Question

我有包含多个字典的列表形式的JSON数据。我想创建一个新字典，其值与旧字典相同，但键不同。我正在使用原始键并删除下划线以及大写每个单词。问题是，当我创建新字典时，它不会在相同的庄园中创建字典（请参见下面的代码输出）。

如何创建具有与旧词典相同的结构但具有经过修改的键的新词典？

JSON数据

contacts_to_parse = [
    {
        "contact_order": "1",
        "contact_type": "Other",
        "contact_role": "role here",
        "contact_text": "contact text here"
    },
    {
        "contact_order": "2",
        "contact_type": "Mail Alias",
        "contact_role": "role here",
        "contact_text": "contact text here"
    },
    {
        "contact_order": "3",
        "contact_type": "Mail Alias",
        "contact_role": "role here",
        "contact_text": "contact text here"
    },
    {
        "contact_order": "5",
        "contact_type": "Other",
        "contact_role": "role here",
        "contact_text": "contact text here"
    },
    {
        "contact_order": "7",
        "contact_type": "Employee",
        "contact_role": "role here",
        "contact_text": "contact text here"
    },
    {
        "contact_order": "8",
        "contact_type": "Employee",
        "contact_role": "role here",
        "contact_text": "contact text here"
    },
    {
        "contact_order": "9",
        "contact_type": "Other",
        "contact_role": "role here",
        "contact_text": "contact text here"
    }
]

到目前为止的

代码：

def check(contacts):
    parsed_keys = []
    new_dict = {}
    for contact in contacts:
        for key, val in contact.items():
            rm_underscore_capitalize = key.replace('_', ' ').title()
            new_dict[rm_underscore_capitalize] = val
            if rm_underscore_capitalize in new_dict:
                parsed_keys.append(new_dict)
                new_dict = dict()

    return parsed_keys


if __name__ == '__main__':
    con = check(contacts_to_parse)
    print(con)

当前输出：

[{'Contact Order': '1'}, {'Contact Type': 'Other'}, {'Contact Role': 'role here'}, {'Contact Text': 'contact text here'}, {'Contact Order': '2'}, {'Contact Type': 'Mail Alias'}, {'Contact Role': 'role here'}, {'Contact Text': 'contact text here'}, {'Contact Order': '3'}, {'Contact Type': 'Mail Alias'}, {'Contact Role': 'role here'}, {'Contact Text': 'contact text here'}, {'Contact Order': '5'}, {'Contact Type': 'Other'}, {'Contact Role': 'role here'}, {'Contact Text': 'contact text here'}, {'Contact Order': '7'}, {'Contact Type': 'Employee'}, {'Contact Role': 'role here'}, {'Contact Text': 'contact text here'}, {'Contact Order': '8'}, {'Contact Type': 'Employee'}, {'Contact Role': 'role here'}, {'Contact Text': 'contact text here'}, {'Contact Order': '9'}, {'Contact Type': 'Other'}, {'Contact Role': 'role here'}, {'Contact Text': 'contact text here'}]

Answer 1

您需要将内循环结束附近的一些处理拉到外循环。您试图遍历字典列表，然后遍历键以更新'ban-eur.kd_kd_ban-eur<sd,td>'，但是您却经常重新定义new_dict。试试这个：

new_dict

Answer 2

您的逻辑略有偏离。您想将条件移至第二个for循环之外：

def check(contacts):
    parsed_keys = []
    new_dict = {}
    for contact in contacts:
        for key, val in contact.items():
            rm_underscore_capitalize = key.replace('_', ' ').title()
            new_dict[rm_underscore_capitalize] = val
        if rm_underscore_capitalize in new_dict:
            parsed_keys.append(new_dict)
            new_dict = dict()

    return parsed_keys

返回：

[
{'Contact Order': '1', 'Contact Type': 'Other',      'Contact Role': 'role here', 'Contact Text': 'contact text here'}, 
{'Contact Order': '2', 'Contact Type': 'Mail Alias', 'Contact Role': 'role here', 'Contact Text': 'contact text here'}, 
{'Contact Order': '3', 'Contact Type': 'Mail Alias', 'Contact Role': 'role here', 'Contact Text': 'contact text here'}, 
{'Contact Order': '5', 'Contact Type': 'Other',      'Contact Role': 'role here', 'Contact Text': 'contact text here'}, 
{'Contact Order': '7', 'Contact Type': 'Employee',   'Contact Role': 'role here', 'Contact Text': 'contact text here'}, 
{'Contact Order': '8', 'Contact Type': 'Employee',   'Contact Role': 'role here', 'Contact Text': 'contact text here'}, 
{'Contact Order': '9', 'Contact Type': 'Other',      'Contact Role': 'role here', 'Contact Text': 'contact text here'}
]

Answer 3

您期望这样吗？我试图使其变得简单。

contacts_to_parse = [
    {
        "contact_order": "1",
        "contact_type": "Other",
        "contact_role": "role here",
        "contact_text": "contact text here"
    },
    {
        "contact_order": "2",
        "contact_type": "Mail Alias",
        "contact_role": "role here",
        "contact_text": "contact text here"
    }
]


def check(contacts):
    result = []
    for item in contacts:
        result.append({})
        for key, val in item.items():
            result[-1][key.replace('_', ' ').title()] = val
    return result


print(check(contacts_to_parse))

结果：

[
    {
        'Contact Order': '1',
        'Contact Type': 'Other',
        'Contact Role': 'role here',
        'Contact Text': 'contact text here'
    },
    {
        'Contact Order': '2',
        'Contact Type': 'Mail Alias',
        'Contact Role': 'role here',
        'Contact Text': 'contact text here'
    }
]

Answer 4

您可以使用列表推导和字典推导来简化代码：

def check(contacts):
    return [_check_contact(contact) for contact in contacts]

def _check_contact(contact):
    return {key.replace('_', ' ').title(): value for key, value in contact.items()}

from pprint import pprint
pprint (check(contacts_to_parse))

结果：

[{'Contact Order': '1',
  'Contact Role': 'role here',
  'Contact Text': 'contact text here',
  'Contact Type': 'Other'},
 {'Contact Order': '2',
  'Contact Role': 'role here',
  'Contact Text': 'contact text here',
  'Contact Type': 'Mail Alias'},
 {'Contact Order': '3',
  'Contact Role': 'role here',
  'Contact Text': 'contact text here',
  'Contact Type': 'Mail Alias'},
 {'Contact Order': '5',
  'Contact Role': 'role here',
  'Contact Text': 'contact text here',
  'Contact Type': 'Other'},
 {'Contact Order': '7',
  'Contact Role': 'role here',
  'Contact Text': 'contact text here',
  'Contact Type': 'Employee'},
 {'Contact Order': '8',
  'Contact Role': 'role here',
  'Contact Text': 'contact text here',
  'Contact Type': 'Employee'},
 {'Contact Order': '9',
  'Contact Role': 'role here',
  'Contact Text': 'contact text here',
  'Contact Type': 'Other'}]

Answer 5

您可以利用列表和字典理解：

contacts_to_parse = [
    {
        "contact_order": "1",
        "contact_type": "Other",
        "contact_role": "role here",
        "contact_text": "contact text here"
    },
    {
        "contact_order": "2",
        "contact_type": "Mail Alias",
        "contact_role": "role here",
        "contact_text": "contact text here"
    },
]

from pprint import pprint

pprint([{k.replace('_', ' ').title():v for k, v in contact.items()} for contact in contacts_to_parse])

打印：

[{'Contact Order': '1',
  'Contact Role': 'role here',
  'Contact Text': 'contact text here',
  'Contact Type': 'Other'},
 {'Contact Order': '2',
  'Contact Role': 'role here',
  'Contact Text': 'contact text here',
  'Contact Type': 'Mail Alias'}]

修改字典键并分配给新字典

5 个答案: