递归父子节点数据

数据样式

idnamep_id1学校A02专业113专业214专业315专业416班级1127班级1228班级213

目的:打印每个节点路径,层级

代码:

# -*- coding: utf-8 -*- from operator import attrgetter from collections import namedtuple d_list = [['1','学校A','0'],['2','专业1','1'],['3','专业2','1'],['4','专业3','1'],['5','专业4','1'],['6','班级11','2'],['7','班级12','2'],['8','班级21','3']] node_dict = {} node = namedtuple('node', 'id, name, p_id') for i in d_list: node_dict[i[0]] = node(*i) def traversal(node_dict, id): if node_dict[id].p_id == '0': return [node_dict[id]] else: return traversal(node_dict, node_dict[id].p_id) + [node_dict[id]] c_list = ['1','2','3','4','5','6','7','8'] for i in c_list: res = traversal(node_dict, i) path = ','.join(map(attrgetter('id'), res)) level = path.count(',') print(i, node_dict[i].name, level, path)

打印结果:

1 学校A 0 1 2 专业1 1 1,2 3 专业2 1 1,3 4 专业3 1 1,4 5 专业4 1 1,5 6 班级11 2 1,2,6 7 班级12 2 1,2,7 8 班级21 2 1,3,8

核心代码参考以下文章: https://bbs.csdn.net/topics/392271500