如何使用bonferroni校正结果从回归结果中删除特征？

Question

我使用

实现了回归模型

formula= "cost ~ C(state) + group_size + C(homeowner) + car_age + C(car_value) + 
risk_factor + age_oldest + age_youngest + C(married_couple) + c_previous + 
duration_previous + C(a) + C(b) + C(c) + C(d) + C(e) + C(f) + C(g)"

model_a = smf.ols(formula = formula, data = train).fit()
model_a.summary()

拟合回归模型后，我使用

进行了 bonferroni 校正

smt.multipletests(model_a.pvalues, alpha=0.05, method='bonferroni', is_sorted=False, 
returnsorted=False)

我得到以下结果：

(array([ True, False,  True,  True,  True,  True,  True, False,  True,
     True,  True, False,  True,  True,  True,  True, False, False,
    False, False,  True, False,  True,  True,  True,  True,  True,
     True,  True, False,  True,  True, False,  True,  True, False,
     True,  True,  True,  True,  True,  True,  True,  True, False,
     True,  True,  True, False, False, False, False, False, False,
     True,  True,  True,  True,  True, False,  True, False,  True,
    False,  True,  True,  True,  True]),
 array([0.00000000e+00, 1.00000000e+00, 1.45352365e-03, 2.14422252e-21,
    5.68726115e-13, 4.81466313e-12, 1.22517937e-05, 3.36565323e-01,
    4.81396354e-45, 1.51138583e-05, 4.27572151e-04, 1.00000000e+00,
    5.91690245e-10, 2.62041907e-16, 3.12129589e-18, 9.88879325e-13,
    1.00000000e+00, 1.00000000e+00, 1.00000000e+00, 6.85853188e-01,
    8.94886169e-07, 1.00000000e+00, 3.55801455e-12, 5.35987286e-54,
    7.77655333e-03, 5.45090922e-04, 5.15690091e-03, 7.40791788e-04,
    1.24797586e-07, 1.00000000e+00, 2.91991310e-04, 1.75502703e-07,
    1.00000000e+00, 2.57023089e-26, 2.34824045e-10, 1.00000000e+00,
    2.79360586e-87, 5.26115182e-09, 4.94812967e-08, 3.36073545e-07,
    5.06333547e-07, 4.44900552e-07, 1.06078148e-05, 1.42866234e-03,
    1.00000000e+00, 3.72074539e-10, 1.38294896e-74, 1.39540646e-69,
    1.00000000e+00, 1.00000000e+00, 1.00000000e+00, 1.00000000e+00,
    1.00000000e+00, 1.00000000e+00, 2.78538149e-18, 3.74576314e-22,
    1.12111501e-19, 1.14698339e-04, 9.34411232e-18, 1.00000000e+00,
    4.10430857e-02, 1.00000000e+00, 5.35030644e-23, 1.00000000e+00,
    7.61651080e-20, 9.49735915e-56, 7.90523832e-66, 8.15390766e-94]),
 0.0007540287301109894,
 0.0007352941176470588)

我想使用这些数组来删除 model_a 中为 False 的特征并创建一个新模型“train_simplified”。

我正在使用以下手动方法，但我想知道是否有更有效的方法。

train_simplified = train.drop(train.columns[[0, 1, 2, 4, 10, 16, 25, 27, 28, 30, 36, 38, 
41, 44, 47, 55, 61, 62, 63, 64, 65, 66, 67, 68, 69, 75, 78]], axis=1)

Answer 1

您可以使用 Pandas

loc

仅选择

model_a

中的

True

功能。

.loc[] 主要基于标签，但也可以与布尔数组一起使用。

train = pd.DataFrame(np.random.rand(5,68))
          0         1         2         3  ...        63        64        65        66        67
0  0.637557  0.887213  0.472215  0.119594  ...  0.908266  0.239562  0.144895  0.489453  0.985650
1  0.242055  0.672136  0.761620  0.237638  ...  0.649633  0.849223  0.657613  0.568309  0.093675
2  0.367716  0.265202  0.243990  0.973011  ...  0.465598  0.542645  0.286541  0.590833  0.030500
3  0.037348  0.822601  0.360191  0.127061  ...  0.070569  0.642419  0.026511  0.585776  0.940230
4  0.575474  0.388170  0.643288  0.458253  ...  0.091206  0.494420  0.057559  0.549529  0.441531

[5 rows x 68 columns]

keep_columns = np.array([ # array from smt.multipletests
    True, False,  True,  True,  True,  True,  True, False,  True,
    True,  True, False,  True,  True,  True,  True, False, False,
    False, False,  True, False,  True,  True,  True,  True,  True,
    True,  True, False,  True,  True, False,  True,  True, False,
    True,  True,  True,  True,  True,  True,  True,  True, False,
    True,  True,  True, False, False, False, False, False, False,
    True,  True,  True,  True,  True, False,  True, False,  True,
    False,  True,  True,  True,  True])
np.sum(keep_columns) # 47 (keep 47 columns)

train_simplified = train.loc[:,keep_columns]

train_simplified

的输出

          0         2         3         4  ...        62        64        65        66        67
0  0.637557  0.472215  0.119594  0.713245  ...  0.278646  0.239562  0.144895  0.489453  0.985650
1  0.242055  0.761620  0.237638  0.728216  ...  0.746491  0.849223  0.657613  0.568309  0.093675
2  0.367716  0.243990  0.973011  0.393098  ...  0.035942  0.542645  0.286541  0.590833  0.030500
3  0.037348  0.360191  0.127061  0.522243  ...  0.162934  0.642419  0.026511  0.585776  0.940230
4  0.575474  0.643288  0.458253  0.545617  ...  0.789618  0.494420  0.057559  0.549529  0.441531

[5 rows x 47 columns]