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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 requests import 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 unicode character that represents it visually. Some bytes have their appearance preserved because 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 these bytes, 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 characters that "look nice", either in their original form, or a funny shifted character like 'Ā', or 'Ġ', etc. """ # the 188 integers that render fine in their original form and need no shifting bs = 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 loop n = 0 for b in range(2**8): if b not in bs: # if this byte is "ugly" then map it to the next available "nice" character bs.append(b) cs.append(2**8+n) n += 1 cs = [chr(n) for n in cs] d = dict(zip(bs, cs)) return d def 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 = char return pairs class Encoder: def __init__(self, encoder, bpe_merges): # byte encoder/decoder self.byte_encoder = bytes_to_unicode() self.byte_decoder = {v:k for k, v in self.byte_encoder.items()} # bpe token encoder/decoder self.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 ab self.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-whitespace so this will consume whitespace characters in a sequence but exclude the last whitespace in that sequence. that last whitespace has the opportunity to then match the optional ' ?' in earlier patterns. - '\s+': 1+ whitespace characters, intended probably to catch a full trailing sequence of whitespaces at end of string So 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 tokens up 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 efficiency if 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 tuple pairs = get_pairs(word) # get all bigrams if not pairs: return token while True: # 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 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 # 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 words try: j = word.index(first, i) new_word.extend(word[i:j]) i = j except: new_word.extend(word[i:]) break # if this occurence is also followed by second, then merge them into one if word[i] == first and i < len(word)-1 and word[i+1] == second: new_word.append(first+second) i += 2 else: new_word.append(word[i]) i += 1 # 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) # 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 character word = ' '.join(word) # cache the result and return self.cache[token] = word return word def 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 integers for 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'') object token_bytes = token.encode('utf-8') # translate all bytes to their unicode string representation and flatten token_translated = ''.join(self.byte_encoder[b] for b in token_bytes) # perform all the applicable bpe merges according to self.bpe_ranks token_merged = self.bpe(token_translated).split(' ') # translate all bpe tokens to integers token_ix = [self.encoder[bpe_token] for bpe_token in token_merged] # extend our running list of all output integers bpe_idx.extend(token_ix) return bpe_idx def 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 out def 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 tokens tokens_merged = [self.decoder[token] for token in bpe_idx] # inverse the byte encoder, e.g. recovering 'Ġ' -> ' ', and get the bytes tokens_flat = ''.join(tokens_merged) tokens_bytes = bytearray([self.byte_decoder[c] for c in tokens_flat]) # recover the full utf-8 string text = tokens_bytes.decode('utf-8', errors='replace') return text def 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/Decoder and 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 index encoder_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 ab 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]] assert len(bpe_merges) == 50000 # 50,000 merged tokens # construct the Encoder object and return enc = 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 interface assert return_tensors == 'pt' # single string input for now, in the future potentially a list of strings assert isinstance(text, str) # encode and create a "batch dimension" of 1 idx = [self.encoder.encode(text)] # wrap into PyTorch tensor out = torch.tensor(idx, dtype=torch.long) return out def decode(self, idx): # ensure a simple 1D tensor for now assert idx.ndim == 1 # decode indices to text text = 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 tuple pairs = get_pairs(word) # get all bigrams if not pairs: return token while True: # 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 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 # 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 words try: j = word.index(first, i) new_word.extend(word[i:j]) i = j except: new_word.extend(word[i:]) break # if this occurence is also followed by second, then merge them into one if word[i] == first and i < len(word)-1 and word[i+1] == second: new_word.append(first+second) i += 2 else: new_word.append(word[i]) i += 1 # 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) # 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 character word = ' '.join(word) # cache the result and return self.cache[token] = word return 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 = char
return 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 words try: j = word.index(first, i) new_word.extend(word[i:j]) i = j except: new_word.extend(word[i:]) break # if this occurence is also followed by second, then merge them into one if word[i] == first and i < len(word)-1 and word[i+1] == second: new_word.append(first+second) i += 2 else: 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
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
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