mysql – 在一次查询调用中计算同一表中多列的中位数

2023年2月5日 181次阅读

StackOverflow救援！我需要在一次查询调用中一次找到五列的中位数.

下面的中值计算适用于单个列,但组合使用时,“rownum”的多次使用会抛出查询.如何更新此功能以适用于多列？谢谢.这是创建一个Web工具,非营利组织可以将其财务指标与用户定义的对等组进行比较.

SELECT t1_wages.totalwages_pctoftotexp AS median_totalwages_pctoftotexp
FROM (

SELECT @rownum := @rownum +1 AS  `row_number` , d_wages.totalwages_pctoftotexp
FROM data_990_c3 d_wages, (

SELECT @rownum :=0
)r_wages
WHERE totalwages_pctoftotexp >0
ORDER BY d_wages.totalwages_pctoftotexp
) AS t1_wages, (

SELECT COUNT( * ) AS total_rows
FROM data_990_c3 d_wages
WHERE totalwages_pctoftotexp >0
) AS t2_wages
WHERE 1 
AND t1_wages.row_number = FLOOR( total_rows /2 ) +1

--- [that was one median, below is another] ---

SELECT t1_solvent.solvent_days AS median_solvent_days
FROM (

SELECT @rownum := @rownum +1 AS  `row_number` , d_solvent.solvent_days
FROM data_990_c3 d_solvent, (

SELECT @rownum :=0
)r_solvent
WHERE solvent_days >0
ORDER BY d_solvent.solvent_days
) AS t1_solvent, (

SELECT COUNT( * ) AS total_rows
FROM data_990_c3 d_solvent
WHERE solvent_days >0
) AS t2_solvent
WHERE 1 
AND t1_solvent.row_number = FLOOR( total_rows /2 ) +1

[这两个 – 总共有五个我最终需要立即找到中位数]

最佳答案 这种事情在MySQL的脖子上是一个巨大的痛苦.如果你要做这个统计排名工作的吨位,你可能明智地使用免费的Oracle Express Edition或postgreSQL.它们都具有MEDIAN(值)聚合函数,这些函数可以是内置函数,也可以作为扩展函数使用.这是一个小小的方形展示.
http://sqlfiddle.com/#!4/53de8/6/0

但你没有问过这件事.

在MySQL中,您的基本问题是@rownum等变量的范围.您还有一个透视问题：也就是说,您需要将查询的行转换为列.

让我们先解决枢轴问题.你要做的是创建几个大胖查询的联合.例如：

SELECT 'median_wages' AS tag, wages AS value
  FROM (big fat query making median wages) A
 UNION
SELECT 'median_volunteer_hours' AS tag, hours AS value
  FROM (big fat query making median volunteer hours) B
 UNION
SELECT 'median_solvent_days' AS tag, days AS value
  FROM (big fat query making median solvency days) C

所以这是您在标签/值对表中的结果.您可以像这样转动该表,以获得每行中具有值的一行.

SELECT SUM( CASE tag WHEN 'median_wages' THEN value ELSE 0 END 
          ) AS median_wages, 
SELECT SUM( CASE tag WHEN 'median_volunteer_hours' THEN value ELSE 0 END
          ) AS median_volunteer_hours, 
SELECT SUM( CASE tag WHEN 'median_solvent_days' THEN value ELSE 0 END 
          ) AS median_solvent_days
FROM (
    /* the above gigantic UNION query */
 ) Q

这就是你将行(从本例中的UNION查询)转移到列的方式.这是关于该主题的教程. http://www.artfulsoftware.com/infotree/qrytip.php?id=523

现在我们需要处理中值计运算符查询.你问题中的代码看起来很不错.我没有您的数据,因此我很难对其进行评估.

但是你需要避免重用@rownum变量.在你的一个查询中调用@ rownum1,在下一个查询中调用@ rownum2,依此类推.这是一个简单的sql小提琴只做其中一个. http://sqlfiddle.com/#!2/2f770/1/0

现在让我们建立它,做两个不同的中位数.这是小提琴http://sqlfiddle.com/#!2/2f770/2/0,这里是UNION查询.请注意,union查询的后半部分使用@ rownum2而不是@rownum.

最后,这是带有旋转的完整查询. http://sqlfiddle.com/#!2/2f770/13/0

 SELECT SUM( CASE tag WHEN 'Boston' THEN value ELSE 0 END ) AS Boston,
           SUM( CASE tag WHEN 'Bronx' THEN value ELSE 0 END ) AS Bronx   
   FROM (
 SELECT 'Boston' AS tag, pop AS VALUE
  FROM (
        SELECT @rownum := @rownum +1 AS  `row_number` , pop
          FROM pops, 
        (SELECT @rownum :=0)r
          WHERE pop >0 AND city = 'Boston'
          ORDER BY pop
        ) AS ordered_rows, 
        ( 
         SELECT COUNT( * ) AS total_rows
           FROM pops
          WHERE pop >0 AND city = 'Boston'
        ) AS rowcount
  WHERE ordered_rows.row_number = FLOOR( total_rows /2 ) +1
  UNION ALL
 SELECT 'Bronx' AS tag, pop AS VALUE
  FROM (
        SELECT @rownum2 := @rownum2 +1 AS  `row_number` , pop
          FROM pops, 
        (SELECT @rownum2 :=0)r
          WHERE pop >0 AND city = 'Bronx'
          ORDER BY pop
        ) AS ordered_rows, 
        ( 
         SELECT COUNT( * ) AS total_rows
           FROM pops
          WHERE pop >0 AND city = 'Bronx'
        ) AS rowcount
  WHERE ordered_rows.row_number = FLOOR( total_rows /2 ) +1
) D

这只是两个中位数.你需要五个.我认为很容易证明,在单个查询中,这种中值计算在MySQL中是非常难以做到的.

点赞