5.决策树特征重要性判别算法python实现_python decisiontree回归怎么看feature importance

特征重要性算法

项目链接：https://github.com/Wchenguang/gglearn/blob/master/DecisionTree/李航机器学习讲解/FeatureImportance.ipynb

信息增益法 公式

• 熵的定义：

  • 属性 $y$ 的熵，表示特征的不确定性：
    $$P\left(Y=y_j\right)=p_j,i=1,2,\cdots ,n$$
    $$H(Y)=-\sum _{j=1}^n{p}_j\log p_j$$

• 条件熵的定义：

  • 在 $x$ 已知的情况下，$y$ 的不确定性
    $$P\left(X=x_i,Y=y_j\right)=p_{ij},i=1,2,\cdots ,n;j=1,2,\cdots ,m$$
    $$H(Y|X)=\sum _{i=1}^n{p}_{i}H\left(Y|X=x_i\right)$$

• 信息增益计算流程

  1. 计算特征A对数据集D的熵，即计算$y$ 的熵
     $$H(D)=-\sum _{k=1}^K\frac{\left|C_k\right|}{|D|}{\log }_2\frac{\left|C_k\right|}{|D|}$$
  2. 计算$ x $不同取值的情况下，$ y $的熵
     $$H(D|A)=\sum _{i=1}^n\frac{\left|D_i\right|}{|D|}H\left(D_i\right)=-\sum _{i=1}^n\frac{\left|D_i\right|}{|D|}\sum _{k=1}^K\frac{\left|D_{ik}\right|}{\left|D_i\right|}{\log }_2\frac{\left|D_{ik}\right|}{\left|D_i\right|}$$
  3. 做差计算增益
     $$g(D,A)=H(D)-H(D|A)$$

import numpy as np
import math
'''
熵的计算
'''
def entropy(y_values):
    e = 0
    unique_vals = np.unique(y_values)
    for val in unique_vals:
        p = np.sum(y_values == val)/len(y_values)
        e += (p * math.log(p, 2))
    return -1 * e

'''
条件熵的计算
'''
def entropy_condition(x_values, y_values):
    ey = entropy(y_values)
    ey_condition = 0
    xy = np.hstack((x_values, y_values))
    unique_x = np.unique(x_values)
    for x_val in unique_x:
        px = np.sum(x_values == x_val) / len(x_values)
        xy_condition_x = xy[np.where(xy[:, 0] == x_val)]
        ey_condition_x = entropy(xy_condition_x[:, 1])
        ey_condition += (px * ey_condition_x)
    return ey - ey_condition

'''
信息增益比：摒弃了选择取值多的特征为重要特征的缺点
'''
def entropy_condition_ratio(x_values, y_values):
    return entropy_condition(x_values, y_values) / entropy(x_values)
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• 以书中P62页的例子作为测试，以下分别为A1， A2的信息增益

xy = np.array([[0,0,0,0,0,1,1,1,1,1,2,2,2,2,2], [0,0,1,1,0,0,0,1,0,0,0,0,1,1,0], [0,0,0,1,0,0,0,1,1,1,1,1,0,0,0], 
             [0,1,1,0,0,0,1,1,2,2,2,1,1,2,0], [0,0,1,1,0,0,0,1,1,1,1,1,1,1,0]]).T
#A1
print(entropy_condition(xy[:, 0].reshape(-1, 1), 
                        xy[:, -1].reshape(-1, 1)))
#A2
print(entropy_condition(xy[:, 1].reshape(-1, 1), 
                        xy[:, -1].reshape(-1, 1)))

#A3
print(entropy_condition(xy[:, 2].reshape(-1, 1), 
                        xy[:, -1].reshape(-1, 1)))

#A4
print(entropy_condition(xy[:, 3].reshape(-1, 1), 
                        xy[:, -1].reshape(-1, 1)))

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• 与书中结果相合

xy = np.array([[0,0,0,0,0,1,1,1,1,1,2,2,2,2,2], [0,0,1,1,0,0,0,1,0,0,0,0,1,1,0], [0,0,0,1,0,0,0,1,1,1,1,1,0,0,0], 
             [0,1,1,0,0,0,1,1,2,2,2,1,1,2,0], [0,0,1,1,0,0,0,1,1,1,1,1,1,1,0]]).T
#A1
print(entropy_condition_ratio(xy[:, 0].reshape(-1, 1), 
                        xy[:, -1].reshape(-1, 1)))
#A2
print(entropy_condition_ratio(xy[:, 1].reshape(-1, 1), 
                        xy[:, -1].reshape(-1, 1)))

#A3
print(entropy_condition_ratio(xy[:, 2].reshape(-1, 1), 
                        xy[:, -1].reshape(-1, 1)))

#A4
print(entropy_condition_ratio(xy[:, 3].reshape(-1, 1), 
                        xy[:, -1].reshape(-1, 1)))

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基尼指数 公式

$$\mathrm{Gini}(p)=\sum _{k=1}^K{p}_k\left(1-p_k\right)=1-\sum _{k=1}^K{p}_k^2$$
$$\mathrm{Gini}(D)=1-\sum _{k=1}^K{\left(\frac{\left|C_k\right|}{|D|}\right)}^2$$
$$\mathrm{Gini}(D,A)=\frac{\left|D_1\right|}{|D|}\mathrm{Gini}\left(D_1\right)+\frac{\left|D_2\right|}{|D|}\mathrm{Gini}\left(D_2\right)$$

'''
基尼指数计算
'''
def gini(y_values):
    g = 0
    unique_vals = np.unique(y_values)
    for val in unique_vals:
        p = np.sum(y_values == val)/len(y_values)
        g += (p * p)
    return 1 - g

'''
按照x取值的基尼指数的计算
'''
def gini_condition(x_values, y_values):
    g_condition = {}
    xy = np.hstack((x_values, y_values))
    unique_x = np.unique(x_values)
    for x_val in unique_x:
        xy_condition_x = xy[np.where(xy[:, 0] == x_val)]
        xy_condition_notx = xy[np.where(xy[:, 0] != x_val)]
        g_condition[x_val] = len(xy_condition_x)/len(x_values) * gini(xy_condition_x[:, 1]) + len(xy_condition_notx)/len(x_values) * gini(xy_condition_notx[:, 1])
    return g_condition
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xy = np.array([[0,0,0,0,0,1,1,1,1,1,2,2,2,2,2], [0,0,1,1,0,0,0,1,0,0,0,0,1,1,0], [0,0,0,1,0,0,0,1,1,1,1,1,0,0,0], 
             [0,1,1,0,0,0,1,1,2,2,2,1,1,2,0], [0,0,1,1,0,0,0,1,1,1,1,1,1,1,0]]).T
#A1
print(gini_condition(xy[:, 0].reshape(-1, 1), 
                        xy[:, -1].reshape(-1, 1)))
#A2
print(gini_condition(xy[:, 1].reshape(-1, 1), 
                        xy[:, -1].reshape(-1, 1)))

#A3
print(gini_condition(xy[:, 2].reshape(-1, 1), 
                        xy[:, -1].reshape(-1, 1)))

#A4
print(gini_condition(xy[:, 3].reshape(-1, 1), 
                        xy[:, -1].reshape(-1, 1)))

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• 与书中p71相符，选择最小的特征及$x$取值作为最优特征及分切点。
• 其实选取基尼指数最小，即选择在哪个特征下以及该特征取哪个值的情况下，$y$的不确定性最小

特征重要性的对比

以随机森林算法进行特征重要性计算，以书中数据为例

from sklearn.ensemble import RandomForestClassifier

rf = RandomForestClassifier(random_state=42).fit(xy[:, :-1], xy[:, -1])

print(rf. feature_importances_)

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• 总体上相符