PHP/MySQL - “最高评级”算法

发布于 2024-08-04 11:27:39 字数 390 浏览 6 评论 0原文

所以我只是建立了一个星级评级系统，并尝试提出一种算法来列出“最高评级”的项目。为简单起见，以下是专栏：

item_name
average_rating (a decimal from 1 to 5)
num_votes

我试图确定票数和评级之间的“最佳点”。例如...

评分为（4.6 / 20 票）的项目在列表中应高于评分为（5.0 / 2 票）的项目
评分为（2.5 / 100 票）的项目应低于评分为（4.5 / 2 票）的项目）

换句话说，num_votes 是“热门”的一个因素。

有人知道一种非常擅长确定这个“最佳位置”的算法吗？

提前致谢。

原文

So I just built a star-rating system and and trying to come up with an algorithm to list the "Top Rated" items. For simplicity, here are the columns:

item_name
average_rating (a decimal from 1 to 5)
num_votes

I'm trying to determine the "sweet spot" between number of votes and rating. For example...

An item rated (4.6 / 20 votes) should be higher on the list than an item that's (5.0 / 2 votes)
An item rated (2.5 / 100 votes) should be below an item that's (4.5 / 2 votes)

So in other words, num_votes plays a factor in what's "Top".

Anyone know of an algorithm that is pretty good at determining this "sweet spot"?

Thanks in advance.

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久隐师 2024-08-11 11:27:39

这是另一种从统计角度来看听起来不错的方法： http://www.thebroth.com/blog/118 /贝叶斯评级

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予囚 2024-08-11 11:27:39

问题是，4.6/20 的评级应比 5.0/2 高多少...

一个想法是不要考虑没有至少 x 票的项目。

另一个想法是用“中等”票来填补。确定最少 10 票。 5.0/2 必须填满 8 个 2.5 的虚拟选票

5.0/2 意味着 2 个 5.0 的选票，加上 8 个 2.5 的选票，您将得到 30/10 -> 3.0 ;)

现在，您必须决定一个项目至少应有多少票。对于已经获得最低票数的，进行直接比较。

4.5/20 > 4.4/100
5.0/2  < 3.1/20  (as 5.0/2 is, as we calculated, 3.0/10)

The question is, how much higher the 4.6/20 shall be rated than the 5.0/2...

An idea not to take items in consideration that do not have at least x votes.

Another idea is to fill up with "medium" votes. Decide that 10votes shall be the minimum. The 5.0/2 must be filled with 8 virtual votes of 2.5

5.0/2 means 2 votes with 5.0, add 8 with 2.5 you'll get 30/10 -> 3.0 ;)

Now, you have to decide how many votes an item shall at least have. For those that already have the minimum votes, a direct comparation shall be done.

4.5/20 > 4.4/100
5.0/2  < 3.1/20  (as 5.0/2 is, as we calculated, 3.0/10)

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薄荷梦 2024-08-11 11:27:39

您给每 10 票的权重为 1，那么 20 票的权重为该项目 2。
那么如果该物品的重量为 0，它将比平均值减少 0.5

4.6/20 = 20/10: 2 weight
5.0/2 = 2/10: 0 weight

(4.6 * 0.02) + 4.6 = 4.692
(5.0 * 0.00) + 5.0 = 5 - 0.5 = 4.5

2.5/100 = 100/10: 10 weight
4.5/2 = 2/10: 0 weight

(2.5 * 0.1) + 2.5 = 2.75
(4.5 * 0.0) + 4.5 = 4.5 - 0.5 = 4

How about you give each 10 votes a weight of 1 so 20 votes gives the item 2 weight.
Then if the item has 0 weight it will loose 0.5 from the average

4.6/20 = 20/10: 2 weight
5.0/2 = 2/10: 0 weight

(4.6 * 0.02) + 4.6 = 4.692
(5.0 * 0.00) + 5.0 = 5 - 0.5 = 4.5

2.5/100 = 100/10: 10 weight
4.5/2 = 2/10: 0 weight

(2.5 * 0.1) + 2.5 = 2.75
(4.5 * 0.0) + 4.5 = 4.5 - 0.5 = 4

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