Python 替换我的功能的更好方法?

Python 替换我的功能的更好方法?,python,json,pandas,dataframe,python-3.6,Python,Json,Pandas,Dataframe,Python 3.6,我附加了一个json数据链接供下载- 目前,我已经编写了以下函数,用于将每一级的子数据放入一个组合数据框中- def get_children(catMapping): level4 = json_normalize(catMapping['SuccessResponse']['Body'], ['children', 'children', 'children', 'children', ['childr

我附加了一个json数据链接供下载-

目前,我已经编写了以下函数,用于将每一级的子数据放入一个组合数据框中-

def get_children(catMapping):
    level4 = json_normalize(catMapping['SuccessResponse']['Body'],
                                        ['children', 'children', 'children', 'children', ['children']])
    level3 = json_normalize(catMapping['SuccessResponse']['Body'],
                                        ['children', 'children', 'children', ['children']])
                        ['children', 'children', ['children']])
    level1 = json_normalize(catMapping['SuccessResponse']['Body'],
                                        ['children', ['children']])
    level0 = json_normalize(catMapping['SuccessResponse']['Body'],
                                        ['children'])

    combined = pd.concat([level0, level1, level2, level3,level4])
    combined = combined.reset_index(drop=True)

    return combined
看起来这不是推荐的方法,但我无法编写一个可以遍历每个级别的函数


你能帮我找到更好的函数吗?

这里有一个递归迭代所有项的函数:

import pandas as pd
import ast

with open(r"data.json", "r") as f:
    data = ast.literal_eval(f.read())

def nest_iter(items):
    for item in items:
        children_ids = [o["categoryId"] for o in item["children"]]
        ret_item = item.copy()
        ret_item["children"] = children_ids
        yield ret_item
        yield from nest_iter(item["children"])

df = pd.DataFrame(nest_iter(data['SuccessResponse']['Body']))
结果是:

      categoryId                        children   leaf         name    var
....
4970    10001244                              []   True     Business  False
4971    10001245                              []   True       Casual  False
4972    10001246                              []   True      Fashion  False
4973    10001247                              []   True       Sports  False
4974        7756  [7761, 7758, 7757, 7759, 7760]  False        Women  False
4975        7761                              []   True  Accessories  False
4976        7758                              []   True     Business  False
4977        7757                              []   True       Casual  False
4978        7759                              []   True      Fashion  False
4979        7760                              []   True       Sports  False

嗯,你为什么要用
literal\u eval
而不是
json
包呢?哦,我明白了,数据不是真正的json。单引号而不是双引号。谢谢@HYRY。这很有帮助。我将对此进行测试,并很快分享结果。