Question

所以我有一个排行榜，我每天使用以下查询有效地获取每个用户的分数：

SELECT DATE(a.time) as time, a.userid, SUM(activity_weight) as weight

FROM activity_entries a INNER JOIN users1 u ON u.id = a.userid 

WHERE competitionId = '$competitionId' GROUP BY a.userid, DATE(time) 

ORDER BY time ASC

我想知道，在比赛的整个过程中，找到“最稳定”的表演者，即每天平均成绩最高的用户，将会是一种有效的方式。

非常感谢！

编辑：对此进行测试，但遇到问题：

SELECT a.userid,
DATE(a.time) as time,
AVG(activity_weight) AS daily_average,

(SELECT a.userid, DATE(a.time) as time,
AVG(AVG(daily_average)) as topAverage
FROM activity_entries a INNER JOIN users1 u ON u.id = a.userid
WHERE competitionId = '$competitionId'
)

FROM activity_entries a INNER JOIN users1 u ON u.id = a.userid
WHERE competitionId = '$competitionId'
GROUP BY userid, time

Answer 1

完全基于您提供的内容，并假设activity_weight是单个事件上特定用户的得分，然后从以下内容开始：

SELECT userid,
       time,
       AVG(activity_weight) AS daily_average
FROM activity_entries
GROUP BY userid, time

此查询返回每个用户的平均每日得分。现在，我们需要定义“最一致”。如果可接受的定义是“每日平均值的最高平均值”，则将上述查询的内部联接作为子查询，选择AVG(daily_average)。

如果“最一致”需要检查标准偏差，或者比平均值更复杂的数学，那么你应该在前端程序中而不是在数据库中这样做。

编辑：尝试此查询（或在此处使用它：SQLFiddle link）：

SELECT davg.userid,
   SUM(davg.daily_average*davg.num_of_activities)/SUM(davg.num_of_activities) AS weighted_total_average
FROM (
  SELECT userid,
         time,
         AVG(activity_weight) AS daily_average,
     COUNT(*) AS num_of_activities
  FROM activity_entries
  GROUP BY userid, time
) AS davg
GROUP BY davg.userid
ORDER BY AVG(davg.daily_average) DESC

数学的实际细节供您决定，但这里是一个加权平均值的例子，其中竞争者参加大量比赛的日子比懒惰的日子更有价值。

如何找到“最一致”的表演者MYSQL

1 个答案: