如何从每个节点提取sklearn决策树规则来满足布尔条件？

Question

此问题类似于（可能是简单的扩展）此处链接的问题：

How to extract sklearn decision tree rules to pandas boolean conditions?

以上环节的解决方案综合如下：

First of all let's use the scikit documentation on decision tree structure to get information about the tree that was constructed:

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

We then define two recursive functions. The first one will find the path from the tree's root to create a specific node (all the leaves in our case). The second one will write the specific rules used to create a node using its creation path:

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

Finally, we use those two functions to first store the path of creation of each leaf. And then to store the rules used to create each leaf :

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)

With the data you gave the output is:

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']> 9.100000381469`727)",  
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)  "}

Since the rules are strings, you can't directly call them using df[rules[3]], you have to use the eval function like so df[eval(rules[3])]

上面发布的解决方案在为每个终止节点找到路径方面非常有效。我想知道是否有可能将每个节点（叶子和终止节点）的路径存储为与上面链接中完全相同的输出格式（字典/列表格式）

谢谢

Answer 1

所以我找到了一个解决问题的方法（虽然我不认为这是最好的/最有效的方法），但它也不是我问题的直接答案（我没有存储每个节点的路径-只是创建一个函数来解析存储的信息）。它是上述解决方案的第二部分，允许您为要查找的特定节点提取子集数据

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)]

此函数在包含要查找的节点id的路径中运行。然后，它将根据节点数拆分规则，创建逻辑以基于一个特定节点对数据帧进行子集划分