Question

我有一个MySQL表格，我从一个游戏中收集了一个数字列表，用户必须点击一个按钮才能接近100。

我以为我会使用AVG()，但这显然不会起作用，因为当你有0和200时，它看起来好像你得了一个完美的100。

我需要的是每一轮（有10个）计算与数字相差100的距离并计算它们。

Example:
User A: 98 + 102 + 102 = 6 (something) Loser!
User B: 95 + 100 + 100 = 5 (something) Winner!

我有什么，但显然有缺陷：

SELECT user_id, AVG(score) AS Average, (100-AVG(score)) AS Difference, 
DATE(playtime) AS Playdate 
FROM playtable
GROUP BY user_id, DATE(playtime) 
ORDER BY Playdate DESC, user_id

Answer 1

这应该这样做

SELECT user_id, SUM( ABS(100-score) ) as cumulative_error,
  DATE(playtime) AS Playdate 
FROM playtable
GROUP BY user_id, DATE(playtime) 
ORDER BY Playdate DESC, user_id

cumulative_error将包含累积的差值

Answer 2

您可以修复查询以返回绝对值，以查看分数与给定值的距离（在您的情况下为100）

SELECT user_id, AVG(score) AS Average, SUM(ABS(100 - score)) AS Difference, 
DATE(playtime) AS Playdate 
FROM playtable
GROUP BY user_id, DATE(playtime) 
ORDER BY Playdate DESC, user_id

如何计算最接近100的平均数？

2 个答案: