我正在对 2 个类(0 和 1)的一些测试数据实现朴素贝叶斯的 10 倍交叉验证。 我按照以下步骤操作并收到错误。
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, 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))
另外,如何计算该模型的 R 方或 AUC
数据集:有 10000 条记录,20 个特征和 Binary 类。
最佳答案
NaiveBayes 是一个分类器,因此将 Y 转换为因子或 bool 值是解决该问题的正确方法。您最初的公式使用了分类器工具,但使用了数值,因此 R 很困惑。
就 R 方而言,该指标仅针对回归问题而不是分类问题计算。为了评估分类问题,还有其他指标,例如精度和召回率。
有关这些指标的更多信息,请参阅维基百科链接: http://en.wikipedia.org/wiki/Binary_classification
关于r - 使用 R 对朴素贝叶斯进行 10 倍交叉验证时出现回归错误的模型类型错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/23357855/