Sometimes you might need to convert a list to dict object for some better and fast operation. Let’s see how to convert a list into a dictionary of none values. Here we will find three methods of doing this.
Method #1: Using zip()and dict
Python3
# Python code to demonstrate# converting list into dictionary with none values# using zip() and dictionary# initializing listini_list = [1, 2, 3, 4, 5]# printing initialized listprint ("initial list", str(ini_list))# Converting list into dictionary using zip() and dictionaryres = dict(zip(ini_list, [None]*len(ini_list)))# printing final resultprint ("final dictionary", str(res)) |
Output:
initial list [1, 2, 3, 4, 5]
final dictionary {1: None, 2: None, 3: None, 4: None, 5: None}
Method #2: Using dict
Python3
# Python code to demonstrate converting # list into dictionary with none values# using dict()# initializing listini_list = [1, 2, 3, 4, 5]# printing initialized listprint ("initial list", str(ini_list))# Converting list into dict()res = dict.fromkeys(ini_list)# printing final resultprint ("final dictionary", str(res)) |
Output:
initial list [1, 2, 3, 4, 5]
final dictionary {1: None, 2: None, 3: None, 4: None, 5: None}
Method #3: Using dict comprehension
Python3
# Python code to demonstrate converting # list into dictionary with none values# using dict comprehension# initializing listini_list = [1, 2, 3, 4, 5]# printing initialized listprint ("initial list", str(ini_list))# Converting list into dict()res = {key: None for key in ini_list}# printing final resultprint ("final dictionary", str(res)) |
Output:
initial list [1, 2, 3, 4, 5]
final dictionary {1: None, 2: None, 3: None, 4: None, 5: None}
Method #4: using enumerate()
Python3
# Initialize a listini_list = [1, 2, 3, 4, 5]# Print the initial listprint("initial list", str(ini_list))# Use a dictionary comprehension and the enumerate() function# to convert the list into a dictionary with None valuesres = {k: None for i, k in enumerate(ini_list)}# Print the final dictionaryprint("final dictionary", str(res))#this code is contributed Vinay Pinjala. |
Output
initial list [1, 2, 3, 4, 5]
final dictionary {1: None, 2: None, 3: None, 4: None, 5: None}
Time Complexity:O(N)
Auxiliary Space :O(N)
Method #5: Using a loop and the setdefault() method of the dictionary
Steps:
- Initialize an empty dictionary.
- Iterate through each element of the list using a loop and for each element, and use the setdefault() method of the dictionary to add a new key-value pair with the element as the key and None as the value. If the key already exists in the dictionary, setdefault() does not add a new key-value pair and returns the existing value.
- The result is the dictionary with the key-value pairs from the list, where all values are None.
Python3
# Python code to demonstrate converting list# into dictionary with none values# using a loop and setdefault()# initializing listini_list = [1, 2, 3, 4, 5]# printing initialized listprint ("initial list", str(ini_list))# Converting list into dictionary using# a loop and setdefault()res = {}for element in ini_list: res.setdefault(element, None)# printing final resultprint ("final dictionary", str(res)) |
Output
initial list [1, 2, 3, 4, 5]
final dictionary {1: None, 2: None, 3: None, 4: None, 5: None}
Time complexity: O(n), where n is the length of the list ini_list.
Auxiliary space: O(n), to store the dictionary with the key-value pairs.
Method #6: Using pop() function and a while loop to create the dictionary with None values
- Initialize the input list, ini_list.
- Initialize an empty dictionary, res.
- Loop over the range of indices of ini_list in ascending order:
- Use pop(0) to remove the first element from ini_list.
- Add the removed element as a key to the res dictionary with a value of None.
- Print the resulting res dictionary.
Python3
# initialize the input listini_list = [1, 2, 3, 4, 5]# initialize an empty dictionaryres = {}# printing initialized listprint("initial list", str(ini_list))# loop over the range of indices of# ini_list in ascending orderfor i in range(len(ini_list)): res[ini_list.pop(0)] = None# printing final resultprint("final dictionary", str(res)) |
Output
{1: None, 2: None, 3: None, 4: None, 5: None}
Time complexity: O(n), where n is the length of the input list.
Space complexity: O(n), where n is the length of the input list.
Method #7: Using map and lambda functions:
- We start by initializing a list of integers ini_list.
- We print the initialized list using the print() function.
- We create a new dictionary by calling map() function, which applies a lambda function to each element of the ini_list and returns a new list of tuples with each tuple consisting of an element of ini_list and None.
- We convert the list of tuples into a dictionary by calling the dict() function on the result from step 3.
- We print the final dictionary using the print() function.
Python3
# Python code to demonstrate# converting list into dictionary with none values# using map() and lambda functions# initializing listini_list = [1, 2, 3, 4, 5]# printing initialized listprint("initial list:", ini_list)# Converting list into dictionary using map() and lambda functionsres = dict(map(lambda x: (x, None), ini_list))# printing final resultprint("final dictionary:", res) |
Output
initial list: [1, 2, 3, 4, 5]
final dictionary: {1: None, 2: None, 3: None, 4: None, 5: None}
Time complexity: O(n), where n is the length of the input list.
Space complexity: O(n), where n is the length of the input list.
