BPE算法详解

Byte Pair Encoding

在NLP模型中，输入一般是一个句子，例如"I went to New York last week."，一句话中包含不少单词（token）。传统的作法是将这些单词以空格进行分隔，例如['i', 'went', 'to', 'New', 'York', 'last', 'week']。然而这种作法存在不少问题，例如模型没法经过old, older, oldest之间的关系学到smart, smarter, smartest之间的关系。若是咱们能使用将一个token分红多个subtokens，上面的问题就能很好的解决。本文将详述目前比较经常使用的subtokens算法——BPE（Byte-Pair Encoding）

如今性能比较好一些的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）。具体来讲，例如['loved', 'loving', 'loves']这三个单词。其实自己的语义都是"爱"的意思，可是若是咱们以词为单位，那它们就算不同的词，在英语中不一样后缀的词很是的多，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算法经过训练，可以把上面的3个单词拆分红["lov","ed","ing","es"]几部分，这样能够把词的自己的意思和时态分开，有效的减小了词表的数量。算法流程以下：

1. 设定最大subwords个数 $V$
2. 将全部单词拆分为单个字符，并在最后添加一个中止符</w>，同时标记出该单词出现的次数。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}
3. 统计每个连续字节对的出现频率，选择最高频者合并成新的subword
4. 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}
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出现最频繁的字节对是**e和s**，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 es t </w>': 6, 'w i d es t </w>': 3}
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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}
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出现最频繁的字节对是**est和</w>**，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}
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出现最频繁的字节对是**l和o**，共出现了5+2=7次，所以将它们合并

{'lo w </w>': 5, 'lo w e r </w>': 2, 'n e w est</w>': 6, 'w i d est</w>': 3}
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出现最频繁的字节对是**lo和w**，共出现了5+2=7次，所以将它们合并

{'low </w>': 5, 'low e r </w>': 2, 'n e w est</w>': 6, 'w i d est</w>': 3}
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......继续迭代直到达到预设的subwords词表大小或下一个最高频的字节对出现频率为1。这样咱们就获得了更加合适的词表，这个词表可能会出现一些不是单词的组合，可是其自己有意义的一种形式

中止符</w>的意义在于表示subword是词后缀。举例来讲：st不加</w>能够出如今词首，如st ar；加了</w>代表改字词位于词尾，如wide st</w>，两者意义大相径庭

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('==========')
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输出以下

==========
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
==========
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编码和解码

编码

在以前的算法中，咱们已经获得了subword的词表，对该词表按照字符个数由多到少排序。编码时，对于每一个单词，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，若是有，则该token是表示单词的tokens之一

咱们从最长的token迭代到最短的token，尝试将每一个单词中的子字符串替换为token。 最终，咱们将迭代全部tokens，并将全部子字符串替换为tokens。 若是仍然有子字符串没被替换但全部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>"]
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解码

将全部的tokens拼在一块儿便可，例如

# 编码序列
["the</w>", "high", "est</w>", "moun", "tain</w>"]

# 解码序列
"the</w> highest</w> mountain</w>"
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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>'))
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输出以下

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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Reference

• 3 subword algorithms help to improve your NLP model performance
• Tokenizers: How machines read
• Overview of tokenization algorithms in NLP
• 一文读懂BERT中的WordPiece
• Byte Pair Encoding
• 深刻理解NLP Subword算法：BPE、WordPiece、ULM