python之我如何在 scikit-learn 中使用 k 折交叉验证来获得每折的精确召回率

假设我有这种情况:

from sklearn import model_selection 
from sklearn.linear_model import LogisticRegression 
 
kfold = model_selection.KFold(n_splits=5, random_state=7) 
acc_per_fold = model_selection.cross_val_score(LogisticRegression(), 
               x_inputs, np.ravel(y_response), cv=kfold, scoring='accuracy') 

我还能从 model_selection.cross_val_score() 得到什么？有没有办法查看每个实际折叠内部发生的情况？我可以获得每次折叠的精确召回率吗？预测值？如何使用经过训练的折叠模型对看不见的数据进行预测？

请您参考如下方法:

您可以使用 cross_validate 函数来查看每次折叠中发生的情况。

import numpy as np 
from sklearn.datasets import make_classification 
from sklearn.model_selection import cross_validate 
from sklearn.linear_model import LogisticRegression 
from sklearn.metrics import accuracy_score, confusion_matrix, recall_score, roc_auc_score, precision_score 
 
X, y = make_classification( 
    n_classes=2, class_sep=1.5, weights=[0.9, 0.1], 
    n_features=20, n_samples=1000, random_state=10 
) 
clf = LogisticRegression(class_weight="balanced") 
scoring = {'accuracy': 'accuracy', 
           'recall': 'recall', 
           'precision': 'precision', 
           'roc_auc': 'roc_auc'} 
cross_val_scores = cross_validate(clf, X, y, cv=3, scoring=scoring) 

输出如下，

{'fit_time': array([ 0.        ,  0.        ,  0.01559997]), 
 'score_time': array([ 0.01559997,  0.        ,  0.        ]), 
 'test_accuracy': array([ 0.9251497 ,  0.95808383,  0.93674699]), 
 'test_precision': array([ 0.59183673,  0.70833333,  0.63636364]), 
 'test_recall': array([ 0.85294118,  1.        ,  0.84848485]), 
 'test_roc_auc': array([ 0.96401961,  0.99343137,  0.96787271]), 
 'train_accuracy': array([ 0.96096096,  0.93693694,  0.95209581]), 
 'train_precision': array([ 0.73033708,  0.62376238,  0.69148936]), 
 'train_recall': array([ 0.97014925,  0.94029851,  0.95588235]), 
 'train_roc_auc': array([ 0.99426906,  0.98509954,  0.99223039])} 

那么第一次折叠发生了什么？

FOLD, METRIC = (0, 'test_precision') 
cross_val_scores[METRIC][FOLD] 

精度得分是否稳定？

np.std(cross_val_scores[METRIC])