我发现了这个问题：

from nltk.translate.bleu_score import corpus_bleu, SmoothingFunction

hyp = ['a', 'a', 'a', 'a', 'a', 'a', 'a']
ref = [['a', 'a', 'a', 'd', 'a', 'a', 'a']]
print(corpus_bleu([ref], [hyp], (0, 0, 0, 1), smoothing_function=SmoothingFunction().method4))

hyp = ['a', 'b', 'b', 'b', 'b', 'b', 'b']
ref = [['a', 'a', 'a', 'd', 'a', 'a', 'a']]
print(corpus_bleu([ref], [hyp], (0, 0, 0, 1), smoothing_function=SmoothingFunction().method4))

输出是：

0.17954959837224665
0.17954959837224665

很奇怪为什么它们的4-gram平滑精度相同，所以我查看了https://github.com/nltk/nltk/blob/develop/nltk/translate/bleu_score.py#L576-L590 的代码。

def method4(self, p_n, references, hypothesis, hyp_len, *args, **kwargs):
        """
        Smoothing method 4:
        Shorter translations may have inflated precision values due to having
        smaller denominators; therefore, we give them proportionally
        smaller smoothed counts. Instead of scaling to 1/(2^k), Chen and Cherry
        suggests dividing by 1/ln(len(T)), where T is the length of the translation.
        """
        for i, p_i in enumerate(p_n):
            if p_i.numerator == 0 and hyp_len != 0:
                incvnt = i + 1 * self.k / math.log(
                    hyp_len
                )  # Note that this K is different from the K from NIST.
                p_n[i] = 1 / incvnt
        return p_n

p_n[i] 是一个分数对象，一旦 p_n[i].numberator == 0 ,它就会变成一个浮点数常量 p_n[i] = 1 / (i + self.k / math.log(hyp_len)) 。
我找到了原始论文，它说：

分子(确切地说，匹配的n-gram计数)应该是 1 / (i + self.k / math.log(hyp_len)) ,但分母应该保持不变！

我大致浏览了一下其他的平滑函数，似乎Smoothing 4-7也有相同的问题......我认为应该立即修复，因为我已经发现一些论文的结果因为 nltk 而错误......