bleu全称为Bilingual Evaluation Understudy（双语评估替换），是2002年提出的用于评估机器翻译效果的一种方法，这种方法简单朴素、短平快、易于理解。因为其效果还算说得过去，因此被广泛迁移到自然语言处理的各种评估任务中。这种方法可以说是：山上无老虎，猴子称大王。时无英雄遂使竖子成名。蜀中无大将，廖化做先锋。

问题描述

首先，对bleu算法建立一个直观的印象。
有两类问题：
1、给定一个句子和一个候选句子集，求bleu值，此问题称为sentence_bleu
2、给定一堆句子和一堆候选句子集，求bleu值，此问题称为corpus_bleu

机器翻译得到的句子称为candidate，候选句子集称为references。
计算方式就是计算candidate和references的公共部分。公共部分越多，说明翻译结果越好。

给定一个句子和一个候选句子集计算bleu值

bleu考虑1，2，3，4共4个n-gram，可以给每个n-gram指定权重。

对于n-gram：

• 对candidate和references分别分词（n-gram分词）
• 统计candidate和references中每个word的出现频次
• 对于candidate中的每个word，它的出现频次不能大于references中最大出现频次
  这一步是为了整治形如the the the the the这样的candidate，因为the在candidate中出现次数太多了，导致分值为1。为了限制这种不正常的candidate，使用正常的references加以约束。
• candidate中每个word的出现频次之和除以总的word数，即为得分score
• score乘以句子长度惩罚因子即为最终的bleu分数
  这一步是为了整治短句子，比如candidate只有一个词：the，并且the在references中出现过，这就导致得分为1。也就是说，有些人因为怕说错而保持沉默。

bleu的发展不是一蹴而就的，很多人为了修正bleu，不断发现bleu的漏洞并提出解决方案。从bleu的发展历程上，我们可以学到如何设计规则整治badcase。

最后，对于1-gram，2-gram，3-gram的组合，应该采用几何平均，也就是s1^w1*s2^2*s3^w3，而不是算术平均w1*s1+w2*s2+w3*s3。

from collections import Counter import numpy as np from nltk.translate import bleu_score def bp(references, candidate): # brevity penality,句子长度惩罚因子 ind = np.argmin([abs(len(i) - len(candidate)) for i in references]) if len(references[ind]) < len(candidate): return 1 scale = 1 - (len(candidate) / len(references[ind])) return np.e ** scale def parse_ngram(sentence, gram): # 把一个句子分成n-gram return [sentence[i:i + gram] for i in range(len(sentence) - gram + 1)] # 此处一定要注意+1，否则会少一个gram def sentence_bleu(references, candidate, weight): bp_value = bp(references, candidate) s = 1 for gram, wei in enumerate(weight): gram = gram + 1 # 拆分n-gram ref = [parse_ngram(i, gram) for i in references] can = parse_ngram(candidate, gram) # 统计n-gram出现次数 ref_counter = [Counter(i) for i in ref] can_counter = Counter(can) # 统计每个词在references中的出现次数 appear = sum(min(cnt, max(i.get(word, 0) for i in ref_counter)) for word, cnt in can_counter.items()) score = appear / len(can) # 每个score的权值不一样 s *= score ** wei s *= bp_value # 最后的分数需要乘以惩罚因子 return s references = [ "the dog jumps high", "the cat runs fast", "dog and cats are good friends" ] candidate = "the d o g jump s hig" weights = [0.25, 0.25, 0.25, 0.25] print(sentence_bleu(references, candidate, weights)) print(bleu_score.sentence_bleu(references, candidate, weights)) 

一个corpus是由多个sentence组成的，计算corpus_bleu并非求sentence_bleu的均值，而是一种略微复杂的计算方式，可以说是没什么道理的狂想曲。

corpus_bleu

一个文档包含3个句子，句子的分值分别为a1/b1，a2/b2，a3/b3。
那么全部句子的分值为：(a1+a2+a3)/(b1+b2+b3)

惩罚因子也是一样：三个句子的长度分别为l1,l2,l3，对应的最接近的reference分别为k1,k2,k3。那么相当于bp(l1+l2+l3,k1+k2+k3)。

也就是说：对于corpus_bleu不是单纯地对sentence_bleu求均值，而是基于更统一的一种方法。

from collections import Counter import numpy as np from nltk.translate import bleu_score def bp(references_len, candidate_len): if references_len < candidate_len: return 1 scale = 1 - (candidate_len / references_len) return np.e ** scale def parse_ngram(sentence, gram): return [sentence[i:i + gram] for i in range(len(sentence) - gram + 1)] def corpus_bleu(references_list, candidate_list, weights): candidate_len = sum(len(i) for i in candidate_list) reference_len = 0 for candidate, references in zip(candidate_list, references_list): ind = np.argmin([abs(len(i) - len(candidate)) for i in references]) reference_len += len(references[ind]) s = 1 for index, wei in enumerate(weights): up = 0 # 分子 down = 0 # 分母 gram = index + 1 for candidate, references in zip(candidate_list, references_list): # 拆分n-gram ref = [parse_ngram(i, gram) for i in references] can = parse_ngram(candidate, gram) # 统计n-gram出现次数 ref_counter = [Counter(i) for i in ref] can_counter = Counter(can) # 统计每个词在references中的出现次数 appear = sum(min(cnt, max(i.get(word, 0) for i in ref_counter)) for word, cnt in can_counter.items()) up += appear down += len(can) s *= (up / down) ** wei return bp(reference_len, candidate_len) * s references = [ [ "the dog jumps high", "the cat runs fast", "dog and cats are good friends"], [ "ba ga ya", "lu ha a df", ] ] candidate = ["the d o g jump s hig", 'it is too bad'] weights = [0.25, 0.25, 0.25, 0.25] print(corpus_bleu(references, candidate, weights)) print(bleu_score.corpus_bleu(references, candidate, weights)) 

如果你用的NLTK版本是3.2，发布时间是2016年3月份，那么计算corpus_bleu时有一处bug。NLTK在2016年10月份已经修复了此处bug。对于句子分值的求和，NLTK代码中是使用Fraction，Fraction会自动对分子和分母进行化简，导致求和的时候计算错误。

简化代码

在计算sentence_bleu和corpus_bleu过程中，许多步骤都是相似的、可以合并的。精简后的代码如下：

from collections import Counter import numpy as np from nltk.translate import bleu_score def bp(references_len, candidate_len): return np.e ** (1 - (candidate_len / references_len)) if references_len > candidate_len else 1 def nearest_len(references, candidate): return len(references[np.argmin([abs(len(i) - len(candidate)) for i in references])]) def parse_ngram(sentence, gram): return [sentence[i:i + gram] for i in range(len(sentence) - gram + 1)] def appear_count(references, candidate, gram): ref = [parse_ngram(i, gram) for i in references] can = parse_ngram(candidate, gram) # 统计n-gram出现次数 ref_counter = [Counter(i) for i in ref] can_counter = Counter(can) # 统计每个词在references中的出现次数 appear = sum(min(cnt, max(i.get(word, 0) for i in ref_counter)) for word, cnt in can_counter.items()) return appear, len(can) def corpus_bleu(references_list, candidate_list, weights): candidate_len = sum(len(i) for i in candidate_list) reference_len = sum(nearest_len(references, candidate) for candidate, references in zip(candidate_list, references_list)) bp_value = bp(reference_len, candidate_len) s = 1 for index, wei in enumerate(weights): up = 0 # 分子 down = 0 # 分母 gram = index + 1 for candidate, references in zip(candidate_list, references_list): appear, total = appear_count(references, candidate, gram) up += appear down += total s *= (up / down) ** wei return bp_value * s def sentence_bleu(references, candidate, weight): bp_value = bp(nearest_len(references, candidate), len(candidate)) s = 1 for gram, wei in enumerate(weight): gram = gram + 1 appear, total = appear_count(references, candidate, gram) score = appear / total # 每个score的权值不一样 s *= score ** wei # 最后的分数需要乘以惩罚因子 return s * bp_value if __name__ == '__main__': references = [ [ "the dog jumps high", "the cat runs fast", "dog and cats are good friends"], [ "ba ga ya", "lu ha a df", ] ] candidate = ["the d o g jump s hig", 'it is too bad'] weights = [0.25, 0.25, 0.25, 0.25] print(corpus_bleu(references, candidate, weights)) print(bleu_score.corpus_bleu(references, candidate, weights)) print(sentence_bleu(references[0], candidate[0], weights)) print(bleu_score.sentence_bleu(references[0], candidate[0], weights)) 

参考资料

https://cloud.tencent.com/developer/article/
https://en.wikipedia.org/wiki/BLEU
理解bleu
https://www.jianshu.com/p/15c22fadcba5