使用 Python Pandas 将嵌套的 json 解析为 csv

Question

我有一个以下格式的 json：

{"MainName":[{"col1":"12345","col2":"False","col3":"190809","SubName1":{"col4":30.00,"SubName2":{"col5":"19703","col6":"USD"}},"col7":"7372267","SubName3":[{"col8":"345337","col9":"PC"}],"col10":"10265","col11":"29889004","col12":"calculated","col13":"9218","SubName4":{"col14":1,"SubName5":{"col15":"1970324","col16":"integer"}},"col17":"434628","col18":"2020-02-06T13:47:40.000-0800","col19":"754878037","SubName6":{"col20":30.00,"SubName7":{"col21":"19703248","col22":"USD"}}},{"col1":"12345","col2":"False","col3":"190809","SubName1":{"col4":30.00,"SubName2":{"col5":"19703","col6":"USD"}},"col7":"7372267","SubName3":[{"col8":"345337","col9":"PC"}],"col10":"10265","col11":"29889004","col12":"calculated","col13":"9218","SubName4":{"col14":1,"SubName5":{"col15":"1970324","col16":"integer"}},"col17":"434628","col18":"2020-02-06T13:47:40.000-0800","col19":"754878037","SubName6":{"col20":30.00,"SubName7":{"col21":"19703248","col22":"USD"}}}],"skip":0,"top":2,"next":"/v1/APIName?skip=2&top=2"}

我想把它转换成以下格式的csv：

MainName_col1,MainName_col2,MainName_col3,MainName_SubName1_col4,MainName_SubName1_SubName2_col5,MainName_SubName1_SubName2_col6,MainName_col7,MainName_SubName3_col8,MainName_SubName3_col9,MainName_col10,MainName_col11,MainName_col12,MainName_col13,MainName_SubName4_col14,MainName_SubName4_SubName5_col15,MainName_SubName4_SubName5_col16,MainName_col17,MainName_col18,MainName_col19,MainName_SubName6_col20,MainName_SubName6_SubName7_col21,MainName_SubName6_SubName7_col22
12345,False,190809,30.0,19703,USD,7372267,345337,PC,10265,29889004,calculated,9218,1,1970324,integer,434628,2020-02-06T13:47:40.000-0800,754878037,30.0,19703248,USD
12345,False,190809,30.0,19703,USD,7372267,345337,PC,10265,29889004,calculated,9218,2,123453,integer,434628,2020-02-06T13:47:40.000-0800,754878037,30.0,19703248,USD

请帮帮我。

Answer 1

使用下面的函数来扁平化你的 JSON 数据。

dc = {"MainName":[{"col1":"12345","col2":False,"col3":"190809","SubName1":{"col4":30.00,"SubName2":{"col5":"19703","col6":"USD"}},"col7":"7372267","SubName3":[{"col8":"345337","col9":"PC"}],"col10":"10265","col11":"29889004","col12":"calculated","col13":"9218","SubName4":{"col14":1,"SubName5":{"col15":"1970324","col16":"integer"}},"col17":"434628","col18":"2020-02-06T13:47:40.000-0800","col19":"754878037","SubName6":{"col20":30.00,"SubName7":{"col21":"19703248","col22":"USD"}}}],"skip":0,"top":1,"next":"/v1/APIName?skip=1&top=1"}
def flatten(root: str, dict_obj: dict):
    flat = {}
    for i in dict_obj.keys():
        val = dict_obj[i]
        if not isinstance(val, dict) and not isinstance(val, list):
            flat[f'{root}_{i}'] = val
        else:
            if isinstance(val, list):
                val = val[-1]
            flat.update(flatten(f'{root}_{i}', val))
    return flat
flatten('MainName', dc['MainName'][0])

它会给你预期的输出。然后按照你想要的方式使用它。

{'MainName_col1': '12345',
 'MainName_col2': False,
 'MainName_col3': '190809',
 'MainName_SubName1_col4': 30.0,
 'MainName_SubName1_SubName2_col5': '19703',
 'MainName_SubName1_SubName2_col6': 'USD',
 'MainName_col7': '7372267',
 'MainName_SubName3_col8': '345337',
 'MainName_SubName3_col9': 'PC',
 'MainName_col10': '10265',
 'MainName_col11': '29889004',
 'MainName_col12': 'calculated',
 'MainName_col13': '9218',
 'MainName_SubName4_col14': 1,
 'MainName_SubName4_SubName5_col15': '1970324',
 'MainName_SubName4_SubName5_col16': 'integer',
 'MainName_col17': '434628',
 'MainName_col18': '2020-02-06T13:47:40.000-0800',
 'MainName_col19': '754878037',
 'MainName_SubName6_col20': 30.0,
 'MainName_SubName6_SubName7_col21': '19703248',
 'MainName_SubName6_SubName7_col22': 'USD'}

Answer 2

据我了解，您的 dc 如下所示

dc = {"MainName":[{"col1":"12345","col2":"False","col3":"190809","SubName1":{"col4":30.00,"SubName2":{"col5":"19703","col6":"USD"}},"col7":"7372267","SubName3":[{"col8":"345337","col9":"PC"}],"col10":"10265","col11":"29889004","col12":"calculated","col13":"9218","SubName4":{"col14":1,"SubName5":{"col15":"1970324","col16":"integer"}},"col17":"434628","col18":"2020-02-06T13:47:40.000-0800","col19":"754878037","SubName6":{"col20":30.00,"SubName7":{"col21":"19703248","col22":"USD"}}},{"col1_a":"12345XX","col2_b":"False","col3_c":"190809","SubName1":{"col4_d":30.00,"SubName2":{"col5_e":"19703","col6_f":"USD"}},"col7_g":"7372267","SubName3":[{"col8_h":"345337","col9":"PC"}],"col10_i":"10265","col11_j":"29889004","col12_k":"calculated","col13_l":"9218","SubName4":{"col14_m":1,"SubName5":{"col15_n":"1970324","col16_o":"integer"}},"col17_p":"434628","col18_q":"2020-02-06T13:47:40.000-0800","col19_r":"754878037","SubName6":{"col20_s":30.00,"SubName7":{"col21_t":"19703248","col22_u":"USDZZ"}}}],"skip":0,"top":2,"next":"/v1/APIName?skip=2&top=2"}

我用上面的答案把所有东西都拼成了一个对象

def flatten(root: str, dict_obj: dict):
    flat = {}
    for i in dict_obj.keys():
        val = dict_obj[i]
        if not isinstance(val, dict) and not isinstance(val, list):
            flat[f'{root}_{i}'] = val
        else:
            if isinstance(val, list):
                val = val[-1]
            flat.update(flatten(f'{root}_{i}', val))
    return flat

keys_list  = []
values_list = []
for i in range(len(dc['MainName'])):  
  result = flatten('MainName', dc['MainName'][i])
  keys_list.append(list(result.keys()))
  values_list.append(list(result.values()))

for k in keys_list:
    for res in k:
      guestFile = open("sample.csv","a")
      guestFile.write(res)
      guestFile.write(",")
      guestFile.close()

for v in values_list:
    for res in v:
      guestFile = open("sample.csv","a")
      guestFile.write(str(res))
      guestFile.write(",")
      guestFile.close()

查看我的代码athttps://repl.it/@TamilselvanLaks/jsontocsvmul

Note: Use the 'run' button to run the program, left side you can see sample.csv 

there you can see all keys as like you want 

请告诉我我的回答符合您的期望