你可以使用下面的代码:

OldDict={'a':'v1', 'b':'v2', 'c':'v3'}

OldKey=['a','b','c']
NewKey=['A','B','C']

def DictKeyChanger(dict,OldKey,NewKey):
    ListAllKey=list(dict.keys())
    for x in range(0,len(NewKey)):
        dict[NewKey[x]]=dict[OldKey[x]] if OldKey[x] in ListAllKey else None
    for x in ListAllKey:
        dict.pop(x)
    return dict

NewDict=DictKeyChanger(OldDict,OldKey,NewKey)
print(NewDict)#===>>{'A': 'v1', 'B': 'v2', 'C': 'v3'}

注:

列表OldKey和列表NewKey的长度必须相等。 列表OldKey的长度必须等于listNewKey，如果该键在OldKey中不存在，则将'noexis'改为如下所示。

例子:

OldDict={'a':'v1', 'b':'v2', 'c':'v3'}
OldKey=['a','b','c','noexis','noexis']
NewKey=['A','B','C','D','E']
NewDict=DictKeyChanger(OldDict,OldKey,NewKey)
print(NewDict)#===>>{'A': 'v1', 'B': 'v2', 'C': 'v3', 'D': None, 'E': None}

我想出了这个函数，它不会改变原来的字典。这个函数也支持字典列表。

import functools
from typing import Union, Dict, List


def rename_dict_keys(
    data: Union[Dict, List[Dict]], old_key: str, new_key: str
):
    """
    This function renames dictionary keys

    :param data:
    :param old_key:
    :param new_key:
    :return: Union[Dict, List[Dict]]
    """
    if isinstance(data, dict):
        res = {k: v for k, v in data.items() if k != old_key}
        try:
            res[new_key] = data[old_key]
        except KeyError:
            raise KeyError(
                "cannot rename key as old key '%s' is not present in data"
                % old_key
            )
        return res
    elif isinstance(data, list):
        return list(
            map(
                functools.partial(
                    rename_dict_keys, old_key=old_key, new_key=new_key
                ),
                data,
            )
        )
    raise ValueError("expected type List[Dict] or Dict got '%s' for data" % type(data))

Python 3.6(以后?)我会用下面的一句话

test = {'a': 1, 'old': 2, 'c': 3}
old_k = 'old'
new_k = 'new'
new_v = 4  # optional

print(dict((new_k, new_v) if k == old_k else (k, v) for k, v in test.items()))

生产

{'a': 1, 'new': 4, 'c': 3}

可能值得注意的是，如果没有打印语句，ipython控制台/jupyter笔记本将按照它们选择的顺序显示字典…

假设你想把键k3重命名为k4:

temp_dict = {'k1':'v1', 'k2':'v2', 'k3':'v3'}
temp_dict['k4']= temp_dict.pop('k3')

@helloswift123我喜欢你的功能。下面是在一次调用中重命名多个键的修改:

def rename(d, keymap):
    """
    :param d: old dict
    :type d: dict
    :param keymap: [{:keys from-keys :values to-keys} keymap]
    :returns: new dict
    :rtype: dict
    """
    new_dict = {}
    for key, value in zip(d.keys(), d.values()):
        new_key = keymap.get(key, key)
        new_dict[new_key] = d[key]
    return new_dict

我结合了上面帖子中的一些答案，并提出了下面的解决方案。虽然它很简单，但它可以用作从字典中进行更复杂的键更新的构建块。

test_dict = {'a': 1, 'b': 2, 'c': 3}
print(test_dict)
# {'a': 1, 'b': 2, 'c': 3}
prefix = 'up'
def dict_key_update(json_file):    
    new_keys = []
    old_keys = []
    for i,(key,value) in enumerate(json_file.items()):
        old_keys.append(key)
        new_keys.append(str(prefix) + key) # i have updated by adding a prefix to the 
        # key
    for old_key, new_key in zip(old_keys,new_keys):
        print('old {}, new {}'.format(old_key, new_key))
        if new_key!=old_key:  
           json_file[new_key] = json_file.pop(old_key)
     return json_file

test_dict = dict_key_update(test_dict)
print(test_dict)
# {'upa': 1, 'upb': 2, 'upc': 3}