对python 树状嵌套结构的实现思路详解

yipeiwu_com5年前Python基础

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))

以上这篇对python 树状嵌套结构的实现思路详解就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持【听图阁-专注于Python设计】。

相关文章

python实现Dijkstra算法的最短路径问题

python实现Dijkstra算法的最短路径问题

迪杰斯特拉(Dijkstra)算法主要是针对没有负值的有向图,求解其中的单一起点到其他顶点的最短路径算法。 1 算法原理 迪杰斯特拉(Dijkstra)算法是一个按照路径长度递增的次序产...

TensorFlow实现iris数据集线性回归

TensorFlow实现iris数据集线性回归

本文将遍历批量数据点并让TensorFlow更新斜率和y截距。这次将使用Scikit Learn的内建iris数据集。特别地,我们将用数据点(x值代表花瓣宽度,y值代表花瓣长度)找到最优...

如何用Python制作微信好友个性签名词云图

如何用Python制作微信好友个性签名词云图

前言 上次查看了微信好友的位置信息,想了想,还是不过瘾,于是就琢磨起了把微信好友的个性签名拿到,然后分词,接着分析词频,最后弄出词云图来。 1.环境说明 Win10 系统下 Pyt...

Python设计模式之MVC模式简单示例

Python设计模式之MVC模式简单示例

本文实例讲述了Python设计模式之MVC模式。分享给大家供大家参考,具体如下: 一.简单介绍 mvc模式  the  model-view-controller p...

详解python中的文件与目录操作

详解python中的文件与目录操作 一 获得当前路径 1、代码1 >>>import os >>>print('Current directo...