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/1042161
https://en.wikipedia.org/wiki/BLEU
https://blog.csdn.net/qq_31584157/article/details/77709454
https://www.jianshu.com/p/15c22fadcba5