如何对r中的LDA进行重采样？

Question

我认为使用bootstrap可以重新采样我的LDA，但我不确定。另外，如果bootstrap可以工作，我不确定如何在r中编写bootstrap。下面是我的LDA代码：

library('MASS') 
n=nrow(iris)
train = sample(n ,size = floor(n*0.75), replace = F)
train.species =Species[train]
test.species=Species[-train]
lda.fit = lda(Species~. , data=iris, subset=train)

Answer 1

下面的代码使用boot() library在iris数据集上使用LDA执行引导，以获得LD1和LD2系数的标准误差。此外，代码的初始部分显示了不具有相同系数的自举LDA拟合。

# Library
library(MASS)
library(boot)

# Get data
data(iris)
names(iris) <- gsub("\\.", "", names(iris)) #remove dots from column names

# Split data into train and test sets
train_index <-sample(seq(nrow(iris)),floor(dim(iris)[1]*0.75))
train <- iris[train_index,]
test <- iris[-train_index,]
test_Y <- test[, c('Species')]
test_X <- subset(test, select=-c(Species))

#### LDA without bootstrap: 
# Fit LDA to train data: 
lda.fit = lda(Species ~ . , data=train)
lda.fit
# Predict test_Y based on lda.fit above
lda.pred <- predict(lda.fit, test_X)
lda.class <- lda.pred$class
# Confusion matrix
table(lda.class, test_Y) 


#### LDA with bootstrap: 
# Fit LDA to train data: to get standard errors for coefficients
set.seed(1)
boot.fn <- function(data,index){ 
  return(coefficients(lda(Species ~ SepalLength + SepalWidth + PetalLength + PetalWidth, data=data, subset=index)))
}
# Call boot(): This returns LD1 and LD2 for each predictor
boot(train, boot.fn, 1000)
# NOTE: Here, in Bootstrap Statistics output, t1* to t4* are LD1 coefficients and t5* to t8* are LD2 coefficients