你可以使用iff/else字典理解。此方法允许您在一行中替换任意数量的键，并且不需要更改所有键。

key_map_dict = {'a':'apple','c':'cat'}
d = {'a':1,'b':2,'c':3}
d = {(key_map_dict[k] if k in key_map_dict else k):v  for (k,v) in d.items() }

返回{“苹果”:1、“b”:2,“猫”:3}

我在下面写了这个函数，您可以将当前键名的名称更改为新名称。

def change_dictionary_key_name(dict_object, old_name, new_name):
    '''
    [PARAMETERS]: 
        dict_object (dict): The object of the dictionary to perform the change
        old_name (string): The original name of the key to be changed
        new_name (string): The new name of the key
    [RETURNS]:
        final_obj: The dictionary with the updated key names
    Take the dictionary and convert its keys to a list.
    Update the list with the new value and then convert the list of the new keys to 
    a new dictionary
    '''
    keys_list = list(dict_object.keys())
    for i in range(len(keys_list)):
        if (keys_list[i] == old_name):
            keys_list[i] = new_name

    final_obj = dict(zip(keys_list, list(dict_object.values()))) 
    return final_obj

假设一个JSON，你可以调用它，并通过以下行重命名它:

data = json.load(json_file)
for item in data:
    item = change_dictionary_key_name(item, old_key_name, new_key_name)

在这里可以找到从列表到字典键的转换:https://www.geeksforgeeks.org/python-ways-to-change-keys-in-dictionary/

对于熊猫，你可以有这样的东西，

from pandas import DataFrame
df = DataFrame([{"fruit":"apple", "colour":"red"}])
df.rename(columns = {'fruit':'fruit_name'}, inplace = True)
df.to_dict('records')[0]
>>> {'fruit_name': 'apple', 'colour': 'red'}

我还没有看到确切的答案:

dict['key'] = value

您甚至可以对对象属性执行此操作。 通过这样做，将它们编入字典:

dict = vars(obj)

然后你可以像操作字典一样操作对象属性:

dict['attribute'] = value

我只是要帮我妻子做一些python类的事情，所以我写了这段代码来告诉她如何做。正如标题所示，它只替换键名。这是非常罕见的，你必须替换一个键名，并保持字典的顺序完整，但我还是想分享，因为这篇文章是当你搜索它时谷歌返回的，即使它是一个非常老的线程。

代码:

dictionary = {
    "cat": "meow",
    "dog": "woof",
    "cow": "ding ding ding",
    "goat": "beh"
}


def countKeys(dictionary):
    num = 0
    for key, value in dictionary.items():
        num += 1
    return num


def keyPosition(dictionary, search):
    num = 0
    for key, value in dictionary.items():
        if key == search:
            return num
        num += 1


def replaceKey(dictionary, position, newKey):
    num = 0
    updatedDictionary = {}
    for key, value in dictionary.items():
        if num == position:
            updatedDictionary.update({newKey: value})
        else:
            updatedDictionary.update({key: value})
        num += 1
    return updatedDictionary


for x in dictionary:
    print("A", x, "goes", dictionary[x])
    numKeys = countKeys(dictionary)

print("There are", numKeys, "animals in this list.\n")
print("Woops, that's not what a cow says...")

keyPos = keyPosition(dictionary, "cow")
print("Cow is in the", keyPos, "position, lets put a fox there instead...\n")
dictionary = replaceKey(dictionary, keyPos, "fox")

for x in dictionary:
    print("A", x, "goes", dictionary[x])

输出:

A cat goes meow
A dog goes woof
A cow goes ding ding ding
A goat goes beh
There are 4 animals in this list.

Woops, that's not what a cow says...
Cow is in the 2 position, lets put a fox there instead...

A cat goes meow
A dog goes woof
A fox goes ding ding ding
A goat goes beh

这个函数获得一个字典，另一个字典指定如何重命名键;它返回一个新的字典，带有重命名的键:

def rekey(inp_dict, keys_replace):
    return {keys_replace.get(k, k): v for k, v in inp_dict.items()}

测试:

def test_rekey():
    assert rekey({'a': 1, "b": 2, "c": 3}, {"b": "beta"}) == {'a': 1, "beta": 2, "c": 3}