Byte Pair Encoding（BPE）算法
BPE算法是Transformer中构建词表的方法，大致分为如下几个步骤：
- 将语料中的文本切分为字符
- 统计高频共现二元组
- 将共现频率最高的二元组合并加入词表
- 重复上述第二和第三直到词表规模达到预先设置的数量，或没有可以合并的二元组为止
以GPT-2中BPE相关的代码为例对代码进行整理
完整代码如下所示
"""
BPE算法:字节对编码算法,将任意UTF-8字符串转换为整数索引序列,方便后续的神经网络运算。bpe is short for Byte Pair Encoder. It translates arbitrary utf-8 strings into
sequences of integers, where each integer represents small chunks of commonly
occuring characters. This implementation is based on openai's gpt2 encoder.py:
https://github.com/openai/gpt-2/blob/master/src/encoder.py
but was mildly modified because the original implementation is a bit confusing.
I also tried to add as many comments as possible, my own understanding of what's
going on.
"""import os
import json
import regex as re
import requestsimport torch# -----------------------------------------------------------------------------def bytes_to_unicode():"""将字节(8bit->2**8->256个)转换为unicode表示的字符。有些字节表示的字符太"丑"了,比如chr(0)为'\x00',OpenAI选择进行额外的转换。Every possible byte (really an integer 0..255) gets mapped by OpenAI to a unicodecharacter that represents it visually. Some bytes have their appearance preservedbecause they don't cause any trouble. These are defined in list bs. For example:chr(33) returns "!", so in the returned dictionary we simply have d[33] -> "!".However, chr(0), for example, is '\x00', which looks ugly. So OpenAI maps thesebytes, into new characters in a range where chr() returns a single nice character.So in the final dictionary we have d[0] -> 'Ā' instead, which is just chr(0 + 2**8).In particular, the space character is 32, which we can see by ord(' '). Instead,this function will shift space (32) by 256 to 288, so d[32] -> 'Ġ'.So this is just a simple one-to-one mapping of bytes 0..255 into unicode charactersthat "look nice", either in their original form, or a funny shifted characterlike 'Ā', or 'Ġ', etc."""# the 188 integers that render fine in their original form and need no shiftingbs = list(range(ord("!"), ord("~")+1))+list(range(ord("¡"), ord("¬")+1))+list(range(ord("®"), ord("ÿ")+1))cs = bs[:] # all integers b in bs will simply map to chr(b) in the output dict# now get the representations of the other 68 integers that do need shifting# each will get mapped chr(256 + n), where n will grow from 0...67 in the loopn = 0for b in range(2**8):if b not in bs:# if this byte is "ugly" then map it to the next available "nice" characterbs.append(b)cs.append(2**8+n)n += 1cs = [chr(n) for n in cs]d = dict(zip(bs, cs))return ddef get_pairs(word):"""获取一个单词中所有可能的字符二元组Return all bigrams as a set of tuples, of consecutive elements in the iterable word."""pairs = set()prev_char = word[0]for char in word[1:]:pairs.add((prev_char, char))prev_char = charreturn pairsclass Encoder:def __init__(self, encoder, bpe_merges):# byte encoder/decoderself.byte_encoder = bytes_to_unicode()self.byte_decoder = {v:k for k, v in self.byte_encoder.items()}# bpe token encoder/decoderself.encoder = encoder # 将字符串转换为整数索引self.decoder = {v:k for k,v in self.encoder.items()} # 将整数索引转换为字符串# bpe merge list that defines the bpe "tree", of tuples (a,b) that are to merge to token abself.bpe_ranks = dict(zip(bpe_merges, range(len(bpe_merges))))# the splitting pattern used for pre-tokenization# Should haved added re.IGNORECASE so BPE merges can happen for capitalized versions of contractions <-- original openai comment"""ok so what is this regex looking for, exactly?python re reference: https://docs.python.org/3/library/re.html- the vertical bars | is OR, so re.findall will chunkate text as the pieces match, from left to right- '\'s' would split up things like Andrej's -> (Andrej, 's)- ' ?\p{L}': optional space followed by 1+ unicode code points in the category "letter"- ' ?\p{N}': optional space followed by 1+ unicode code points in the category "number"- ' ?[^\s\p{L}\p{N}]+': optional space, then 1+ things that are NOT a whitespace, letter or number- '\s+(?!\S)': 1+ whitespace characters (e.g. space or tab or etc) UNLESS they are followed by non-whitespaceso this will consume whitespace characters in a sequence but exclude the last whitespace inthat sequence. that last whitespace has the opportunity to then match the optional ' ?' inearlier patterns.- '\s+': 1+ whitespace characters, intended probably to catch a full trailing sequence of whitespaces at end of stringSo TLDR:- we are special casing a few common apostrophe constructs ('s, 't, 're, ...) and making those into separate tokens- we then separate out strings into consecutive chunks of 1) letters, 2) numbers, 3) non-letter-numbers, 4) whitespaces"""self.pat = re.compile(r"""'s|'t|'re|'ve|'m|'ll|'d| ?\p{L}+| ?\p{N}+| ?[^\s\p{L}\p{N}]+|\s+(?!\S)|\s+""") # 预先使用一些正则表达式提前将字符串切分，例如将字符串划分为连续的字母、数字、空格和其他字符。包括一些英文的规则。self.cache = {}def bpe(self, token):"""对每个预先切分出来的token进行进一步的bpe切分,切分主要依赖于预先统计的bpe_ranks;bpe_ranks: 从大规模语料中统计的bi-gram共现频率this function uses self.bpe_ranks to iteratively merge all the possible bpe tokensup the tree. token is a string of one individual 'word' (after regex tokenization)and after byte encoding, e.g. 'Ġthere'."""# token is a string of one individual 'word', after byte encoding, e.g. 'Ġthere'# memoization, for efficiencyif token in self.cache: # cache缓存加速bpe算法return self.cache[token]word = tuple(token) # individual characters that make up the token, in a tuplepairs = get_pairs(word) # get all bigramsif not pairs:return tokenwhile True:# find the next lowest rank bigram that can be mergedbigram = min(pairs, key = lambda pair: self.bpe_ranks.get(pair, float('inf'))) # 优先合并共现频率高的二元组if bigram not in self.bpe_ranks: # 如果剩下的二元组共现频率过低break # no more bigrams are eligible to be mergedfirst, second = bigram# we will now replace all occurences of (first, second) in the list of current# words into one merged token first_second, in the output list new_wordsnew_word = []i = 0while i < len(word): # 合并二元组(考虑多次出现的情况)# find the next occurence of first in the sequence of current wordstry:j = word.index(first, i)new_word.extend(word[i:j])i = jexcept:new_word.extend(word[i:])break# if this occurence is also followed by second, then merge them into oneif word[i] == first and i < len(word)-1 and word[i+1] == second:new_word.append(first+second)i += 2else:new_word.append(word[i])i += 1# all occurences of (first, second) have been merged to first_secondnew_word = tuple(new_word)word = new_wordif len(word) == 1:breakelse:pairs = get_pairs(word)# concat all words into a string, and use ' ' as the separator. Note that# by now all characters have been byte encoded, guaranteeing that ' ' is# not used in the actual data and is a 'special' delimiter characterword = ' '.join(word)# cache the result and returnself.cache[token] = wordreturn worddef encode(self, text):""" 字符串序列转整数索引序列string goes in, list of integers comes out"""bpe_idx = []# pre-tokenize the input text into string tokens (words, roughly speaking)tokens = re.findall(self.pat, text) # 预先使用正则表达式粗糙切分# process each token into BPE integersfor token in tokens: # 每个token内部使用bpe不断合并二元组# encode the token as a bytes (b'') objecttoken_bytes = token.encode('utf-8')# translate all bytes to their unicode string representation and flattentoken_translated = ''.join(self.byte_encoder[b] for b in token_bytes)# perform all the applicable bpe merges according to self.bpe_rankstoken_merged = self.bpe(token_translated).split(' ')# translate all bpe tokens to integerstoken_ix = [self.encoder[bpe_token] for bpe_token in token_merged]# extend our running list of all output integersbpe_idx.extend(token_ix)return bpe_idxdef encode_and_show_work(self, text):""" debugging function, same as encode but returns all intermediate work """bpe_idx = []parts = []tokens = re.findall(self.pat, text)for token in tokens:token_bytes = token.encode('utf-8')token_translated = ''.join(self.byte_encoder[b] for b in token_bytes)token_merged = self.bpe(token_translated).split(' ')token_ix = [self.encoder[bpe_token] for bpe_token in token_merged]bpe_idx.extend(token_ix)parts.append({'token': token,'token_bytes': token_bytes,'token_translated': token_translated,'token_merged': token_merged,'token_ix': token_ix,})out = {'bpe_idx': bpe_idx, # the actual output sequence'tokens': tokens, # result of pre-tokenization'parts': parts, # intermediates for each token part}return outdef decode(self, bpe_idx):""" 整数索引序列恢复成字符串序列list of integers comes in, string comes out """# inverse map the integers to get the tokenstokens_merged = [self.decoder[token] for token in bpe_idx]# inverse the byte encoder, e.g. recovering 'Ġ' -> ' ', and get the bytestokens_flat = ''.join(tokens_merged)tokens_bytes = bytearray([self.byte_decoder[c] for c in tokens_flat])# recover the full utf-8 stringtext = tokens_bytes.decode('utf-8', errors='replace')return textdef get_file(local_file, remote_file):""" downloads remote_file to local_file if necessary """if not os.path.isfile(local_file):print(f"downloading {remote_file} to {local_file}")response = requests.get(remote_file)open(local_file, "wb").write(response.content)def get_encoder():"""从OpenAI官方的GPT-2分词器cache文件初始化Returns an instance of the GPT BPE Encoder/Decoderand handles caching of "database" files."""home_dir = os.path.expanduser('~')cache_dir = os.path.join(home_dir, '.cache', 'mingpt')os.makedirs(cache_dir, exist_ok=True)# load encoder.json that has the raw mappings from token -> bpe indexencoder_local_file = os.path.join(cache_dir, 'encoder.json')encoder_remote_file = 'https://openaipublic.blob.core.windows.net/gpt-2/models/124M/encoder.json'get_file(encoder_local_file, encoder_remote_file)with open(encoder_local_file, 'r') as f:encoder = json.load(f)assert len(encoder) == 50257 # 256 individual byte tokens, 50,000 merged tokens, and 1 special <|endoftext|> token# load vocab.bpe that contains the bpe merges, i.e. the bpe tree structure# in the form tuples (a, b), that indicate that (a, b) is to be merged to one token abvocab_local_file = os.path.join(cache_dir, 'vocab.bpe')vocab_remote_file = 'https://openaipublic.blob.core.windows.net/gpt-2/models/124M/vocab.bpe'get_file(vocab_local_file, vocab_remote_file)with open(vocab_local_file, 'r', encoding="utf-8") as f:bpe_data = f.read()# light postprocessing: strip the version on first line and the last line is a blankbpe_merges = [tuple(merge_str.split()) for merge_str in bpe_data.split('\n')[1:-1]]assert len(bpe_merges) == 50000 # 50,000 merged tokens# construct the Encoder object and returnenc = Encoder(encoder, bpe_merges)return enc# -----------------------------------------------------------------------------class BPETokenizer:""" PyTorch-aware class that wraps the Encoder above """def __init__(self):self.encoder = get_encoder()def __call__(self, text, return_tensors='pt'):# PyTorch only; here because we want to match huggingface/transformers interfaceassert return_tensors == 'pt'# single string input for now, in the future potentially a list of stringsassert isinstance(text, str)# encode and create a "batch dimension" of 1idx = [self.encoder.encode(text)]# wrap into PyTorch tensorout = torch.tensor(idx, dtype=torch.long)return outdef decode(self, idx):# ensure a simple 1D tensor for nowassert idx.ndim == 1# decode indices to texttext = self.encoder.decode(idx.tolist())return text
从Encoder类中bpe方法出发，理解BPE的全过程，以下为bpe方法代码：
def bpe(self, token):# cache缓存加速bpe算法if token in self.cache: return self.cache[token]word = tuple(token) # individual characters that make up the token, in a tuplepairs = get_pairs(word) # get all bigramsif not pairs:return tokenwhile True:# find the next lowest rank bigram that can be mergedbigram = min(pairs, key = lambda pair: self.bpe_ranks.get(pair, float('inf'))) # 优先合并共现频率高的二元组if bigram not in self.bpe_ranks: # 如果剩下的二元组共现频率过低break # no more bigrams are eligible to be mergedfirst, second = bigram# we will now replace all occurences of (first, second) in the list of current# words into one merged token first_second, in the output list new_wordsnew_word = []i = 0while i < len(word): # 合并二元组(考虑多次出现的情况)# find the next occurence of first in the sequence of current wordstry:j = word.index(first, i)new_word.extend(word[i:j])i = jexcept:new_word.extend(word[i:])break# if this occurence is also followed by second, then merge them into oneif word[i] == first and i < len(word)-1 and word[i+1] == second:new_word.append(first+second)i += 2else:new_word.append(word[i])i += 1# all occurences of (first, second) have been merged to first_secondnew_word = tuple(new_word)word = new_wordif len(word) == 1:breakelse:pairs = get_pairs(word)# concat all words into a string, and use ' ' as the separator. Note that# by now all characters have been byte encoded, guaranteeing that ' ' is# not used in the actual data and is a 'special' delimiter characterword = ' '.join(word)# cache the result and returnself.cache[token] = wordreturn word
以下是对bpe方法代码分块进行解读：
"""
在Encoder类中初始化一个缓存空间，在每次对token进行bpe操作时先验证缓存空间中是否包含，若有包含则直接结束。
"""
# cache缓存加速bpe算法
if token in self.cache: return self.cache[token]
"""
将输入bpe方法的token进行切分，此时输入的token是一个已将文本切分后的单词，使用tuple对单词中所有字符进行拆分形成一个包含token中所有字符的元组。
"""
word = tuple(token) # individual characters that make up the token, in a tuple
"""
使用get_pairs函数通过对已经拆分好的token字符元组获取所有可能的字符二元组
"""
pairs = get_pairs(word) # get all bigrams
"""
输入的word是token中所有字符的有序元组，从元组中的第一个字符开始，每两个相邻的字符组成一个二元组
"""
def get_pairs(word):pairs = set()prev_char = word[0]for char in word[1:]:pairs.add((prev_char, char))prev_char = charreturn pairs
"""
判断输入的token是否产生了二元组，若没有产生二元组则结束
"""
if not pairs:return token
"""
找到生成的二元组中共现频率最高的，其中使用bpe_ranks获得二元组频率排名，通过排名找到排名最小也就是频率最高的二元组
"""
# find the next lowest rank bigram that can be merged
bigram = min(pairs, key = lambda pair: self.bpe_ranks.get(pair, float('inf'))) # 优先合并共现频率高的二元组
"""
形成二元组对应共现频率的字典，其中bpe_merges是从已经统计好的文件中读取二元组频率数据
"""
self.bpe_ranks = dict(zip(bpe_merges, range(len(bpe_merges))))
"""
读取的文件中每行是一个二元组，行号即为频率，行号越小频率越高
"""
vocab_local_file = os.path.join(cache_dir, 'vocab.bpe')
vocab_remote_file = 'https://openaipublic.blob.core.windows.net/gpt-2/models/124M/vocab.bpe'
get_file(vocab_local_file, vocab_remote_file)
with open(vocab_local_file, 'r', encoding="utf-8") as f:bpe_data = f.read()
# light postprocessing: strip the version on first line and the last line is a blank
bpe_merges = [tuple(merge_str.split()) for merge_str in bpe_data.split('\n')[1:-1]]
"""
bpe_ranks中不存在的频率过低的二元组直接跳过，first代表二元组中的第一个字符，second代表二元组中第二个字符
"""
if bigram not in self.bpe_ranks: # 如果剩下的二元组共现频率过低break # no more bigrams are eligible to be merged
first, second = bigram
"""
此部分代码是将token中所有的字符和最高频率二元组加入到new_word列表中
"""
# we will now replace all occurences of (first, second) in the list of current
# words into one merged token first_second, in the output list new_words
new_word = []
i = 0
while i < len(word): # 合并二元组(考虑多次出现的情况)# find the next occurence of first in the sequence of current wordstry:j = word.index(first, i)new_word.extend(word[i:j])i = jexcept:new_word.extend(word[i:])break# if this occurence is also followed by second, then merge them into oneif word[i] == first and i < len(word)-1 and word[i+1] == second:new_word.append(first+second)i += 2else:new_word.append(word[i])i += 1
"""
如果新生成的字符只有一个则直接退出，如果有多个则获得新的字符对继续执行
"""
# all occurences of (first, second) have been merged to first_second
new_word = tuple(new_word)
word = new_word
if len(word) == 1:break
else:pairs = get_pairs(word)
"""
最后将字符通过空格连接为一个字符串，并存入缓存中
"""
word = ' '.join(word)# cache the result and return
self.cache[token] = word
注
本文以GPT-2中的BPE代码为例，主要记录了其中Encoder类里的bpe方法相关代码的阅读笔记
