forked from filipg/aitech-eks-pub
303 lines
8.3 KiB
Plaintext
303 lines
8.3 KiB
Plaintext
{
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"cells": [
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{
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"cell_type": "markdown",
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"id": "cathedral-newark",
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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": "heard-clinton",
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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": "preceding-impression",
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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 w m$^2$ ($x_1 = 32.3$) $w_1 = 7000$\n",
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"\n",
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"* liczby pokoi ($x_2 = 3$) $w_2 = -30000$\n",
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" \n",
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"* nr piętra ($x_3 = 4$) \n",
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"\n",
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"* wieku ($x_4 = 13$) $w_3 = -1000$\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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"* wielkość miasta\n",
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"\n",
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"* gęstość zaludnienia\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? ($x_7 = 1$) $w_7 = 5000$\n",
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"\n",
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" * czy jest grzyb? ($x_8 = 0$) $w_8 = -40000$\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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" * czy to Kraków ($x_{10} = 0$)\n",
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" \n",
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" * czy to Katowice ($x_{11} = 0$)\n",
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" \n",
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" * czy obok budynku jest parking \n",
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" \n",
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" * czy w budynku jest parking\n",
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"\n",
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"* zakodowany opis \n",
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"\n",
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" * $(x_{12}, x_{|V|+12})$ - wektor tf-idf \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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"$$ y = w_0 + w_1x_1 + \\ldots + w_kx_k = w_0 + \\sum_{j=1}^{k} w_jx_j = w_0 + \\vec{w}\\vec{x}$$\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": "confused-increase",
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"metadata": {},
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"source": [
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"## Kilka spostrzeżeń\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": "freelance-controversy",
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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": "divine-medium",
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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": "supreme-tennessee",
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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": "markdown",
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"id": "seasonal-syndication",
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"metadata": {},
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"source": [
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"### Przykład \n",
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"\n",
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"Wyzwanie RetroC2 - odgadywanie roku dla krótkiego tekstu (1814-2013), <https://gonito.net/challenge/retroc2>.\n",
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" \n",
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"Lista słów (obcięta do 7 znaków) z największą/najmniejszymi wagami. \n",
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"\n",
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"```\n",
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"wzbudze -0.08071490\n",
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"paczka -0.08000180\n",
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"szarpi -0.05906200\n",
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"spadoch -0.05784140\n",
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"rzymsko -0.05466660\n",
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"sosnowy -0.05162170\n",
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"dębowyc -0.04778910\n",
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"nawinię -0.04649400\n",
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"odmówie -0.04522140\n",
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"zacisko -0.04480620\n",
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"funkcją -0.04479500\n",
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"werben -0.04423350\n",
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"nieumyś -0.04415200\n",
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"wodomie -0.04351570\n",
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"szczote -0.04313390\n",
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"exekucy -0.04297940\n",
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"listew -0.04214090\n",
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"daley -0.04145400\n",
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"metro -0.04080110\n",
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"wyjąwsz -0.04078060\n",
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"salda -0.04042050\n",
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"tkach -0.04020180\n",
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"cetnar -0.03999050\n",
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"zgóry -0.03855980\n",
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"belek -0.03833100\n",
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"formier -0.03805890\n",
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"wekslu -0.03796510\n",
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"odmową -0.03753760\n",
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"\n",
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"odwadni 0.04662140\n",
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"dozując 0.04672770\n",
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"wyników 0.04744650\n",
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"sprawst 0.04746330\n",
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"jakub 0.04750710\n",
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"ścieran 0.04791070\n",
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"wrodzon 0.04799800\n",
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"koryguj 0.04843560\n",
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"odnotow 0.04854360\n",
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"tłumiąc 0.04917320\n",
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"leasing 0.04963200\n",
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"ecznej 0.04994810\n",
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"2013r 0.05009500\n",
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"kompens 0.05049060\n",
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"comarch 0.05058620\n",
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"pojazde 0.05078540\n",
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"badanyc 0.05340480\n",
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"kontakc 0.05377990\n",
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"sygnali 0.05601120\n",
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"piasta 0.05658670\n",
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"2000r 0.05716820\n",
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"stropni 0.06123470\n",
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"oszone 0.06124600\n",
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"zamonto 0.06424310\n",
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"……….. 0.06498500\n",
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"kumulat 0.06596770\n",
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"faktura 0.07313080\n",
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"wielost 0.09677770\n",
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"wielomi 0.12307300\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": "encouraging-martial",
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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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"nbformat": 4,
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"nbformat_minor": 5
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