如何将sklearn决策树规则提取到熊猫布尔条件?

时间:2019-05-28 02:09:24

标签: pandas machine-learning scikit-learn decision-tree

关于如何提取sklearn决策树规则的帖子like this太多了,但我找不到有关使用熊猫的任何信息。

this data and model为例,如下

# Create Decision Tree classifer object
clf = DecisionTreeClassifier(criterion="entropy", max_depth=3)

# Train Decision Tree Classifer
clf = clf.fit(X_train,y_train)

结果:

enter image description here

预期:

关于此示例,有8条规则。

从左到右,请注意数据帧为df

r1 = (df['glucose']<=127.5) & (df['bmi']<=26.45) & (df['bmi']<=9.1)
……
r8 =  (df['glucose']>127.5) & (df['bmi']>28.15) & (df['glucose']>158.5)

我不掌握提取sklearn决策树规则的代码。得到大熊猫布尔条件后,它将帮助我为每个规则计算样本和其他度量。因此,我必须为熊猫的布尔条件提取规则。

3 个答案:

答案 0 :(得分:14)

首先,让我们在决策树结构上使用scikit documentation来获取有关构造的树的信息:

n_nodes = clf.tree_.node_count
children_left = clf.tree_.children_left
children_right = clf.tree_.children_right
feature = clf.tree_.feature
threshold = clf.tree_.threshold

然后,我们定义两个递归函数。第一个将找到从树的根部开始的路径,以创建一个特定的节点(在本例中为所有叶子)。第二个将使用其创建路径编写用于创建节点的特定规则:

def find_path(node_numb, path, x):
        path.append(node_numb)
        if node_numb == x:
            return True
        left = False
        right = False
        if (children_left[node_numb] !=-1):
            left = find_path(children_left[node_numb], path, x)
        if (children_right[node_numb] !=-1):
            right = find_path(children_right[node_numb], path, x)
        if left or right :
            return True
        path.remove(node_numb)
        return False


def get_rule(path, column_names):
    mask = ''
    for index, node in enumerate(path):
        #We check if we are not in the leaf
        if index!=len(path)-1:
            # Do we go under or over the threshold ?
            if (children_left[node] == path[index+1]):
                mask += "(df['{}']<= {}) \t ".format(column_names[feature[node]], threshold[node])
            else:
                mask += "(df['{}']> {}) \t ".format(column_names[feature[node]], threshold[node])
    # We insert the & at the right places
    mask = mask.replace("\t", "&", mask.count("\t") - 1)
    mask = mask.replace("\t", "")
    return mask

最后,我们使用这两个函数来首先存储每个叶子的创建路径。然后存储用于创建每个叶子的规则:

# Leaves
leave_id = clf.apply(X_test)

paths ={}
for leaf in np.unique(leave_id):
    path_leaf = []
    find_path(0, path_leaf, leaf)
    paths[leaf] = np.unique(np.sort(path_leaf))

rules = {}
for key in paths:
    rules[key] = get_rule(paths[key], pima.columns)

根据您提供的数据,输出为:

rules =
{3: "(df['insulin']<= 127.5) & (df['bp']<= 26.450000762939453) & (df['bp']<= 9.100000381469727)  ",
 4: "(df['insulin']<= 127.5) & (df['bp']<= 26.450000762939453) & (df['bp']> 9.100000381469727)  ",
 6: "(df['insulin']<= 127.5) & (df['bp']> 26.450000762939453) & (df['skin']<= 27.5)  ",
 7: "(df['insulin']<= 127.5) & (df['bp']> 26.450000762939453) & (df['skin']> 27.5)  ",
 10: "(df['insulin']> 127.5) & (df['bp']<= 28.149999618530273) & (df['insulin']<= 145.5)  ",
 11: "(df['insulin']> 127.5) & (df['bp']<= 28.149999618530273) & (df['insulin']> 145.5)  ",
 13: "(df['insulin']> 127.5) & (df['bp']> 28.149999618530273) & (df['insulin']<= 158.5)  ",
 14: "(df['insulin']> 127.5) & (df['bp']> 28.149999618530273) & (df['insulin']> 158.5)  "}

由于规则是字符串,因此无法使用df[rules[3]]直接调用它们,因此必须像df[eval(rules[3])]这样使用eval函数

答案 1 :(得分:2)

现在您可以使用export_text。

from sklearn.tree import export_text

r = export_text(loan_tree, feature_names=(list(X_train.columns)))
print(r)

来自sklearn

的完整示例
from sklearn.datasets import load_iris
from sklearn.tree import DecisionTreeClassifier
from sklearn.tree import export_text
iris = load_iris()
X = iris['data']
y = iris['target']
decision_tree = DecisionTreeClassifier(random_state=0, max_depth=2)
decision_tree = decision_tree.fit(X, y)
r = export_text(decision_tree, feature_names=iris['feature_names'])
print(r)

答案 2 :(得分:-1)

我想出了解决此问题的另一种方法(vlemaistre发布的第二部分),该方法允许用户遍历任何节点并根据pandas布尔条件对数据进行子集化。

node_id = 3

def datatree_path_summarystats(node_id):
    for k, v in paths.items():
        if node_id in v:
            d = k,v

    ruleskey = d[0]
    numberofsteps = sum(map(lambda x : x<node_id, d[1]))

    for k, v in rules.items():
        if k == ruleskey:
            b = k,v

    stringsubset = b[1]

    datasubset = "&".join(stringsubset.split('&')[:numberofsteps])
    return datasubset

datasubset = datatree_path_summarystats(node_id)

df[eval(datasubset)]

此功能贯穿包含您要查找的节点ID的路径。然后,它将基于该节点数拆分规则,从而创建基于该特定节点对数据帧进行子集化的逻辑。

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