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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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"# Linear regression\n",
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"\n",
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"The linear regression is a training procedure based on a linear model. The model makes a prediction by simply computing a weighted sum of the input features, plus a constant term called the bias term (also called the intercept term):\n",
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"\n",
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"$$ \\hat{y}=\\theta_0 + \\theta_1 x_1 + \\theta_2 x_2 + \\cdots + \\theta_n x_n$$\n",
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"\n",
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"This can be writen more easy by using vector notation form for $m$ values. Therefore, the model will become:\n",
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"\n",
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"$$ \n",
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" \\begin{bmatrix}\n",
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" \\hat{y}^0 \\\\ \n",
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" \\hat{y}^1\\\\\n",
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" \\hat{y}^2\\\\\n",
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" \\vdots \\\\\n",
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" \\hat{y}^m\n",
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" \\end{bmatrix}\n",
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" =\n",
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" \\begin{bmatrix}\n",
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" 1 & x_1^0 & x_2^0 & \\cdots &x_n^0\\\\\n",
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" 1 & x_1^1 & x_2^1 & \\cdots & x_n^1\\\\\n",
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" \\vdots & \\vdots &\\vdots & \\cdots & \\vdots\\\\\n",
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" 1 & x_1^m & x_2^m & \\cdots & x_n^m\n",
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" \\end{bmatrix}\n",
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"\n",
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" \\begin{bmatrix}\n",
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" \\theta_0 \\\\\n",
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" \\theta_1 \\\\\n",
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" \\theta_2 \\\\\n",
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" \\vdots \\\\\n",
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" \\theta_n\n",
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" \\end{bmatrix}\n",
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"$$\n",
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"\n",
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"Resulting:\n",
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"\n",
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"$$\\hat{y}= h_\\theta(x) = x \\theta $$\n",
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"\n",
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"**Now that we have our mode, how do we train it?**\n",
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"\n",
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"Please, consider that training the model means adjusting the parameters to reduce the error or minimizing the cost function. The most common performance measure of a regression model is the Mean Square Error (MSE). Therefore, to train a Linear Regression model, you need to find the value of θ that minimizes the MSE:\n",
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"\n",
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"$$ MSE(X,h_\\theta) = \\frac{1}{m} \\sum_{i=1}^{m} \\left(\\hat{y}^{(i)}-y^{(i)} \\right)^2$$\n",
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"\n",
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"\n",
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"$$ MSE(X,h_\\theta) = \\frac{1}{m} \\sum_{i=1}^{m} \\left( x^{(i)}\\theta-y^{(i)} \\right)^2$$\n",
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"\n",
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"$$ MSE(X,h_\\theta) = \\frac{1}{m} \\left( x\\theta-y \\right)^T \\left( x\\theta-y \\right)$$\n"
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]
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},
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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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"# The normal equation\n",
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"\n",
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"To find the value of $\\theta$ that minimizes the cost function, there is a closed-form solution that gives the result directly. This is called the **Normal Equation**; and can be find it by derivating the *MSE* equation as a function of $\\theta$ and making it equals to zero:\n",
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"\n",
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"\n",
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"$$\\hat{\\theta} = (X^T X)^{-1} X^{T} y $$\n",
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"\n",
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"$$ Temp = \\theta_0 + \\theta_1 * t $$\n"
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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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{
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"data": {
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"text/html": [
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"<div>\n",
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"<style scoped>\n",
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" .dataframe tbody tr th:only-of-type {\n",
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" vertical-align: middle;\n",
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" }\n",
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"\n",
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" .dataframe tbody tr th {\n",
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" vertical-align: top;\n",
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" }\n",
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"\n",
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" .dataframe thead th {\n",
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" text-align: right;\n",
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" }\n",
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"</style>\n",
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"<table border=\"1\" class=\"dataframe\">\n",
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" <thead>\n",
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" <tr style=\"text-align: right;\">\n",
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" <th></th>\n",
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" <th>0</th>\n",
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" </tr>\n",
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" </thead>\n",
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" <tbody>\n",
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" <tr>\n",
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" <th>0</th>\n",
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" <td>24.218</td>\n",
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" </tr>\n",
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" <tr>\n",
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" <th>1</th>\n",
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" <td>23.154</td>\n",
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" </tr>\n",
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" <tr>\n",
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" <th>2</th>\n",
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" <td>24.347</td>\n",
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" </tr>\n",
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" <tr>\n",
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" <th>3</th>\n",
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" <td>24.411</td>\n",
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" </tr>\n",
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" <tr>\n",
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" <th>4</th>\n",
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" <td>24.411</td>\n",
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" </tr>\n",
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" <tr>\n",
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" <th>...</th>\n",
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" <td>...</td>\n",
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" </tr>\n",
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" <tr>\n",
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" <th>295</th>\n",
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" <td>46.357</td>\n",
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" </tr>\n",
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" <tr>\n",
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" <th>296</th>\n",
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" <td>46.551</td>\n",
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" </tr>\n",
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" <tr>\n",
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" <th>297</th>\n",
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" <td>46.519</td>\n",
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" </tr>\n",
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" <tr>\n",
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" <th>298</th>\n",
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" <td>46.551</td>\n",
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" </tr>\n",
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" <tr>\n",
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" <th>299</th>\n",
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" <td>46.583</td>\n",
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" </tr>\n",
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" </tbody>\n",
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"</table>\n",
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"<p>300 rows × 1 columns</p>\n",
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"</div>"
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],
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"text/plain": [
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" 0\n",
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"0 24.218\n",
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"1 23.154\n",
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"2 24.347\n",
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"3 24.411\n",
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"4 24.411\n",
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".. ...\n",
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"295 46.357\n",
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"296 46.551\n",
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"297 46.519\n",
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"298 46.551\n",
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"299 46.583\n",
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"\n",
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"[300 rows x 1 columns]"
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]
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},
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"execution_count": 1,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"import pandas as pd\n",
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"df = pd.read_csv('data.csv')\n",
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"df"
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]
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},
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"metadata": {},
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"outputs": [
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{
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"ename": "NameError",
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"evalue": "name 'df' is not defined",
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"output_type": "error",
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"traceback": [
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"\u001b[0;31m---------------------------------------------------------------------------\u001b[0m",
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"\u001b[0;31mNameError\u001b[0m Traceback (most recent call last)",
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"Cell \u001b[0;32mIn[1], line 1\u001b[0m\n\u001b[0;32m----> 1\u001b[0m \u001b[38;5;28mprint\u001b[39m(\u001b[43mdf\u001b[49m)\n",
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"\u001b[0;31mNameError\u001b[0m: name 'df' is not defined"
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]
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}
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],
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"source": []
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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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"display_name": ".venv",
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"language": "python",
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"name": "python3"
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"pygments_lexer": "ipython3",
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"version": "3.12.5"
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