# -*- coding: utf-8 -*-
# Natural Language Toolkit: BLEU Score
#
# Copyright (C) 2001-2015 NLTK Project
# Authors: Chin Yee Lee, Hengfeng Li, Ruxin Hou, Calvin Tanujaya Lim
# Contributors: Dmitrijs Milajevs, Liling Tan
# URL: <http://nltk.org/>
# For license information, see LICENSE.TXT
"""BLEU score implementation."""
from __future__ import division
import math
from fractions import Fraction
from collections import Counter
from nltk.util import ngrams
def sentence_bleu(references, hypothesis, weights=(0.25, 0.25, 0.25, 0.25),
smoothing_function=None):
"""
Calculate BLEU score (Bilingual Evaluation Understudy) from
Papineni, Kishore, Salim Roukos, Todd Ward, and Wei-Jing Zhu. 2002.
"BLEU: a method for automatic evaluation of machine translation."
In Proceedings of ACL. http://www.aclweb.org/anthology/P02-1040.pdf
>>> hypothesis1 = ['It', 'is', 'a', 'guide', 'to', 'action', 'which',
... 'ensures', 'that', 'the', 'military', 'always',
... 'obeys', 'the', 'commands', 'of', 'the', 'party']
>>> hypothesis2 = ['It', 'is', 'to', 'insure', 'the', 'troops',
... 'forever', 'hearing', 'the', 'activity', 'guidebook',
... 'that', 'party', 'direct']
>>> reference1 = ['It', 'is', 'a', 'guide', 'to', 'action', 'that',
... 'ensures', 'that', 'the', 'military', 'will', 'forever',
... 'heed', 'Party', 'commands']
>>> reference2 = ['It', 'is', 'the', 'guiding', 'principle', 'which',
... 'guarantees', 'the', 'military', 'forces', 'always',
... 'being', 'under', 'the', 'command', 'of', 'the',
... 'Party']
>>> reference3 = ['It', 'is', 'the', 'practical', 'guide', 'for', 'the',
... 'army', 'always', 'to', 'heed', 'the', 'directions',
... 'of', 'the', 'party']
>>> sentence_bleu([reference1, reference2, reference3], hypothesis1) # doctest: +ELLIPSIS
0.5045...
>>> sentence_bleu([reference1, reference2, reference3], hypothesis2) # doctest: +ELLIPSIS
0.3969...
The default BLEU calculates a score for up to 4grams using uniform
weights. To evaluate your translations with higher/lower order ngrams,
use customized weights. E.g. when accounting for up to 6grams with uniform
weights:
>>> weights = (0.1666, 0.1666, 0.1666, 0.1666, 0.1666)
>>> sentence_bleu([reference1, reference2, reference3], hypothesis1, weights)
0.45838627164939455
:param references: reference sentences
:type references: list(list(str))
:param hypothesis: a hypothesis sentence
:type hypothesis: list(str)
:param weights: weights for unigrams, bigrams, trigrams and so on
:type weights: list(float)
:return: The sentence-level BLEU score.
:rtype: float
"""
# Calculates the brevity penalty.
# *hyp_len* is referred to as *c* in Papineni et. al. (2002)
hyp_len = len(hypothesis)
# *closest_ref_len* is referred to as *r* variable in Papineni et. al. (2002)
closest_ref_len = _closest_ref_length(references, hyp_len)
bp = _brevity_penalty(closest_ref_len, hyp_len)
# Calculates the modified precision *p_n* for each order of ngram.
p_n = [_modified_precision(references, hypothesis, i)
for i, _ in enumerate(weights, start=1)]
# Smoothen the modified precision.
# Note: smooth_precision() converts values into float.
if smoothing_function:
p_n = smoothing_function(p_n, references=references,
hypothesis=hypothesis, hyp_len=hyp_len)
# Calculates the overall modified precision for all ngrams.
# By sum of the product of the weights and the respective *p_n*
s = (w * math.log(p_i) if p_i else 0
for w, p_i in zip(weights, p_n))
sum_s = math.fsum(s)
if sum_s == 0 and all(p_n) == 0:
return 0
return bp * math.exp(sum_s)
def corpus_bleu(list_of_references, hypotheses, weights=(0.25, 0.25, 0.25, 0.25),
smoothing_function=None):
"""
Calculate a single corpus-level BLEU score (aka. system-level BLEU) for all
the hypotheses and their respective references.
Instead of averaging the sentence level BLEU scores (i.e. marco-average
precision), the original BLEU metric (Papineni et al. 2002) accounts for
the micro-average precision (i.e. summing the numerators and denominators
for each hypothesis-reference(s) pairs before the division).
>>> hyp1 = ['It', 'is', 'a', 'guide', 'to', 'action', 'which',
... 'ensures', 'that', 'the', 'military', 'always',
... 'obeys', 'the', 'commands', 'of', 'the', 'party']
>>> ref1a = ['It', 'is', 'a', 'guide', 'to', 'action', 'that',
... 'ensures', 'that', 'the', 'military', 'will', 'forever',
... 'heed', 'Party', 'commands']
>>> ref1b = ['It', 'is', 'the', 'guiding', 'principle', 'which',
... 'guarantees', 'the', 'military', 'forces', 'always',
... 'being', 'under', 'the', 'command', 'of', 'the', 'Party']
>>> ref1c = ['It', 'is', 'the', 'practical', 'guide', 'for', 'the',
... 'army', 'always', 'to', 'heed', 'the', 'directions',
... 'of', 'the', 'party']
>>> hyp2 = ['he', 'read', 'the', 'book', 'because', 'he', 'was',
... 'interested', 'in', 'world', 'history']
>>> ref2a = ['he', 'was', 'interested', 'in', 'world', 'history',
... 'because', 'he', 'read', 'the', 'book']
>>> list_of_references = [[ref1a, ref1b, ref1c], [ref2a]]
>>> hypotheses = [hyp1, hyp2]
>>> corpus_bleu(list_of_references, hypotheses) # doctest: +ELLIPSIS
0.5520...
The example below show that corpus_bleu() is different from averaging
sentence_bleu() for hypotheses
>>> score1 = sentence_bleu([ref1a, ref1b, ref1c], hyp1)
>>> score2 = sentence_bleu([ref2a], hyp2)
>>> (score1 + score2) / 2 # doctest: +ELLIPSIS
0.6223...
:param references: a corpus of lists of reference sentences, w.r.t. hypotheses
:type references: list(list(list(str)))
:param hypotheses: a list of hypothesis sentences
:type hypotheses: list(list(str))
:param weights: weights for unigrams, bigrams, trigrams and so on
:type weights: list(float)
:return: The corpus-level BLEU score.
:rtype: float
"""
p_numerators = Counter() # Key = ngram order, and value = no. of ngram matches.
p_denominators = Counter() # Key = ngram order, and value = no. of ngram in ref.
hyp_lengths, ref_lengths = 0, 0
assert len(list_of_references) == len(hypotheses), "The number of hypotheses and their reference(s) should be the same"
# Iterate through each hypothesis and their corresponding references.
for references, hypothesis in zip(list_of_references, hypotheses):
# For each order of ngram, calculate the numerator and
# denominator for the corpus-level modified precision.
for i, _ in enumerate(weights, start=1):
p_i = _modified_precision(references, hypothesis, i)
p_numerators[i] += p_i.numerator
p_denominators[i] += p_i.denominator
# Calculate the hypothesis length and the closest reference length.
# Adds them to the corpus-level hypothesis and reference counts.
hyp_len = len(hypothesis)
hyp_lengths += hyp_len
ref_lengths += _closest_ref_length(references, hyp_len)
# Calculate corpus-level brevity penalty.
bp = _brevity_penalty(ref_lengths, hyp_lengths)
# Collects the various precision values for the different ngram orders.
p_n = [Fraction(p_numerators[i], p_denominators[i])
for i, _ in enumerate(weights, start=1)]
# Smoothen the modified precision.
# Note: smooth_precision() converts values into float.
if smoothing_function:
p_n = smoothing_function(p_n, references=references,
hypothesis=hypothesis, hyp_len=hyp_len)
# Calculates the overall modified precision for all ngrams.
# By sum of the product of the weights and the respective *p_n*
s = (w * math.log(p_i) if p_i else 0
for w, p_i in zip(weights, p_n))
return bp * math.exp(math.fsum(s))
def _modified_precision(references, hypothesis, n):
"""
Calculate modified ngram precision.
The normal precision method may lead to some wrong translations with
high-precision, e.g., the translation, in which a word of reference
repeats several times, has very high precision.
This function only returns the Fraction object that contains the numerator
and denominator necessary to calculate the corpus-level precision.
To calculate the modified precision for a single pair of hypothesis and
references, cast the Fraction object into a float.
The famous "the the the ... " example shows that you can get BLEU precision
by duplicating high frequency words.
>>> reference1 = 'the cat is on the mat'.split()
>>> reference2 = 'there is a cat on the mat'.split()
>>> hypothesis1 = 'the the the the the the the'.split()
>>> references = [reference1, reference2]
>>> float(_modified_precision(references, hypothesis1, n=1)) # doctest: +ELLIPSIS
0.2857...
In the modified n-gram precision, a reference word will be considered
exhausted after a matching hypothesis word is identified, e.g.
>>> reference1 = ['It', 'is', 'a', 'guide', 'to', 'action', 'that',
... 'ensures', 'that', 'the', 'military', 'will',
... 'forever', 'heed', 'Party', 'commands']
>>> reference2 = ['It', 'is', 'the', 'guiding', 'principle', 'which',
... 'guarantees', 'the', 'military', 'forces', 'always',
... 'being', 'under', 'the', 'command', 'of', 'the',
... 'Party']
>>> reference3 = ['It', 'is', 'the', 'practical', 'guide', 'for', 'the',
... 'army', 'always', 'to', 'heed', 'the', 'directions',
... 'of', 'the', 'party']
>>> hypothesis = 'of the'.split()
>>> references = [reference1, reference2, reference3]
>>> float(_modified_precision(references, hypothesis, n=1))
1.0
>>> float(_modified_precision(references, hypothesis, n=2))
1.0
An example of a normal machine translation hypothesis:
>>> hypothesis1 = ['It', 'is', 'a', 'guide', 'to', 'action', 'which',
... 'ensures', 'that', 'the', 'military', 'always',
... 'obeys', 'the', 'commands', 'of', 'the', 'party']
>>> hypothesis2 = ['It', 'is', 'to', 'insure', 'the', 'troops',
... 'forever', 'hearing', 'the', 'activity', 'guidebook',
... 'that', 'party', 'direct']
>>> reference1 = ['It', 'is', 'a', 'guide', 'to', 'action', 'that',
... 'ensures', 'that', 'the', 'military', 'will',
... 'forever', 'heed', 'Party', 'commands']
>>> reference2 = ['It', 'is', 'the', 'guiding', 'principle', 'which',
... 'guarantees', 'the', 'military', 'forces', 'always',
... 'being', 'under', 'the', 'command', 'of', 'the',
... 'Party']
>>> reference3 = ['It', 'is', 'the', 'practical', 'guide', 'for', 'the',
... 'army', 'always', 'to', 'heed', 'the', 'directions',
... 'of', 'the', 'party']
>>> references = [reference1, reference2, reference3]
>>> float(_modified_precision(references, hypothesis1, n=1)) # doctest: +ELLIPSIS
0.9444...
>>> float(_modified_precision(references, hypothesis2, n=1)) # doctest: +ELLIPSIS
0.5714...
>>> float(_modified_precision(references, hypothesis1, n=2)) # doctest: +ELLIPSIS
0.5882352941176471
>>> float(_modified_precision(references, hypothesis2, n=2)) # doctest: +ELLIPSIS
0.07692...
:param references: A list of reference translations.
:type references: list(list(str))
:param hypothesis: A hypothesis translation.
:type hypothesis: list(str)
:param n: The ngram order.
:type n: int
:return: BLEU's modified precision for the nth order ngram.
:rtype: Fraction
"""
counts = Counter(ngrams(hypothesis, n))
if not counts:
return Fraction(0)
max_counts = {}
for reference in references:
reference_counts = Counter(ngrams(reference, n))
for ngram in counts:
max_counts[ngram] = max(max_counts.get(ngram, 0), reference_counts[ngram])
clipped_counts = dict((ngram, min(count, max_counts[ngram]))
for ngram, count in counts.items())
numerator = sum(clipped_counts.values())
denominator = sum(counts.values())
return Fraction(numerator, denominator)
def _closest_ref_length(references, hyp_len):
"""
This function finds the reference that is the closest length to the
hypothesis. The closest reference length is referred to as *r* variable
from the brevity penalty formula in Papineni et. al. (2002)
:param references: A list of reference translations.
:type references: list(list(str))
:param hypothesis: The length of the hypothesis.
:type hypothesis: int
:return: The length of the reference that's closest to the hypothesis.
:rtype: int
"""
ref_lens = (len(reference) for reference in references)
closest_ref_len = min(ref_lens, key=lambda ref_len:
(abs(ref_len - hyp_len), ref_len))
return closest_ref_len
def _brevity_penalty(closest_ref_len, hyp_len):
"""
Calculate brevity penalty.
As the modified n-gram precision still has the problem from the short
length sentence, brevity penalty is used to modify the overall BLEU
score according to length.
An example from the paper. There are three references with length 12, 15
and 17. And a concise hypothesis of the length 12. The brevity penalty is 1.
>>> reference1 = list('aaaaaaaaaaaa') # i.e. ['a'] * 12
>>> reference2 = list('aaaaaaaaaaaaaaa') # i.e. ['a'] * 15
>>> reference3 = list('aaaaaaaaaaaaaaaaa') # i.e. ['a'] * 17
>>> hypothesis = list('aaaaaaaaaaaa') # i.e. ['a'] * 12
>>> references = [reference1, reference2, reference3]
>>> hyp_len = len(hypothesis)
>>> closest_ref_len = _closest_ref_length(references, hyp_len)
>>> _brevity_penalty(closest_ref_len, hyp_len)
1.0
In case a hypothesis translation is shorter than the references, penalty is
applied.
>>> references = [['a'] * 28, ['a'] * 28]
>>> hypothesis = ['a'] * 12
>>> hyp_len = len(hypothesis)
>>> closest_ref_len = _closest_ref_length(references, hyp_len)
>>> _brevity_penalty(closest_ref_len, hyp_len)
0.2635971381157267
The length of the closest reference is used to compute the penalty. If the
length of a hypothesis is 12, and the reference lengths are 13 and 2, the
penalty is applied because the hypothesis length (12) is less then the
closest reference length (13).
>>> references = [['a'] * 13, ['a'] * 2]
>>> hypothesis = ['a'] * 12
>>> hyp_len = len(hypothesis)
>>> closest_ref_len = _closest_ref_length(references, hyp_len)
>>> _brevity_penalty(closest_ref_len, hyp_len) # doctest: +ELLIPSIS
0.9200...
The brevity penalty doesn't depend on reference order. More importantly,
when two reference sentences are at the same distance, the shortest
reference sentence length is used.
>>> references = [['a'] * 13, ['a'] * 11]
>>> hypothesis = ['a'] * 12
>>> hyp_len = len(hypothesis)
>>> closest_ref_len = _closest_ref_length(references, hyp_len)
>>> bp1 = _brevity_penalty(closest_ref_len, hyp_len)
>>> hyp_len = len(hypothesis)
>>> closest_ref_len = _closest_ref_length(reversed(references), hyp_len)
>>> bp2 = _brevity_penalty(closest_ref_len, hyp_len)
>>> bp1 == bp2 == 1
True
A test example from mteval-v13a.pl (starting from the line 705):
>>> references = [['a'] * 11, ['a'] * 8]
>>> hypothesis = ['a'] * 7
>>> hyp_len = len(hypothesis)
>>> closest_ref_len = _closest_ref_length(references, hyp_len)
>>> _brevity_penalty(closest_ref_len, hyp_len) # doctest: +ELLIPSIS
0.8668...
>>> references = [['a'] * 11, ['a'] * 8, ['a'] * 6, ['a'] * 7]
>>> hypothesis = ['a'] * 7
>>> hyp_len = len(hypothesis)
>>> closest_ref_len = _closest_ref_length(references, hyp_len)
>>> _brevity_penalty(closest_ref_len, hyp_len)
1.0
:param hyp_len: The length of the hypothesis for a single sentence OR the
sum of all the hypotheses' lengths for a corpus
:type hyp_len: int
:param closest_ref_len: The length of the closest reference for a single
hypothesis OR the sum of all the closest references for every hypotheses.
:type closest_reference_len: int
:return: BLEU's brevity penalty.
:rtype: float
"""
if hyp_len > closest_ref_len:
return 1
else:
return math.exp(1 - closest_ref_len / hyp_len)
class SmoothingFunction:
"""
This is an implementation of the smoothing techniques
for segment-level BLEU scores that was presented in
Boxing Chen and Collin Cherry (2014) A Systematic Comparison of
Smoothing Techniques for Sentence-Level BLEU. In WMT14.
http://acl2014.org/acl2014/W14-33/pdf/W14-3346.pdf
"""
def __init__(self, epsilon=0.1, alpha=5, k=5):
"""
This will initialize the parameters required for the various smoothing
techniques, the default values are set to the numbers used in the
experiments from Chen and Cherry (2014).
>>> hypothesis1 = ['It', 'is', 'a', 'guide', 'to', 'action', 'which', 'ensures',
... 'that', 'the', 'military', 'always', 'obeys', 'the',
... 'commands', 'of', 'the', 'party']
>>> reference1 = ['It', 'is', 'a', 'guide', 'to', 'action', 'that', 'ensures',
... 'that', 'the', 'military', 'will', 'forever', 'heed',
... 'Party', 'commands']
>>> chencherry = SmoothingFunction()
>>> print (sentence_bleu([reference1], hypothesis1)) # doctest: +ELLIPSIS
0.4118...
>>> print (sentence_bleu([reference1], hypothesis1, smoothing_function=chencherry.method0)) # doctest: +ELLIPSIS
0.4118...
>>> print (sentence_bleu([reference1], hypothesis1, smoothing_function=chencherry.method1)) # doctest: +ELLIPSIS
0.4118...
>>> print (sentence_bleu([reference1], hypothesis1, smoothing_function=chencherry.method2)) # doctest: +ELLIPSIS
0.4576...
>>> print (sentence_bleu([reference1], hypothesis1, smoothing_function=chencherry.method3)) # doctest: +ELLIPSIS
0.4118...
>>> print (sentence_bleu([reference1], hypothesis1, smoothing_function=chencherry.method4)) # doctest: +ELLIPSIS
0.4118...
>>> print (sentence_bleu([reference1], hypothesis1, smoothing_function=chencherry.method5)) # doctest: +ELLIPSIS
0.4905...
>>> print (sentence_bleu([reference1], hypothesis1, smoothing_function=chencherry.method6)) # doctest: +ELLIPSIS
0.1801...
>>> print (sentence_bleu([reference1], hypothesis1, smoothing_function=chencherry.method7)) # doctest: +ELLIPSIS
0.4905...
:param epsilon: the epsilon value use in method 1
:type epsilon: float
:param alpha: the alpha value use in method 6
:type alpha: int
:param k: the k value use in method 4
:type k: int
"""
self.epsilon = epsilon
self.alpha = alpha
self.k = k
def method0(self, p_n, *args, **kwargs):
""" No smoothing. """
return p_n
def method1(self, p_n, *args, **kwargs):
"""
Smoothing method 1: Add *epsilon* counts to precision with 0 counts.
"""
return [(p_i.numerator + self.epsilon)/ p_i.denominator
if p_i.numerator == 0 else p_i for p_i in p_n]
def method2(self, p_n, *args, **kwargs):
"""
Smoothing method 2: Add 1 to both numerator and denominator from
Chin-Yew Lin and Franz Josef Och (2004) Automatic evaluation of
machine translation quality using longest common subsequence and
skip-bigram statistics. In ACL04.
"""
return [Fraction(p_i.numerator + 1, p_i.denominator + 1) for p_i in p_n]
def method3(self, p_n, *args, **kwargs):
"""
Smoothing method 3: NIST geometric sequence smoothing
The smoothing is computed by taking 1 / ( 2^k ), instead of 0, for each
precision score whose matching n-gram count is null.
k is 1 for the first 'n' value for which the n-gram match count is null/
For example, if the text contains:
- one 2-gram match
- and (consequently) two 1-gram matches
the n-gram count for each individual precision score would be:
- n=1 => prec_count = 2 (two unigrams)
- n=2 => prec_count = 1 (one bigram)
- n=3 => prec_count = 1/2 (no trigram, taking 'smoothed' value of 1 / ( 2^k ), with k=1)
- n=4 => prec_count = 1/4 (no fourgram, taking 'smoothed' value of 1 / ( 2^k ), with k=2)
"""
incvnt = 1 # From the mteval-v13a.pl, it's referred to as k.
for i, p_i in enumerate(p_n):
if p_i == 0:
p_n[i] = 1 / (2**incvnt * p_i.denominator)
incvnt+=1
return p_n
def method4(self, p_n, references, hypothesis, hyp_len):
"""
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.
"""
incvnt = 1
for i, p_i in enumerate(p_n):
if p_i == 0 and hyp_len != 0:
p_n[i] = incvnt * self.k / math.log(hyp_len) # Note that this K is different from the K from NIST.
incvnt+=1
return p_n
def method5(self, p_n, references, hypothesis, hyp_len):
"""
Smoothing method 5:
The matched counts for similar values of n should be similar. To a
calculate the n-gram matched count, it averages the n−1, n and n+1 gram
matched counts.
"""
m = {}
# Requires an precision value for an addition ngram order.
p_n_plus1 = p_n + [_modified_precision(references, hypothesis, 5)]
m[-1] = p_n[0] + 1
for i, p_i in enumerate(p_n):
p_n[i] = (m[i-1] + p_i + p_n_plus1[i+1]) / 3
m[i] = p_n[i]
return p_n
def method6(self, p_n, references, hypothesis, hyp_len):
"""
Smoothing method 6:
Interpolates the maximum likelihood estimate of the precision *p_n* with
a prior estimate *pi0*. The prior is estimated by assuming that the ratio
between pn and pn−1 will be the same as that between pn−1 and pn−2.
"""
for i, p_i in enumerate(p_n):
if i in [1,2]: # Skips the first 2 orders of ngrams.
continue
else:
pi0 = 0 if p_n[i-2] == 0 else p_n[i-1]**2 / p_n[i-2]
# No. of ngrams in translation.
l = sum(1 for _ in ngrams(hypothesis, i+1))
p_n[i] = (p_i + self.alpha * pi0) / (l + self.alpha)
return p_n
def method7(self, p_n, references, hypothesis, hyp_len):
"""
Smoothing method 6:
Interpolates the maximum likelihood estimate of the precision *p_n* with
a prior estimate *pi0*. The prior is estimated by assuming that the ratio
between pn and pn−1 will be the same as that between pn−1 and pn−2.
"""
p_n = self.method4(p_n, references, hypothesis, hyp_len)
p_n = self.method5(p_n, references, hypothesis, hyp_len)
return p_n