deux premiers algorithmes sous forme de Jupyter Notebook
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exercices.ipynb
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exercices.ipynb
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"cells": [
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{
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"cell_type": "markdown",
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"id": "0f3d617c",
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"metadata": {},
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"source": [
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"# TP Word Embedding\n",
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"\n",
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"## Bag of Words\n",
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"Un sac de mots (ou *Bag of Words* en anglais, parfois abbrévié *BOW*) est un description d'un ensemble de mot sous forme d'un vecteur où l'ordre des mots ne rentre pas en compte.\n",
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"\n",
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"### Term Frequency\n",
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"L'idée de Term Frequency est d'effectué un simple compte du nombre d'occurence (ou de la fréquence) du nombre de mots dans le corpus.\n",
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"\n",
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"Soit un vocabulaire $V$ dans un corps $C$ contenant $D$ documents.\n",
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"Soit $w$ un mot dans un document $d \\in C$.\n",
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"\n",
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"Alors $TF(C)$ est une matrice de taille $|V|\\times|D|$ tel que\n",
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"\n",
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"$$ TF(C)_{ij} = \\frac{\\text{# words $i$ in document $j$}}{|V|} $$"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 1,
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"id": "a1445527",
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"metadata": {},
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"outputs": [],
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"source": [
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"document_1 = \"le chat mange la souris\"\n",
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"document_2 = \"le chien regarde le canard\"\n",
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"document_3 = \"le canard regarde le chat\"\n",
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"\n",
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"corpus = (document_1, document_2, document_3)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 2,
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"id": "6c989264",
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"{'le': 5, 'chat': 2, 'mange': 1, 'la': 1, 'souris': 1, 'chien': 1, 'regarde': 2, 'canard': 2}\n"
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]
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}
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],
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"source": [
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"# construction du vocabulaire\n",
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"vocabulary = []\n",
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"for d in corpus:\n",
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" for w in d.split(\" \"):\n",
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" if w not in vocabulary:\n",
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" vocabulary.append(w)\n",
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" \n",
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"# calcul d'un histogramme simple sur le corpus\n",
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"\n",
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"# intialisation du dictionnaire\n",
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"freq = dict()\n",
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"for v in vocabulary:\n",
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" freq[v] = 0\n",
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"\n",
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"# compte des fréquences\n",
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"for d in corpus:\n",
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" for w in d.split(\" \"):\n",
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" freq[w] += 1 \n",
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" \n",
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"print(freq)"
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]
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},
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{
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"cell_type": "markdown",
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"id": "34fa1346",
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"metadata": {},
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"source": [
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"problèmes liés avec cette approche :\n",
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"* indépendence au document (pousse les mots fréquents comme \"le\" vers le dessus alors qu'ils ne sont pas informatifs sémantiquements)\n",
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"* pas de prise en compte de la case (majusucle / miniscule)\n",
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"* simpliste"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 3,
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"id": "5fc408eb",
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"[[1. 1. 1. 1. 1. 0. 0. 0.]\n",
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" [2. 0. 0. 0. 0. 1. 1. 1.]\n",
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" [2. 1. 0. 0. 0. 0. 1. 1.]]\n"
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]
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}
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],
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"source": [
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"# calcul d'un histogramme par document\n",
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"import numpy as np\n",
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"\n",
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"V = len(vocabulary)\n",
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"D = len(corpus)\n",
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"\n",
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"tf_idf = np.zeros([D, V])\n",
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"\n",
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"for i, d in enumerate(corpus):\n",
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" for w in d.split(\" \"):\n",
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" j = vocabulary.index(w)\n",
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" tf_idf[i,j] += 1\n",
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" \n",
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"print(tf_idf)"
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]
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},
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{
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"cell_type": "markdown",
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"id": "771b997f",
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"metadata": {},
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"source": [
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"problèmes liés avec cette approche :\n",
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"* résoud uniquement le premier problème cité précedement\n",
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"* devrait être une implémentation en matrice creuse (*sparse matrix*) car va en pratique contenir beaucoup de zéros pour un vocabulaire grand\n",
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"\n",
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"Heuresement des implémentations existantes comme dans `scikit learn` permettent de résoudre ces problèmes techniques."
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3 (ipykernel)",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.9.7"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 5
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}
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