use*_*438 61 algorithm weighted-average rating-system
我需要计算亚马逊网站上的五星评级.我已经做了足够的搜索来找到最好的算法,但我无法得到正确的答案.例如,如果这些是评级
5 star - 252
4 star - 124
3 star - 40
2 star - 29
1 star - 33
Run Code Online (Sandbox Code Playgroud)
共478条评论
亚马逊计算出这是"4.1星,5星".谁能告诉我这个数字是如何得出的?我只是通过平均而无法做到这一点.
Bli*_*ndy 141
这是一个加权平均值,您可以将每个评级与其获得的投票数量进行权衡:
(5*252 + 4*124 + 3*40 + 2*29 + 1*33) / (252+124+40+29+33) = 4.11 and change
Run Code Online (Sandbox Code Playgroud)
小智 11
如果您从头开始计算总体评级,那么此公式将对您有所帮助.
式
((Overall Rating * Total Rating) + new Rating) / (Total Rating + 1)
Run Code Online (Sandbox Code Playgroud)
例
假设你到目前为止没有评级,那么公式就像,到目前为止整体评价为"0".总评分"0",给定评级为"4"
((0*0)+4)/1 = 4
Run Code Online (Sandbox Code Playgroud)
如果总体评分是"4.11"总评分是"478"并且由一个用户给出的新评级是"2"
然后公式就好
((4.11 * 478)+ 2)/479 // 479 is increment of new rating from 478
Run Code Online (Sandbox Code Playgroud)
Asa*_*sad 11
一个更好的方法来做到这一点,
rating = (sum_of_rating * 5)/sum_of_max_rating_of_user_count
Run Code Online (Sandbox Code Playgroud)
例子:
total users rated: 6
sum_of_max_rating_of_user_count: 6 x 5 = 30
sum_of_rating: 25
rating = (25 * 5) / 30
Run Code Online (Sandbox Code Playgroud)
完毕!
Jef*_*eff 10
在evanmiller.org上有一篇非常好的关于这个主题的文章.他介绍并讨论了一些方法的优缺点,并提出了一种数学上可靠的加权和计算投票方式.
http://evanmiller.org/ranking-items-with-star-ratings.html
Blindy的回复非常有用,这是基于此的PHP代码。有些可能会有用。根据OP的示例,结果将为4.11:
$ratings = array(
5 => 252,
4 => 124,
3 => 40,
2 => 29,
1 => 33
);
function calcAverageRating($ratings) {
$totalWeight = 0;
$totalReviews = 0;
foreach ($ratings as $weight => $numberofReviews) {
$WeightMultipliedByNumber = $weight * $numberofReviews;
$totalWeight += $WeightMultipliedByNumber;
$totalReviews += $numberofReviews;
}
//divide the total weight by total number of reviews
$averageRating = $totalWeight / $totalReviews;
return $averageRating;
}
Run Code Online (Sandbox Code Playgroud)
如何构建上面的$ ratings数组
示例伪代码,但是假设您有一个名为“ ratings”的表和名为“ rating”的列,则该伪代码应能解释在数据库中存储信息时如何构建$ ratings数组。在这种情况下,这是1次加入,您需要进行4次加入才能获得所有评分,但这应该可以帮助您开始:
SELECT count(c1.rating) as one_star, count(c2.rating) as two_star
FROM ratings c1
LEFT OUTER JOIN
ratings c2
ON
c1.id = c2.id
WHERE
c1.rating = 1
AND
c2.rating = 2
Run Code Online (Sandbox Code Playgroud)
评论中建议的另一种方法
SELECT SUM(rating = 1) AS one_s ,SUM(rating = 2) AS two_s ,SUM(rating = 3) as three_s FROM reviews where product_id = 9
Run Code Online (Sandbox Code Playgroud)