{
"cells": [
{
"cell_type": "markdown",
"id": "f4bdb240-9099-411b-a859-daaf3269c900",
"metadata": {},
"source": [
"## Implémentation de l'algorithme BPE\n",
"L'objectif de ce premier TP est de mettre en place l'algorithme de tokenization BPE. Pour rappel le principe consiste à rassembler les mots ou ``tokens'' apparaissant le plus de fois successivement.\n",
"\n",
"Par exemple si l'on considère le corpus contenant les mots de la table suivante (avec le nombre d’occurrences de chaque mot) :\n",
"\n",
"| mots | occurence |\n",
"|------|-----------|\n",
"| manger | 2 |\n",
"| voter | 3 |\n",
"| lent | 1 |\n",
"| lentement | 2 |\n",
"\n",
"Et que les \"tokens\" initiaux sont les lettres de l'alphabet alors, c'est le sufixe \"er\" qui sera ajouté à la liste des sous-mots (tokens) dans un premier temps.\n",
"\n",
"Les étapes pour réaliser l'algorithme BPE sont les suivantes :\n",
"1. Télécharger un corpus de textes (ici une page wikipedia)\n",
"2. Découper le texte en mots (en utilisant le caractère \"espace\") et compter le nombre d’occurrences de chaque mot\n",
"3. Initialiser le dictionnaire de mots avec les tokens initiaux (les lettres de l’alphabet)\n",
"4. Faire tourner l'algorithme BPE (apprendre le vocabulaire)\n",
"5. Tester la décomposition en tokens sur des phrases choisies (on applique les règles de fusion)"
]
},
{
"cell_type": "markdown",
"id": "ee88b07e-4de8-436e-9dc7-f405c69c9042",
"metadata": {},
"source": [
"### Étape 1: Télécharger un corpus"
]
},
{
"cell_type": "code",
"execution_count": 2,
"id": "5352d48f-27ca-41ad-9265-0e128578a5ec",
"metadata": {},
"outputs": [],
"source": [
"import urllib.request\n",
"import re\n",
"import numpy as np\n",
"import collections\n",
"import json\n",
"url_request = 'https://fr.wikipedia.org/w/api.php?format=json&action=query&prop=extracts&explaintext&redirects=1&titles=Gr%C3%A8ce_antique'\n",
"raw_page = urllib.request.urlopen(url_request)\n",
"json_page = json.load(raw_page)\n",
"\n",
"with open('wikitest.json', 'w') as f:\n",
" json.dump(json_page, f)"
]
},
{
"cell_type": "markdown",
"id": "f8b6b7a2-74b4-4ba2-aa4b-93885a286fb2",
"metadata": {},
"source": [
"### Étape 2 : Découper le texte en mots\n",
"\n",
"Pour découper le texte en mot nous utiliserons la regex suivante ```r'(\\b[^\\s]+\\b)'```. Pour compter les mots nous utiliserons l'objet Counter de python. \n",
"1. Stocker dans **count_words** chaque mot ainsi que le nombre d’occurrences\n",
"2. En regardant la documentation donnez les 10 mots les plus fréquents (vous les stockerez dans most_commons_words)."
]
},
{
"cell_type": "code",
"execution_count": 3,
"id": "6d7341ff-61bf-4a5a-9d0d-4ae8fd8797ce",
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"['de', 'la', 'et', 'des', 'les', 'à', 'le', 'en', 'qui', 'dans']"
]
},
"execution_count": 3,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"from collections import Counter\n",
"\n",
"corpus = list(json_page['query']['pages'].values())[0]['extract']\n",
"word_regex = re.compile(r'(\\b[^\\s]+\\b)')\n",
"words = word_regex.findall(corpus)\n",
"\n",
"count_words = collections.Counter(words)\n",
"most_commons_words = [k for k, v in count_words.most_common(10)]\n",
"\n",
"most_commons_words"
]
},
{
"cell_type": "markdown",
"id": "7a7a38b8-97f5-4851-b642-f7b40c023491",
"metadata": {},
"source": [
"### Étape 3 : Initialiser le dictionnaire de mots avec les tokens initiaux (les lettre de l'aplhabet)\n",
"\n",
"Créer le vocabulaire initial dans la variable vocab. Combien de tokens initiaux avez-vous ?"
]
},
{
"cell_type": "code",
"execution_count": 4,
"id": "901f7ea5-9d71-4f8c-95bf-32b098c3241c",
"metadata": {},
"outputs": [],
"source": [
"vocab = list({char for word in count_words.keys() for char in word })\n",
"vocab.sort()"
]
},
{
"cell_type": "markdown",
"id": "efd79ab8-b1b1-4802-ac80-39b12cb5917f",
"metadata": {},
"source": [
"### Étape 4 : Apprendre le tokenizer\n",
"Pour apprendre le tokenizer nous avons besoins de plusieurs fonctions:\n",
"1. Une fonction pour calculer la fréquence de chacune des pairs de ``tokens''\n",
"2. Une fonction pour fusionner un paire\n",
"\n",
"Plusieurs variables sont nécessaires:\n",
"1. **vocab** contenant le vocabulaire courant\n",
"2. **merge_rules** contenant toutes les règles de fusion (un dictionnaire contenant comme clef un couple de tokens à fusionner et le résultat de la fusion des tokens). Par exemple : {('e', 's'), 'es', ('en', 't') :'ent'}\n",
"3. **splits** Un dictionnaire contenant le découpage courant du corpus avec pour clef le mot et comme valeur la liste des \"tokens\"\n",
"\n"
]
},
{
"cell_type": "code",
"execution_count": 6,
"id": "8357b6bd-9414-4263-9bf5-ef249614cb04",
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"{'La': ['L', 'a'],\n",
" 'Grèce': ['G', 'r', 'è', 'c', 'e'],\n",
" 'antique': ['a', 'n', 't', 'i', 'q', 'u', 'e'],\n",
" 'est': ['e', 's', 't'],\n",
" 'une': ['u', 'n', 'e']}"
]
},
"execution_count": 6,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"# En première étape splits contient les mots décomposer en caractères\n",
"splits = {word: [c for c in word] for word in count_words.keys()}\n",
"{k: splits[k] for k in list(splits.keys())[:5]}"
]
},
{
"cell_type": "markdown",
"id": "722df20d-fae1-41f1-96bc-fecb6b10a0ca",
"metadata": {},
"source": [
"#### Calculer la fréquences des pairs de tokens\n",
"Créer un fonction **compute_pair_freqs** qui étant donné les mots décomposés en tokens (dictionnaire splits) et la fréquence des mots retourne la fréquence de chaque couple de tokens (attention seulement les sous-mots successifs). "
]
},
{
"cell_type": "code",
"execution_count": 7,
"id": "cb995e01-74f4-432f-9ddb-5c676afc32c8",
"metadata": {},
"outputs": [],
"source": [
"def compute_pair_freqs(splits, word_freqs):\n",
" pair_freqs = {}\n",
" for word, freq in word_freqs.items():\n",
" split = splits[word]\n",
" if len(split) == 1:\n",
" continue\n",
" for i in range(len(split) - 1):\n",
" pair = (split[i], split[i + 1])\n",
" if(pair not in pair_freqs):\n",
" pair_freqs[pair] = 0\n",
" pair_freqs[pair] += freq\n",
" return pair_freqs"
]
},
{
"cell_type": "code",
"execution_count": 8,
"id": "d806c606-8cd3-43ff-8e14-aea74f9d4172",
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"{('L', 'a'): 188,\n",
" ('G', 'r'): 266,\n",
" ('r', 'è'): 279,\n",
" ('è', 'c'): 316,\n",
" ('c', 'e'): 1342}"
]
},
"execution_count": 8,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"pair_freqs = compute_pair_freqs(splits, count_words)\n",
"{k: pair_freqs[k] for k in list(pair_freqs.keys())[:5]}"
]
},
{
"cell_type": "markdown",
"id": "72e05d19-df89-4b0b-a2ad-ba5af727b1e5",
"metadata": {},
"source": [
"#### Retrouver la paire la plus fréquente et fusionner une pair \n",
"1. Créer une fonction **most_frequent(pair_freqs)** retournant la paire de token la plus fréquente.\n",
"2. Créer une fonction **merge_pair()** qui étant donnée une paire, l'objet splits retourne la nouvelle séparation en token des données (splits))"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "c608944a",
"metadata": {},
"outputs": [],
"source": []
},
{
"cell_type": "code",
"execution_count": 9,
"id": "9b60759e-0bb4-4969-bc86-ee39fd3dc7bb",
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"(('e', 's'), 6895)"
]
},
"execution_count": 9,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"def most_frequent(pair_freqs):\n",
" return max(pair_freqs.items(), key=lambda x: x[1])\n",
"most_frequent(pair_freqs)\n"
]
},
{
"cell_type": "code",
"execution_count": 10,
"id": "dd44467a-326b-499c-9995-8661fd102bec",
"metadata": {},
"outputs": [],
"source": [
"def merge_pair(a, b, splits):\n",
" for word in splits.keys():\n",
" split = splits[word]\n",
" if len(split) == 1:\n",
" continue\n",
" i = 0\n",
" while i < len(split) - 1:\n",
" if split[i] == a and split[i + 1] == b:\n",
" split = split[:i] + [a + b] + split[i + 2 :]\n",
" else:\n",
" i += 1\n",
" splits[word] = split\n",
" return splits"
]
},
{
"cell_type": "code",
"execution_count": 11,
"id": "462376c4-6abc-4d4c-8bb6-a7e4d5bb96bb",
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"['g', 'r', 'e', 'c', 'q', 'u', 'es']\n"
]
}
],
"source": [
"new_splits = merge_pair(*most_frequent(pair_freqs)[0], splits)\n",
"print(new_splits['grecques'])"
]
},
{
"cell_type": "markdown",
"id": "1a9cdfeb-ade8-4ac6-9cd5-9049090ca142",
"metadata": {},
"source": [
"#### Appliquer l'algorithme jusqu'a obtenir la taille du vocabulaire souhaitée\n",
"Créer un objet BPE qui prend en argument un corpus, un nombre de mots et applique l'algorithme BPE. L'algorithme stocke dans l'attribut vocab le vcocabulaire final et dans merge_rule les règles de fusion."
]
},
{
"cell_type": "code",
"execution_count": 13,
"id": "c0367f37-9b40-4ae7-8ec6-ca7ab86ae687",
"metadata": {},
"outputs": [],
"source": [
"class BPE:\n",
" def __init__(self, corpus, vocabulary_size=500):\n",
" super().__init__()\n",
" self.word_regex = re.compile(r'(\\b[^\\s]+\\b)')\n",
" words = self.word_regex.findall(corpus)\n",
"\n",
" # counting words\n",
" count_words = collections.Counter(words)\n",
" # create initial vocab\n",
" self.vocab = list({char for word in count_words.keys() for char in word })\n",
" self.vocab.sort()\n",
" # create the initial split\n",
" splits = {word: [c for c in word] for word in count_words.keys()}\n",
" # initialise merge_rules\n",
" self.merge_rules = {}\n",
" while len(self.vocab) < vocabulary_size:\n",
" pair_freqs = compute_pair_freqs(splits, count_words)\n",
" # I considered the format (('e', 's'), 6848) for best_pair\n",
" best_pair = most_frequent(pair_freqs)\n",
" \n",
" splits = merge_pair(*best_pair[0], splits)\n",
"\n",
" self.merge_rules[best_pair[0]] = best_pair[0][0] + best_pair[0][1]\n",
" self.vocab.append(''.join(best_pair[0]))\n",
"\n",
" def tokenize(self, text):\n",
" words = self.word_regex.findall(text)\n",
" splits = [[l for l in word] for word in words]\n",
" for pair, merge in self.merge_rules.items():\n",
" for idx, split in enumerate(splits):\n",
" i = 0\n",
" while i < len(split) - 1:\n",
" if split[i] == pair[0] and split[i + 1] == pair[1]:\n",
" split = split[:i] + [merge] + split[i + 2 :]\n",
" else:\n",
" i += 1\n",
" splits[idx] = split\n",
" \n",
" return sum(splits, [])\n"
]
},
{
"cell_type": "code",
"execution_count": 14,
"id": "c282a2a1-f23c-4b22-b858-35a0049250c1",
"metadata": {
"scrolled": true
},
"outputs": [],
"source": [
"my_bpe = BPE(corpus)"
]
},
{
"cell_type": "code",
"execution_count": 15,
"id": "b225b59b-6103-4b3f-b776-ea2b71714d64",
"metadata": {
"scrolled": true
},
"outputs": [
{
"data": {
"text/plain": [
"['culture', 'grecques', 'dévelop', 'p', 'ée', 'en', 'Grèce']"
]
},
"execution_count": 15,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"texte = '''culture grecques développée en Grèce '''\n",
"my_bpe.tokenize(texte)[:12]"
]
},
{
"cell_type": "markdown",
"id": "a6f478a3-5f52-43c4-9b48-0e912304985a",
"metadata": {},
"source": [
"#### Testez en modifiant les paramètres ou le corpus\n",
"Tester l'algorithme avec différents hyper-paramètres ou données"
]
},
{
"cell_type": "code",
"execution_count": 39,
"id": "e33bdef2-8392-44f8-a846-81249fd60ac9",
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"{('e', 's'): 'es',\n",
" ('n', 't'): 'nt',\n",
" ('q', 'u'): 'qu',\n",
" ('r', 'e'): 're',\n",
" ('o', 'n'): 'on',\n",
" ('d', 'e'): 'de',\n",
" ('l', 'e'): 'le',\n",
" ('l', 'a'): 'la',\n",
" ('t', 'i'): 'ti',\n",
" ('i', 's'): 'is',\n",
" ('e', 'nt'): 'ent',\n",
" ('e', 'n'): 'en'}"
]
},
"execution_count": 39,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"{k: my_bpe.merge_rules[k] for k in list(my_bpe.merge_rules.keys())[:12]}"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "7513567a-5237-4dad-9500-bf2cded9b8a2",
"metadata": {},
"outputs": [],
"source": []
},
{
"cell_type": "code",
"execution_count": null,
"id": "2ccb41e3",
"metadata": {},
"outputs": [],
"source": []
}
],
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