我正在使用 scikit-learn 运行一堆模型来解决分类问题。这是应该完成所有运行的代码:for model_name, classifier, param_grid, cv, cv_name in tqdm(zip(model_names, classifiers, param_grids, cvs, cv_names)): pipeline = Pipeline(steps=[('preprocessor', preprocessor), ('classifier', classifier)]) train_and_score_model(model_name, pipeline, param_grid, cv=cv)我的问题是,如何保留train_and_score_model函数的输出?它返回一个 cv 对象,即一个模型。我试图做但我认为不正确的是创建一个列表cv_names = ['dm_cv', 'lr_cv', 'knn_cv', 'svm_cv', 'dt_cv', 'rf_cv', 'nb_cv']并将每个列表设置为 for 循环运行。那是cv_namefor 循环头中的迭代器。我不认为那是对的,因为我不会设置字符串而不是变量吗?就像,我真正应该拥有的是cv_names = [dm_cv, lr_cv, knn_cv, svm_cv, dt_cv, rf_cv, nb_cv],但我不认为我可以拥有这样的列表。我想到的另一种方法是将每个模型保存在字典中,其中的键是我上面概述的列表中的元素。我不知道我是否可以将模型作为字典值。这是我目前在 for 循环中运行的笨重、重复的代码:pipeline = Pipeline(steps=[('preprocessor', preprocessor), ('classifier', classifier_dm)])dm_cv = train_and_score_model('Dummy Model', pipeline, param_grid_dm)pipeline = Pipeline(steps=[('preprocessor', preprocessor), ('classifier', classifier_lr)])lr_cv = train_and_score_model('Logistic Regression', pipeline, param_grid_lr)pipeline = Pipeline(steps=[('preprocessor', preprocessor), ('classifier', classifier_knn)])knn_cv = train_and_score_model('K Nearest Neighbors', pipeline, param_grid_knn)pipeline = Pipeline(steps=[('preprocessor', preprocessor), ('classifier', classifier_svm)])svm_cv = train_and_score_model('Support Vector Machine', pipeline, param_grid_svm)pipeline = Pipeline(steps=[('preprocessor', preprocessor), ('classifier', classifier_dt)])dt_cv = train_and_score_model('Decision Tree', pipeline, param_grid_dt)
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您可以创建一个字典,其中包含分类器名称及其信息(即对象和参数网格)的映射:
models_list = {'Logistic Regression': (classifier_lr, param_grid_lr),
'K Nearest Neighbours': (classifier_knn, param_grid_knn)}
遍历字典中的每个键值对并构建您的管道:
model_cvs = {}
for model_name, model_info in models_list.items():
pipeline = Pipeline(steps=[('preprocessor', preprocessor),
('classifier', model_info[0])])
model_cvs[model_name] = train_and_score_model(model_name, pipeline, model_info[1])
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