215 lines
5.8 KiB
Plaintext
215 lines
5.8 KiB
Plaintext
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{
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"cells": [
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{
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"cell_type": "markdown",
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"id": "continent-intermediate",
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"metadata": {},
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"source": [
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"# Regresja liniowa\n",
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"\n",
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"Regresja liniowa jest prosta...\n",
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"\n",
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"\n",
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"\n",
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"... dosłownie — dopasuj prostą $y = ax + b$ do punktów\n",
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"\n",
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"Należy odgadnąć $a$ i $b$ tak, aby zminimalizować błąd\n",
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"kwadratowy, tj. wartość:\n",
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"\n",
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"$$\\sum_{i=1}^n (y_i - (ax_i + b))^2$$\n",
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"\n",
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"\n"
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]
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},
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{
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"cell_type": "markdown",
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"id": "original-speed",
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"metadata": {},
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"source": [
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"Regresje liniowa (jednej zmiennej) jest łatwa do rozwiązania — wystarczy podstawić do wzoru!\n",
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"\n",
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"$$\\hat{b} = \\frac{ \\sum_{i=1}^{n}{x_i y_i} - \\frac{1}{n} \\sum_{i=1}^n x_i\n",
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" \\sum_{j=1}^n y_j}{ \\sum_{i=1}^n {x_i^2} - \\frac{1}{n} (\\sum_{i=1}^n\n",
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" x_i)^2 }$$\n",
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"\n",
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"$$\\hat{a} = \\bar{y} - \\hat{b}\\,\\bar{x}$$\n",
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"\n",
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"\n",
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"Na przykład dla mieszkań: $b =$ -30809.203 zł, $a =$ 5733.693 zł/m$^2$.\n",
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"\n",
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"\n",
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"\n"
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]
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},
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{
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"cell_type": "markdown",
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"id": "significant-relaxation",
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"metadata": {},
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"source": [
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"## Regresja wielu zmiennych\n",
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"\n",
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"W praktyce mamy do czynienia z **wielowymiarową** regresją\n",
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"liniową.\n",
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"\n",
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"Cena mieszkań może być prognozowana na podstawie:\n",
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"\n",
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"* powierzchni ($x_1 = 32.3$) \n",
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"\n",
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"* liczby pokoi ($x_2 = 3$)\n",
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" \n",
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"* piętra ($x_3 = 4$)\n",
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"\n",
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"* wieku ($x_4 = 13$)\n",
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"\n",
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"* odległości od Dworca Centralnego w Warszawie ($x_5 = 371.3$)\n",
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"\n",
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"* cech zerojedynkowych:\n",
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"\n",
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" * czy wielka płyta? ($x_6 = 0$)\n",
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"\n",
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" * czy jest jacuzzi? ($x_7 = 1$)\n",
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"\n",
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" * czy jest grzyb? ($x_8 = 0$)\n",
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"\n",
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" * czy to Kielce? ($x_9 = 1$)\n",
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"\n",
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"* ...\n",
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"\n",
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"... więc uogólniamy na wiele ($k$) wymiarów:\n",
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"\n",
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"$$ y = w_0 + w_1x_1 + \\ldots + w_kx_k = w_0 + \\sum_{j=1}^{k} w_jx_j $$\n",
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"\n",
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"gdzie:\n",
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"\n",
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"* $x_1,\\dots,x_k$ -- zmienne, na podstawie których zgadujemy\n",
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"\n",
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"* $w_0, w_1,\\dots,w_k$ -- wagi modelu (do wymyślenia na\n",
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" podstawie przykładów)\n",
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"\n",
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"* $y$ -- odgadywana wartość\n",
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"\n",
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"Też istnieje wzór ładny wzór na wyliczenie wektora wag!\n",
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"\n",
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"$$\\mathbf{w} = (\\mathbf{X}^{\\rm T}\\mathbf{X})^{-1} \\mathbf{X}^{\\rm T}\\mathbf{y}$$\n",
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"\n",
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"... niestety odwracanie macierzy nie jest tanie :("
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]
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},
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{
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"cell_type": "markdown",
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"id": "ordinary-appendix",
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"metadata": {},
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"source": [
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"## Kilka sporzeżeń\n",
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"\n",
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"Regresja liniowa to najprostszy możliwy model:\n",
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"\n",
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"* im czegoś więcej na wejściu, tym proporcjonalnie (troszkę) więcej/mniej na wyjściu\n",
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"\n",
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"* nic prostszego nie da się wymyślić (funkcja stała??)\n",
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"\n",
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"* niestety model liniowy czasami kompletnie nie ma sensu (np. wzrost człowieka w\n",
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" stosunku do wieku)\n"
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]
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},
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{
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"cell_type": "markdown",
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"id": "egyptian-austria",
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"metadata": {},
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"source": [
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"## Uczenie\n",
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"\n",
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"A jak nauczyć się wag z przykładów?\n",
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"\n",
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"* wzór (z odwracaniem macierzy) — problematyczny\n",
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"\n",
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"### Metoda gradientu prostego\n",
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"\n",
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"\n",
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"\n",
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"Schodź wzdłuż lokalnego spadku funkcji błędu.\n",
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"\n",
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"Tak więc w praktyce zamiast podstawiać do wzoru lepiej się uczyć iteracyjnie -\n",
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" metodą **gradientu prostego** (ang. _gradient descent_).\n",
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"\n",
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"1. Zacznij od byle jakich wag $w_i$ (np. wylosuj)\n",
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"2. Weź losowy przykład uczący $x_1,\\dots,x_n$, $y$.\n",
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"3. Oblicz wyjście $\\hat{y}$ na podstawie $x_1,\\dots,x_n$.\n",
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"4. Oblicz funkcję błędu między $y$ a $\\hat{y}$.\n",
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"5. Zmodyfikuj lekko wagi $(w_i)$ w kierunku spadku funkcji błędu.\n",
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"6. Jeśli błąd jest duży, idź do 2.\n",
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"\n",
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"Modyfikacja wag:\n",
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"\n",
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"$$w_i := w_i - x_i (\\hat{y} - y) \\eta$$\n",
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"\n",
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"gdzie $\\eta$ to **współczynnik uczenia** _learning rate_.\n"
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]
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},
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{
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"cell_type": "markdown",
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"id": "exact-train",
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"metadata": {},
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"source": [
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"## Ewaluacja regresji\n",
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"\n",
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"To miary błędu (im mniej, tym lepiej!)}\n",
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"\n",
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"### Błąd bezwzględny (Mean Absolute Error, MAE)\n",
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"\n",
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"$$\\frac{1}{n}\\sum_{i=1}^n |\\hat{y}_i - y_i| $$\n",
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"\n",
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"### Mean Squared Error (MSE)\n",
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"\n",
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"$$\\frac{1}{n}\\sum_{i=1}^n (\\hat{y}_i - y_i)^2$$\n",
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"\n",
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"### Root Mean Squared Error (RMSE)\n",
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"\n",
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"$$\\sqrt{\\frac{1}{n}\\sum_{i=1}^n (\\hat{y}_i - y_i)^2}$$\n",
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"\n"
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]
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},
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{
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"cell_type": "markdown",
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"id": "selective-agriculture",
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"metadata": {},
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"source": [
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"## Regresja liniowa dla tekstu\n",
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"\n",
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"Czym jest wektor $\\vec{x} = (x_1,\\dots,x_n)$? Wiemy, np. reprezentacja tf-idf (być z trikiem z haszowaniem, Word2vec etc.).\n",
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"\n",
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"\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": null,
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"id": "numerous-limitation",
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"metadata": {},
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"outputs": [],
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"source": []
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.9.2"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 5
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}
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