如何使用可变标签获取特征重要性答案

【问题标题】：How to get features importances with variable labels如何使用可变标签获取特征重要性
【发布时间】：2021-03-23 00:48:52
【问题描述】：

我正在训练一个决策树回归器，但是当我得到特征重要性时，只有值来了。

有谁也知道如何获取带有变量名称的数据框？

下面是代码的主要部分：

num_pipeline = Pipeline([
    ('imputer', SimpleImputer(strategy="median")),
    ('std_scaler', StandardScaler()),
])

cat_pipeline = Pipeline([
    ('imputer', SimpleImputer(strategy="most_frequent")),
    ('oneHot', OneHotEncoder(handle_unknown='ignore')),
])

num_attribs = x_train.select_dtypes(include=np.number).columns.tolist()
cat_attribs = x_train.select_dtypes(include='object').columns.tolist()

full_pipeline = ColumnTransformer([
    ("num", num_pipeline, num_attribs),
    ("cat", cat_pipeline, cat_attribs),
])

train_prepared = full_pipeline.fit_transform(x_train)

param_grid = {'max_leaf_nodes': list(range(2, 100)), 'min_samples_split': [2, 3, 4], 'max_depth': list(range(3, 20))}

dtr = DecisionTreeRegressor()
grid_search = GridSearchCV(dtr, param_grid, cv=5, scoring='neg_mean_squared_error', verbose=1, return_train_score=True, n_jobs=-1)
grid_search = grid_search.fit(train_prepared, y_train)

grid_search.best_estimator_.feature_importances_

这是 feature_importances_ 的输出：

array([2.59182901e-03, 5.08807106e-04, 1.46808641e-03, 2.20756886e-03,
       1.48878361e-01, 5.65411415e-03, 5.16351699e-03, 9.37444882e-03,
       0.00000000e+00, 7.19228983e-03, 1.00581364e-03, 1.05073934e-03,
       2.63424620e-03, 9.41587243e-03, 7.22742602e-02, 0.00000000e+00,
       2.41075666e-03, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00,
       0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 1.12861715e-02,
       3.39987538e-03, 5.27924849e-04, 2.20562317e-03, 4.14808367e-03,
       5.82557008e-04, 1.40134963e-03, 0.00000000e+00, 0.00000000e+00,
       1.08351677e-03, 0.00000000e+00, 0.00000000e+00, 1.58022433e-03,
       0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 2.79779634e-02,
       5.94436576e-01, 3.72725666e-02, 1.11665462e-03, 2.39049915e-03,
       0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 1.15314788e-03,
       0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00,
       0.00000000e+00,...])

【问题讨论】：

标签： python machine-learning scikit-learn feature-selection

【解决方案1】：

虽然您不能直接调用方法从模型中获取标签，但它们与x_train 的方式相同且索引方式相同，因此您可以通过以下方式获取名称：

x_train.select_dtypes(include=np.number).columns

或者你可以创建一个字典，例如：

feature_importances = {x_train.select_dtypes(include=np.number).columns[x]:grid_search.best_estimator_.feature_importances_[x] for x in range(len(grid_search.best_estimator_.feature_importances_))}

【讨论】：

感谢您的回答。我正在运行字典方法，它给出了一个错误“索引 82 超出轴 0 的范围，大小为 82”。会是什么呢？如有必要，我可以将完整的代码放在帖子中。
抱歉耽搁了，是的，我会很感激有关 x_train 的更多信息，可能会有所帮助。
嘿！有了你的提示，我设法解决了这个问题。这是解决方案：'feats = {} # 保存 feature_name 的字典：feature_importance for feature,importance in zip(x_train.columns, grid_search.best_estimator_.feature_importances_): feats[feature] = important #添加名称/值对重要性= pd.DataFrame.from_dict(feats, orient='index').rename(columns={0: 'Gini-importance'}).sort_values(by='Gini-importance', ascending=False) 重要性[:10] '