如何把如下这样一个 Json tree 转换成如何 dataframe

[ariera]

import json

str_json = '{"id":"S0","label":"部门","child":[{"id":"S1","label":"管理层","child":[]},{"id":"S2","label":"人力资源部","child":[{"id":"S21","label":"招聘组","child":[{"id":"S210","label":"上海组","child":[{"id":"S2101","label":"上海其他","child":[]},{"id":"S2102","label":"陆家嘴","child":[]}]},{"id":"S211","label":"深圳组","child":[]},{"id":"S212","label":"北京组","child":[]}]},{"id":"S22","label":"考核组","child":[]},{"id":"S23","label":"制度组","child":[]}]},{"id":"S3","label":"行政管理部","child":[{"id":"S31","label":"后勤组","child":[]},{"id":"S31","label":"食堂组","child":[]}]},{"id":"S4","label":"信息技术部","child":[{"id":"S41","label":"运行部","child":[{"id":"S411","label":"运维组","child":[]}]}]}]}'
js = json.loads(str_json)

https://imgur.com/OwfYUme

以上只是个示例，实际中最深可能有 6 层。 将上图的 js 转成如下 dataframe （ id 列在第一列也行），求赐教

					id
部门	管理层				S1
部门	人力资源部	招聘组	上海组	上海其他	S2101
部门	人力资源部	招聘组	上海组	陆家嘴	S2102
部门	人力资源部	招聘组	深圳组		S211
部门	人力资源部	招聘组	北京组		S212
部门	人力资源部	考核组			S22
部门	人力资源部	制度组			S23
部门	行政管理部	后勤组			S31
部门	行政管理部	食堂组			S32
部门	行政管理部	设施组			S33
部门	信息技术部	运行部	运维组		S411

[Donahue]

import json
from typing import List
import pandas as pd

class People():
def __init__(self, js_data, label_prefix:List=None):
self.id = js_data['id']

if label_prefix == None:
self.label = [js_data['label']]
else:
self.label = label_prefix + [js_data['label']]

self.child = [People(child_js, self.label) for child_js in js_data['child']]

def get_label_i(self, i:int):
if i >= len(self.label):
return ''
else:
return self.label[i]

def get_all_instance(p:People):
result = []
result.append(p)
for child in p.child:
result += get_all_instance(child)
return result



str_json = '{"id":"S0","label":"部门","child":[{"id":"S1","label":"管理层","child":[]},{"id":"S2","label":"人力资源部","child":[{"id":"S21","label":"招聘组","child":[{"id":"S210","label":"上海组","child":[{"id":"S2101","label":"上海其他","child":[]},{"id":"S2102","label":"陆家嘴","child":[]}]},{"id":"S211","label":"深圳组","child":[]},{"id":"S212","label":"北京组","child":[]}]},{"id":"S22","label":"考核组","child":[]},{"id":"S23","label":"制度组","child":[]}]},{"id":"S3","label":"行政管理部","child":[{"id":"S31","label":"后勤组","child":[]},{"id":"S31","label":"食堂组","child":[]}]},{"id":"S4","label":"信息技术部","child":[{"id":"S41","label":"运行部","child":[{"id":"S411","label":"运维组","child":[]}]}]}]}'
js = json.loads(str_json)

root_instance = People(js)
all_instance = get_all_instance(root_instance)

df = pd.DataFrame()
df['id'] = [i.id for i in all_instance]

max_depth_people = max(all_instance, key=lambda x: len(x.label))
label_depth = len(max_depth_people.label)

for label_num in range(label_depth):
df[f"label{label_num}"] = [i.get_label_i(label_num) for i in all_instance]

print(df)

[Donahue]

ps: 不要吐槽代码写得烂 hh

[Donahue]

就是递归建立对象，然后把所有对象保存到列表里

[fkdog]

本质上就是树结构的遍历算法。
BFS 请用队列
DFS 请用递归 or 栈

[ariera]

@Donahue #1

[ariera]

@Donahue #1 好像无法运行，可能是缩进有问题。呜呜呜.... 能不能用 三个点``` 圈一下代码

[Donahue]

https://pastebin.com/a86cyvPG

[Donahue]

上面弄错了
https://pastebin.com/D1rbnbBW

[ariera]

@Donahue #8 赞！