加权平均和评级

发布于 2024-08-28 01:06:50 字数 244 浏览 7 评论 0原文

数学不是我的强项,我在这里不知所措。

基本上,我需要的只是一个简单的公式,该公式将给出 1 到 5 级的加权评级。如果票数很少,那么它们的影响力就会较小,并且评级会更接近平均水平(在这种情况下,我希望它为 3,而不是所有其他评级的平均值)。

我尝试了一些不同的贝叶斯实现,但这些都没有成功。我相信我正在寻找的图形表示可以显示为:

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    /
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干杯

Maths isn't my strong point and I'm at a loss here.

Basically, all I need is a simple formula that will give a weighted rating on a scale of 1 to 5. If there are very few votes, they carry less influence and the rating pressess more towards the average (in this case I want it to be 3, not the average of all other ratings).

I've tried a few different bayesian implementations but these haven't worked out. I believe the graphical representation I am looking for could be shown as:

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    /
___/

Cheers

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评论(2

不必在意 2024-09-04 01:06:50

我会这样做

1*num(1) + 2*num(2) + 3*num(3) + 4*num(4) + 5*num(5) + A*3
-----------------------------------------------------------
      num(1) + num(2) + num(3) + num(4) + num(5) + A

其中 num(i) 是 i 的投票数。
A是一个参数。我无法告诉你它的确切价值。这取决于你所说的“几票”是什么意思。一般来说,A 的高值意味着您需要很多选票才能获得与 3 不同的平均值,A 的低值意味着您需要很少的选票才能获得与 3 不同的值。

如果您将 5 视为“很少的选票”,那么您可以取 A= 5.

在这个解决方案中,我只是假设每个产品以 A 票投 3 票而不是无票开始。

希望有帮助。

I'd do this this way

1*num(1) + 2*num(2) + 3*num(3) + 4*num(4) + 5*num(5) + A*3
-----------------------------------------------------------
      num(1) + num(2) + num(3) + num(4) + num(5) + A

Where num(i) is number of votes for i.
A is a parameter. I can't tell You exact value of it. It depends on what do You mean by "few votes". In general high value of A means that You need many votes to get average different than 3, low value of A means You need few votes to get different value than 3.

If You consider 5 as "few votes" then You can take A=5.

In this solution I just assume that each product starts with A votes for 3 instead of no votes.

Hope it helps.

油饼 2024-09-04 01:06:50
(sum(ratings) / number(ratings)) * min(number(ratings), 10)/max(number(ratings), 10)

第一部分是非标准化平均评级。随着个人评分的数量增加到 10,第二部分将慢慢地将评分提高到 5。这个问题还不够清楚,我无法提供更好的答案,但我相信上面的公式可能是您可以开始并适应的东西随你走。不用说,您必须检查是否有任何评级(不要除以零)。

(sum(ratings) / number(ratings)) * min(number(ratings), 10)/max(number(ratings), 10)

The first part is the un-normalized average rating. The second part will slowly increase the rating towards 5 as the number of individual ratings grows to 10. The question isn't clear enough for me to provide a better answer, but I believe the above formula might be something you can start with and adapt as you go. It goes without saying that you have to check if there are any ratings at all (not to divide by zero).

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