Python – 使用给定的列表创建嵌套字典

给定一个列表和字典，将列表的每个元素与字典的每个项目映射，形成嵌套字典作为值。

Input : test_dict = {‘Gfg’ : 4, ‘best’ : 9}, test_list = [8, 2]
Output : {8: {‘Gfg’: 4}, 2: {‘best’: 9}}
Explanation : Index-wise key-value pairing from list [8] to dict {‘Gfg’ : 4} and so on.

Input : test_dict = {‘Gfg’ : 4}, test_list = [8]
Output : {8: {‘Gfg’: 4}}
Explanation : Index-wise key-value pairing from list [8] to dict {‘Gfg’ : 4}.

编程需要懂一点英语

方法 #1：使用循环 + zip()

这是可以执行此任务的方式之一。在此，我们使用 zip() 组合两个列表，并使用循环来进行压缩键的迭代和字典构造。

Python3

# Python3 code to demonstrate working of 
# Nested Dictionary with List
# Using loop + zip()
  
# initializing dictionary and list
test_dict = {'Gfg' : 4, 'is' : 5, 'best' : 9} 
test_list = [8, 3, 2]
  
# printing original dictionary and list
print("The original dictionary is : " + str(test_dict))
print("The original list is : " + str(test_list))
  
# using zip() and loop to perform 
# combining and assignment respectively.
res = {}
for key, ele in zip(test_list, test_dict.items()):
    res[key] = dict([ele])
          
# printing result 
print("The mapped dictionary : " + str(res))

Python3

# Python3 code to demonstrate working of 
# Nested Dictionary with List
# Using dictionary comprehension + zip()
  
# initializing dictionary and list
test_dict = {'Gfg' : 4, 'is' : 5, 'best' : 9} 
test_list = [8, 3, 2]
  
# printing original dictionary and list
print("The original dictionary is : " + str(test_dict))
print("The original list is : " + str(test_list))
  
# zip() and dictionary comprehension mapped in one liner to solve
res = {idx: {key : test_dict[key]} for idx, key in zip(test_list, test_dict)}
          
# printing result 
print("The mapped dictionary : " + str(res))

输出

The original dictionary is : {'Gfg': 4, 'is': 5, 'best': 9}
The original list is : [8, 3, 2]
The mapped dictionary : {8: {'Gfg': 4}, 3: {'is': 5}, 2: {'best': 9}}

方法 #2：使用字典理解 + zip()

这是可以执行此任务的另一种方式。在此，我们执行与上述方法类似的任务，但在一个内衬中使用字典理解

Python3

# Python3 code to demonstrate working of 
# Nested Dictionary with List
# Using dictionary comprehension + zip()
  
# initializing dictionary and list
test_dict = {'Gfg' : 4, 'is' : 5, 'best' : 9} 
test_list = [8, 3, 2]
  
# printing original dictionary and list
print("The original dictionary is : " + str(test_dict))
print("The original list is : " + str(test_list))
  
# zip() and dictionary comprehension mapped in one liner to solve
res = {idx: {key : test_dict[key]} for idx, key in zip(test_list, test_dict)}
          
# printing result 
print("The mapped dictionary : " + str(res))

输出

The original dictionary is : {'Gfg': 4, 'is': 5, 'best': 9}
The original list is : [8, 3, 2]
The mapped dictionary : {8: {'Gfg': 4}, 3: {'is': 5}, 2: {'best': 9}}