如何获得R中交叉验证数据集的F1、精度和召回率

我有两个数据集

train <- read.csv("train.csv")
test  <- read.csv("test.csv")

测试集中的数据如下所示

> str(train)
'data.frame':   891 obs. of  12 variables:
 $ PassengerId: int  1 2 3 4 5 6 7 8 9 10 ...
 $ Survived   : Factor w/ 2 levels "0","1": 1 2 2 2 1 1 1 1 2 2 ...
 $ Pclass     : int  3 1 3 1 3 3 1 3 3 2 ...
 $ Name       : Factor w/ 891 levels "Abbing, Mr. Anthony",..: 109 191 358 
 277 16 559 520 629 417 581 ...
 $ Sex        : Factor w/ 2 levels "female","male": 2 1 1 1 2 2 2 2 1 1 ...
 $ Age        : num  22 38 26 35 35 NA 54 2 27 14 ...
 $ SibSp      : int  1 1 0 1 0 0 0 3 0 1 ...
 $ Parch      : int  0 0 0 0 0 0 0 1 2 0 ...
 $ Ticket     : Factor w/ 681 levels "110152","110413",..: 524 597 670 50 473 276 86 396 345 133 ...
 $ Fare       : num  7.25 71.28 7.92 53.1 8.05 ...
 $ Cabin      : Factor w/ 148 levels "","A10","A14",..: NA 83 NA 57 NA NA 131 NA NA NA ...
 $ Embarked   : Factor w/ 4 levels "","C","Q","S": 4 2 4 4 4 3 4 4 4 2 ...

> str(test)
'data.frame':   418 obs. of  11 variables:
 $ PassengerId: int  892 893 894 895 896 897 898 899 900 901 ...
 $ Pclass     : int  3 3 2 3 3 3 3 2 3 3 ...
 $ Name       : Factor w/ 418 levels "Abbott, Master. Eugene Joseph",..: 210 
409 273 414 182 370 85 58 5 104 ...
 $ Sex        : Factor w/ 2 levels "female","male": 2 1 2 2 1 2 1 2 1 2 ...
 $ Age        : num  34.5 47 62 27 22 14 30 26 18 21 ...
 $ SibSp      : int  0 1 0 0 1 0 0 1 0 2 ...
 $ Parch      : int  0 0 0 0 1 0 0 1 0 0 ...
 $ Ticket     : Factor w/ 363 levels "110469","110489",..: 153 222 74 148 
 139 262 159 85 101 270 ...
 $ Fare       : num  7.83 7 9.69 8.66 12.29 ...
 $ Cabin      : Factor w/ 77 levels "","A11","A18",..: 1 1 1 1 1 1 1 1 1 1 ...
 $ Embarked   : Factor w/ 3 levels "C","Q","S": 2 3 2 3 3 3 2 3 1 3 ... 

我使用decison树作为分类器。我想使用10倍交叉验证来训练和评估训练集。 为此，我使用胡萝卜包装

library(caret)
tc <- trainControl("cv",10)
rpart.grid <- expand.grid(.cp=0.2)

(train.rpart <- train(  Survived ~ Pclass + Sex + Age + SibSp + Parch + Fare 
                       + Embarked, 
                       data=train, 
                       method="rpart",
                       trControl=tc,
                       na.action = na.omit,
                       tuneGrid=rpart.grid))

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我的问题是如何以类似的方式找到10倍交叉验证数据集的精确度、召回率和F1？

当前方法将生存结果解读为整数，这导致rpart执行回归而不是分类。最好重新编码到因子级别

精度、召回率和F1等评估指标可通过奇妙的confusionMatrix函数获得

library(caret)
train <- read.csv("train.csv")
test  <- read.csv("test.csv")
tc <- trainControl("cv",10)
rpart.grid <- expand.grid(.cp=0.2)

# Convert variable interpreted as integer to factor
train$Survived <- as.factor(train$Survived)

(train.rpart <- train(  Survived ~ Pclass + Sex + Age + SibSp + Parch + Fare 
                    + Embarked, 
                    data=train, 
                    method="rpart",
                    trControl=tc,
                    na.action = na.omit,
                    tuneGrid=rpart.grid))
# Predict
pred <- predict(train.rpart, train) 

# Produce confusion matrix from prediction and data used for training
cf <- confusionMatrix(pred, train.rpart$trainingData$.outcome, mode = "everything")
print(cf)

library(caret)
train <- read.csv("train.csv")
test  <- read.csv("test.csv")
tc <- trainControl("cv",10)
rpart.grid <- expand.grid(.cp=0.2)

# Convert variable interpreted as integer to factor
train$Survived <- as.factor(train$Survived)

(train.rpart <- train(  Survived ~ Pclass + Sex + Age + SibSp + Parch + Fare 
                    + Embarked, 
                    data=train, 
                    method="rpart",
                    trControl=tc,
                    na.action = na.omit,
                    tuneGrid=rpart.grid))
# Predict
pred <- predict(train.rpart, train) 

# Produce confusion matrix from prediction and data used for training
cf <- confusionMatrix(pred, train.rpart$trainingData$.outcome, mode = "everything")
print(cf)