根据不同选民人数调整投票

发布于 2024-12-10 20:45:44 字数 555 浏览 0 评论 0原文

我有一个 1 到 5 的投票系统,我正在尝试找出找到最受欢迎的投票项目的最佳方法,同时考虑到可能的总投票数。为了获得总票数,我将“1”票计为-3,“2”票计为-2,“3”票计为+1,“4”票计为+2,“5”票计为+3,因此,“1”票将抵消“5”票,反之亦然。

对于这个例子,假设我们有 3 部电影在 3 个不同规模的剧院放映。

电影1:800个座位 / 电影2:400个座位 / 电影3:180个座位

在某种程度上,我们根据座位限制了总票数,所以我想为电影提供一种方式在小剧院里看电影,以免被大剧院里的电影淹没。较大的剧院很可能会有更多的票数,从而导致总分更高。


编辑10/18:

好吧,希望我能更好地解释这一点。我正在为一个电影节工作,我们正在对电影节中每部电影的首映进行投票。因此,根据每个影院的规模,每部电影的票数从 0 到最大。我正在寻找 3 个类别中最受欢迎的电影:叙事片、纪录片、短片。我所说的受欢迎程度是指最高平均票数和票数的组合。

看来我正在寻找加权平均值,对较大剧院的投票给予较小的权重,而对较小剧院的投票给予较大的权重,以平衡情况。

I have a 1 to 5 voting system and i'm trying to figure out the best way to find the most popular item voted on, taking into consideration the total possible number of votes cast. To get a vote total, i'm counting "1" votes as -3, "2" votes as -2, "3" votes as +1, "4" votes as +2, "5" votes as +3, so a "1" vote would cancel out a "5" vote and vice versa.

For this example, say we have 3 films playing in 3 different size theaters.

Film 1: 800 seats / Film 2: 400 seats / Film 3: 180 seats

In a way, we're limiting the total amount of votes based on seats, so I would like a way for the film in the smaller theater to not get automatically overwhelmed by the film in the larger theater. It's likely that there will be more votes cast in the larger theater, resulting in a higher total score.


Edit 10/18:

Alright, hopefully I can explain this better. I'm working for a film festival, and we're balloting the first screening of each film in the fest. Therefore, each film will have from 0 to a maximum number of votes based on the size of each theater. I'm looking to find the most popular film in 3 categories: narrative, documentary, short film. By popular I mean a combination of highest average vote and number of votes.

It seems like a weighted average is what i'm looking for, giving less weight to votes from a bigger theater and more weight to votes from a smaller theater to even things out.

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诗化ㄋ丶相逢 2024-12-17 20:45:45

您正在使用加权平均值

而不是仅仅相加并除以元素总数(算术平均值):

 a + b + c
 ---------
     3

您要为每个元素添加权重,因为它们并非全部均匀分布:

 w1*a + w2*b + w3*c
 ------------------
         3

在您的情况下,权重可能是这样的:

# of people in current theater
--------------------------------
# of people in all the theaters

让我们尝试一个测试用例:

Theater 1: 100 people       (rating: 1)
Theater 2: 1,000,000 people (rating: 5)

Average = (100 / (100 + 1000000)) * 1 + (1000000/(100 + 1000000)) * 5
          -----------------------------------------------------------
                                      2
        = 2.49980002

You're working with weighted averages.

Instead of just adding up and dividing by the total number of elements (arithmetic mean):

 a + b + c
 ---------
     3

You are adding weights to each element, as they are not all evenly distributed:

 w1*a + w2*b + w3*c
 ------------------
         3

In your case, the weights could be this:

# of people in current theater
--------------------------------
# of people in all the theaters

Let's try a test case:

Theater 1: 100 people       (rating: 1)
Theater 2: 1,000,000 people (rating: 5)

Average = (100 / (100 + 1000000)) * 1 + (1000000/(100 + 1000000)) * 5
          -----------------------------------------------------------
                                      2
        = 2.49980002
秋日私语 2024-12-17 20:45:45

好吧,根据您的目标,听起来您对某种加权平均值感兴趣。

继续你的电影例子,在我看来,你似乎在试图评价这些电影有多“好”。为此,您不想在最终决定中过多考虑任何特定电影的观看次数。然而,你必须考虑到这一点,因为一部只被观看了 5 次且平均评分为 +2.7 的电影的可信度比一部观看次数为 10,000 次并获得相同评分的电影要低得多。

您可能会考虑干脆不将某部电影包含在结果中,除非该电影具有最低票数。

Well, depending on your goals it sounds like you are interested in some sort of weighted average.

Continuing your film example, it sounds to me like you are trying to rate how "good" the films are. To do this, you don't want to factor the number of views of any particular film too highly into the final determination. However, you have to take it into account somewhat since a film that only got viewed 5 times and had an average rating of +2.7 has much less credibility than a film with 10,000 views getting the same rating.

You might consider simply not including a film in the results unless it has a minimum number of votes.

暖风昔人 2024-12-17 20:45:45

鉴于 {1,2,3,4,5} 上的投票均匀(均匀)分布,您的电影的预期评分为 0.2。这是因为投票 {1 和 5} 相互抵消,{2 和 4} 也是如此。但投票 3 的预期值为 1/5 = 0.2。因此,如果人们以相同的概率给出 {1,2,3,4,5} 的评分,那么您会期望一部电影(无论有多少人观看)的平均评分接近 0.2。

因此,我认为对您来说最好的选择是将收到的所有分数相加,然后除以看过每部电影的人数。这应该可以很好地猜测人们对这部电影的情绪,因为发行的平均值不应该仅仅因为更多的人观看这部电影而变得更大。

如果我是你,我还会建议在你的最终结果中添加一个小的惩罚项,以考虑到有些人甚至不想去看电影的事实。如果很多人一开始就不想看这部电影,但看过这部电影的大约 5 个人给了它 5 分,这并不能使它成为一部好电影,不是吗?

因此,我建议的最终解决方案是:将您所描述的所有点相加,然后除以去电影院的总人数。虽然不完美(无论完美意味着什么),但它应该让您了解人们喜欢什么和不喜欢什么。这本质上意味着选择不看电影的人将总分加零,但仍然影响平均值,因为最终结果除以更大的数字。

Given a uniform (even) distribution of votes across {1,2,3,4,5}, the expected rating of your film is 0.2. This is because the the votes {1 and 5} cancel eachother out, as do {2 and 4}. But the vote 3 has an expected value of 1/5 = 0.2. So if people give a rating of {1,2,3,4,5} with equal probability, then you would expect a film (no matter how many people see it) to have an average rating close to 0.2.

So I think the best option for you would be to add up all the scores received and simply divide by the number of people who have seen each film. This should be a good guess at people's sentiment toward the film as the average of the distribution should not get larger simply because more people see the film.

If I were you, I would also suggest adding a small penalty term to your final result, to take into account the fact that some people didn't even want to go see the movie. If lots of people didn't want to see the movie in the first place, but the 5 or so people that saw it gave it a 5* rating, that doesn't make it a good movie, does it?

So a final solution I would recommend: Add up all the points as you have described, and divide by the total number of people who have gone to the cinema. While not perfect (whatever perfect means), it should give you some indication of what people like and don't like. This essentially means people who chose not to see a movie are adding zero to the points total, but still affect the average because the end result is divided by a larger number.

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