R-caret包(rpart):构建分类树
我花了好几天的时间来使用插入符号包执行分类树。 问题是我的因素变量。我生成了树,但当我尝试使用最佳模型对测试样本进行预测时,它失败了,因为训练函数为我的因子变量创建了假人,然后预测函数无法在测试集中找到这些新创建的假人。我应该如何处理这个问题 我的代码如下:R-caret包(rpart):构建分类树,r,rpart,r-caret,R,Rpart,R Caret,我花了好几天的时间来使用插入符号包执行分类树。 问题是我的因素变量。我生成了树,但当我尝试使用最佳模型对测试样本进行预测时,它失败了,因为训练函数为我的因子变量创建了假人,然后预测函数无法在测试集中找到这些新创建的假人。我应该如何处理这个问题 我的代码如下: install.packages("caret", dependencies = c("Depends", "Suggests")) library(caret)
install.packages("caret", dependencies = c("Depends", "Suggests"))
library(caret)
db=data.frame(read.csv ("db.csv", head=TRUE, sep=";", na.strings ="?"))
fix(db)
db$defaillance=factor(db$defaillance)
db$def=ifelse(db$defaillance==0,"No","Yes")
db$def=factor(db$def)
db$defaillance=NULL
db$canal=factor(db$canal)
db$sect_isodev=factor(db$sect_isodev)
db$sect_risq=factor(db$sect_risq)
#delete zero variance predictors
nzv <- nearZeroVar(db[,-78])
db_new <- db[,-nzv]
inTrain <- createDataPartition(y = db_new$def, p = .75, list = FALSE)
training <- db_new[inTrain,]
testing <- db_new[-inTrain,]
str(training)
str(testing)
dim(training)
dim(testing)
然后我的代码是这样的:
fitControl <- trainControl(method = "repeatedcv",
number = 10,
repeats = 10,
classProbs = TRUE,
summaryFunction = twoClassSummary)
#CART1
set.seed(1234)
tree1 = train (def~.,
training,
method = "rpart",
tuneLength=20,
metric="ROC",
trControl = fitControl)
在这儿吗
RNTB 38.397731
sect_isodev1 6.742289
sect_isodev3 4.005016
sect_isodev8 2.520850
sect_risq3 9.909127
sect_risq4 6.737908
sect_risq5 3.085714
SOLV 73.067539
TRES 47.906884
sect_isodev2 0.000000
sect_isodev4 0.000000
sect_isodev5 0.000000
sect_isodev6 0.000000
sect_isodev7 0.000000
sect_isodev9 0.000000
sect_risq0 0.000000
sect_risq1 0.000000
sect_risq2 0.000000
下面是错误:
model.tree1据我所知,有两个问题:
- R无法为
找到合适的tree1$finalModel
函数,该函数应为predict
,因为predict.rpart
属于tree1$finalModel
类。我也犯了这个错误,不幸的是,我不知道根本原因。这也是R不接受rpart
的原因type=“class”
会接受它predict.rpart
- 为
函数提供一个公式,而不是x和y对象,会导致以后找不到类似train
的变量的问题sect_isodev1
str
)复制错误并从rpart
显式调用predict.rpart
后,我的做法是:
tree1 = train (y = training$def,
x = training[, -which(colnames(training) == "def")],
method = "rpart",
tuneLength=20,
metric="ROC",
trControl = fitControl)
summary(tree1$finalModel)
# This still results in Error: could not find function "predict.rpart":
model.tree1 <- predict.rpart(tree1$finalModel, newdata = testing)
# Explicitly calling predict.rpart from the rpart package works:
rpart:::predict.rpart(object = tree1$finalModel,
newdata = testing,
type = "class")
tree1=训练(y=训练$def,
x=训练[,-其中(colnames(training)=“def”)],
method=“rpart”,
tuneLength=20,
metric=“ROC”,
trControl=fitControl)
摘要(tree1$finalModel)
#这仍然会导致错误:找不到函数“predict.rpart”:
model.tree1除非有很好的理由,否则不要将predict.rpart
与train$finalModel
一起使用。rpart
对象没有;我不知道train
所做的任何事情,包括预处理。它可能不会给你正确的答案。毕竟,您可能正在使用train
来避免细节,所以让预测。train
做这项工作
马克斯
编辑-
关于type=“class”
和type=“prob”
位
predict.rpart
默认生成类概率。尽管rpart
是最早的包之一,但这是非典型的,因为大多数默认情况下都生成类
predict.train
默认生成类,您必须使用type=“prob”
获取概率
RNTB 38.397731
sect_isodev1 6.742289
sect_isodev3 4.005016
sect_isodev8 2.520850
sect_risq3 9.909127
sect_risq4 6.737908
sect_risq5 3.085714
SOLV 73.067539
TRES 47.906884
sect_isodev2 0.000000
sect_isodev4 0.000000
sect_isodev5 0.000000
sect_isodev6 0.000000
sect_isodev7 0.000000
sect_isodev9 0.000000
sect_risq0 0.000000
sect_risq1 0.000000
sect_risq2 0.000000
predict(rpartTune$finalModel, newdata, type = "class")
tree1 = train (y = training$def,
x = training[, -which(colnames(training) == "def")],
method = "rpart",
tuneLength=20,
metric="ROC",
trControl = fitControl)
summary(tree1$finalModel)
# This still results in Error: could not find function "predict.rpart":
model.tree1 <- predict.rpart(tree1$finalModel, newdata = testing)
# Explicitly calling predict.rpart from the rpart package works:
rpart:::predict.rpart(object = tree1$finalModel,
newdata = testing,
type = "class")