如何更改Python字典中条目的键?
当前回答
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。
其他回答
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}
在python 2.7及更高版本中,您可以使用字典理解: 这是我在使用DictReader读取CSV时遇到的一个例子。用户已经在所有列名后面加上了':'
ori_dict ={“key1:”:1、“key2:”:2,“key3:”:3}
在键中去掉后面的':':
Corrected_dict = {k.replace(':', "): v for k, v in ori_dict.items()}
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/
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
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