @sql1;
execute statement1;
set @v2 := 0;
set @sql2 := concat('select lat_n into @v2 from station order by lat_n asc limit 1 offset ', ceil((@rows + 1)/2) - 1);
prepare statement2 from @sql2;
execute statement2;
select (@v1 + @v2)/2;
