随机森林 (Random Forest)

最后发布时间:2023-04-23 14:10:21 浏览量:

在随机森林方法中，创建了大量决策树。每个观察结果都被输入到每个决策树中。每个观察的最常见结果用作最终输出。新的观察结果被输入到所有树中，并对每个分类模型进行多数表决。

数据准备

加载R包

library(randomForest)

这里使用R语言的自带数据集iris，选择versicolor和virginica两种类型用于二分类问题

data(iris)
iris <- dplyr::filter(iris,Species %in% c("versicolor","virginica"))
iris$Species<-as.factor(as.character(iris$Species))
head(iris)

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使用训练集建立模型，并查看训练集的混淆矩阵和ROC曲线

set.seed(2022)
iris.rf<-randomForest(Species ~ ., data=iris,ntree=10,importance=T,proximity=T)
print(iris.rf)

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查看混淆矩阵

data.frame(predicted =iris.rf$predicted,
        group = iris$Species,
        stringsAsFactors = F)|> table()

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data.frame(predicted =predict(iris.rf, newdata = iris,type="response"),
        group = iris$Species,
        stringsAsFactors = F)|> table()

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绘制ROC曲线只需要简单的下面一行代码

library(pROC)
plot.roc(iris$Species,iris.rf$votes[,2])

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下面的ROC曲线添加了图例和颜色

library(RColorBrewer)
mycol<-brewer.pal(11, "Spectral")
x <- plot.roc(iris$Species,iris.rf$votes[,2],ylim=c(0,1),xlim=c(1,0),
                smooth=FALSE, 
                ci=TRUE, 
                col=mycol[1],
                lwd=2, 
                legacy.axes=T)
                
ci.lower <- round(as.numeric(x$ci[1]),3) 
ci.upper <- round(as.numeric(x$ci[3]),3) 

legend.name <- paste("AUC:",round(as.numeric(x$auc),3),paste(ci.lower,ci.upper,sep="-"))

legend("bottomright", 
         legend=legend.name,
         col = mycol[1],
         lwd = 2,
         bty="n")

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使用训练集建立模型，并查看测试集的混淆矩阵和ROC曲线

将数据集划分成70%的训练集和30%的测试集

library(caTools)
set.seed(2022)
sample <- sample.split(iris$Species, SplitRatio = 0.7)
train  <- subset(iris, sample == TRUE)
test   <- subset(iris, sample == FALSE)

建立二分类器的随机扥林模型

iris.rf<-randomForest(Species ~ ., data=train,ntree=10)
print(iris.rf)
iris.rf$importance

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查看混淆矩阵

data.frame(predicted =predict(iris.rf, newdata = test,type="response"),
        group = test$Species,
        stringsAsFactors = F)|> table()

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绘制ROC曲线

library(pROC)
plot.roc(test$Species,predict(iris.rf, newdata = test,type="vote")[,2])

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使用交叉验证建立随机森林的二分类模型

将数据集划分为大小相同的k份
每一次将其中一份作为测试集，剩余的k-1份作为训练集
以k次测试结果的平均值作为最终的测试误差

cv_kfold <- function (data,k=10,seed=2022){
  n_row <- nrow(data)
  n_foldmarkers <- rep(1:k, ceiling(n_row/k))[1:n_row]
  set.seed(seed)
  n_foldmarkers <- sample(n_foldmarkers)
  k_fold <- lapply(1:k,function (i){
    (1:n_row)[n_foldmarkers==i]
  })
  return(k_fold)
}

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res <- data.frame()
for (i in 1:length(kfolds)){
    curr_fold <- kfolds[[i]]
    train_set <- iris[-curr_fold,]
    test_set <- iris[curr_fold,]
    predicted_train <-randomForest(Species ~ ., data=train_set,ntree=10)    
    train_accuracy <- sum(train_set$Species ==predict(predicted_train, newdata = train_set,type="response")) / length(train_set$Species)
    test_accuracy <- sum(test_set$Species ==predict(predicted_train, newdata = test_set,type="response") )/length(test_set$Species)
    res <- rbind(res,data.frame(train_accuracy=train_accuracy, test_accuracy=test_accuracy))
}
res

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Mean Decrease Accuracy and Mean Decrease Gini

https://stackoverflow.com/questions/55036416/why-is-mean-decrease-gini-in-random-forest-dependent-on-population-size/55055245#55055245

varImpPlot(model.rf)

model.rf$importance

set.seed(1)

n <- 1000
# There are three classes in equal proportions
a <- rep(c(-10,0,10), each = n)
# One feature is useless
b <- rnorm(3*n)
# The other feature is highly predictive but we need at least two splits
c <- rnorm(3*n, a)
data <- data.frame(a = as.factor(a), b = b, c = c)
head(data)

rf3 <- randomForest(data = data, a ~ b + c, importance = TRUE,
                    ntree = 1000, mtry = 2, maxnodes = 2)
rf3$importance

rf3 <- randomForest(data = data, a ~ b + c, importance = TRUE,
                    ntree = 1000, mtry = 2, maxnodes = 3)
rf3$importance

关于ROC曲线上的每一个点

library(caTools)
set.seed(2022)
sample <- sample.split(iris$Species, SplitRatio = 0.9)
train  <- subset(iris, sample == TRUE)
test   <- subset(iris, sample == FALSE)
iris.rf<-randomForest(Species ~ ., data=train,ntree=10)
prob <- predict(iris.rf, newdata = test,type="vote")
predicted <- predict(iris.rf, newdata = test,type="response")
cbind(test,predicted, prob)