怎么在python中实现一个树状嵌套结构-创新互联
怎么在python中实现一个树状嵌套结构？针对这个问题，这篇文章详细介绍了相对应的分析和解答，希望可以帮助更多想解决这个问题的小伙伴找到更简单易行的方法。
Python是一种跨平台的、具有解释性、编译性、互动性和面向对象的脚本语言，其最初的设计是用于编写自动化脚本，随着版本的不断更新和新功能的添加，常用于用于开发独立的项目和大型项目。
原始数据
原始数据大致是这样子的：
每条数据中的四个数据分别是 当前节点名称，节点描述（指代一些需要的节点属性),源节点（即最顶层节点），父节点（当前节点上一层节点）。
datas = [ ["root", "根节点", "root", None], ["node1", "一级节点1", "root", "root"], ["node2", "一级节点2", "root", "root"], ["node11", "二级节点11", "root", "node1"], ["node12", "二级节点12", "root", "node1"], ["node21", "二级节点21", "root", "node2"], ["node22", "二级节点22", "root", "node2"], ]
节点类
抽象封装出一个节点类：
class Node(object): def __init__(self, name: str, desc, parent: str, children: list): """ 初始化 :param name: :param desc: :param parent: :param children: """ self.name = name self.desc = desc self.parent = parent self.children = children def get_nodes(self): """ 获取该节点下的全部结构字典 """ d = dict() d['name'] = self.name d['desc'] = self.desc d['parent'] = self.parent children = self.get_children() if children: d['children'] = [child.get_nodes() for child in children] return d def get_children(self): """ 获取该节点下的全部节点对象 """ return [n for n in nodes if n.parent == self.name] def __repr__(self): return self.name
将原始数据转换为节点对象
nodes = list() for data in datas: node = Node(data[0], data[1], data[-1], []) nodes.append(node)
为各个节点建立联系
for node in nodes: children_names = [data[0] for data in datas if data[-1] == node.name] children = [node for node in nodes if node.name in children_names] node.children.extend(children)
测试
root = nodes[0] print(root) tree = root.get_nodes() print(json.dumps(tree, indent=4))
运行结果：
原始数据也可以是字典的形式：
### fork_tool.py
import json
class Node(object):
def __init__(self, **kwargs):
"""
初始化
:param nodes: 树的全部节点对象
:param kwargs: 当前节点参数
"""
self.forked_id = kwargs.get("forked_id")
self.max_drawdown = kwargs.get("max_drawdown")
self.annualized_returns = kwargs.get("annualized_returns")
self.create_time = kwargs.get("create_time")
self.desc = kwargs.get("desc")
self.origin = kwargs.get("origin")
self.parent = kwargs.get("parent")
self.children = kwargs.get("children", [])
def get_nodes(self, nodes):
"""
获取该节点下的全部结构字典，即建立树状联系
"""
d = dict()
d['forked_id'] = self.forked_id
d['max_drawdown'] = self.max_drawdown
d['annualized_returns'] = self.annualized_returns
d['create_time'] = self.create_time
d['desc'] = self.desc
d['origin'] = self.origin
d['parent'] = self.parent
children = self.get_children(nodes)
if children:
d['children'] = [child.get_nodes(nodes) for child in children]
return d
def get_children(self, nodes):
"""
获取该节点下的全部节点对象
"""
return [n for n in nodes if n.parent == self.forked_id]
# def __repr__(self):
# return str(self.desc)
def process_datas(datas):
"""
处理原始数据
:param datas:
:return:
"""
# forked_infos.append({"forked_id": str(forked_strategy.get("_id")),
# "max_drawdown": max_drawdown,
# "annualized_returns": annualized_returns,
# "create_time": create_time, # 分支创建时间
# "desc": desc,
# "origin": origin,
# "parent": parent,
# "children": [],
# })
nodes = []
# 构建节点列表集
for data in datas:
node = Node(**data)
nodes.append(node)
# 为各个节点对象建立类 nosql 结构的联系
for node in nodes:
children_ids = [data["forked_id"] for data in datas if data["parent"] == node.forked_id]
children = [node for node in nodes if node.forked_id in children_ids]
node.children.extend(children)
return nodes
test_datas = [
{'annualized_returns': 0.01,
'children': [],
'create_time': 1562038393,
'desc': 'root',
'forked_id': '5d1ad079e86117f3883f361e',
'max_drawdown': 0.01,
'origin': None,
'parent': None},
{'annualized_returns': 0.314,
'children': [],
'create_time': 1562060612,
'desc': 'level1',
'forked_id': '5d1b2744b264566d3f3f3632',
'max_drawdown': 0.2,
'origin': '5d1ad079e86117f3883f361e',
'parent': '5d1ad079e86117f3883f361e'},
{'annualized_returns': 0.12,
'children': [],
'create_time': 1562060613,
'desc': 'level11',
'forked_id': '5d1b2745e86117f3883f3632',
'max_drawdown': None,
'origin': '5d1ad079e86117f3883f361e',
'parent': '5d1b2744b264566d3f3f3632'},
{'annualized_returns': 0.09,
'children': [],
'create_time': 1562060614,
'desc': 'level12',
'forked_id': '5d1b2746b264566d3f3f3633',
'max_drawdown': None,
'origin': '5d1ad079e86117f3883f361e',
'parent': '5d1b2744b264566d3f3f3632'},
{'annualized_returns': None,
'children': [],
'create_time': 1562060614,
'desc': 'level2',
'forked_id': '5d1b2746e86117f3883f3633',
'max_drawdown': None,
'origin': '5d1ad079e86117f3883f361e',
'parent': '5d1ad079e86117f3883f361e'},
{'annualized_returns': None,
'children': [],
'create_time': 1562060627,
'desc': 'level21',
'forked_id': '5d1b2753b264566d3f3f3635',
'max_drawdown': None,
'origin': '5d1ad079e86117f3883f361e',
'parent': '5d1b2746e86117f3883f3633'},
{'annualized_returns': None,
'children': [],
'create_time': 1562060628,
'desc': 'level211',
'forked_id': '5d1b2754b264566d3f3f3637',
'max_drawdown': None,
'origin': '5d1ad079e86117f3883f361e',
'parent': '5d1b2753b264566d3f3f3635'},
{'annualized_returns': None,
'children': [],
'create_time': 1562060640,
'desc': 'level212',
'forked_id': '5d1b2760e86117f3883f3634',
'max_drawdown': None,
'origin': '5d1ad079e86117f3883f361e',
'parent': '5d1b2753b264566d3f3f3635'},
]
if __name__ == "__main__":
nodes = process_datas(test_datas)
info = nodes[0].get_nodes(nodes)
print(json.dumps(info, indent=4))关于怎么在python中实现一个树状嵌套结构问题的解答就分享到这里了，希望以上内容可以对大家有一定的帮助，如果你还有很多疑惑没有解开，可以关注创新互联行业资讯频道了解更多相关知识。
分享名称：怎么在python中实现一个树状嵌套结构-创新互联
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