90 lines
2.2 KiB
Plaintext
90 lines
2.2 KiB
Plaintext
{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"\n",
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"<div class=\"alert alert-block alert-info\">\n",
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"<h1> Ekstrakcja informacji </h1>\n",
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"<h2> 3. <i>tfidf (1)</i> [ćwiczenia]</h2> \n",
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"<h3> Jakub Pokrywka (2021)</h3>\n",
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"</div>\n",
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"\n",
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""
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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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"metadata": {},
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"outputs": [],
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"source": [
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"def word_to_index(word):\n",
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" vec = np.zeros(len(vocabulary))\n",
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" if word in vocabulary:\n",
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" idx = vocabulary.index(word)\n",
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" vec[idx] = 1\n",
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" else:\n",
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" vec[-1] = 1\n",
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" return vec"
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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": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"def tf(document):\n",
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" document_vector = None\n",
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" for word in document:\n",
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" if document_vector is None:\n",
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" document_vector = word_to_index(word)\n",
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" else:\n",
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" document_vector += word_to_index(word)\n",
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" return document_vector"
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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": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"def similarity(query, document):\n",
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" numerator = np.sum(query * document)\n",
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" denominator = np.sqrt(np.sum(query*query)) * np.sqrt(np.sum(document*document)) \n",
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" return numerator / denominator"
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]
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}
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],
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"metadata": {
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"author": "Jakub Pokrywka",
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"email": "kubapok@wmi.amu.edu.pl",
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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"name": "python3"
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},
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"lang": "pl",
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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": "3.8.3"
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},
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"subtitle": "3.tfidf (1)[ćwiczenia]",
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"title": "Ekstrakcja informacji",
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"year": "2021"
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},
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"nbformat": 4,
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"nbformat_minor": 4
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}
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