What is a better way to sort by a 5 star rating?

问题

I\'m trying to sort a bunch of products by customer ratings using a 5 star system. The site I\'m setting this up for does not have a lot of ratings and continue to add new products so it will usually have a few products with a low number of ratings.

I tried using average star rating but that algorithm fails when there is a small number of ratings.

Example a product that has 3x 5 star ratings would show up better than a product that has 100x 5 star ratings and 2x 2 star ratings.

Shouldn\'t the second product show up higher because it is statistically more trustworthy because of the larger number of ratings?

回答1:

Prior to 2015, the Internet Movie Database (IMDb) publicly listed the formula used to rank their Top 250 movies list. To quote:

The formula for calculating the Top Rated 250 Titles gives a true Bayesian estimate:
weighted rating (WR) = (v ÷ (v+m)) × R + (m ÷ (v+m)) × C
where:

R = average for the movie (mean)

v = number of votes for the movie

m = minimum votes required to be listed in the Top 250 (currently 25000)

C = the mean vote across the whole report (currently 7.0)

For the Top 250, only votes from regular voters are considered.

It's not so hard to understand. The formula is:

rating = (v / (v + m)) * R +
         (m / (v + m)) * C;

Which can be mathematically simplified to:

rating = (R * v + C * m) / (v + m);

The variables are:

R – The item's own rating. R is the average of the item's votes. (For example, if an item has no votes, its R is 0. If someone gives it 5 stars, R becomes 5. If someone else gives it 1 star, R becomes 3, the average of [1, 5]. And so on.)
C – The average item's rating. Find the R of every single item in the database, including the current one, and take the average of them; that is C. (Suppose there are 4 items in the database, and their ratings are [2, 3, 5, 5]. C is 3.75, the average of those numbers.)
v – The number of votes for an item. (To given another example, if 5 people have cast votes on an item, v is 5.)
m – The tuneable parameter. The amount of "smoothing" applied to the rating is based on the number of votes (v) in relation to m. Adjust m until the results satisfy you. And don't misinterpret IMDb's description of m as "minimum votes required to be listed" – this system is perfectly capable of ranking items with less votes than m.

All the formula does is: add m imaginary votes, each with a value of C, before calculating the average. In the beginning, when there isn't enough data (i.e. the number of votes is dramatically less than m), this causes the blanks to be filled in with average data. However, as votes accumulates, eventually the imaginary votes will be drowned out by real ones.

In this system, votes don't cause the rating to fluctuate wildly. Instead, they merely perturb it a bit in some direction.

When there are zero votes, only imaginary votes exist, and all of them are C. Thus, each item begins with a rating of C.

回答2:

Evan Miller shows a Bayesian approach to ranking 5-star ratings:

where

nk is the number of k-star ratings,
sk is the "worth" (in points) of k stars,
N is the total number of votes
K is the maximum number of stars (e.g. K=5, in a 5-star rating system)
z_alpha/2 is the 1 - alpha/2 quantile of a normal distribution. If you want 95% confidence (based on the Bayesian posterior distribution) that the actual sort criterion is at least as big as the computed sort criterion, choose z_alpha/2 = 1.65.

In Python, the sorting criterion can be calculated with

def starsort(ns):
    """
    http://www.evanmiller.org/ranking-items-with-star-ratings.html
    """
    N = sum(ns)
    K = len(ns)
    s = list(range(K,0,-1))
    s2 = [sk**2 for sk in s]
    z = 1.65
    def f(s, ns):
        N = sum(ns)
        K = len(ns)
        return sum(sk*(nk+1) for sk, nk in zip(s,ns)) / (N+K)
    fsns = f(s, ns)
    return fsns - z*math.sqrt((f(s2, ns)- fsns**2)/(N+K+1))

For example, if an item has 60 five-stars, 80 four-stars, 75 three-stars, 20 two-stars and 25 one-stars, then its overall star rating would be about 3.4:

x = (60, 80, 75, 20, 25)
starsort(x)
# 3.3686975120774694

and you can sort a list of 5-star ratings with

sorted([(60, 80, 75, 20, 25), (10,0,0,0,0), (5,0,0,0,0)], key=starsort, reverse=True)
# [(10, 0, 0, 0, 0), (60, 80, 75, 20, 25), (5, 0, 0, 0, 0)]

This shows the effect that more ratings can have upon the overall star value.

You'll find that this formula tends to give an overall rating which is a bit lower than the overall rating reported by sites such as Amazon, Ebay or Wal-mart particularly when there are few votes (say, less than 300). This reflects the higher uncertainy that comes with fewer votes. As the number of votes increases (into the thousands) all overall these rating formulas should tend to the (weighted) average rating.

Since the formula only depends on the frequency distribution of 5-star ratings for the item itself, it is easy to combine reviews from multiple sources (or, update the overall rating in light of new votes) by simply adding the frequency distributions together.

Unlike the IMDb formula, this formula does not depend on the average score across all items, nor an artificial minimum number of votes cutoff value.

Moreover, this formula makes use of the full frequency distribution -- not just the average number of stars and the number of votes. And it makes sense that it should since an item with ten 5-stars and ten 1-stars should be treated as having more uncertainty than (and therefore not rated as highly as) an item with twenty 3-star ratings:

In [78]: starsort((10,0,0,0,10))
Out[78]: 2.386028063783418

In [79]: starsort((0,0,20,0,0))
Out[79]: 2.795342687927806

The IMDb formula does not take this into account.

回答3:

See this page for a good analysis of star-based rating systems, and this one for a good analysis of upvote-/downvote- based systems.

For up and down voting you want to estimate the probability that, given the ratings you have, the "real" score (if you had infinite ratings) is greater than some quantity (like, say, the similar number for some other item you're sorting against).

See the second article for the answer, but the conclusion is you want to use the Wilson confidence. The article gives the equation and sample Ruby code (easily translated to another language).

回答4:

You could sort by median instead of arithmetic mean. In this case both examples have a median of 5, so both would have the same weight in a sorting algorithm.

You could use a mode to the same effect, but median is probably a better idea.

If you want to assign additional weight to the product with 100 5-star ratings, you'll probably want to go with some kind of weighted mode, assigning more weight to ratings with the same median, but with more overall votes.

回答5:

Well, depending on how complex you want to make it, you could have ratings additionally be weighted based on how many ratings the person has made, and what those ratings are. If the person has only made one rating, it could be a shill rating, and might count for less. Or if the person has rated many things in category a, but few in category b, and has an average rating of 1.3 out of 5 stars, it sounds like category a may be artificially weighed down by the low average score of this user, and should be adjusted.

But enough of making it complex. Let’s make it simple.

Assuming we’re working with just two values, ReviewCount and AverageRating, for a particular item, it would make sense to me to look ReviewCount as essentially being the “reliability” value. But we don’t just want to bring scores down for low ReviewCount items: a single one-star rating is probably as unreliable as a single 5 star rating. So what we want to do is probably average towards the middle: 3.

So, basically, I’m thinking of an equation something like X * AverageRating + Y * 3 = the-rating-we-want. In order to make this value come out right we need X+Y to equal 1. Also we need X to increase in value as ReviewCount increases...with a review count of 0, x should be 0 (giving us an equation of “3”), and with an infinite review count X should be 1 (which makes the equation = AverageRating).

So what are X and Y equations? For the X equation want the dependent variable to asymptotically approach 1 as the independent variable approaches infinity. A good set of equations is something like: Y = 1/(factor^RatingCount) and (utilizing the fact that X must be equal to 1-Y) X = 1 – (1/(factor^RatingCount)

Then we can adjust "factor" to fit the range that we're looking for.

I used this simple C# program to try a few factors:

        // We can adjust this factor to adjust our curve.
        double factor = 1.5;  

        // Here's some sample data
        double RatingAverage1 = 5;
        double RatingCount1 = 1;

        double RatingAverage2 = 4.5;
        double RatingCount2 = 5;

        double RatingAverage3 = 3.5;
        double RatingCount3 = 50000; // 50000 is not infinite, but it's probably plenty to closely simulate it.

        // Do the calculations
        double modfactor = Math.Pow(factor, RatingCount1);
        double modRating1 = (3 / modfactor)
            + (RatingAverage1 * (1 - 1 / modfactor));

        double modfactor2 = Math.Pow(factor, RatingCount2);
        double modRating2 = (3 / modfactor2)
            + (RatingAverage2 * (1 - 1 / modfactor2));

        double modfactor3 = Math.Pow(factor, RatingCount3);
        double modRating3 = (3 / modfactor3)
            + (RatingAverage3 * (1 - 1 / modfactor3));

        Console.WriteLine(String.Format("RatingAverage: {0}, RatingCount: {1}, Adjusted Rating: {2:0.00}", 
            RatingAverage1, RatingCount1, modRating1));
        Console.WriteLine(String.Format("RatingAverage: {0}, RatingCount: {1}, Adjusted Rating: {2:0.00}",
            RatingAverage2, RatingCount2, modRating2));
        Console.WriteLine(String.Format("RatingAverage: {0}, RatingCount: {1}, Adjusted Rating: {2:0.00}",
            RatingAverage3, RatingCount3, modRating3));

        // Hold up for the user to read the data.
        Console.ReadLine();

So you don’t bother copying it in, it gives this output:

RatingAverage: 5, RatingCount: 1, Adjusted Rating: 3.67
RatingAverage: 4.5, RatingCount: 5, Adjusted Rating: 4.30
RatingAverage: 3.5, RatingCount: 50000, Adjusted Rating: 3.50

Something like that? You could obviously adjust the "factor" value as needed to get the kind of weighting you want.

回答6:

If you just need a fast and cheap solution that will mostly work without using a lot of computation here's one option (assuming a 1-5 rating scale)

SELECT Products.id, Products.title, avg(Ratings.score), etc
FROM
Products INNER JOIN Ratings ON Products.id=Ratings.product_id
GROUP BY 
Products.id, Products.title
ORDER BY (SUM(Ratings.score)+25.0)/(COUNT(Ratings.id)+20.0) DESC, COUNT(Ratings.id) DESC

By adding in 25 and dividing by the total ratings + 20 you're basically adding 10 worst scores and 10 best scores to the total ratings and then sorting accordingly.

This does have known issues. For example, it unfairly rewards low-scoring products with few ratings (as this graph demonstrates, products with an average score of 1 and just one rating score a 1.2 while products with an average score of 1 and 1k+ ratings score closer to 1.05). You could also argue it unfairly punishes high-quality products with few ratings.

This chart shows what happens for all 5 ratings over 1-1000 ratings: http://www.wolframalpha.com/input/?i=Plot3D%5B%2825%2Bxy%29/%2820%2Bx%29%2C%7Bx%2C1%2C1000%7D%2C%7By%2C0%2C6%7D%5D

You can see the dip upwards at the very bottom ratings, but overall it's a fair ranking, I think. You can also look at it this way:

http://www.wolframalpha.com/input/?i=Plot3D%5B6-%28%2825%2Bxy%29/%2820%2Bx%29%29%2C%7Bx%2C1%2C1000%7D%2C%7By%2C0%2C6%7D%5D

If you drop a marble on most places in this graph, it will automatically roll towards products with both higher scores and higher ratings.

回答7:

Obviously, the low number of ratings puts this problem at a statistical handicap. Never the less...

A key element to improving the quality of an aggregate rating is to "rate the rater", i.e. to keep tabs of the ratings each particular "rater" has supplied (relative to others). This allows weighing their votes during the aggregation process.

Another solution, more of a cope out, is to supply the end-users with a count (or a range indication thereof) of votes for the underlying item.

回答8:

One option is something like Microsoft's TrueSkill system, where the score is given by mean - 3*stddev, where the constants can be tweaked.

回答9:

After look for a while, I choose the Bayesian system. If someone is using Ruby, here a gem for it:

https://github.com/wbotelhos/rating

回答10:

I'd highly recommend the book Programming Collective Intelligence by Toby Segaran (OReilly) ISBN 978-0-596-52932-1 which discusses how to extract meaningful data from crowd behaviour. The examples are in Python, but its easy enough to convert.

来源：https://stackoverflow.com/questions/1411199/what-is-a-better-way-to-sort-by-a-5-star-rating

标签

sorting

statistics

user-experience

rating

bayesian