如何獲得F1中的交叉驗證數據集的精度和調用

Question

我有兩個數據集。

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

列車組中的數據如下所示。

> 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 ... 

測試集中的數據如下所示。

> 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樹作爲分類器。我想使用10倍交叉驗證來訓練和評估火車組。 爲此，我正在使用胡蘿蔔包。

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)) 

從這裏，我能夠獲得交叉驗證準確性的值。

712 samples 
    7 predictor 
    2 classes: '0', '1' 

No pre-processing 
Resampling: Cross-Validated (10 fold) 
Summary of sample sizes: 641, 641, 640, 640, 641, 641, ... 
Resampling results: 

    Accuracy Kappa  
    0.7794601 0.5334528 

Tuning parameter 'cp' was held constant at a value of 0.2 

我的問題是如何找到準確率，召回和F1用於以類似的方式設置的10倍交叉驗證的數據？

Answer 1

當前的方法將整個生存結果讀爲整數，這導致rpart執行迴歸而不是分類。更好地重新編碼到一個因子水平。

評估指標如精度，召回率和F1可通過美妙的confusionMatrix函數獲得。

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)