Prediction using Linear Regression.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Simplest Linear Regression Prediction"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Define a function \"train\" to take parameters for predictors (x) and target (y) and \"fit\" it"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [],
   "source": [
    "def train(x,y):\n",
    "    from sklearn.linear_model import LinearRegression\n",
    "    model = LinearRegression().fit(x,y)\n",
    "    return model"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Predict the output (y_hat) for a given value of predictor (x_new)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[3.664]\n"
     ]
    }
   ],
   "source": [
    "import numpy as np\n",
    "p = np.array([0,10,20,30,40]).reshape(-1,1) # to reshae it to (row, col) as it has to be 2D array\n",
    "#print(p.shape)\n",
    "#print(p)\n",
    "q = np.array([2.3, 4.5, 5.6, 6.7, 7.8]) # 1D array \n",
    "model = train(p,q)\n",
    "x_new = 7\n",
    "y_hat = model.predict(x_new)\n",
    "print(y_hat)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Plot to see the spread of data"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "<matplotlib.collections.PathCollection at 0x2095bae2400>"
      ]
     },
     "execution_count": 11,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "data": {
      "image/png": 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2TaanNvznG6ytI/d0vjJzMjP3ZOZe1vrqwcz8ZXo1V02fld1wdvYm1s6wPwP8UdN5NuS6nLVdLo8BTzSZDfg8a2+jj7P2LuXXWFtX+xfgm9XXi/sk1z8A88DjrJXlpT3O9LOsvU19HDhS3W5qer5OkavR+aqy/QQwV2U4CvxxNX458AjwNPBF4I19kOnBar6OAp+h2nnSxA34Of5/V0lP5spPTkpSYfplqUSSVJPFLUmFsbglqTAWtyQVxuKWpMJY3JJUGItbkgpjcUtSYf4P2hU9d5gmzK4AAAAASUVORK5CYII=\n",
      "text/plain": [
       "<Figure size 432x288 with 1 Axes>"
      ]
     },
     "metadata": {
      "needs_background": "light"
     },
     "output_type": "display_data"
    }
   ],
   "source": [
    "import matplotlib.pyplot as plt\n",
    "plt.scatter(p,q)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "coefficient of determination (𝑅²) with .score() called on model"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "R SQUARE =  0.9729729729729729\n"
     ]
    }
   ],
   "source": [
    "r_sq = model.score(p,q)\n",
    "print(\"R SQUARE = \", r_sq)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The code below illustrates how to get 𝑏₀ and 𝑏₁. You can notice that .intercept_ is a scalar, while .coef_ is an array.\n",
    "\n",
    "The value 𝑏₀ = 5.63 (approximately) illustrates that your model predicts the response 2.7399999999998 when 𝑥 is zero. The value 𝑏₁ = 0.132 means that the predicted response rises by 0.54 when 𝑥 is increased by one."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Intercept (b0) =  2.7399999999999998\n",
      "Slope (b1) =  [0.132]\n"
     ]
    }
   ],
   "source": [
    "print(\"Intercept (b0) = \", model.intercept_)\n",
    "print(\"Slope (b1) = \", model.coef_)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
 ],
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