如何更改Python字典中条目的键?
当前回答
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())
其他回答
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
流行与新鲜
>>>a = {1:2, 3:4}
>>>a[5] = a.pop(1)
>>>a
{3: 4, 5: 2}
>>>
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/
d = {1:2,3:4}
假设我们想要改变列表元素p=['a', 'b']的键值。 下面的代码可以做到:
d=dict(zip(p,list(d.values())))
我们得到
{'a': 2, 'b': 4}
如果你想更改所有的键:
d = {'x':1, 'y':2, 'z':3}
d1 = {'x':'a', 'y':'b', 'z':'c'}
In [10]: dict((d1[key], value) for (key, value) in d.items())
Out[10]: {'a': 1, 'b': 2, 'c': 3}
如果你想改变单键: 你可以选择上面的任何一个建议。
推荐文章
- 证书验证失败:无法获得本地颁发者证书
- 当使用pip3安装包时,“Python中的ssl模块不可用”
- 无法切换Python与pyenv
- Python if not == vs if !=
- 如何从scikit-learn决策树中提取决策规则?
- 为什么在Mac OS X v10.9 (Mavericks)的终端中apt-get功能不起作用?
- 将旋转的xtick标签与各自的xtick对齐
- 为什么元组可以包含可变项?
- 如何合并字典的字典?
- 如何创建类属性?
- 不区分大小写的“in”
- 在Python中获取迭代器中的元素个数
- 解析日期字符串并更改格式
- 使用try和。Python中的if
- 如何在Python中获得所有直接子目录