R 中不带插入符号的交叉验证

发布于 2025-01-10 04:25:41 字数 824 浏览 0 评论 0原文

由于我使用的模型（fastNaiveBayes 包）不在 caret 包的内置库中，因此我尝试在 R 中进行 k 折交叉验证而不使用 caret 包。有人有解决办法吗？

编辑：这是迄今为止我学到的如何在没有插入符号的情况下进行简历的代码。我非常确定这里出了问题。

library(fastNaiveBayes)


k<- 10
outs <- NULL
proportion <- 0.8
for (i in 1:10)
{
  split <- sample(1:nrow(data), round(proportion*nrow(data)))
  traindata <- data[split,]
  testdata <- data[-split,]

y <- traindata$Label

x <- traindata[,0 - 15:ncol(traindata)]

model <- fnb.train(x, y=y, priors = NULL, laplace=0, 
               distribution = fnb.detect_distribution(x, nrows = nrow(x)))
model




test1 <- testdata[,0 - 15:ncol(testdata)]
pred <- predict(model, newdata = test1)


cm<- table(testdata$Label, pred)


print(confusionMatrix(cm))

}

它给了我 10 个不同的结果，我认为这不是交叉验证应该如何工作的。我是一名入门级 R 学习者，非常感谢从中获得启发

原文

Since the model (package fastNaiveBayes) that I am using is not in the built-in library of the caret package, I am trying to make a k-fold cross validation in R without using the caret package. Does anyone have a solution to this?

Edit:
Here is my code so far from what I learned on how to do cv without caret. I am very certain something is wrong here.

library(fastNaiveBayes)


k<- 10
outs <- NULL
proportion <- 0.8
for (i in 1:10)
{
  split <- sample(1:nrow(data), round(proportion*nrow(data)))
  traindata <- data[split,]
  testdata <- data[-split,]

y <- traindata$Label

x <- traindata[,0 - 15:ncol(traindata)]

model <- fnb.train(x, y=y, priors = NULL, laplace=0, 
               distribution = fnb.detect_distribution(x, nrows = nrow(x)))
model




test1 <- testdata[,0 - 15:ncol(testdata)]
pred <- predict(model, newdata = test1)


cm<- table(testdata$Label, pred)


print(confusionMatrix(cm))

}

It gave me 10 different results and I think that's not how it cross validation is supposed to work. I'm an entry-level R learner and I appreciate so much to receive enlightenment from this

分享到QQ

分享到微博