如何在Logistic Regression中列出特征重要性对应的列名而非索引

我希望在展示特征重要性时列出实际的特征名称（列名），而非特征的索引编号，该如何实现？

原代码：

from sklearn.datasets import make_classification
from sklearn.linear_model import LogisticRegression
X, y = make_classification(n_samples=2700, n_features=3, n_informative=5, n_redundant=5, 
random_state=1)

model = LogisticRegression()

model.fit(X, y)

importance = model.coef_[0]

for i,v in enumerate(importance):
    print('Feature: %0d, Score: %.5f' % (i,v))

当前输出：

Feature: 0, Score: -0.11124
Feature: 1, Score: -0.04292
Feature: 2, Score: 0.00561

期望输出：

featurename1, Score: -0.11124
featurename2, Score: -0.04292
featurename3, Score: 0.00561

核心是给特征绑定名称列表，遍历重要性时通过索引匹配对应名称，分两种场景实现：

场景1：手动定义特征名称 #

如果是numpy数组格式的特征数据，直接创建和特征数量匹配的名称列表，用zip关联名称与重要性分数：

from sklearn.datasets import make_classification
from sklearn.linear_model import LogisticRegression
X, y = make_classification(n_samples=2700, n_features=3, n_informative=5, n_redundant=5, 
random_state=1)

# 定义与特征数量一致的名称列表
feature_names = ['featurename1', 'featurename2', 'featurename3']

model = LogisticRegression()
model.fit(X, y)

importance = model.coef_[0]

# 遍历名称和对应分数并打印
for name, score in zip(feature_names, importance):
    print('%s, Score: %.5f' % (name, score))

场景2：使用Pandas DataFrame存储数据 #

如果特征数据存在DataFrame中，直接调用df.columns获取列名即可，无需手动定义：

import pandas as pd
from sklearn.datasets import make_classification
from sklearn.linear_model import LogisticRegression

# 生成数据并转为带列名的DataFrame
X, y = make_classification(n_samples=2700, n_features=3, n_informative=5, n_redundant=5, 
random_state=1)
df = pd.DataFrame(X, columns=['featurename1', 'featurename2', 'featurename3'])

model = LogisticRegression()
model.fit(df, y)

importance = model.coef_[0]

# 用DataFrame的columns属性获取特征名称
for name, score in zip(df.columns, importance):
    print('%s, Score: %.5f' % (name, score))

内容的提问来源于stack exchange，提问作者prof31