如何显示每个交叉验证折叠的混淆矩阵和报告(调用,精度,fmeasure)

问题描述 投票:0回答:1

我试图在python中执行10倍交叉验证。我知道如何计算混淆矩阵和分裂测试报告(例如分开80%训练和20%测试)。但问题是我不知道如何计算混淆矩阵和每个折叠的报告,例如fold-10时,我只知道平均准确度的代码。

python machine-learning scikit-learn cross-validation confusion-matrix
1个回答
2
投票

这是一个可重复的例子,其中包含乳腺癌数据和简单的3倍CV:

from sklearn.datasets import load_breast_cancer
from sklearn.tree import DecisionTreeClassifier
from sklearn.metrics import confusion_matrix, classification_report
from sklearn.model_selection import KFold

X, y = load_breast_cancer(return_X_y=True)
n_splits = 3
kf = KFold(n_splits=n_splits, shuffle=True)
model = DecisionTreeClassifier()

for train_index, val_index in kf.split(X):
    model.fit(X[train_index], y[train_index])
    pred = model.predict(X[val_index])
    print(confusion_matrix(y[val_index], pred))
    print(classification_report(y[val_index], pred))

结果是3个混淆矩阵和分类报告,每个CV折叠一个:

[[ 63   9]
 [ 10 108]]
              precision    recall  f1-score   support

           0       0.86      0.88      0.87        72
           1       0.92      0.92      0.92       118

   micro avg       0.90      0.90      0.90       190
   macro avg       0.89      0.90      0.89       190
weighted avg       0.90      0.90      0.90       190

[[ 66   8]
 [  6 110]]
              precision    recall  f1-score   support

           0       0.92      0.89      0.90        74
           1       0.93      0.95      0.94       116

   micro avg       0.93      0.93      0.93       190
   macro avg       0.92      0.92      0.92       190
weighted avg       0.93      0.93      0.93       190

[[ 59   7]
 [  8 115]]
              precision    recall  f1-score   support

           0       0.88      0.89      0.89        66
           1       0.94      0.93      0.94       123

   micro avg       0.92      0.92      0.92       189
   macro avg       0.91      0.91      0.91       189
weighted avg       0.92      0.92      0.92       189
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