如何在 R 中获得交叉验证数据集的 F1、精确度和召回率答案

【问题标题】：How to get F1,Precision and Recall for a Cross Validated Data Set in R如何在 R 中获得交叉验证数据集的 F1、精确度和召回率
【发布时间】：2017-11-18 23:44:28
【问题描述】：

我有两个数据集。

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

训练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 ...

测试集中的数据如下所示。

> 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 ...

我使用决策树作为分类器。我想使用 10 折交叉验证来训练和评估训练集。为此，我正在使用胡萝卜包。

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))

从这里，我可以获得交叉验证准确性的值。

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

我的问题是如何以类似的方式找到 10 倍交叉验证数据集的准确率、召回率和 F1？

【问题讨论】：

标签： r decision-tree cross-validation kaggle precision-recall

【解决方案1】：

当前方法将生存结果读取为整数，这导致 rpart 执行回归而不是分类。最好重新编码到因子水平。

精确度、召回率和 F1 等评估指标可通过出色的混淆矩阵函数获得。

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)

【讨论】：