python - 我怎样才能改进这个方程，使负面投票多于正面投票的项目返回更有用的威尔逊分数

Question

使用此处描述的 Wilson Score 方程 http://www.evanmiller.org/how-not-to-sort-by-average-rating.html ，我正在对我评分的项目进行排序。但是，如果某个项目有 1 票反对票(0 票赞成票)，则它会返回与有 1000 票反对票(0 票赞成票)的项目相同的分数(即 0 分)。

我想要么允许威尔逊分数为负，以克服这个缺点，要么也许有人建议其他解决方案。

无论如何，我不知道如何改变这个方程/函数:

def ci_lower_bound(pos, n, confidence):
    if n==0: return 0

    z = 1.96
    phat = 1.0*pos/n

    score = (phat + z*z/(2*n) - z*math.sqrt((phat*(1-phat)+z*z/(4*n))/n))/(1+z*z/n)
    return score

其中pos是正面评分的数量，n是评分的总数，confidence是指统计置信水平。

Answer 1

从逻辑上讲，您的评分系统必须处理以下情况:

+----------+----------+------------+---------------+
| Positive | Negative | Any Votes? | Wilson Score? |
+----------+----------+------------+---------------+
| N        | N        | N          | Y, = 0        |
| Y        | Y        | Y          | Y             |
| Y        | N        | Y          | Y             |
| N        | Y        | Y          | N             |
+----------+----------+------------+---------------+

The missing item being when you have 0 positive votes and more than 0 negative votes, as you note.

As you have both positive and negative scores at the time, why not follow your own idea and create a negative Wilson Score to deal with this, remembering that the square root of a negative number is complex.

To get around complexity assume that negative votes are positive. You then calculate how "liked" a negatively scored item is and multiple this by -1 to turn it into how disliked it is.

import math

def ci_lower_bound(pos, n, neg=0):

    if n == 0:
        return 0

    # Cannot calculate the square-root of a negative number
    if pos == 0:
        votes, use_neg = neg, True
    else:
        votes, use_neg = pos, False


    # Confidence
    z = 1.96

    phat = 1.0 * votes / n

    # Calculate how confident we are that this is bad or good.
    score = (phat + z*z/(2*n) - z * math.sqrt((phat*(1-phat)+z*z/(4*n))/n))/(1+z*z/n)

    # This relationship is defined above.
    # Multiply by -1 to return a negative confidence.
    if use_neg:
        return -1 * score

    return score