如何更改Python字典中条目的键?
当前回答
完整解决方案的示例
声明一个json文件,其中包含你想要的映射
{
"old_key_name": "new_key_name",
"old_key_name_2": "new_key_name_2",
}
加载它
with open("<filepath>") as json_file:
format_dict = json.load(json_file)
创建此函数来使用映射格式化字典
def format_output(dict_to_format,format_dict):
for row in dict_to_format:
if row in format_dict.keys() and row != format_dict[row]:
dict_to_format[format_dict[row]] = dict_to_format.pop(row)
return dict_to_format
其他回答
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'}}
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
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
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