Question

我正在尝试使用 GridSearchCV 实现决策树分类器。实施后，我尝试访问 cv_results_.mean_train_score ，但出现关键错误。

    tuned_parameters = [{'max_depth': [1, 5, 10, 25, 50, 75, 100, 150, 250, 500, 750, 1000], 
                         'min_samples_split' : [5, 10, 25, 50, 75, 150, 250, 500]}] 
    cv_timeSeries = TimeSeriesSplit(n_splits=4).split(X_train)
    base_estimator = DecisionTreeClassifier(class_weight='balanced') 
    gsearch_cv = GridSearchCV(estimator=base_estimator, 
                              param_grid=tuned_parameters, 
                              cv=cv_timeSeries, 
                              scoring='roc_auc', 
                              n_jobs=-1)
    gsearch_cv.fit(X_train, y_train)

当我尝试访问gsearch_cv的所有键时，找不到字典键mean_train_score。

Answer 1

能否请您发布产生错误的代码？

mean_train_score是cv_results_的键，因此要获取他的值，您应该：

gsearch_cv = GridSearchCV(estimator=base_estimator, 
                          param_grid=tuned_parameters, 
                          cv=cv_timeSeries, 
                          scoring='roc_auc', 
                          return_train_score=True,
                          n_jobs=-1)
gsearch_cv.fit(X_train, y_train)
gsearch_cv.cv_results_['mean_train_score']

您可以在sklearn页面https://scikit-learn.org/stable/modules/generated/sklearn.model_selection.GridSearchCV.html

中找到完整的示例

Answer 2

在GridSearchCV中的参数后面添加参数

GridSearchCV（return_train_score = True）

Answer 3

尝试以下更改：

gsearch_cv = GridSearchCV(estimator=base_estimator,
                          param_grid=tuned_parameters,
                          cv=cv_timeSeries,
                          scoring='roc_auc',
                          return_train_score=True,
                          n_jobs=-1,
                          return_train_score=True)

从sklearn.model_selection.GridSearchCV获取密钥错误

3 个答案: