Python | Initializing dictionary with empty lists

In python one usually comes across situations in which one has to use dictionary for storing the lists. But in those cases, one usually checks for first element and then creates a list corresponding to key when it comes. But its always wanted a method to initialize the dict. keys with list. Let’s discuss certain ways to achieve this particular task.

Method #1 : Using Dictionary comprehension
This is most sought of method to do this initialization. In this method, we create the no. of keys we require and then initialize the empty list as we keep on creating the keys, so as to facilitate the append operation afterwards without an error.

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# Python3 code to demonstrate 
# to initialize dictionary with list 
# using dictionary comprehension
  
# using dictionary comprehension to construct
new_dict = {new_list: [] for new_list in range(4)}
      
# printing result
print ("New dictionary with empty lists as keys : " + str(new_dict))

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Output :

New dictionary with empty lists as keys : {0: [], 1: [], 2: [], 3: []}

Method #2 : Using fromkeys()
fromkeys() can be used to perform this by specifying the additional empty list as argument and the range of elements which need to be the key of dictionary being made.

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# Python3 code to demonstrate 
# to initialize dictionary with list 
# using fromkeys()
  
# using fromkeys() to construct
new_dict = dict.fromkeys(range(4), [])
      
# printing result
print ("New dictionary with empty lists as keys : " + str(new_dict))

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Output :

New dictionary with empty lists as keys : {0: [], 1: [], 2: [], 3: []}

Method #3 : Using defaultdict
This is most pythonic way and error free way to use any key without initialization of its value, it has to be told the type of default container of all its keys and then evaluates the operations and structures accordingly.

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# Python3 code to demonstrate 
# to initialize dictionary with list 
# using defauldict
from collections import defaultdict
  
# initializing dict with lists
new_dict = defaultdict(list)
  
# performing append
# shows no error
new_dict[0].append('GeeksforGeeks')
      
# printing result
print ("New dictionary created : " + str(dict(new_dict)))

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Output :

New dictionary created : {0: ['GeeksforGeeks']}


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