使用R对朴素贝叶斯进行10倍交叉验证时回归误差的错误模型类型
我正在对一些包含2个类(0和1)的测试数据实施10倍的朴素贝叶斯交叉验证。 我遵循下面的步骤并得到了错误使用R对朴素贝叶斯进行10倍交叉验证时回归误差的错误模型类型,r,machine-learning,R,Machine Learning,我正在对一些包含2个类(0和1)的测试数据实施10倍的朴素贝叶斯交叉验证。 我遵循下面的步骤并得到了错误 data(testdata) attach(testdata) X <- subset(testdata, select=-Class) Y <- Class library(e1071) naive_bayes <- naiveBayes(X,Y) library(caret) library(klaR) nb_cv <- train(X, Y, me
data(testdata)
attach(testdata)
X <- subset(testdata, select=-Class)
Y <- Class
library(e1071)
naive_bayes <- naiveBayes(X,Y)
library(caret)
library(klaR)
nb_cv <- train(X, Y, method = "nb", trControl = trainControl(method = "cv", number = 10))
## Error:
## Error in train.default(X, Y, method = "nb", trControl = trainControl(number = 10)) :
## wrong model type for regression
dput(testdata)
structure(list(Feature.1 = 6.534088, Feature.2 = -19.050915,
Feature.3 = 7.599378, Feature.4 = 5.093594, Feature.5 = -22.15166,
Feature.6 = -7.478444, Feature.7 = -59.534652, Feature.8 = -1.587918,
Feature.9 = -5.76889, Feature.10 = 95.810563, Feature.11 = 49.124086,
Feature.12 = -21.101489, Feature.13 = -9.187984, Feature.14 = -10.53006,
Feature.15 = -3.782506, Feature.16 = -10.805074, Feature.17 = 34.039509,
Feature.18 = 5.64245, Feature.19 = 19.389724, Feature.20 = 16.450196,
Class = 1L), .Names = c("Feature.1", "Feature.2", "Feature.3",
"Feature.4", "Feature.5", "Feature.6", "Feature.7", "Feature.8",
"Feature.9", "Feature.10", "Feature.11", "Feature.12", "Feature.13",
"Feature.14", "Feature.15", "Feature.16", "Feature.17", "Feature.18",
"Feature.19", "Feature.20", "Class"), class = "data.frame", row.names = c(NA,
-1L))
数据(testdata)
附加(测试数据)
X它在改变标签向量Y后工作,bayes是一个分类器,因此将Y转换为因子或布尔值是解决该问题的正确方法。您最初的公式使用的是分类器工具,但使用的是数值,因此R混淆了
就R平方而言,同样地,度量只针对回归问题而不是分类问题进行计算。要评估分类问题,还有其他指标,如精确度和召回率
有关这些指标的更多信息,请参阅维基百科链接:
添加到您的结构中
colClasses=c("Class"="character")
如果您想获得帮助,请dput(testdata)
,谢谢David。添加了带有1条记录的dput(testdata)。它在将类标签从(1,0)更改为(是,否)后工作。它在更改标签向量Y后也工作