R中的Caret和KNN:预测函数给出错误 [英] Caret and KNN in R: predict function gives error

问题描述

我尝试使用R中的插入符号包通过简化的KNN模型进行预测.即使在这里非常简单的可重现示例中，它总是会产生相同的错误:

I try to predict with a simplified KNN model using the caret package in R. It always gives the same error, even in the very simple reproducible example here:

library(caret)
set.seed(1)

#generate training dataset "a" 
n = 10000
a = matrix(rnorm(n*8,sd=1000000),nrow = n)
y = round(runif(n))
a = cbind(y,a)
a = as.data.frame(a)
a[,1] = as.factor(a[,1])
colnames(a) = c("y",paste0("V",1:8))

#estimate simple KNN model
ctrl <- trainControl(method="none",repeats = 1)
knnFit <- train(y ~ ., data = a, method = "knn", trControl = ctrl, preProcess = c("center","scale"),  tuneGrid = data.frame(k = 10))

#predict on the training dataset (=useless, but should work)
knnPredict <- predict(knnFit,newdata = a,  type="prob")

这给

[.data.frame中的错误(out，，obsLevels，drop = FALSE): 未定义的列已选择

Error in [.data.frame(out, , obsLevels, drop = FALSE) : undefined columns selected

在没有目标变量y的情况下定义更现实的测试数据集"b".

Defining a more realistic test dataset "b" without the target variable y...

#generate test dataset
b =  matrix(rnorm(n*8,sd=1000000),nrow = n) 
b = as.data.frame(b)
colnames(b) = c(paste0("V",1:8))

#predict on the test datase
knnPredict <- predict(knnFit,newdata = b,  type="prob")

给出相同的错误

[.data.frame中的错误(out，，obsLevels，drop = FALSE): 未定义的列已选择

Error in [.data.frame(out, , obsLevels, drop = FALSE) : undefined columns selected

我知道这些列很重要，但是在这里它们是相同的.这是怎么了谢谢！

I know that the columnames are important, but here they are identical. What is wrong here? Thanks!

推荐答案

问题是您的y变量.当您要求班级概率时，训练和/或预测函数会将其放入每个班级都有一列的数据框中.如果因子级别不是有效的变量名称，则会自动更改它们(例如，"0"变为"X0").另请参见此帖子.

The problem is your y variable. When you are asking for the class probabilities, the train and / or the predict function puts them into a data frame with a column for each class. If the factor levels are not valid variable names, they are automatically changed (e.g. "0" becomes "X0"). See also this post.

如果您在代码中更改此行，则该行应该有效:

If you change this line in your code it should work:

a[,1] = factor(a[,1], labels = c("no", "yes"))

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