使用 Pandas 将嵌套的 CSV 转换为嵌套的 JSON

时间：2021-06-09 15:04:19

标签： json python-3.x pandas dataframe csv

我有一个这样的数据框

org.iden.account,org.iden.id,adress.city,adress.country,person.name.fullname,person.gender,person.birthYear,subs.id,subs.subs1.birthday,subs.subs1.org.address.country,subs.subs1.org.address.strret1,subs.org.buyer.email.address,subs.org.buyer.phone.number
account123,id123,riga,latvia,laura,female,1990,subs123,1990-12-14T00:00:00Z,latvia,street 1,email1@myorg.com|email2@sanoma.com,+371401234567
account123,id000,riga,latvia,laura,female,1990,subs456,1990-12-14T00:00:00Z,latvia,street 1,email1@myorg.com,+371401234567
account123,id456,riga,latvia,laura,female,1990,subs789,1990-12-14T00:00:00Z,latvia,street 1,email1@myorg.com,+371401234567

我需要根据由点（.）分隔的列将其转换为嵌套的 JSON。所以对于第一行，预期的结果应该是

{
    "org": {
        "iden": {
            "account":  "account123",
            "id": "id123"
        }
    },
    "address": {
        "city": "riga",
        "country": "country"
    },
    "person": {
        "name": {
            "fullname": laura,
        },
        "gender": "female",
        "birthYear": 1990
    },
    "subs": {
        "id": "subs123",
        "subs1": {
            "birthday": "1990-12-14T00:00:00Z",
            "org": {
                "address": {
                    "country": "latvia",
                    "street1": "street 1"
                }
            }
        },
        "org": {
            "buyer": {
                "email": {
                    "address": "email1@myorg.com|email2@sanoma.com"
                },
            "phone": {
                "number": "+371401234567"
                }
            }
        }
    }

}

然后当然是所有记录作为列表。我曾尝试使用简单的熊猫 .to_json() 但它没有帮助，我得到以下没有我需要的嵌套结构。

[{"org.iden.account":"account123","org.iden.id":"id123","adress.city":"riga","adress.country":"latvia","person.name.fullname":"laura","person.gender":"female","person.birthYear":1990,"subs.id":"subs123","subs.subs1.birthday":"1990-12-14T00:00:00Z","subs.subs1.org.address.country":"latvia","subs.subs1.org.address.strret1":"street 1","subs.org.buyer.email.address":"email1@myorg.com|email2@sanoma.com","subs.org.buyer.phone.number":371401234567},{"org.iden.account":"account123","org.iden.id":"id000","adress.city":"riga","adress.country":"latvia","person.name.fullname":"laura","person.gender":"female","person.birthYear":1990,"subs.id":"subs456","subs.subs1.birthday":"1990-12-14T00:00:00Z","subs.subs1.org.address.country":"latvia","subs.subs1.org.address.strret1":"street 1","subs.org.buyer.email.address":"email1@myorg.com","subs.org.buyer.phone.number":371407654321},{"org.iden.account":"account123","org.iden.id":"id456","adress.city":"riga","adress.country":"latvia","person.name.fullname":"laura","person.gender":"female","person.birthYear":1990,"subs.id":"subs789","subs.subs1.birthday":"1990-12-14T00:00:00Z","subs.subs1.org.address.country":"latvia","subs.subs1.org.address.strret1":"street 1","subs.org.buyer.email.address":"email1@myorg.com","subs.org.buyer.phone.number":371407654321}]

对此的任何帮助将不胜感激！

2 个答案:

答案 0 :(得分：1)

def df_to_json(row):
    tree = {}
    for item in row.index:
        t = tree
        for part in item.split('.'):
            prev, t = t, t.setdefault(part, {})
        prev[part] = row[item]
    return tree

>>> df.apply(df_to_json, axis='columns').tolist()

[{'org': {'iden': {'account': 'account123', 'id': 'id123'}},
  'adress': {'city': 'riga', 'country': 'latvia'},
  'person': {'name': {'fullname': 'laura'},
   'gender': 'female',
   'birthYear': 1990},
  'subs': {'id': 'subs123',
   'subs1': {'birthday': '1990-12-14T00:00:00Z',
    'org': {'address': {'country': 'latvia', 'strret1': 'street 1'}}},
   'org': {'buyer': {'email': {'address': 'email1@myorg.com|email2@sanoma.com'},
     'phone': {'number': 371401234567}}}}},
 {'org': {'iden': {'account': 'account123', 'id': 'id000'}},
  'adress': {'city': 'riga', 'country': 'latvia'},
  'person': {'name': {'fullname': 'laura'},
   'gender': 'female',
   'birthYear': 1990},
  'subs': {'id': 'subs456',
   'subs1': {'birthday': '1990-12-14T00:00:00Z',
    'org': {'address': {'country': 'latvia', 'strret1': 'street 1'}}},
   'org': {'buyer': {'email': {'address': 'email1@myorg.com'},
     'phone': {'number': 371401234567}}}}},
 {'org': {'iden': {'account': 'account123', 'id': 'id456'}},
  'adress': {'city': 'riga', 'country': 'latvia'},
  'person': {'name': {'fullname': 'laura'},
   'gender': 'female',
   'birthYear': 1990},
  'subs': {'id': 'subs789',
   'subs1': {'birthday': '1990-12-14T00:00:00Z',
    'org': {'address': {'country': 'latvia', 'strret1': 'street 1'}}},
   'org': {'buyer': {'email': {'address': 'email1@myorg.com'},
     'phone': {'number': 371401234567}}}}}]

答案 1 :(得分：0)

假设您的 POST /my_source_index/_clone/my_target_index 结构看起来像这样

json

您可以将它嵌套在 json_data = [ { "org.iden.account": "account123", "org.iden.id": "id123", "adress.city": "riga", "adress.country": "latvia", "person.name.fullname": "laura", "person.gender": "female", "person.birthYear": 1990, "subs.id": "subs123", "subs.subs1.birthday": "1990-12-14T00:00:00Z", "subs.subs1.org.address.country": "latvia", "subs.subs1.org.address.strret1": "street 1", "subs.org.buyer.email.address": "email1@myorg.com|email2@sanoma.com", "subs.org.buyer.phone.number": 371401234567 }, { "org.iden.account": "account123", "org.iden.id": "id000", "adress.city": "riga", "adress.country": "latvia", "person.name.fullname": "laura", "person.gender": "female", "person.birthYear": 1990, "subs.id": "subs456", "subs.subs1.birthday": "1990-12-14T00:00:00Z", "subs.subs1.org.address.country": "latvia", "subs.subs1.org.address.strret1": "street 1", "subs.org.buyer.email.address": "email1@myorg.com", "subs.org.buyer.phone.number": 371407654321 }, { "org.iden.account": "account123", "org.iden.id": "id456", "adress.city": "riga", "adress.country": "latvia", "person.name.fullname": "laura", "person.gender": "female", "person.birthYear": 1990, "subs.id": "subs789", "subs.subs1.birthday": "1990-12-14T00:00:00Z", "subs.subs1.org.address.country": "latvia", "subs.subs1.org.address.strret1": "street 1", "subs.org.buyer.email.address": "email1@myorg.com", "subs.org.buyer.phone.number": 371407654321 } ] 的基础上 dict。

dict

此函数采用未嵌套的 def nestify(unnested): nested = dict() for k, v in unnested.items(): current_dict = nested parts = k.split('.') for i in parts[:-1]: if i not in current_dict: current_dict[i] = dict() current_dict = current_dict[i] current_dict[parts[-1]] = v return nested 之一，遍历键并将值分配给最终深度。

评论版

dict

然后您可以在推导式中对 def nestify(unnested): # this will be our return value nested = dict() for k, v in unnested.items(): # current_dict is the current dict were operating on # gets reset to the base dict on each unnested key current_dict = nested parts = k.split('.') # only create dicts up to the final period # for example, current_dict is the base # and creates an empty dict under the org key # then current_dict is under the org key # and creates an empty dict under the iden key # then current_dict is under the iden key for i in parts[:-1]: # no reason to create an empty dict if it was # already created for a prior key if i not in current_dict: current_dict[i] = dict() current_dict = current_dict[i] # assign the value of the unnested dict # to each final current_dict # for example, the final part of the first key is "account" # so rather than assign an empty dict, assign it "account123" current_dict[parts[-1]] = v return nested json_data 的每个元素调用它。

list

完整代码：

nested = [nestify(i) for i in json_data]

输出：

json_data = [
    {
        "org.iden.account": "account123",
        "org.iden.id": "id123",
        "adress.city": "riga",
        "adress.country": "latvia",
        "person.name.fullname": "laura",
        "person.gender": "female",
        "person.birthYear": 1990,
        "subs.id": "subs123",
        "subs.subs1.birthday": "1990-12-14T00:00:00Z",
        "subs.subs1.org.address.country": "latvia",
        "subs.subs1.org.address.strret1": "street 1",
        "subs.org.buyer.email.address": "email1@myorg.com|email2@sanoma.com",
        "subs.org.buyer.phone.number": 371401234567
    },
    {
        "org.iden.account": "account123",
        "org.iden.id": "id000",
        "adress.city": "riga",
        "adress.country": "latvia",
        "person.name.fullname": "laura",
        "person.gender": "female",
        "person.birthYear": 1990,
        "subs.id": "subs456",
        "subs.subs1.birthday": "1990-12-14T00:00:00Z",
        "subs.subs1.org.address.country": "latvia",
        "subs.subs1.org.address.strret1": "street 1",
        "subs.org.buyer.email.address": "email1@myorg.com",
        "subs.org.buyer.phone.number": 371407654321
    },
    {
        "org.iden.account": "account123",
        "org.iden.id": "id456",
        "adress.city": "riga",
        "adress.country": "latvia",
        "person.name.fullname": "laura",
        "person.gender": "female",
        "person.birthYear": 1990,
        "subs.id": "subs789",
        "subs.subs1.birthday": "1990-12-14T00:00:00Z",
        "subs.subs1.org.address.country": "latvia",
        "subs.subs1.org.address.strret1": "street 1",
        "subs.org.buyer.email.address": "email1@myorg.com",
        "subs.org.buyer.phone.number": 371407654321
    }
]


def nestify(unnested):
    nested = dict()
    for k, v in unnested.items():
        current_dict = nested
        parts = k.split('.')
        for i in parts[:-1]:
            if i not in current_dict:
                current_dict[i] = dict()
            current_dict = current_dict[i]
        current_dict[parts[-1]] = v
    return nested

nested = [nestify(i) for i in json_data]
print(nested)
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