用于创建包含字典列表中所有键的子字典的Python程序
给定字典列表，我们的任务是创建一个包含所有键的新字典列表，如果没有，则将 None 分配给每个字典的键并持久化。
例子：
Input : test_list = [{‘gfg’ : 3, ‘is’ : 7}, {‘gfg’ : 3, ‘is’ : 1, ‘best’ : 5}, {‘gfg’ : 8}]
Output : [{‘is’: 7, ‘best’: None, ‘gfg’: 3}, {‘is’: 1, ‘best’: 5, ‘gfg’: 3}, {‘is’: None, ‘best’: None, ‘gfg’: 8}]
Explanation : The items with “is” and “best” are added to all lists, whereever missing as None if no values populated.
Input : test_list = [{‘gfg’ : 3}, {‘gfg’ : 3, ‘best’ : 5}, {‘gfg’ : 8}]
Output : [{‘best’: None, ‘gfg’: 3}, {‘best’: 5, ‘gfg’: 3}, {‘best’: None, ‘gfg’: 8}]
Explanation : The items with “best” are added to all lists, whereever missing as None if no values populated.
方法 #1：使用set() + chain.from_iterable() + get() +列表理解
在此，我们使用 set() 和 chain.from_iterable() 执行获取所有必需密钥的任务。下一步是使用列表理解和 get() 更新所有未找到键的字典。
Python3
# Python3 code to demonstrate working of
# Ensure all keys in dictionary list
# Using set() + chain.from_iterable() + get() + list comprehension
from itertools import chain
# initializing list
test_list = [{'gfg' : 3, 'is' : 7},
{'gfg' : 3, 'is' : 1, 'best' : 5},
{'gfg' : 8}]
# printing original list
print("The original list is : " + str(test_list))
# extracting all keys
all_keys = set(chain.from_iterable(test_list))
# assigning None using get() if key's value is not found
res = [dict((key, sub.get(key, None)) for key in all_keys) for sub in test_list]
# printing result
print("Reformed dictionaries list : " + str(res))Python3
# Python3 code to demonstrate working of
# Ensure all keys in dictionary list
# Using set() + chain.from_iterable() + update()
from itertools import chain
# initializing list
test_list = [{'gfg' : 3, 'is' : 7},
{'gfg' : 3, 'is' : 1, 'best' : 5},
{'gfg' : 8}]
# printing original list
print("The original list is : " + str(test_list))
# extracting all keys
all_keys = set(chain.from_iterable(test_list))
# assigning None using update() if key is not found
for sub in test_list:
sub.update({key: None for key in all_keys if key not in sub})
# printing result
print("Reformed dictionaries list : " + str(test_list))输出：
The original list is : [{‘gfg’: 3, ‘is’: 7}, {‘gfg’: 3, ‘is’: 1, ‘best’: 5}, {‘gfg’: 8}]
Reformed dictionaries list : [{‘gfg’: 3, ‘best’: None, ‘is’: 7}, {‘gfg’: 3, ‘best’: 5, ‘is’: 1}, {‘gfg’: 8, ‘best’: None, ‘is’: None}]
方法 #2：使用 set() + chain.from_iterable() + update()
其中，字典中所有键的更新和检查是使用 update() 完成的，其余所有功能保持相似。
蟒蛇3
# Python3 code to demonstrate working of
# Ensure all keys in dictionary list
# Using set() + chain.from_iterable() + update()
from itertools import chain
# initializing list
test_list = [{'gfg' : 3, 'is' : 7},
{'gfg' : 3, 'is' : 1, 'best' : 5},
{'gfg' : 8}]
# printing original list
print("The original list is : " + str(test_list))
# extracting all keys
all_keys = set(chain.from_iterable(test_list))
# assigning None using update() if key is not found
for sub in test_list:
sub.update({key: None for key in all_keys if key not in sub})
# printing result
print("Reformed dictionaries list : " + str(test_list))
输出：
The original list is : [{‘gfg’: 3, ‘is’: 7}, {‘gfg’: 3, ‘is’: 1, ‘best’: 5}, {‘gfg’: 8}]
Reformed dictionaries list : [{‘gfg’: 3, ‘best’: None, ‘is’: 7}, {‘gfg’: 3, ‘best’: 5, ‘is’: 1}, {‘gfg’: 8, ‘best’: None, ‘is’: None}]