在python中解析没有明确键名的嵌套json

Question

我有一个嵌套的 json ，如下所示。我正在尝试从下面的 json 中获取城市和标签的值。

   {

      "5c7a3dd5-29e4-4f5b-9dc8-d2d451906cf1": {

        "id": "1d14eba1-de9e-4e53-9e9e-619432fd815b",

        "opinion_id": "3063e6ff-1483-4daf-8e80-41afa43b6348",

        "paragraph_id": "e6d419fe-5017-47f6-acf1-0e74034aed4a",

        "position": 7,

        "city": "Berlin.",

        "label": "Capital"

      },

      "0c0adcad-992f-4c30-aa58-7bf539f8d649": {

        "case_id": "6f6e70f7-3257-4f26-a8fa-8cf0e5f97920",

        "opinion_id": "71b435ff-112c-470b-a36b-c552e54a5605",

        "paragraph_id": "e6d419fe-5017-47f6-acf1-0e74034aed4a",

        "position": 3,

        "city": "Frankfurt.",

        "label": "Economic Capital"

      },

      "de968cdf-d865-4049-9f7d-f9a68be27432": {

        "case_id": "2d749a58-412c-4bc8-b268-7d5f7bec3d9a",

        "opinion_id": "1836c972-d6f7-420b-ba33-8f863cfca482",

        "paragraph_id": "e6d419fe-5017-47f6-acf1-0e74034aed4a",

        "position": 4,

        "city": "Paris.",

        "label": "Capital"

      }

    }

我尝试的是循环遍历 json 并将值提取到字典列表中。它给我带来了一个关键错误。我有点明白了。如果外部json的键相同，那么我可以使用text = jsonData['KEY'][i]["city"]

with open(json_file_name) as input:

    jsonData = json.load(input)

    JSONLength = len(jsonData)

    data = []

    classes = dict()

    for i in range(0, JSONLength):

        text = jsonData[i]["city"]

        label = jsonData[i]["label"]

        print(text)

        print(label)

        data += [(classes[label], text)]

Answer 1

迭代字典的值：with open(json_file_name) as input:     jsonData = json.load(input)     data = [(values['city'], values['label']) for values in jsonData.values()]         print(data)出去：[('Berlin.', 'Capital'), ('Frankfurt.', 'Economic Capital'), ('Paris.', 'Capital')]

Answer 2

尝试data =   {      "5c7a3dd5-29e4-4f5b-9dc8-d2d451906cf1": {        "id": "1d14eba1-de9e-4e53-9e9e-619432fd815b",        "opinion_id": "3063e6ff-1483-4daf-8e80-41afa43b6348",        "paragraph_id": "e6d419fe-5017-47f6-acf1-0e74034aed4a",        "position": 7,        "city": "Berlin.",        "label": "Capital"      },      "0c0adcad-992f-4c30-aa58-7bf539f8d649": {        "case_id": "6f6e70f7-3257-4f26-a8fa-8cf0e5f97920",        "opinion_id": "71b435ff-112c-470b-a36b-c552e54a5605",        "paragraph_id": "e6d419fe-5017-47f6-acf1-0e74034aed4a",        "position": 3,        "city": "Frankfurt.",        "label": "Economic Capital"      },      "de968cdf-d865-4049-9f7d-f9a68be27432": {        "case_id": "2d749a58-412c-4bc8-b268-7d5f7bec3d9a",        "opinion_id": "1836c972-d6f7-420b-ba33-8f863cfca482",        "paragraph_id": "e6d419fe-5017-47f6-acf1-0e74034aed4a",        "position": 4,        "city": "Paris.",        "label": "Capital"      }    }    city_n_label = [(x['city'],x['label']) for x in data.values()]print(city_n_label)输出[('Berlin.', 'Capital'), ('Frankfurt.', 'Economic Capital'), ('Paris.', 'Capital')]

Answer 3

尝试这个with open(json_file_name) as input:    jsonData = json.load(input)    for k, v in jsonData.items():        print(k) # prints dynamic key        print(v) # prints its corresponding obj        print(v['city']) # will print city of the particular key        print(v['label']) # will print label of the particular key