Bosques Aleatorios
Miguel Angel Lopez
2025-08-20
Teoría
El Bosque Aleatorio es un algoritmo de aprendizaje automatico que combina el resultado de multiples arboles de decision para llegar a un resultado optimo.
Ejemplo 1 . Melbourne
En esta base de datos tenemos los precios de mas de 13,000 casas en la ciudad de Melbourne:
Instalar paquetes y llamar librerias
library(tidyverse)## ── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ──
## ✔ dplyr 1.1.4 ✔ readr 2.1.5
## ✔ forcats 1.0.0 ✔ stringr 1.5.1
## ✔ ggplot2 3.5.2 ✔ tibble 3.3.0
## ✔ lubridate 1.9.4 ✔ tidyr 1.3.1
## ✔ purrr 1.1.0
## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag() masks stats::lag()
## ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errorslibrary(rpart)
library(randomForest)## randomForest 4.7-1.2
## Type rfNews() to see new features/changes/bug fixes.
##
## Attaching package: 'randomForest'
##
## The following object is masked from 'package:dplyr':
##
## combine
##
## The following object is masked from 'package:ggplot2':
##
## marginlibrary(rpart.plot)
library(modelr)
library(caret)## Loading required package: lattice
##
## Attaching package: 'caret'
##
## The following object is masked from 'package:purrr':
##
## liftCargar Base de datos
df <- read.csv("/Users/miguel/Desktop/melbourne.csv")Entender Base de datos
summary(df)## Suburb Address Rooms Type
## Length:13580 Length:13580 Min. : 1.000 Length:13580
## Class :character Class :character 1st Qu.: 2.000 Class :character
## Mode :character Mode :character Median : 3.000 Mode :character
## Mean : 2.938
## 3rd Qu.: 3.000
## Max. :10.000
##
## Price Method SellerG Date
## Min. : 85000 Length:13580 Length:13580 Length:13580
## 1st Qu.: 650000 Class :character Class :character Class :character
## Median : 903000 Mode :character Mode :character Mode :character
## Mean :1075684
## 3rd Qu.:1330000
## Max. :9000000
##
## Distance Postcode Bedroom2 Bathroom
## Min. : 0.00 Min. :3000 Min. : 0.000 Min. :0.000
## 1st Qu.: 6.10 1st Qu.:3044 1st Qu.: 2.000 1st Qu.:1.000
## Median : 9.20 Median :3084 Median : 3.000 Median :1.000
## Mean :10.14 Mean :3105 Mean : 2.915 Mean :1.534
## 3rd Qu.:13.00 3rd Qu.:3148 3rd Qu.: 3.000 3rd Qu.:2.000
## Max. :48.10 Max. :3977 Max. :20.000 Max. :8.000
##
## Car Landsize BuildingArea YearBuilt
## Min. : 0.00 Min. : 0.0 Min. : 0 Min. :1196
## 1st Qu.: 1.00 1st Qu.: 177.0 1st Qu.: 93 1st Qu.:1940
## Median : 2.00 Median : 440.0 Median : 126 Median :1970
## Mean : 1.61 Mean : 558.4 Mean : 152 Mean :1965
## 3rd Qu.: 2.00 3rd Qu.: 651.0 3rd Qu.: 174 3rd Qu.:1999
## Max. :10.00 Max. :433014.0 Max. :44515 Max. :2018
## NA's :62 NA's :6450 NA's :5375
## CouncilArea Lattitude Longtitude Regionname
## Length:13580 Min. :-38.18 Min. :144.4 Length:13580
## Class :character 1st Qu.:-37.86 1st Qu.:144.9 Class :character
## Mode :character Median :-37.80 Median :145.0 Mode :character
## Mean :-37.81 Mean :145.0
## 3rd Qu.:-37.76 3rd Qu.:145.1
## Max. :-37.41 Max. :145.5
##
## Propertycount
## Min. : 249
## 1st Qu.: 4380
## Median : 6555
## Mean : 7454
## 3rd Qu.:10331
## Max. :21650
## str(df)## 'data.frame': 13580 obs. of 21 variables:
## $ Suburb : chr "Abbotsford" "Abbotsford" "Abbotsford" "Abbotsford" ...
## $ Address : chr "85 Turner St" "25 Bloomburg St" "5 Charles St" "40 Federation La" ...
## $ Rooms : int 2 2 3 3 4 2 3 2 1 2 ...
## $ Type : chr "h" "h" "h" "h" ...
## $ Price : num 1480000 1035000 1465000 850000 1600000 ...
## $ Method : chr "S" "S" "SP" "PI" ...
## $ SellerG : chr "Biggin" "Biggin" "Biggin" "Biggin" ...
## $ Date : chr "3/12/2016" "4/02/2016" "4/03/2017" "4/03/2017" ...
## $ Distance : num 2.5 2.5 2.5 2.5 2.5 2.5 2.5 2.5 2.5 2.5 ...
## $ Postcode : num 3067 3067 3067 3067 3067 ...
## $ Bedroom2 : num 2 2 3 3 3 2 4 2 1 3 ...
## $ Bathroom : num 1 1 2 2 1 1 2 1 1 1 ...
## $ Car : num 1 0 0 1 2 0 0 2 1 2 ...
## $ Landsize : num 202 156 134 94 120 181 245 256 0 220 ...
## $ BuildingArea : num NA 79 150 NA 142 NA 210 107 NA 75 ...
## $ YearBuilt : num NA 1900 1900 NA 2014 ...
## $ CouncilArea : chr "Yarra" "Yarra" "Yarra" "Yarra" ...
## $ Lattitude : num -37.8 -37.8 -37.8 -37.8 -37.8 ...
## $ Longtitude : num 145 145 145 145 145 ...
## $ Regionname : chr "Northern Metropolitan" "Northern Metropolitan" "Northern Metropolitan" "Northern Metropolitan" ...
## $ Propertycount: num 4019 4019 4019 4019 4019 ...df<- na.omit(df)Arbol de decisión
library(rpart)
arbol <- rpart(Price ~ Rooms + Distance + Bedroom2 + Bathroom + Car +
Landsize + BuildingArea + Propertycount, data = df)
plot(arbol, uniform = TRUE)
text(arbol, cex = 0.5)
predict(arbol,head(df))## 2 3 5 7 8 10
## 561850.9 1159988.6 1159988.6 2422139.7 833730.6 561850.9head(df$Price)## [1] 1035000 1465000 1600000 1876000 1636000 1097000prueba_arbol <- head(df)
# MAE: error cuadrado promedio (Ventaja: Mismas unidades)
mae_arbol <- mae(arbol, prueba_arbol)Bosque Aleatorio
set.seed(123)
renglones_entrenamiento <- createDataPartition (df$Price, p=0.8,
list=FALSE)
entrenamiento <- df[renglones_entrenamiento, ]
prueba <- df[-renglones_entrenamiento, ]
rf <- randomForest(Price ~ Rooms + Distance + Bedroom2 + Bathroom + Car +
Landsize + BuildingArea + Propertycount, data = entrenamiento, ntree=500, mtry=3, importance =TRUE)
resultado_entrenamiento <- predict(rf,entrenamiento)
resultado_prueba <- predict(rf,prueba)
mae_rf <- mae(rf,prueba)
resultados <- tibble(Modelo = c("Arbol de decisión", "Bosque Aleatorio"),
MAE = c(mae_arbol, mae_rf))Ejercicio 1. Redimiento Automotriz
Importar basde de datos
df2 <-mtcarsEntender Base de datos
df2 <- mtcars
summary(df2)## mpg cyl disp hp
## Min. :10.40 Min. :4.000 Min. : 71.1 Min. : 52.0
## 1st Qu.:15.43 1st Qu.:4.000 1st Qu.:120.8 1st Qu.: 96.5
## Median :19.20 Median :6.000 Median :196.3 Median :123.0
## Mean :20.09 Mean :6.188 Mean :230.7 Mean :146.7
## 3rd Qu.:22.80 3rd Qu.:8.000 3rd Qu.:326.0 3rd Qu.:180.0
## Max. :33.90 Max. :8.000 Max. :472.0 Max. :335.0
## drat wt qsec vs
## Min. :2.760 Min. :1.513 Min. :14.50 Min. :0.0000
## 1st Qu.:3.080 1st Qu.:2.581 1st Qu.:16.89 1st Qu.:0.0000
## Median :3.695 Median :3.325 Median :17.71 Median :0.0000
## Mean :3.597 Mean :3.217 Mean :17.85 Mean :0.4375
## 3rd Qu.:3.920 3rd Qu.:3.610 3rd Qu.:18.90 3rd Qu.:1.0000
## Max. :4.930 Max. :5.424 Max. :22.90 Max. :1.0000
## am gear carb
## Min. :0.0000 Min. :3.000 Min. :1.000
## 1st Qu.:0.0000 1st Qu.:3.000 1st Qu.:2.000
## Median :0.0000 Median :4.000 Median :2.000
## Mean :0.4062 Mean :3.688 Mean :2.812
## 3rd Qu.:1.0000 3rd Qu.:4.000 3rd Qu.:4.000
## Max. :1.0000 Max. :5.000 Max. :8.000str(df2)## 'data.frame': 32 obs. of 11 variables:
## $ mpg : num 21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
## $ cyl : num 6 6 4 6 8 6 8 4 4 6 ...
## $ disp: num 160 160 108 258 360 ...
## $ hp : num 110 110 93 110 175 105 245 62 95 123 ...
## $ drat: num 3.9 3.9 3.85 3.08 3.15 2.76 3.21 3.69 3.92 3.92 ...
## $ wt : num 2.62 2.88 2.32 3.21 3.44 ...
## $ qsec: num 16.5 17 18.6 19.4 17 ...
## $ vs : num 0 0 1 1 0 1 0 1 1 1 ...
## $ am : num 1 1 1 0 0 0 0 0 0 0 ...
## $ gear: num 4 4 4 3 3 3 3 4 4 4 ...
## $ carb: num 4 4 1 1 2 1 4 2 2 4 ...df2 <- na.omit(df2)Arbol de decisión
# Árbol de decisión SIN las variables vs y am (objetivo: mpg)
library(rpart)
arbol2 <- rpart(mpg ~ . - vs - am, data = df2)
plot(arbol2, uniform = TRUE)
text(arbol2, cex = 0.5)
# Predicción y comparación rápida
predict(arbol2, head(df2))## Mazda RX4 Mazda RX4 Wag Datsun 710 Hornet 4 Drive
## 18.26429 18.26429 26.66364 18.26429
## Hornet Sportabout Valiant
## 18.26429 18.26429head(df2$mpg)## [1] 21.0 21.0 22.8 21.4 18.7 18.1prueba_arbol2 <- head(df2)
# MAE: error absoluto medio (mismas unidades que mpg)
mae_arbol2 <- mean(abs(prueba_arbol2$mpg - predict(arbol2, prueba_arbol2)))
mae_arbol2## [1] 2.178463Bosque Aleatorio
set.seed(123)
# Partición 80/20 usando la variable objetivo mpg
renglones_entrenamiento <- createDataPartition(df2$mpg, p = 0.8, list = FALSE)
entrenamiento <- df2[renglones_entrenamiento, ]
prueba <- df2[-renglones_entrenamiento, ]
# Bosque Aleatorio EXCLUYENDO vs y am
rf2 <- randomForest(mpg ~ . - vs - am,
data = entrenamiento,
ntree = 500, mtry = 3, importance = TRUE)
# Predicciones
pred_train2 <- predict(rf2, entrenamiento)
pred_test2 <- predict(rf2, prueba)
# MAE (definido manualmente)
mae_rf2_train <- mean(abs(entrenamiento$mpg - pred_train2))
mae_rf2_test <- mean(abs(prueba$mpg - pred_test2))
# Si ya calculaste el MAE del árbol para mtcars como 'mae_arbol2', compáralo aquí:
# (de lo contrario, puedes dejar mae_arbol2 <- NA)
# mae_arbol2 <- ... # del árbol arbol2 que hicimos antes
mae_arbol2 <- if (exists("mae_arbol2")) mae_arbol2 else NA_real_
resultados <- tibble(
Modelo = c("Arbol de decisión", "Bosque Aleatorio (test)"),
MAE = c(mae_arbol2, mae_rf2_test)
)
resultados## # A tibble: 2 × 2
## Modelo MAE
## <chr> <dbl>
## 1 Arbol de decisión 2.18
## 2 Bosque Aleatorio (test) 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