平均值计算SQL Server中的加权平均值

Question

有人可以解释一下为什么AVG（）函数给我代码中的加权平均值吗？

SELECT s.stud_id, s.country, SUM(e.paid) AS totalpaid
    INTO #totalpaid 
    FROM oc.students AS s
    JOIN oc.enrollment AS e ON s.stud_id = e.stud_id
GROUP BY s.country ,s.stud_id;

SELECT DISTINCT s.country, ROUND(AVG(t.totalpaid) OVER (PARTITION BY s.country),0) AS avg_country
    FROM #totalpaid t
    JOIN oc.students s ON t.stud_id = s.stud_id
    JOIN oc.enrollment e ON e.stud_id = s.stud_id; 

例如，在马耳他，学生12参加了1门课程并支付了45欧元，学生837参加了7门课程并支付了294欧元。我想对平均值进行简单的（45 + 294）/ 2计算，但是系统的计算结果类似于（1 * 45 + 7 * 294）/ 8。我究竟做错了什么？

Answer 1

因为您要两次加入桌子。

通过将INSERT和SELECT语句放在一起，您的查询等同于：

SELECT
  DISTINCT s.country, 
  ROUND(AVG(t.totalpaid) OVER (PARTITION BY s.country),0) AS avg_country
FROM (
  SELECT s.stud_id, s.country, SUM(e.paid) AS totalpaid
  FROM oc.students AS s
  JOIN oc.enrollment AS e ON s.stud_id = e.stud_id
  GROUP BY s.country ,s.stud_id    
) t
JOIN oc.students s ON t.stud_id = s.stud_id
JOIN oc.enrollment e ON e.stud_id = s.stud_id

您可以清楚地看到表students和enrollment被连接了两次。这将产生偏斜的平均函数。

Answer 2

在第二个查询中，当您将临时表重新连接到enrollment时，它将为每个类生成一行；这就是totalpaid列中多个值的来源。

第二个查询未使用临时表中尚未存在的任何列，因此您根本不需要那些联接。这应该会产生您想要的东西。

SELECT 
  t.country, 
  ROUND(AVG(t.totalpaid) OVER (PARTITION BY t.country),0) AS avg_country
FROM #totalpaid t
GROUP BY 
  t.country;

Answer 3

SELECT s.country, sum(e.paid) / count(DISTINCT s.stud_id) as average
FROM oc.students s 
JOIN oc.enrollment e ON e.stud_id = s.stud_id
GROUP BY s.country; 

Answer 4

与此同时，我找到了解决方案：

 SELECT
      country,
      ROUND(AVG(totalpaid) OVER (PARTITION BY country),0) AS avg_country
    FROM #totalpaid;

超级容易:)