Byte Pair Encoding
在NLP模型中,输入通常是一个句子,例如I went to New York last week.
。传统做法:空格分隔,例如['i', 'went', 'to', 'New', 'York', 'last', 'week']
。
这种做法存在问题:例如模型无法通过 old, older, oldest 之间的关系学到 smart, smarter, smartest 之间的关系。如果我们能将一个token分成多个subtokens,上面的问题就能很好地解决。本文将详述目前比较常用的 subtokens 算法 ——BPE(Byte-Pair Encoding)
现在性能比较好一些的 NLP 模型,例如 GPT、BERT、RoBERTa 等,在数据预处理的时候都会有 WordPiece 的过程,其主要的实现方式就是 BPE(Byte-Pair Encoding)。具体来说,例如 ['loved', 'loving', 'loves']
这三个单词。其实本身的语义都是 "爱" 的意思,但是如果我们以词为单位,那它们就算不一样的词,在英语中不同后缀的词非常的多,就会使得词表变的很大,训练速度变慢,训练的效果也不是太好。
BPE算法通过训练,能够把上面的3个单词拆分成["lov", "ed", "ing", "es"]几个部分,这样可以把词的本身的意思和时态分开,有效的减少了此表的数量。算法流程如下:
- 设定最大subwords个数
- 将所有单词拆分为单个字符,并且在最后添加一个停止符
</w>
,同时标记处该单词出现的次数。例如,"low"
这个单词出现了5次,那么它将会被处理为{'l o w </w>': 5}
- 统计每一个连续字节对的出现频率,选择最高频者合成新的subword
- 重复第3步直到达到第1步设定的subwords词表大小或下一个最高频的字节对出现频率为1
例如:
{'l o w </w>': 5, 'l o w e r </w>': 2, 'n e w e s t </w>': 6, 'w i d e s t </w>': 3}
出现最频繁的字节对是e
和s
,共出现了 6+3 = 9次,因此将它们合并
{'l o w </w>': 5, 'l o w e r </w>': 2, 'n e w es t </w>': 6, 'w i d es t </w>': 3}
出现最频繁的字节对是es
和t
,共出现了6+3=9次,所以将它们合并
{'l o w </w>': 5, 'l o w e r </w>': 2, 'n e w est </w>': 6, 'w i d est </w>': 3}
出现最频繁的字节对是est
和</w>
,共出现了6+3=9次,因此将它们合并
{'l o w </w>': 5, 'l o w e r </w>': 2, 'n e w est</w>': 6, 'w i d est</w>': 3}
出现最频繁的字节对是l
和o
,共出现了5+2 = 7 次,因此将它们合并
{'low </w>': 5, 'low e r </w>': 2, 'n e w est</w>': 6, 'w i d est</w>': 3}
...... 继续迭代直到达到预设的 subwords 词表大小V或下一个最高频的字节对出现频率为 1。这样我们就得到了更加合适的词表,这个词表可能会出现一些不是单词的组合,但是其本身有意义的一种形式
停止符</w>
的意义在于表示subword是词后缀。举例来说:st
不加</w>
可以出现在词首,如st ar
;加了</w>
表明该子词位于词尾,如wide st</w>
,二者意义截然不同。
BPE实现
import re, collections
def get_vocab(filename):
vocab = collections.defaultdict(int)
with open(filename, 'r', encoding='utf-8') as fhand:
for line in fhand:
words = line.strip().split()
for word in words:
vocab[' '.join(list(word)) + ' </w>'] += 1
return vocab
def get_stats(vocab):
pairs = collections.defaultdict(int)
for word, freq in vocab.items():
symbols = word.split()
for i in range(len(symbols)-1):
pairs[symbols[i],symbols[i+1]] += freq
return pairs
def merge_vocab(pair, v_in):
v_out = {}
bigram = re.escape(' '.join(pair))
p = re.compile(r'(?<!\S)' + bigram + r'(?!\S)')
for word in v_in:
w_out = p.sub(''.join(pair), word)
v_out[w_out] = v_in[word]
return v_out
def get_tokens(vocab):
tokens = collections.defaultdict(int)
for word, freq in vocab.items():
word_tokens = word.split()
for token in word_tokens:
tokens[token] += freq
return tokens
vocab = {'l o w </w>': 5, 'l o w e r </w>': 2, 'n e w e s t </w>': 6, 'w i d e s t </w>': 3}
# Get free book from Gutenberg
# wget http://www.gutenberg.org/cache/epub/16457/pg16457.txt
# vocab = get_vocab('pg16457.txt')
print('==========')
print('Tokens Before BPE')
tokens = get_tokens(vocab)
print('Tokens: {}'.format(tokens))
print('Number of tokens: {}'.format(len(tokens)))
print('==========')
num_merges = 5
for i in range(num_merges):
pairs = get_stats(vocab)
if not pairs:
break
best = max(pairs, key=pairs.get)
vocab = merge_vocab(best, vocab)
print('Iter: {}'.format(i))
print('Best pair: {}'.format(best))
tokens = get_tokens(vocab)
print('Tokens: {}'.format(tokens))
print('Number of tokens: {}'.format(len(tokens)))
print('==========')
输出如下
==========
Tokens Before BPE
Tokens: defaultdict(<class 'int'>, {'l': 7, 'o': 7, 'w': 16, '</w>': 16, 'e': 17, 'r': 2, 'n': 6, 's': 9, 't': 9, 'i': 3, 'd': 3})
Number of tokens: 11
==========
Iter: 0
Best pair: ('e', 's')
Tokens: defaultdict(<class 'int'>, {'l': 7, 'o': 7, 'w': 16, '</w>': 16, 'e': 8, 'r': 2, 'n': 6, 'es': 9, 't': 9, 'i': 3, 'd': 3})
Number of tokens: 11
==========
Iter: 1
Best pair: ('es', 't')
Tokens: defaultdict(<class 'int'>, {'l': 7, 'o': 7, 'w': 16, '</w>': 16, 'e': 8, 'r': 2, 'n': 6, 'est': 9, 'i': 3, 'd': 3})
Number of tokens: 10
==========
Iter: 2
Best pair: ('est', '</w>')
Tokens: defaultdict(<class 'int'>, {'l': 7, 'o': 7, 'w': 16, '</w>': 7, 'e': 8, 'r': 2, 'n': 6, 'est</w>': 9, 'i': 3, 'd': 3})
Number of tokens: 10
==========
Iter: 3
Best pair: ('l', 'o')
Tokens: defaultdict(<class 'int'>, {'lo': 7, 'w': 16, '</w>': 7, 'e': 8, 'r': 2, 'n': 6, 'est</w>': 9, 'i': 3, 'd': 3})
Number of tokens: 9
==========
Iter: 4
Best pair: ('lo', 'w')
Tokens: defaultdict(<class 'int'>, {'low': 7, '</w>': 7, 'e': 8, 'r': 2, 'n': 6, 'w': 9, 'est</w>': 9, 'i': 3, 'd': 3})
Number of tokens: 9
==========
编码和解码
编码
在之前的算法中,我们已经得到了 subword 的词表,对该词表按照字符个数由多到少排序。编码时,对于每个单词,遍历排好序的子词词表寻找是否有 token 是当前单词的子字符串,如果有,则该 token 是表示单词的 tokens 之一
我们从最长的token迭代到最短的token,尝试将每个单词中的子字符串替换为token。最终,我们将迭代所有的tokens,并将所有子字符串替换为tokens。 如果仍然有子字符串没被替换但所有token都已迭代完毕,则将剩余的子词替换为特殊token,如<unk>
例如
# 给定单词序列
["the</w>", "highest</w>", "mountain</w>"]
# 排好序的subword表
# 长度 6 5 4 4 4 4 2
["errrr</w>", "tain</w>", "moun", "est</w>", "high", "the</w>", "a</w>"]
# 迭代结果
"the</w>" -> ["the</w>"]
"highest</w>" -> ["high", "est</w>"]
"mountain</w>" -> ["moun", "tain</w>"]
解码
将所有的tokens拼在一起即可,例如
# 编码序列
["the</w>", "high", "est</w>", "moun", "tain</w>"]
# 解码序列
"the</w> highest</w> mountain</w>"
编码和解码实现
import re, collections
def get_vocab(filename):
vocab = collections.defaultdict(int)
with open(filename, 'r', encoding='utf-8') as fhand:
for line in fhand:
words = line.strip().split()
for word in words:
vocab[' '.join(list(word)) + ' </w>'] += 1
return vocab
def get_stats(vocab):
pairs = collections.defaultdict(int)
for word, freq in vocab.items():
symbols = word.split()
for i in range(len(symbols)-1):
pairs[symbols[i],symbols[i+1]] += freq
return pairs
def merge_vocab(pair, v_in):
v_out = {}
bigram = re.escape(' '.join(pair))
p = re.compile(r'(?<!\S)' + bigram + r'(?!\S)')
for word in v_in:
w_out = p.sub(''.join(pair), word)
v_out[w_out] = v_in[word]
return v_out
def get_tokens_from_vocab(vocab):
tokens_frequencies = collections.defaultdict(int)
vocab_tokenization = {}
for word, freq in vocab.items():
word_tokens = word.split()
for token in word_tokens:
tokens_frequencies[token] += freq
vocab_tokenization[''.join(word_tokens)] = word_tokens
return tokens_frequencies, vocab_tokenization
def measure_token_length(token):
if token[-4:] == '</w>':
return len(token[:-4]) + 1
else:
return len(token)
def tokenize_word(string, sorted_tokens, unknown_token='</u>'):
if string == '':
return []
if sorted_tokens == []:
return [unknown_token]
string_tokens = []
for i in range(len(sorted_tokens)):
token = sorted_tokens[i]
token_reg = re.escape(token.replace('.', '[.]'))
matched_positions = [(m.start(0), m.end(0)) for m in re.finditer(token_reg, string)]
if len(matched_positions) == 0:
continue
substring_end_positions = [matched_position[0] for matched_position in matched_positions]
substring_start_position = 0
for substring_end_position in substring_end_positions:
substring = string[substring_start_position:substring_end_position]
string_tokens += tokenize_word(string=substring, sorted_tokens=sorted_tokens[i+1:], unknown_token=unknown_token)
string_tokens += [token]
substring_start_position = substring_end_position + len(token)
remaining_substring = string[substring_start_position:]
string_tokens += tokenize_word(string=remaining_substring, sorted_tokens=sorted_tokens[i+1:], unknown_token=unknown_token)
break
return string_tokens
# vocab = {'l o w </w>': 5, 'l o w e r </w>': 2, 'n e w e s t </w>': 6, 'w i d e s t </w>': 3}
vocab = get_vocab('pg16457.txt')
print('==========')
print('Tokens Before BPE')
tokens_frequencies, vocab_tokenization = get_tokens_from_vocab(vocab)
print('All tokens: {}'.format(tokens_frequencies.keys()))
print('Number of tokens: {}'.format(len(tokens_frequencies.keys())))
print('==========')
num_merges = 10000
for i in range(num_merges):
pairs = get_stats(vocab)
if not pairs:
break
best = max(pairs, key=pairs.get)
vocab = merge_vocab(best, vocab)
print('Iter: {}'.format(i))
print('Best pair: {}'.format(best))
tokens_frequencies, vocab_tokenization = get_tokens_from_vocab(vocab)
print('All tokens: {}'.format(tokens_frequencies.keys()))
print('Number of tokens: {}'.format(len(tokens_frequencies.keys())))
print('==========')
# Let's check how tokenization will be for a known word
word_given_known = 'mountains</w>'
word_given_unknown = 'Ilikeeatingapples!</w>'
sorted_tokens_tuple = sorted(tokens_frequencies.items(), key=lambda item: (measure_token_length(item[0]), item[1]), reverse=True)
sorted_tokens = [token for (token, freq) in sorted_tokens_tuple]
print(sorted_tokens)
word_given = word_given_known
print('Tokenizing word: {}...'.format(word_given))
if word_given in vocab_tokenization:
print('Tokenization of the known word:')
print(vocab_tokenization[word_given])
print('Tokenization treating the known word as unknown:')
print(tokenize_word(string=word_given, sorted_tokens=sorted_tokens, unknown_token='</u>'))
else:
print('Tokenizating of the unknown word:')
print(tokenize_word(string=word_given, sorted_tokens=sorted_tokens, unknown_token='</u>'))
word_given = word_given_unknown
print('Tokenizing word: {}...'.format(word_given))
if word_given in vocab_tokenization:
print('Tokenization of the known word:')
print(vocab_tokenization[word_given])
print('Tokenization treating the known word as unknown:')
print(tokenize_word(string=word_given, sorted_tokens=sorted_tokens, unknown_token='</u>'))
else:
print('Tokenizating of the unknown word:')
print(tokenize_word(string=word_given, sorted_tokens=sorted_tokens, unknown_token='</u>'))
输出如下
Tokenizing word: mountains</w>...
Tokenization of the known word:
['mountains</w>']
Tokenization treating the known word as unknown:
['mountains</w>']
Tokenizing word: Ilikeeatingapples!</w>...
Tokenizating of the unknown word:
['I', 'like', 'ea', 'ting', 'app', 'l', 'es!</w>']
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