Python-从字典创建树

Question

我有一个问题，我必须创建一个函数，从任意数量元素的字典中创建一棵树。字典示例：

D1 = {'name': 'musica',
      'children': [
          {'name': 'rock',
           'children': [
               {'name': 'origini', 'children': []},
               {'name': 'rock&roll', 'children': []},
               {'name': 'hard rock', 'children': []}]},
          {'name': 'jazz',
           'children': [
               {'name': 'origini',
                'children': [{'name': '1900',
                              'children': [{'name': 'origini', 'children': []}]}]},
               {'name': 'ragtime', 'children': []},
               {'name': 'swing', 'children': []}]}]}

D2 = {'name': 'html',
      'children': [
          {'name': 'head',
           'children': [
               {'name': 'meta', 'children': []},
               {'name': 'title', 'children': []},
               {'name': 'style', 'children': []}]},
          {'name': 'body',
           'children': [
               {'name': 'h1', 'children': []},
               {'name': 'section',
                'children': [
                    {'name': 'p',
                     'children': [
                         {'name': 'strong', 'children': []},
                         {'name': 'b', 'children': []},
                         {'name': 'em', 'children': []},
                         {'name': 'i', 'children': []}]},
                    {'name': 'p',
                     'children': [
                         {'name': 'q', 'children': []},
                         {'name': 'code', 'children': []},
                         {'name': 'kbd', 'children': []}]},
                    {'name': 'p',
                     'children': [
                         {'name': 'sup', 'children': []},
                         {'name': 'sub', 'children': []}]},
                    {'name': 'p',
                     'children': [
                         {'name': 'span', 'children': []}]}]},
               {'name': 'footer',
                'children': [
                    {'name': 'a',
                     'children': [
                         {'name': 'strong', 'children': []}]},
                    {'name': 'a',
                     'children': [
                         {'name': 'strong', 'children': []}]}]}]}]}

D3 = {'name': 'Giovanni di Bicci',
      'children': [
          {'name': 'Cosimo il vecchio',
           'children': [
               {'name': 'Piero il gottuso',
                'children': [
                    {'name': 'Lorenzo il magnifico',
                     'children': [
                         {'name': 'Piero II',
                          'children': [
                              {'name': 'Lorenzo II', 'children': []}]},
                         {'name': 'Papa Leone X', 'children': []},
                         {'name': 'Giuliano', 'children': []}]}]},
               {'name': 'Giovanni Carlo', 'children': []}]},
          {'name': 'Lorenzo',
           'children': [
               {'name': 'Pierfrancesco',
                'children': [
                    {'name': 'Lorenzo', 'children': []},
                    {'name': 'Giovanni',
                     'children': [
                         {'name': 'Giovanni dalle Bande Nere',
                          'children': [
                              {'name': 'Lorenzino', 'children': []},
                              {'name': 'Cosimo I',
                               'children': [
                                   {'name': 'Francesco I',
                                    'children': [
                                        {'name': 'Maria', 'children': []}]},
                                   {'name': 'Ferdinando I',
                                    'children': {}}]}]}]}]}]}]}

有什么解决办法吗？

多谢

添加

谢谢大家的回答。

现在我写了完整的练习，以便您可以更好地理解和回答我：实际上我已经实现了用这些方法制作的类 TNode：

class TNode(object):

def __init__(self, name, Sinistra= None, Destra= None):
    self._name = name
    self.Destra = Destra
    self.Sinistra = Sinistra
    self._children = []
    self._copy = []
    self.c = c
    
def add(self, c): 
    self._children.append(c)
 

def children(self):
    self._copy = self._children
    return self._copy
    
def height(self):
    h = 1
    for node in self._children:
        h = max(h, node.height() + 1)
    return h
    
def count(self):
    c = 1
    for node in self._children:
        c += node.count()
    return c

 def count_by_name(self, name):
    lst = []
    if self._name == name: 
        lst += [self]
    for node in self._children:
        lst += node.count_by_name(name)
    return lst
    
def leaves(self):
    leaves_s = []
    if not self._children:
        leaves_s.append(self.c)
    for node in self._children:
        leaves_s.update(node.leaves())
    return len(leaves_s)
    
def paths(self, name):
    paths_s = set()
    if self._name == name:
        paths_s.add((name,))
    for node in self._children:
        for j in node.paths(name):
            paths_s.add((self._name,)+j)
    return paths_s

我还需要创建一个函数 create_tree(d)，它采用代表树的字典“d”，创建具有 TNode 类型节点的相应树并返回根。该函数必须按照“children”键列表中列出的顺序添加子项。

抱歉，如果最初我没有写下所有这些。

我无法创建引用类的函数，该函数从字典创建树。

我使用Python 2.7

谢谢。

Answer 1

您可以使用以下命令创建一棵树defaultdict：

from collections import defaultdict

def Tree():
    return defaultdict(Tree)

然后使用它：

>>> tree = Tree()
>>> tree['house']['car']['red']['hubcap'] = 1950