r - R中混淆矩阵的平均值

Question

我应用了 10 次交叉验证，输出是混淆矩阵的 10 倍，那么如何找到混淆矩阵的倍数平均值？

我的工作正确吗？

这是我的代码:

set.seed(100)
    library(caTools)
    library(caret)
    library(e1071)
    folds<-createFolds(wpdc$outcome, k=10) 
    CV <- lapply(folds, function(x){
      traing_folds=wpdc[-x,]
      test_folds=wpdc[x,]
      dataset_model_nb<-naiveBayes(outcome ~ ., data = traing_folds)
      dataset_predict_nB<-predict(dataset_model_nb, test_folds[-1])
      dataset_table_nB<-table(test_folds[,1],dataset_predict_nB)
      accuracy<-confusionMatrix(dataset_table_nB, positive ="R")
      return(accuracy)
    })

 outcome radius_mean texture_mean perimeter_mean area_mean smoothness_mean compactness_mean concavity_mean concave_points_mean symmetry_mean fractal_dimension_mean radius_se texture_se perimeter_se area_se smoothness_se
1       N       18.02        27.60         117.50    1013.0         0.09489           0.1036         0.1086             0.07055        0.1865                0.06333    0.6249     1.8900        3.972   71.55      0.004433
2       N       17.99        10.38         122.80    1001.0         0.11840           0.2776         0.3001             0.14710        0.2419                0.07871    1.0950     0.9053        8.589  153.40      0.006399
3       N       21.37        17.44         137.50    1373.0         0.08836           0.1189         0.1255             0.08180        0.2333                0.06010    0.5854     0.6105        3.928   82.15      0.006167

Answer 1

我也需要同样的东西，然后按照@Stephen Handerson 的提示，我就是:

1. 定义矩阵列表:
   • rfConfusionMatrices <- list()
2. 将每个矩阵存储在该列表中:
   • RrfConfusionMatrix[[i]] <- confMatrix
3. 使用Reduce函数对矩阵求和并除以折叠:
   • rfConfusionMatrixMean <- Reduce('+', rfConfusionMatrix) / nFolds