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/

流行与新鲜

>>>a = {1:2, 3:4}
>>>a[5] = a.pop(1)
>>>a
{3: 4, 5: 2}
>>> 

如果你有一个复杂的字典，这意味着字典中有一个字典或列表:

myDict = {1:"one",2:{3:"three",4:"four"}}
myDict[2][5] = myDict[2].pop(4)
print myDict

Output
{1: 'one', 2: {3: 'three', 5: 'four'}}

转换字典中的所有键

假设这是你的字典:

>>> sample = {'person-id': '3', 'person-name': 'Bob'}

将示例字典键中的所有破折号转换为下划线:

>>> sample = {key.replace('-', '_'): sample.pop(key) for key in sample.keys()}
>>> sample
>>> {'person_id': '3', 'person_name': 'Bob'}

没有直接的方法做到这一点，但你可以删除然后分配

d = {1:2,3:4}

d[newKey] = d[1]
del d[1]

或者做大量的键改变:

d = dict((changeKey(k), v) for k, v in d.items())

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