{ "cells": [ { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## Uczenie maszynowe – zastosowania\n", "# 2. Regresja liniowa" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## 2.1. Funkcja kosztu" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "subslide" } }, "source": [ "### Zadanie\n", "Znając $x$ – ludność miasta (w dziesiątkach tysięcy mieszkańców),\n", "należy przewidzieć $y$ – dochód firmy transportowej (w dziesiątkach tysięcy dolarów).\n", "\n", "(Dane pochodzą z kursu „Machine Learning”, Andrew Ng, Coursera)." ] }, { "cell_type": "code", "execution_count": 71, "metadata": { "slideshow": { "slide_type": "notes" } }, "outputs": [], "source": [ "# Przydatne importy\n", "\n", "import numpy as np\n", "import matplotlib\n", "import matplotlib.pyplot as plt\n", "import ipywidgets as widgets\n", "import pandas as pd\n", "\n", "%matplotlib inline\n", "%config InlineBackend.figure_format = \"svg\"\n", "\n", "from IPython.display import display, Math, Latex" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "### Wczytanie danych" ] }, { "cell_type": "code", "execution_count": 37, "metadata": { "slideshow": { "slide_type": "fragment" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ " x y\n", "0 6.1101 17.5920\n", "1 5.5277 9.1302\n", "2 8.5186 13.6620\n", "3 7.0032 11.8540\n", "4 5.8598 6.8233\n", "5 8.3829 11.8860\n", "6 7.4764 4.3483\n", "7 8.5781 12.0000\n", "8 6.4862 6.5987\n", "9 5.0546 3.8166\n" ] } ], "source": [ "data = pd.read_csv(\"data01_train.csv\", names=[\"x\", \"y\"])\n", "print(data[:10])" ] }, { "cell_type": "code", "execution_count": 38, "metadata": { "slideshow": { "slide_type": "fragment" } }, "outputs": [], "source": [ "x = data[[\"x\"]].to_numpy().flatten()\n", "y = data[[\"y\"]].to_numpy().flatten()" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "### Hipoteza i parametry modelu" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "notes" } }, "source": [ "Jak przewidzieć $y$ na podstawie danego $x$? W celu odpowiedzi na to pytanie będziemy starać się znaleźć taką funkcję $h(x)$, która będzie najlepiej obrazować zależność między $x$ a $y$, tj. $y \\sim h(x)$.\n", "\n", "Zacznijmy od najprostszego przypadku, kiedy $h(x)$ jest po prostu funkcją liniową." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "subslide" } }, "source": [ "Ogólny wzór funkcji liniowej to\n", "$$ h(x) = a \\, x + b $$" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "notes" } }, "source": [ "Pamiętajmy jednak, że współczynniki $a$ i $b$ nie są w tej chwili dane z góry – naszym zadaniem właśnie będzie znalezienie takich ich wartości, żeby $h(x)$ było „możliwie jak najbliżej” $y$ (co właściwie oznacza to sformułowanie, wyjaśnię potem).\n", "\n", "Poszukiwaną funkcję $h$ będziemy nazywać **funkcją hipotezy**, a jej współczynniki – **parametrami modelu**.\n", "\n", "W teorii uczenia maszynowego parametry modelu oznacza się na ogół grecką literą $\\theta$ z odpowiednimi indeksami, dlatego powyższy wzór opisujący liniową funkcję hipotezy zapiszemy jako\n", "$$ h(x) = \\theta_0 + \\theta_1 x $$\n", "\n", "**Parametry modelu** tworzą wektor, który oznaczymy po prostu przez $\\theta$:" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "subslide" } }, "source": [ "$$ \\theta = \\left[\\begin{array}{c}\\theta_0\\\\ \\theta_1\\end{array}\\right] $$" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "notes" } }, "source": [ "Żeby podkreślić fakt, że funkcja hipotezy zależy od parametrów modelu, będziemy pisać $h_\\theta$ zamiast $h$:" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "$$ h_{\\theta}(x) = \\theta_0 + \\theta_1 x $$" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "notes" } }, "source": [ "Przyjrzyjmy się teraz, jak wyglądają dane, które mamy modelować:" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "Na poniższym wykresie możesz spróbować ręcznie dopasować parametry modelu $\\theta_0$ i $\\theta_1$ tak, aby jak najlepiej modelowały zależność między $x$ a $y$:" ] }, { "cell_type": "code", "execution_count": 39, "metadata": { "slideshow": { "slide_type": "skip" } }, "outputs": [], "source": [ "# Funkcje rysujące wykres kropkowy oraz prostą regresyjną\n", "\n", "def regdots(x, y): \n", " fig = plt.figure(figsize=(16*.6, 9*.6))\n", " ax = fig.add_subplot(111)\n", " fig.subplots_adjust(left=0.1, right=0.9, bottom=0.1, top=0.9)\n", " ax.scatter(x, y, c='r', s=50, label='Dane')\n", " \n", " ax.set_xlabel(u'Wielkość miejscowości [dzies. tys. mieszk.]')\n", " ax.set_ylabel(u'Dochód firmy [dzies. tys. dolarów]')\n", " ax.margins(.05, .05)\n", " plt.ylim(min(y) - 1, max(y) + 1)\n", " plt.xlim(min(x) - 1, max(x) + 1)\n", " return fig\n", "\n", "def regline(fig, fun, theta, x):\n", " ax = fig.axes[0]\n", " x0, x1 = min(x), max(x)\n", " X = [x0, x1]\n", " Y = [fun(theta, x) for x in X]\n", " ax.plot(X, Y, linewidth='2',\n", " label=(r'$y={theta0}{op}{theta1}x$'.format(\n", " theta0=theta[0],\n", " theta1=(theta[1] if theta[1] >= 0 else -theta[1]),\n", " op='+' if theta[1] >= 0 else '-')))\n", "\n", "def legend(fig):\n", " ax = fig.axes[0]\n", " handles, labels = ax.get_legend_handles_labels()\n", " # try-except block is a fix for a bug in Poly3DCollection\n", " try:\n", " fig.legend(handles, labels, fontsize='15', loc='lower right')\n", " except AttributeError:\n", " pass" ] }, { "cell_type": "code", "execution_count": 40, "metadata": { "slideshow": { "slide_type": "subslide" } }, "outputs": [ { "data": { "image/svg+xml": [ "\r\n", "\r\n", "\r\n", "\r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", "\r\n" ], "text/plain": [ "
" ] }, "metadata": { "needs_background": "light" }, "output_type": "display_data" } ], "source": [ "fig = regdots(x,y)\n", "legend(fig)" ] }, { "cell_type": "code", "execution_count": 41, "metadata": { "slideshow": { "slide_type": "notes" } }, "outputs": [], "source": [ "# Hipoteza: funkcja liniowa jednej zmiennej\n", "\n", "def h(theta, x):\n", " return theta[0] + theta[1] * x" ] }, { "cell_type": "code", "execution_count": 42, "metadata": { "slideshow": { "slide_type": "notes" } }, "outputs": [], "source": [ "# Przygotowanie interaktywnego wykresu\n", "\n", "sliderTheta01 = widgets.FloatSlider(min=-10, max=10, step=0.1, value=0, description=r'$\\theta_0$', width=300)\n", "sliderTheta11 = widgets.FloatSlider(min=-5, max=5, step=0.1, value=0, description=r'$\\theta_1$', width=300)\n", "\n", "def slide1(theta0, theta1):\n", " fig = regdots(x, y)\n", " regline(fig, h, [theta0, theta1], x)\n", " legend(fig)" ] }, { "cell_type": "code", "execution_count": 43, "metadata": { "slideshow": { "slide_type": "subslide" } }, "outputs": [ { "data": { "application/vnd.jupyter.widget-view+json": { "model_id": "acd1d08a63d349429522ddb2170c79c4", "version_major": 2, "version_minor": 0 }, "text/plain": [ "interactive(children=(FloatSlider(value=0.0, description='$\\\\theta_0$', max=10.0, min=-10.0), FloatSlider(valu…" ] }, "metadata": {}, "output_type": "display_data" }, { "data": { "text/plain": [ "" ] }, "execution_count": 43, "metadata": {}, "output_type": "execute_result" } ], "source": [ "widgets.interact_manual(slide1, theta0=sliderTheta01, theta1=sliderTheta11)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "notes" } }, "source": [ "Skąd wiadomo, że przewidywania modelu (wartości funkcji $h(x)$) zgadzaja się z obserwacjami (wartości $y$)?\n", "\n", "Aby to zmierzyć wprowadzimy pojęcie funkcji kosztu." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "### Funkcja kosztu" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "notes" } }, "source": [ "Funkcję kosztu zdefiniujemy w taki sposób, żeby odzwierciedlała ona różnicę między przewidywaniami modelu a obserwacjami.\n", "\n", "Jedną z możliwosci jest zdefiniowanie funkcji kosztu jako wartość **błędu średniokwadratowego** (metoda najmniejszych kwadratów, *mean-square error, MSE*).\n", "\n", "My zdefiniujemy funkcję kosztu jako *połowę* błędu średniokwadratowego w celu ułatwienia późniejszych obliczeń (obliczenie pochodnej funkcji kosztu w dalszej części wykładu). Możemy tak zrobić, ponieważ $\\frac{1}{2}$ jest stałą, a pomnożenie przez stałą nie wpływa na przebieg zmienności funkcji." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "subslide" } }, "source": [ "$$ J(\\theta) \\, = \\, \\frac{1}{2m} \\sum_{i = 1}^{m} \\left( h_{\\theta} \\left( x^{(i)} \\right) - y^{(i)} \\right) ^2 $$" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "notes" } }, "source": [ "gdzie $m$ jest liczbą wszystkich przykładów (obserwacji), czyli wielkością zbioru danych uczących.\n", "\n", "W powyższym wzorze sumujemy kwadraty różnic między przewidywaniami modelu ($h_\\theta \\left( x^{(i)} \\right)$) a obserwacjami ($y^{(i)}$) po wszystkich przykładach $i$." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "Teraz nasze zadanie sprowadza się do tego, że będziemy szukać takich parametrów $\\theta = \\left[\\begin{array}{c}\\theta_0\\\\ \\theta_1\\end{array}\\right]$, które minimalizują fukcję kosztu $J(\\theta)$:" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "$$ \\hat\\theta = \\mathop{\\arg\\min}_{\\theta} J(\\theta) $$" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "$$ \\theta \\in \\mathbb{R}^2, \\quad J \\colon \\mathbb{R}^2 \\to \\mathbb{R} $$" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "notes" } }, "source": [ "Proszę zwrócić uwagę, że dziedziną funkcji kosztu jest zbiór wszystkich możliwych wartości parametrów $\\theta$." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "subslide" } }, "source": [ "$$ J(\\theta_0, \\theta_1) \\, = \\, \\frac{1}{2m} \\sum_{i = 1}^{m} \\left( \\theta_0 + \\theta_1 x^{(i)} - y^{(i)} \\right) ^2 $$" ] }, { "cell_type": "code", "execution_count": 44, "metadata": { "slideshow": { "slide_type": "fragment" } }, "outputs": [], "source": [ "def J(h, theta, x, y):\n", " \"\"\"Funkcja kosztu\"\"\"\n", " m = len(y)\n", " return 1.0 / (2 * m) * sum((h(theta, x[i]) - y[i])**2 for i in range(m))" ] }, { "cell_type": "code", "execution_count": 45, "metadata": { "slideshow": { "slide_type": "notes" } }, "outputs": [], "source": [ "# Oblicz wartość funkcji kosztu i pokaż na wykresie\n", "\n", "def regline2(fig, fun, theta, xx, yy):\n", " \"\"\"Rysuj regresję liniową\"\"\"\n", " ax = fig.axes[0]\n", " x0, x1 = min(xx), max(xx)\n", " X = [x0, x1]\n", " Y = [fun(theta, x) for x in X]\n", " cost = J(fun, theta, xx, yy)\n", " ax.plot(X, Y, linewidth=\"2\", \n", " label=(r'$y={theta0}{op}{theta1}x, \\; J(\\theta)={cost:.3}$'.format(\n", " theta0=theta[0],\n", " theta1=(theta[1] if theta[1] >= 0 else -theta[1]),\n", " op='+' if theta[1] >= 0 else '-',\n", " cost=str(cost))))\n", "\n", "sliderTheta02 = widgets.FloatSlider(min=-10, max=10, step=0.1, value=0, description=r'$\\theta_0$', width=300)\n", "sliderTheta12 = widgets.FloatSlider(min=-5, max=5, step=0.1, value=0, description=r'$\\theta_1$', width=300)\n", "\n", "def slide2(theta0, theta1):\n", " fig = regdots(x, y)\n", " regline2(fig, h, [theta0, theta1], x, y)\n", " legend(fig)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "notes" } }, "source": [ "Poniższy interaktywny wykres pokazuje wartość funkcji kosztu $J(\\theta)$. Czy teraz łatwiej jest dobrać parametry modelu?" ] }, { "cell_type": "code", "execution_count": 46, "metadata": { "slideshow": { "slide_type": "subslide" } }, "outputs": [ { "data": { "application/vnd.jupyter.widget-view+json": { "model_id": "dd2af5f4ffa542569abe93e0000369cf", "version_major": 2, "version_minor": 0 }, "text/plain": [ "interactive(children=(FloatSlider(value=0.0, description='$\\\\theta_0$', max=10.0, min=-10.0), FloatSlider(valu…" ] }, "metadata": {}, "output_type": "display_data" }, { "data": { "text/plain": [ "" ] }, "execution_count": 46, "metadata": {}, "output_type": "execute_result" } ], "source": [ "widgets.interact_manual(slide2, theta0=sliderTheta02, theta1=sliderTheta12)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "### Funkcja kosztu jako funkcja zmiennej $\\theta$" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "notes" } }, "source": [ "Funkcja kosztu zdefiniowana jako MSE jest funkcją zmiennej wektorowej $\\theta$, czyli funkcją dwóch zmiennych rzeczywistych: $\\theta_0$ i $\\theta_1$.\n", " \n", "Zobaczmy, jak wygląda jej wykres." ] }, { "cell_type": "code", "execution_count": 47, "metadata": { "slideshow": { "slide_type": "notes" } }, "outputs": [], "source": [ "# Wykres funkcji kosztu dla ustalonego theta_1=1.0\n", "\n", "def costfun(fun, x, y):\n", " return lambda theta: J(fun, theta, x, y)\n", "\n", "def costplot(hypothesis, x, y, theta1=1.0):\n", " fig = plt.figure(figsize=(16*.6, 9*.6))\n", " ax = fig.add_subplot(111)\n", " fig.subplots_adjust(left=0.1, right=0.9, bottom=0.1, top=0.9)\n", " ax.set_xlabel(r'$\\theta_0$')\n", " ax.set_ylabel(r'$J(\\theta)$')\n", " j = costfun(hypothesis, x, y)\n", " fun = lambda theta0: j([theta0, theta1])\n", " X = np.arange(-10, 10, 0.1)\n", " Y = [fun(x) for x in X]\n", " ax.plot(X, Y, linewidth='2', label=(r'$J(\\theta_0, {theta1})$'.format(theta1=theta1)))\n", " return fig\n", "\n", "def slide3(theta1):\n", " fig = costplot(h, x, y, theta1)\n", " legend(fig)\n", "\n", "sliderTheta13 = widgets.FloatSlider(min=-5, max=5, step=0.1, value=1.0, description=r'$\\theta_1$', width=300)" ] }, { "cell_type": "code", "execution_count": 48, "metadata": { "slideshow": { "slide_type": "subslide" } }, "outputs": [ { "data": { "application/vnd.jupyter.widget-view+json": { "model_id": "f7e676eb219045e588a64bb58c520829", "version_major": 2, "version_minor": 0 }, "text/plain": [ "interactive(children=(FloatSlider(value=1.0, description='$\\\\theta_1$', max=5.0, min=-5.0), Button(description…" ] }, "metadata": {}, "output_type": "display_data" }, { "data": { "text/plain": [ "" ] }, "execution_count": 48, "metadata": {}, "output_type": "execute_result" } ], "source": [ "widgets.interact_manual(slide3, theta1=sliderTheta13)" ] }, { "cell_type": "code", "execution_count": 49, "metadata": { "slideshow": { "slide_type": "notes" } }, "outputs": [], "source": [ "# Wykres funkcji kosztu względem theta_0 i theta_1\n", "\n", "from mpl_toolkits.mplot3d import Axes3D\n", "import pylab\n", "\n", "%matplotlib inline\n", "\n", "def costplot3d(hypothesis, x, y, show_gradient=False):\n", " fig = plt.figure(figsize=(16*.6, 9*.6))\n", " ax = fig.add_subplot(111, projection='3d')\n", " fig.subplots_adjust(left=0.0, right=1.0, bottom=0.0, top=1.0)\n", " ax.set_xlabel(r'$\\theta_0$')\n", " ax.set_ylabel(r'$\\theta_1$')\n", " ax.set_zlabel(r'$J(\\theta)$')\n", " \n", " j = lambda theta0, theta1: costfun(hypothesis, x, y)([theta0, theta1])\n", " X = np.arange(-10, 10.1, 0.1)\n", " Y = np.arange(-1, 4.1, 0.1)\n", " X, Y = np.meshgrid(X, Y)\n", " Z = np.array([[J(hypothesis, [theta0, theta1], x, y) \n", " for theta0, theta1 in zip(xRow, yRow)] \n", " for xRow, yRow in zip(X, Y)])\n", " \n", " ax.plot_surface(X, Y, Z, rstride=2, cstride=8, linewidth=0.5,\n", " alpha=0.5, cmap='jet', zorder=0,\n", " label=r\"$J(\\theta)$\")\n", " ax.view_init(elev=20., azim=-150)\n", "\n", " ax.set_xlim3d(-10, 10);\n", " ax.set_ylim3d(-1, 4);\n", " ax.set_zlim3d(-100, 800);\n", "\n", " N = range(0, 800, 20)\n", " plt.contour(X, Y, Z, N, zdir='z', offset=-100, cmap='coolwarm', alpha=1)\n", " \n", " ax.plot([-3.89578088] * 2,\n", " [ 1.19303364] * 2,\n", " [-100, 4.47697137598], \n", " color='red', alpha=1, linewidth=1.3, zorder=100, linestyle='dashed',\n", " label=r'minimum: $J(-3.90, 1.19) = 4.48$')\n", " ax.scatter([-3.89578088] * 2,\n", " [ 1.19303364] * 2,\n", " [-100, 4.47697137598], \n", " c='r', s=80, marker='x', alpha=1, linewidth=1.3, zorder=100, \n", " label=r'minimum: $J(-3.90, 1.19) = 4.48$')\n", " \n", " if show_gradient:\n", " ax.plot([3.0, 1.1],\n", " [3.0, 2.4],\n", " [263.0, 125.0], \n", " color='green', alpha=1, linewidth=1.3, zorder=100)\n", " ax.scatter([3.0],\n", " [3.0],\n", " [263.0], \n", " c='g', s=30, marker='D', alpha=1, linewidth=1.3, zorder=100)\n", "\n", " ax.margins(0,0,0)\n", " fig.tight_layout()" ] }, { "cell_type": "code", "execution_count": 50, "metadata": { 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" ] }, "metadata": { "needs_background": "light" }, "output_type": "display_data" } ], "source": [ "costplot3d(h, x, y)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "notes" } }, "source": [ "Na powyższym wykresie poszukiwane minimum funkcji kosztu oznaczone jest czerwonym krzyżykiem.\n", "\n", "Możemy też zobaczyć rzut powyższego trójwymiarowego wykresu na płaszczyznę $(\\theta_0, \\theta_1)$ poniżej:" ] }, { "cell_type": "code", "execution_count": 51, "metadata": { "slideshow": { "slide_type": "notes" } }, "outputs": [], "source": [ "def costplot2d(hypothesis, x, y, gradient_values=[], nohead=False):\n", " fig = plt.figure(figsize=(16*.6, 9*.6))\n", " ax = fig.add_subplot(111)\n", " fig.subplots_adjust(left=0.1, right=0.9, bottom=0.1, top=0.9)\n", " ax.set_xlabel(r'$\\theta_0$')\n", " ax.set_ylabel(r'$\\theta_1$')\n", " \n", " j = lambda theta0, theta1: costfun(hypothesis, x, y)([theta0, theta1])\n", " X = np.arange(-10, 10.1, 0.1)\n", " Y = np.arange(-1, 4.1, 0.1)\n", " X, Y = np.meshgrid(X, Y)\n", " Z = np.array([[J(hypothesis, [theta0, theta1], x, y) \n", " for theta0, theta1 in zip(xRow, yRow)] \n", " for xRow, yRow in zip(X, Y)])\n", " \n", " N = range(0, 800, 20)\n", " plt.contour(X, Y, Z, N, cmap='coolwarm', alpha=1)\n", "\n", " ax.scatter([-3.89578088], [1.19303364], c='r', s=80, marker='x',\n", " label=r'minimum: $J(-3.90, 1.19) = 4.48$')\n", " \n", " if len(gradient_values) > 0:\n", " prev_theta = gradient_values[0][1]\n", " ax.scatter([prev_theta[0]], [prev_theta[1]],\n", " c='g', s=30, marker='D', zorder=100)\n", " for cost, theta in gradient_values[1:]:\n", " dtheta = [theta[0] - prev_theta[0], theta[1] - prev_theta[1]]\n", " ax.arrow(prev_theta[0], prev_theta[1], dtheta[0], dtheta[1], \n", " color='green', \n", " head_width=(0.0 if nohead else 0.1), \n", " head_length=(0.0 if nohead else 0.2),\n", " zorder=100)\n", " prev_theta = theta\n", " \n", " return fig" ] }, { "cell_type": "code", "execution_count": 52, "metadata": { "slideshow": { "slide_type": "subslide" } }, "outputs": [ { "data": { "image/svg+xml": [ "\r\n", "\r\n", "\r\n", "\r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", "\r\n" ], "text/plain": [ "
" ] }, "metadata": { "needs_background": "light" }, "output_type": "display_data" } ], "source": [ "fig = costplot2d(h, x, y)\n", "legend(fig)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "subslide" } }, "source": [ "### Cechy funkcji kosztu\n", "* $J(\\theta)$ jest funkcją wypukłą\n", "* $J(\\theta)$ posiada tylko jedno minimum lokalne" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## 2.2. Metoda gradientu prostego" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "subslide" } }, "source": [ "### Metoda gradientu prostego\n", "Metoda znajdowania minimów lokalnych." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "Idea:\n", " * Zacznijmy od dowolnego $\\theta$.\n", " * Zmieniajmy powoli $\\theta$ tak, aby zmniejszać $J(\\theta)$, aż w końcu znajdziemy minimum." ] }, { "cell_type": "code", "execution_count": 53, "metadata": { "slideshow": { "slide_type": "subslide" } }, "outputs": [ { "data": { "image/svg+xml": [ "\r\n", "\r\n", "\r\n", "\r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " 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" ] }, "metadata": { "needs_background": "light" }, "output_type": "display_data" } ], "source": [ "costplot3d(h, x, y, show_gradient=True)" ] }, { "cell_type": "code", "execution_count": 54, "metadata": { "slideshow": { "slide_type": "notes" } }, "outputs": [], "source": [ "# Przykładowe wartości kolejnych przybliżeń (sztuczne)\n", "\n", "gv = [[_, [3.0, 3.0]], [_, [2.6, 2.4]], [_, [2.2, 2.0]], [_, [1.6, 1.6]], [_, [0.4, 1.2]]]\n", "\n", "# Przygotowanie interaktywnego wykresu\n", "\n", "sliderSteps1 = widgets.IntSlider(min=0, max=3, step=1, value=0, description='kroki', width=300)\n", "\n", "def slide4(steps):\n", " costplot2d(h, x, y, gradient_values=gv[:steps+1])" ] }, { "cell_type": "code", "execution_count": 55, "metadata": { "slideshow": { "slide_type": "subslide" } }, "outputs": [ { "data": { "application/vnd.jupyter.widget-view+json": { "model_id": "f7b8eaa326d646948a97a700d7c1f681", "version_major": 2, "version_minor": 0 }, "text/plain": [ "interactive(children=(IntSlider(value=0, description='kroki', max=3), Output()), _dom_classes=('widget-interac…" ] }, "metadata": {}, "output_type": "display_data" }, { "data": { "text/plain": [ "" ] }, "execution_count": 55, "metadata": {}, "output_type": "execute_result" } ], "source": [ "widgets.interact(slide4, steps=sliderSteps1)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "### Metoda gradientu prostego\n", "W każdym kroku będziemy aktualizować parametry $\\theta_j$:\n", "\n", "$$ \\theta_j := \\theta_j - \\alpha \\frac{\\partial}{\\partial \\theta_j} J(\\theta) \\quad \\mbox{ dla każdego } j $$" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "Współczynnik $\\alpha$ nazywamy **długością kroku** lub **współczynnikiem szybkości uczenia** (*learning rate*)." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "subslide" } }, "source": [ "$$ \\begin{array}{rcl}\n", "\\dfrac{\\partial}{\\partial \\theta_j} J(\\theta)\n", " & = & \\dfrac{\\partial}{\\partial \\theta_j} \\dfrac{1}{2m} \\displaystyle\\sum_{i = 1}^{m} \\left( h_{\\theta} \\left( x^{(i)} \\right) - y^{(i)} \\right) ^2 \\\\\n", " & = & 2 \\cdot \\dfrac{1}{2m} \\displaystyle\\sum_{i=1}^m \\left( h_\\theta \\left( x^{(i)} \\right) - y^{(i)} \\right) \\cdot \\dfrac{\\partial}{\\partial\\theta_j} \\left( h_\\theta \\left( x^{(i)} \\right) - y^{(i)} \\right) \\\\\n", " & = & \\dfrac{1}{m}\\displaystyle\\sum_{i=1}^m \\left( h_\\theta \\left( x^{(i)} \\right) - y^{(i)} \\right) \\cdot \\dfrac{\\partial}{\\partial\\theta_j} \\left( \\displaystyle\\sum_{i=0}^n \\theta_i x_i^{(i)} - y^{(i)} \\right)\\\\\n", " & = & \\dfrac{1}{m}\\displaystyle\\sum_{i=1}^m \\left( h_\\theta \\left( x^{(i)} \\right) -y^{(i)} \\right) x_j^{(i)} \\\\\n", "\\end{array} $$" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "subslide" } }, "source": [ "Czyli dla regresji liniowej jednej zmiennej:\n", "\n", "$$ h_\\theta(x) = \\theta_0 + \\theta_1x $$\n", "\n", "w każdym kroku będziemy aktualizować:\n", "\n", "$$\n", "\\begin{array}{rcl}\n", "\\theta_0 & := & \\theta_0 - \\alpha \\, \\dfrac{1}{m}\\displaystyle\\sum_{i=1}^m \\left( h_\\theta(x^{(i)})-y^{(i)} \\right) \\\\ \n", "\\theta_1 & := & \\theta_1 - \\alpha \\, \\dfrac{1}{m}\\displaystyle\\sum_{i=1}^m \\left( h_\\theta(x^{(i)})-y^{(i)} \\right) x^{(i)}\\\\ \n", "\\end{array}\n", "$$" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "subslide" } }, "source": [ "###### Uwaga!\n", " * W każdym kroku aktualizujemy *jednocześnie* $\\theta_0$ i $\\theta_1$" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ " * Kolejne kroki wykonujemy aż uzyskamy zbieżność" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "### Metoda gradientu prostego – implementacja" ] }, { "cell_type": "code", "execution_count": 56, "metadata": { "slideshow": { "slide_type": "notes" } }, "outputs": [], "source": [ "# Wyświetlanie macierzy w LaTeX-u\n", "\n", "def LatexMatrix(matrix):\n", " ltx = r'\\left[\\begin{array}'\n", " m, n = matrix.shape\n", " ltx += '{' + (\"r\" * n) + '}'\n", " for i in range(m):\n", " ltx += r\" & \".join([('%.4f' % j.item()) for j in matrix[i]]) + r\" \\\\ \"\n", " ltx += r'\\end{array}\\right]'\n", " return ltx" ] }, { "cell_type": "code", "execution_count": 57, "metadata": { "slideshow": { "slide_type": "fragment" } }, "outputs": [], "source": [ "def gradient_descent(h, cost_fun, theta, x, y, alpha, eps):\n", " current_cost = cost_fun(h, theta, x, y)\n", " log = [[current_cost, theta]] # log przechowuje wartości kosztu i parametrów\n", " m = len(y)\n", " while True:\n", " new_theta = [\n", " theta[0] - alpha/float(m) * sum(h(theta, x[i]) - y[i]\n", " for i in range(m)), \n", " theta[1] - alpha/float(m) * sum((h(theta, x[i]) - y[i]) * x[i]\n", " for i in range(m))]\n", " theta = new_theta # jednoczesna aktualizacja - używamy zmiennej tymaczasowej\n", " prev_cost = current_cost\n", " current_cost = cost_fun(h, theta, x, y)\n", " if current_cost > prev_cost:\n", " print(\"Zbyt duża długość kroku!\")\n", " break\n", " if abs(prev_cost - current_cost) <= eps:\n", " break \n", " log.append([current_cost, theta])\n", " return theta, log" ] }, { "cell_type": "code", "execution_count": 58, "metadata": { "slideshow": { "slide_type": "subslide" } }, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle \\large\\textrm{Wynik:}\\quad \\theta = \\left[\\begin{array}{r}-3.4894 \\\\ 1.1786 \\\\ \\end{array}\\right] \\quad J(\\theta) = 4.7371 \\quad \\textrm{po 22362 iteracjach}$" ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "best_theta, log = gradient_descent(h, J, [0.0, 0.0], x, y, alpha=0.001, eps=0.0000001)\n", "\n", "display(Math(r'\\large\\textrm{Wynik:}\\quad \\theta = ' + \n", " LatexMatrix(np.matrix(best_theta).reshape(2,1)) + \n", " (r' \\quad J(\\theta) = %.4f' % log[-1][0]) \n", " + r' \\quad \\textrm{po %d iteracjach}' % len(log))) " ] }, { "cell_type": "code", "execution_count": 59, "metadata": { "slideshow": { "slide_type": "notes" } }, "outputs": [], "source": [ "# Przygotowanie interaktywnego wykresu\n", "\n", "sliderSteps2 = widgets.IntSlider(min=0, max=500, step=1, value=1, description='kroki', width=300)\n", "\n", "def slide5(steps):\n", " costplot2d(h, x, y, gradient_values=log[:steps+1], nohead=True)" ] }, { "cell_type": "code", "execution_count": 60, "metadata": { "scrolled": true, "slideshow": { "slide_type": "subslide" } }, "outputs": [ { "data": { "application/vnd.jupyter.widget-view+json": { "model_id": "e47db1599f11451781b5f0010049a6bf", "version_major": 2, "version_minor": 0 }, "text/plain": [ "interactive(children=(IntSlider(value=1, description='kroki', max=500), Button(description='Run Interact', sty…" ] }, "metadata": {}, "output_type": "display_data" }, { "data": { "text/plain": [ "" ] }, "execution_count": 60, "metadata": {}, "output_type": "execute_result" } ], "source": [ "widgets.interact_manual(slide5, steps=sliderSteps2)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "### Współczynnik $\\alpha$ (długość kroku)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "notes" } }, "source": [ "Tempo zbieżności metody gradientu prostego możemy regulować za pomocą parametru $\\alpha$, pamiętając, że:" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ " * Jeżeli długość kroku jest zbyt mała, algorytm może działać zbyt wolno." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ " * Jeżeli długość kroku jest zbyt duża, algorytm może nie być zbieżny." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## 2.3. Predykcja wyników" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "notes" } }, "source": [ "Zbudowaliśmy model, dzięki któremu wiemy, jaka jest zależność między dochodem firmy transportowej ($y$) a ludnością miasta ($x$).\n", "\n", "Wróćmy teraz do postawionego na początku wykładu pytania: jak przewidzieć dochód firmy transportowej w mieście o danej wielkości?\n", "\n", "Odpowiedź polega po prostu na zastosowaniu funkcji $h$ z wyznaczonymi w poprzednim kroku parametrami $\\theta$.\n", "\n", "Na przykład, jeżeli miasto ma $536\\,000$ ludności, to $x = 53.6$ (bo dane trenujące były wyrażone w dziesiątkach tysięcy mieszkańców, a $536\\,000 = 53.6 \\cdot 10\\,000$) i możemy użyć znalezionych parametrów $\\theta$, by wykonać następujące obliczenia:\n", "$$ \\hat{y} \\, = \\, h_\\theta(x) \\, = \\, \\theta_0 + \\theta_1 \\, x \\, = \\, 0.0494 + 0.7591 \\cdot 53.6 \\, = \\, 40.7359 $$\n", "\n", "Czyli używając zdefiniowanych wcześniej funkcji:" ] }, { "cell_type": "code", "execution_count": 72, "metadata": { "slideshow": { "slide_type": "subslide" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "59.68111013077243\n" ] } ], "source": [ "example_x = 53.6\n", "predicted_y = h(best_theta, example_x)\n", "print(predicted_y) ## taki jest przewidywany dochód tej firmy transportowej w 536-tysięcznym mieście" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## 2.4. Ewaluacja modelu" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "subslide" } }, "source": [ "Jak ocenić jakość stworzonego przez nas modelu?\n", "\n", " * Trzeba sprawdzić, jak przewidywania modelu zgadzają się z oczekiwaniami!" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "subslide" } }, "source": [ "Czy możemy w tym celu użyć danych, których użyliśmy do wytrenowania modelu?\n", "**NIE!**\n", "\n", " * Istotą uczenia maszynowego jest budowanie modeli/algorytmów, które dają dobre przewidywania dla **nieznanych** danych – takich, z którymi algorytm nie miał jeszcze styczności! Nie sztuką jest przewidywać rzeczy, które juz sie zna.\n", " * Dlatego testowanie/ewaluowanie modelu na zbiorze uczącym mija się z celem i jest nieprzydatne.\n", " * Do ewaluacji modelu należy użyć oddzielnego zbioru danych.\n", " * **Dane uczące i dane testowe zawsze powinny stanowić oddzielne zbiory!**" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "notes" } }, "source": [ "Na wykładzie *5. Dobre praktyki w uczeniu maszynowym* dowiesz się, jak podzielić posiadane dane na zbiór uczący i zbiór testowy.\n", "\n", "Tutaj, na razie, do ewaluacji użyjemy specjalnie przygotowanego zbioru testowego." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "subslide" } }, "source": [ "Jako metrykę ewaluacji wykorzystamy znany nam już błąd średniokwadratowy (MSE):" ] }, { "cell_type": "code", "execution_count": 73, "metadata": {}, "outputs": [], "source": [ "def mse(expected, predicted):\n", " \"\"\"Błąd średniokwadratowy\"\"\"\n", " m = len(expected)\n", " if len(predicted) != m:\n", " raise Exception('Wektory mają różne długości!')\n", " return 1.0 / (2 * m) * sum((expected[i] - predicted[i])**2 for i in range(m))" ] }, { "cell_type": "code", "execution_count": 74, "metadata": { "slideshow": { "slide_type": "subslide" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "3.4988278621350606\n" ] } ], "source": [ "# Wczytwanie danych testowych z pliku za pomocą numpy\n", "\n", "test_data = np.loadtxt('data01_test.csv', delimiter=',')\n", "x_test = test_data[:, 0]\n", "y_test = test_data[:, 1]\n", "\n", "# Obliczenie przewidywań modelu\n", "y_pred = h(best_theta, x_test)\n", "\n", "# Obliczenie MSE na zbiorze testowym (im mniejszy MSE, tym lepiej!)\n", "evaluation_result = mse(y_test, y_pred)\n", "\n", "print(evaluation_result)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "notes" } }, "source": [ "Otrzymana wartość mówi nam o tym, jak dobry jest stworzony przez nas model.\n", "\n", "W przypadku metryki MSE im mniejsza wartość, tym lepiej.\n", "\n", "W ten sposób możemy np. porównywać różne modele." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## 2.5. Regresja liniowa wielu zmiennych" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "notes" } }, "source": [ "Do przewidywania wartości $y$ możemy użyć więcej niż jednej cechy $x$:" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "subslide" } }, "source": [ "### Przykład – ceny mieszkań" ] }, { "cell_type": "code", "execution_count": 75, "metadata": { "slideshow": { "slide_type": "subslide" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "y : price x1: isNew x2: rooms x3: floor x4: location x5: sqrMetres\n", "476118.0 False 3 1 Centrum 78 \n", "459531.0 False 3 2 Sołacz 62 \n", "411557.0 False 3 0 Sołacz 15 \n", "496416.0 False 4 0 Sołacz 14 \n", "406032.0 False 3 0 Sołacz 15 \n", "450026.0 False 3 1 Naramowice 80 \n", "571229.15 False 2 4 Wilda 39 \n", "325000.0 False 3 1 Grunwald 54 \n", "268229.0 False 2 1 Grunwald 90 \n" ] } ], "source": [ "import csv\n", "\n", "reader = csv.reader(open('data02_train.tsv', encoding='utf-8'), delimiter='\\t')\n", "for i, row in enumerate(list(reader)[:10]):\n", " if i == 0:\n", " print(' '.join(['{}: {:8}'.format('x' + str(j) if j > 0 else 'y ', entry)\n", " for j, entry in enumerate(row)]))\n", " else:\n", " print(' '.join(['{:12}'.format(entry) for entry in row]))" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "$$ x^{(2)} = ({\\rm \"False\"}, 3, 2, {\\rm \"Sołacz\"}, 62), \\quad x_3^{(2)} = 2 $$" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "### Hipoteza" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "W naszym przypadku (wybraliśmy 5 cech):\n", "\n", "$$ h_\\theta(x) = \\theta_0 + \\theta_1 x_1 + \\theta_2 x_2 + \\theta_3 x_3 + \\theta_4 x_4 + \\theta_5 x_5 $$" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "W ogólności ($n$ cech):\n", "\n", "$$ h_\\theta(x) = \\theta_0 + \\theta_1 x_1 + \\theta_2 x_2 + \\ldots + \\theta_n x_n $$" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "subslide" } }, "source": [ "Jeżeli zdefiniujemy $x_0 = 1$, będziemy mogli powyższy wzór zapisać w bardziej kompaktowy sposób:\n", "\n", "$$\n", "\\begin{array}{rcl}\n", "h_\\theta(x)\n", " & = & \\theta_0 x_0 + \\theta_1 x_1 + \\theta_2 x_2 + \\ldots + \\theta_n x_n \\\\\n", " & = & \\displaystyle\\sum_{i=0}^{n} \\theta_i x_i \\\\\n", " & = & \\theta^T \\, x \\\\\n", " & = & x^T \\, \\theta \\\\\n", "\\end{array}\n", "$$" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "### Metoda gradientu prostego – notacja macierzowa" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "notes" } }, "source": [ "Metoda gradientu prostego przyjmie bardzo elegancką formę, jeżeli do jej zapisu użyjemy wektorów i macierzy." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "$$\n", "X=\\left[\\begin{array}{cc}\n", "1 & \\left( \\vec x^{(1)} \\right)^T \\\\\n", "1 & \\left( \\vec x^{(2)} \\right)^T \\\\\n", "\\vdots & \\vdots\\\\\n", "1 & \\left( \\vec x^{(m)} \\right)^T \\\\\n", "\\end{array}\\right] \n", "= \\left[\\begin{array}{cccc}\n", "1 & x_1^{(1)} & \\cdots & x_n^{(1)} \\\\\n", "1 & x_1^{(2)} & \\cdots & x_n^{(2)} \\\\\n", "\\vdots & \\vdots & \\ddots & \\vdots\\\\\n", "1 & x_1^{(m)} & \\cdots & x_n^{(m)} \\\\\n", "\\end{array}\\right]\n", "\\quad\n", "\\vec{y} = \n", "\\left[\\begin{array}{c}\n", "y^{(1)}\\\\\n", "y^{(2)}\\\\\n", "\\vdots\\\\\n", "y^{(m)}\\\\\n", "\\end{array}\\right]\n", "\\quad\n", "\\theta = \\left[\\begin{array}{c}\n", "\\theta_0\\\\\n", "\\theta_1\\\\\n", "\\vdots\\\\\n", "\\theta_n\\\\\n", "\\end{array}\\right]\n", "$$" ] }, { "cell_type": "code", "execution_count": 76, "metadata": { "slideshow": { "slide_type": "skip" } }, "outputs": [], "source": [ "# Wersje macierzowe funkcji rysowania wykresów punktowych oraz krzywej regresyjnej\n", "\n", "def hMx(theta, X):\n", " return X * theta\n", "\n", "def regdotsMx(X, y): \n", " fig = plt.figure(figsize=(16*.6, 9*.6))\n", " ax = fig.add_subplot(111)\n", " fig.subplots_adjust(left=0.1, right=0.9, bottom=0.1, top=0.9)\n", " ax.scatter([X[:, 1]], [y], c='r', s=50, label='Dane')\n", " \n", " ax.set_xlabel('Populacja')\n", " ax.set_ylabel('Zysk')\n", " ax.margins(.05, .05)\n", " plt.ylim(y.min() - 1, y.max() + 1)\n", " plt.xlim(np.min(X[:, 1]) - 1, np.max(X[:, 1]) + 1)\n", " return fig\n", "\n", "def reglineMx(fig, fun, theta, X):\n", " ax = fig.axes[0]\n", " x0, x1 = np.min(X[:, 1]), np.max(X[:, 1])\n", " L = [x0, x1]\n", " LX = np.matrix([1, x0, 1, x1]).reshape(2, 2)\n", " ax.plot(L, fun(theta, LX), linewidth='2',\n", " label=(r'$y={theta0:.2}{op}{theta1:.2}x$'.format(\n", " theta0=float(theta[0][0]),\n", " theta1=(float(theta[1][0]) if theta[1][0] >= 0 else float(-theta[1][0])),\n", " op='+' if theta[1][0] >= 0 else '-')))" ] }, { "cell_type": "code", "execution_count": 77, "metadata": { "slideshow": { "slide_type": "notes" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "[[ 1. 3. 1. 78.]\n", " [ 1. 3. 2. 62.]\n", " [ 1. 3. 0. 15.]\n", " [ 1. 4. 0. 14.]\n", " [ 1. 3. 0. 15.]]\n", "(1339, 4)\n", "\n", "[[476118.]\n", " [459531.]\n", " [411557.]\n", " [496416.]\n", " [406032.]]\n", "(1339, 1)\n" ] } ], "source": [ "# Wczytwanie danych z pliku za pomocą numpy – regresja liniowa wielu zmiennych – notacja macierzowa\n", "\n", "import pandas\n", "\n", "data = pandas.read_csv('data02_train.tsv', delimiter='\\t', usecols=['price', 'rooms', 'floor', 'sqrMetres'])\n", "m, n_plus_1 = data.values.shape\n", "n = n_plus_1 - 1\n", "Xn = data.values[:, 1:].reshape(m, n)\n", "\n", "# Dodaj kolumnę jedynek do macierzy\n", "XMx = np.matrix(np.concatenate((np.ones((m, 1)), Xn), axis=1)).reshape(m, n_plus_1)\n", "yMx = np.matrix(data.values[:, 0]).reshape(m, 1)\n", "\n", "print(XMx[:5])\n", "print(XMx.shape)\n", "\n", "print()\n", "\n", "print(yMx[:5])\n", "print(yMx.shape)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "subslide" } }, "source": [ "### Funkcja kosztu – notacja macierzowa" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "$$J(\\theta)=\\dfrac{1}{2|\\vec y|}\\left(X\\theta-\\vec{y}\\right)^T\\left(X\\theta-\\vec{y}\\right)$$ \n" ] }, { "cell_type": "code", "execution_count": 78, "metadata": { "slideshow": { "slide_type": "notes" } }, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle \\Large J(\\theta) = 85104141370.9717$" ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "from IPython.display import display, Math, Latex\n", "\n", "def JMx(theta,X,y):\n", " \"\"\"Wersja macierzowa funkcji kosztu\"\"\"\n", " m = len(y)\n", " J = 1.0 / (2.0 * m) * ((X * theta - y) . T * ( X * theta - y))\n", " return J.item()\n", "\n", "thetaMx = np.matrix([10, 90, -1, 2.5]).reshape(4, 1) \n", "\n", "cost = JMx(thetaMx,XMx,yMx) \n", "display(Math(r'\\Large J(\\theta) = %.4f' % cost))" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "subslide" } }, "source": [ "### Gradient – notacja macierzowa" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "$$\\nabla J(\\theta) = \\frac{1}{|\\vec y|} X^T\\left(X\\theta-\\vec y\\right)$$" ] }, { "cell_type": "code", "execution_count": 79, "metadata": { "slideshow": { "slide_type": "notes" } }, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle \\large \\theta = \\left[\\begin{array}{r}10.0000 \\\\ 90.0000 \\\\ -1.0000 \\\\ 2.5000 \\\\ \\end{array}\\right]\\quad\\large \\nabla J(\\theta) = \\left[\\begin{array}{r}-373492.7442 \\\\ -1075656.5086 \\\\ -989554.4921 \\\\ -23806475.6561 \\\\ \\end{array}\\right]$" ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "from IPython.display import display, Math, Latex\n", "\n", "def dJMx(theta,X,y):\n", " \"\"\"Wersja macierzowa gradientu funckji kosztu\"\"\"\n", " return 1.0 / len(y) * (X.T * (X * theta - y)) \n", "\n", "thetaMx = np.matrix([10, 90, -1, 2.5]).reshape(4, 1) \n", "\n", "display(Math(r'\\large \\theta = ' + LatexMatrix(thetaMx) + \n", " r'\\quad' + r'\\large \\nabla J(\\theta) = ' \n", " + LatexMatrix(dJMx(thetaMx,XMx,yMx))))" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "subslide" } }, "source": [ "### Algorytm gradientu prostego – notacja macierzowa" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "$$ \\theta := \\theta - \\alpha \\, \\nabla J(\\theta) $$" ] }, { "cell_type": "code", "execution_count": 80, "metadata": { "slideshow": { "slide_type": "notes" } }, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle \\large\\textrm{Wynik:}\\quad \\theta = \\left[\\begin{array}{r}17446.2104 \\\\ 86476.7968 \\\\ -1374.8949 \\\\ 2165.0689 \\\\ \\end{array}\\right] \\quad J(\\theta) = 10324864803.0591 \\quad \\textrm{po 374576 iteracjach}$" ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "# Implementacja algorytmu gradientu prostego za pomocą numpy i macierzy\n", "\n", "def GDMx(fJ, fdJ, theta, X, y, alpha, eps):\n", " current_cost = fJ(theta, X, y)\n", " log = [[current_cost, theta]]\n", " while True:\n", " theta = theta - alpha * fdJ(theta, X, y) # implementacja wzoru\n", " current_cost, prev_cost = fJ(theta, X, y), current_cost\n", " if abs(prev_cost - current_cost) <= eps:\n", " break\n", " if current_cost > prev_cost:\n", " print('Długość kroku (alpha) jest zbyt duża!')\n", " break\n", " log.append([current_cost, theta])\n", " return theta, log\n", "\n", "thetaStartMx = np.zeros((n + 1, 1))\n", "\n", "# Zmieniamy wartości alpha (rozmiar kroku) oraz eps (kryterium stopu)\n", "thetaBestMx, log = GDMx(JMx, dJMx, thetaStartMx, \n", " XMx, yMx, alpha=0.0001, eps=0.1)\n", "\n", "######################################################################\n", "display(Math(r'\\large\\textrm{Wynik:}\\quad \\theta = ' + \n", " LatexMatrix(thetaBestMx) + \n", " (r' \\quad J(\\theta) = %.4f' % log[-1][0]) \n", " + r' \\quad \\textrm{po %d iteracjach}' % len(log))) " ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## 2.6. Metoda gradientu prostego w praktyce" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "subslide" } }, "source": [ "### Kryterium stopu" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "notes" } }, "source": [ "Algorytm gradientu prostego polega na wykonywaniu określonych kroków w pętli. Pytanie brzmi: kiedy należy zatrzymać wykonywanie tej pętli?\n", "\n", "W każdej kolejnej iteracji wartość funkcji kosztu maleje o coraz mniejszą wartość.\n", "Parametr `eps` określa, jaka wartość graniczna tej różnicy jest dla nas wystarczająca:\n", "\n", " * Im mniejsza wartość `eps`, tym dokładniejszy wynik, ale dłuższy czas działania algorytmu.\n", " * Im większa wartość `eps`, tym krótszy czas działania algorytmu, ale mniej dokładny wynik." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "Na wykresie zobaczymy porównanie regresji dla różnych wartości `eps`" ] }, { "cell_type": "code", "execution_count": 81, "metadata": { "slideshow": { "slide_type": "subslide" } }, "outputs": [ { "data": { "image/svg+xml": [ "\r\n", "\r\n", "\r\n", "\r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " 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" ] }, "metadata": { "needs_background": "light" }, "output_type": "display_data" } ], "source": [ "# Wczytwanie danych z pliku za pomocą numpy – wersja macierzowa\n", "data = np.loadtxt('data01_train.csv', delimiter=',')\n", "m, n_plus_1 = data.shape\n", "n = n_plus_1 - 1\n", "Xn = data[:, 0:n].reshape(m, n)\n", "\n", "# Dodaj kolumnę jedynek do macierzy\n", "XMx = np.matrix(np.concatenate((np.ones((m, 1)), Xn), axis=1)).reshape(m, n_plus_1)\n", "yMx = np.matrix(data[:, 1]).reshape(m, 1)\n", "\n", "thetaStartMx = np.zeros((2, 1))\n", "\n", "fig = regdotsMx(XMx, yMx)\n", "theta_e1, log1 = GDMx(JMx, dJMx, thetaStartMx, XMx, yMx, alpha=0.01, eps=0.01) # niebieska linia\n", "reglineMx(fig, hMx, theta_e1, XMx)\n", "theta_e2, log2 = GDMx(JMx, dJMx, thetaStartMx, XMx, yMx, alpha=0.01, eps=0.000001) # pomarańczowa linia\n", "reglineMx(fig, hMx, theta_e2, XMx)\n", "legend(fig)" ] }, { "cell_type": "code", "execution_count": 82, "metadata": { "slideshow": { "slide_type": "subslide" } }, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle \\theta_{10^{-2}} = \\left[\\begin{array}{r}0.0531 \\\\ 0.8365 \\\\ \\end{array}\\right]\\quad\\theta_{10^{-6}} = \\left[\\begin{array}{r}-3.4895 \\\\ 1.1786 \\\\ \\end{array}\\right]$" ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "display(Math(r'\\theta_{10^{-2}} = ' + LatexMatrix(theta_e1) +\n", " r'\\quad\\theta_{10^{-6}} = ' + LatexMatrix(theta_e2)))" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "### Długość kroku ($\\alpha$)" ] }, { "cell_type": "code", "execution_count": 83, "metadata": { "slideshow": { "slide_type": "notes" } }, "outputs": [], "source": [ "# Jak zmienia się koszt w kolejnych krokach w zależności od alfa\n", "\n", "def costchangeplot(logs):\n", " fig = plt.figure(figsize=(16*.6, 9*.6))\n", " ax = fig.add_subplot(111)\n", " fig.subplots_adjust(left=0.1, right=0.9, bottom=0.1, top=0.9)\n", " ax.set_xlabel('krok')\n", " ax.set_ylabel(r'$J(\\theta)$')\n", "\n", " X = np.arange(0, 500, 1)\n", " Y = [logs[step][0] for step in X]\n", " ax.plot(X, Y, linewidth='2', label=(r'$J(\\theta)$'))\n", " return fig\n", "\n", "def slide7(alpha):\n", " best_theta, log = gradient_descent(h, J, [0.0, 0.0], x, y, alpha=alpha, eps=0.0001)\n", " fig = costchangeplot(log)\n", " legend(fig)\n", "\n", "sliderAlpha1 = widgets.FloatSlider(min=0.01, max=0.03, step=0.001, value=0.02, description=r'$\\alpha$', width=300)" ] }, { "cell_type": "code", "execution_count": 84, "metadata": { "slideshow": { "slide_type": "subslide" } }, "outputs": [ { "data": { "application/vnd.jupyter.widget-view+json": { "model_id": "68bdd2c08dfc42cc9bf033cb4eb7e213", "version_major": 2, "version_minor": 0 }, "text/plain": [ "interactive(children=(FloatSlider(value=0.02, description='$\\\\alpha$', max=0.03, min=0.01, step=0.001), Button…" ] }, "metadata": {}, "output_type": "display_data" }, { "data": { "text/plain": [ "" ] }, "execution_count": 84, "metadata": {}, "output_type": "execute_result" } ], "source": [ "widgets.interact_manual(slide7, alpha=sliderAlpha1)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "## 2.7. Normalizacja danych" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Normalizacja danych to proces, który polega na dostosowaniu danych wejściowych w taki sposób, żeby ułatwić działanie algorytmowi gradientu prostego.\n", "\n", "Wyjaśnię to na przykladzie." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "Użyjemy danych z „Gratka flats challenge 2017”.\n", "\n", "Rozważmy model $h(x) = \\theta_0 + \\theta_1 x_1 + \\theta_2 x_2$, w którym cena mieszkania prognozowana jest na podstawie liczby pokoi $x_1$ i metrażu $x_2$:" ] }, { "cell_type": "code", "execution_count": 85, "metadata": { "slideshow": { "slide_type": "fragment" } }, "outputs": [ { "data": { "text/html": [ "
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priceroomssqrMetres
0476118.00378
1459531.00362
2411557.00315
3496416.00414
4406032.00315
5450026.00380
6571229.15239
7325000.00354
8268229.00290
9604836.00440
\n", "
" ], "text/plain": [ " price rooms sqrMetres\n", "0 476118.00 3 78\n", "1 459531.00 3 62\n", "2 411557.00 3 15\n", "3 496416.00 4 14\n", "4 406032.00 3 15\n", "5 450026.00 3 80\n", "6 571229.15 2 39\n", "7 325000.00 3 54\n", "8 268229.00 2 90\n", "9 604836.00 4 40" ] }, "execution_count": 85, "metadata": {}, "output_type": "execute_result" } ], "source": [ "# Wczytanie danych przy pomocy biblioteki pandas\n", "import pandas\n", "alldata = pandas.read_csv('data_flats.tsv', header=0, sep='\\t',\n", " usecols=['price', 'rooms', 'sqrMetres'])\n", "alldata[:10]" ] }, { "cell_type": "code", "execution_count": 86, "metadata": { "slideshow": { "slide_type": "notes" } }, "outputs": [], "source": [ "# Funkcja, która pokazuje wartości minimalne i maksymalne w macierzy X\n", "\n", "def show_mins_and_maxs(XMx):\n", " mins = np.amin(XMx, axis=0).tolist()[0] # wartości minimalne\n", " maxs = np.amax(XMx, axis=0).tolist()[0] # wartości maksymalne\n", " for i, (xmin, xmax) in enumerate(zip(mins, maxs)):\n", " display(Math(\n", " r'${:.2F} \\leq x_{} \\leq {:.2F}$'.format(xmin, i, xmax)))" ] }, { "cell_type": "code", "execution_count": 87, "metadata": { "slideshow": { "slide_type": "notes" } }, "outputs": [], "source": [ "# Przygotowanie danych\n", "\n", "import numpy as np\n", "\n", "%matplotlib inline\n", "\n", "data2 = np.matrix(alldata[['rooms', 'sqrMetres', 'price']])\n", "\n", "m, n_plus_1 = data2.shape\n", "n = n_plus_1 - 1\n", "Xn = data2[:, 0:n]\n", "\n", "XMx2 = np.matrix(np.concatenate((np.ones((m, 1)), Xn), axis=1)).reshape(m, n_plus_1)\n", "yMx2 = np.matrix(data2[:, -1]).reshape(m, 1) / 1000.0" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "subslide" } }, "source": [ "Cechy w danych treningowych przyjmują wartości z zakresu:" ] }, { "cell_type": "code", "execution_count": 88, "metadata": { "slideshow": { "slide_type": "fragment" } }, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle 1.00 \\leq x_0 \\leq 1.00$" ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" }, { "data": { "text/latex": [ "$\\displaystyle 2.00 \\leq x_1 \\leq 7.00$" ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" }, { "data": { "text/latex": [ "$\\displaystyle 12.00 \\leq x_2 \\leq 196.00$" ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "show_mins_and_maxs(XMx2)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "subslide" } }, "source": [ "Jak widzimy, $x_2$ przyjmuje wartości dużo większe niż $x_1$.\n", "Powoduje to, że wykres funkcji kosztu jest bardzo „spłaszczony” wzdłuż jednej z osi:" ] }, { "cell_type": "code", "execution_count": 89, "metadata": { "slideshow": { "slide_type": "notes" } }, "outputs": [], "source": [ "def contour_plot(X, y, rescale=10**8):\n", " theta0_vals = np.linspace(-100000, 100000, 100)\n", " theta1_vals = np.linspace(-100000, 100000, 100)\n", "\n", " J_vals = np.zeros(shape=(theta0_vals.size, theta1_vals.size))\n", " for t1, element in enumerate(theta0_vals):\n", " for t2, element2 in enumerate(theta1_vals):\n", " thetaT = np.matrix([1.0, element, element2]).reshape(3,1)\n", " J_vals[t1, t2] = JMx(thetaT, X, y) / rescale\n", " \n", " plt.figure()\n", " plt.contour(theta0_vals, theta1_vals, J_vals.T, np.logspace(-2, 3, 20))\n", " plt.xlabel(r'$\\theta_1$')\n", " plt.ylabel(r'$\\theta_2$')" ] }, { "cell_type": "code", "execution_count": 90, "metadata": { "slideshow": { "slide_type": "fragment" } }, "outputs": [ { "data": { "image/svg+xml": [ "\r\n", "\r\n", "\r\n", "\r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " 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" ] }, "metadata": { "needs_background": "light" }, "output_type": "display_data" } ], "source": [ "contour_plot(XMx2, yMx2, rescale=10**10)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "notes" } }, "source": [ "Jeżeli funkcja kosztu ma kształt taki, jak na powyższym wykresie, to łatwo sobie wyobrazić, że znalezienie minimum lokalnego przy użyciu metody gradientu prostego musi stanowć nie lada wyzwanie: algorytm szybko znajdzie „rynnę”, ale „zjazd” wzdłuż „rynny” w poszukiwaniu minimum będzie odbywał się bardzo powoli.\n", "\n", "Jak temu zaradzić?\n", "\n", "Spróbujemy przekształcić dane tak, żeby funkcja kosztu miała „ładny”, regularny kształt." ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "### Skalowanie" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "Będziemy dążyć do tego, żeby każda z cech przyjmowała wartości w podobnym zakresie.\n", "\n", "W tym celu przeskalujemy wartości każdej z cech, dzieląc je przez wartość maksymalną:\n", "\n", "$$ \\hat{x_i}^{(j)} := \\frac{x_i^{(j)}}{\\max_j x_i^{(j)}} $$" ] }, { "cell_type": "code", "execution_count": 91, "metadata": { "slideshow": { "slide_type": "subslide" } }, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle 1.00 \\leq x_0 \\leq 1.00$" ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" }, { "data": { "text/latex": [ "$\\displaystyle 0.29 \\leq x_1 \\leq 1.00$" ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" }, { "data": { "text/latex": [ "$\\displaystyle 0.06 \\leq x_2 \\leq 1.00$" ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "XMx2_scaled = XMx2 / np.amax(XMx2, axis=0)\n", "\n", "show_mins_and_maxs(XMx2_scaled)" ] }, { "cell_type": "code", "execution_count": 92, "metadata": { "slideshow": { "slide_type": "subslide" } }, "outputs": [ { "data": { "image/svg+xml": [ 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\r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", "\r\n" ], "text/plain": [ "
" ] }, "metadata": { "needs_background": "light" }, "output_type": "display_data" } ], "source": [ "contour_plot(XMx2_scaled, yMx2)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "### Normalizacja średniej" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "fragment" } }, "source": [ "Będziemy dążyć do tego, żeby dodatkowo średnia wartość każdej z cech była w okolicach $0$.\n", "\n", "W tym celu oprócz przeskalowania odejmiemy wartość średniej od wartości każdej z cech:\n", "\n", "$$ \\hat{x_i}^{(j)} := \\frac{x_i^{(j)} - \\mu_i}{\\max_j x_i^{(j)}} $$" ] }, { "cell_type": "code", "execution_count": 93, "metadata": { "slideshow": { "slide_type": "subslide" } }, "outputs": [ { "data": { "text/latex": [ "$\\displaystyle 0.00 \\leq x_0 \\leq 0.00$" ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" }, { "data": { "text/latex": [ "$\\displaystyle -0.10 \\leq x_1 \\leq 0.62$" ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" }, { "data": { "text/latex": [ "$\\displaystyle -0.23 \\leq x_2 \\leq 0.70$" ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "XMx2_norm = (XMx2 - np.mean(XMx2, axis=0)) / np.amax(XMx2, axis=0)\n", "\n", "show_mins_and_maxs(XMx2_norm)" ] }, { "cell_type": "code", "execution_count": 94, "metadata": { "slideshow": { "slide_type": "subslide" } }, "outputs": [ { "data": { "image/svg+xml": [ "\r\n", "\r\n", "\r\n", "\r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", "\r\n" ], "text/plain": [ "
" ] }, "metadata": { "needs_background": "light" }, "output_type": "display_data" } ], "source": [ "contour_plot(XMx2_norm, yMx2)" ] }, { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "notes" } }, "source": [ "Teraz funkcja kosztu ma wykres o bardzo regularnym kształcie – algorytm gradientu prostego zastosowany w takim przypadku bardzo szybko znajdzie minimum funkcji kosztu." ] } ], "metadata": { "celltoolbar": "Slideshow", "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.8.3" }, "livereveal": { "start_slideshow_at": "selected", "theme": "white" } }, "nbformat": 4, "nbformat_minor": 4 }