Python | Initialize dictionary with None values

Sometimes, while working with dictionaries, we might have a utility in which we need to initialize a dictionary with None values so that they can be altered later. This kind of application can occur in cases of memoization in general or competitive programming. Let’s discuss certain ways in which this task can be performed. 

Initialize dictionary with None values Using zip() + repeat()

The combination of these functions can be used to perform this particular task. In this, the None value is attached to the keys repeated using the repeat() by help of zip() 

Python3

# Python3 code to demonstrate working of
# Initialize dictionary with None values
# Using zip() + repeat()
from itertools import repeat

# Using zip() + repeat()
# Initialize dictionary with None values
res = dict(zip(range(10), repeat(None)))

# printing result
print("The dictionary with None values : " + str(res))

Output

The dictionary with None values : {0: None, 1: None, 2: None, 3: None, 4: None, 5: None, 6: None, 7: None, 8: None, 9: None}

Time Complexity: O(n), where n is the length of the list test_dict
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the res list

Initialize dictionary with None values Using fromkeys()

This task can also be performed more efficiently using the inbuilt function of fromkeys() which is tailor-made for this task itself and hence recommended. 

Python3

# Python3 code to demonstrate working of
# Initialize dictionary with None values
# Using fromkeys()

# Using fromkeys()
# Initialize dictionary with None values
res = dict.fromkeys(range(10))

# printing result
print("The dictionary with None values : " + str(res))

Output

The dictionary with None values : {0: None, 1: None, 2: None, 3: None, 4: None, 5: None, 6: None, 7: None, 8: None, 9: None}

Initialize dictionary with None values Using dict() and enumerate

In this method, we use the built-in function enumerate() which returns an iterator that produces tuples containing the index and the value of the elements in an iterable. We use repeat(None, 10) which returns an iterator that repeats the value None 10 times. We then pass these tuples to the dict() constructor to create the dictionary with keys being the indexes and values being None.

Python3

# Python3 code to demonstrate working of
# Initialize dictionary with None values
# Using dict() and enumerate

from itertools import repeat

# Using dict() and enumerate
# Initialize dictionary with None values
res = dict(enumerate(repeat(None, 10)))

# printing result
print("The dictionary with None values : " + str(res))
# This code is contributed by Edula Vinay Kumar Reddy

Output

The dictionary with None values : {0: None, 1: None, 2: None, 3: None, 4: None, 5: None, 6: None, 7: None, 8: None, 9: None}

Time complexity: O(n)
Auxiliary Space: O(n)

Initialize dictionary with None values Using a loop and update ()

Algorithm:

1. Initialize variable n to 10.
2. Initialize an empty dictionary res.
3. Start a loop for n times: a. Update the dictionary res with a key-value pair where key is i and value is None.
4. Print the dictionary res.

Python3

n = 10
res = {}
for i in range(n):
    res.update({i: None})
print("The dictionary with None values : " + str(res))

Output

The dictionary with None values : {0: None, 1: None, 2: None, 3: None, 4: None, 5: None, 6: None, 7: None, 8: None, 9: None}

Time complexity: O(n) The loop iterates n times, where n is the size of the input, so the time complexity is O(n).
Auxiliary Space: O(n) The dictionary res stores n key-value pairs, so the space complexity is O(n).

Initialize dictionary with None values Using a dictionary comprehension

Step-by-step approach:

• The code uses a dictionary comprehension to create a new dictionary named ‘res‘.
• The dictionary comprehension iterates over a range of 10 numbers using the ‘range()‘ function.
• For each number in the range, a new key-value pair is created in the ‘res‘ dictionary. The key is the current number from the range, and the value is set to None.
• Finally, the resulting dictionary is printed using the ‘print()‘ function.

Below is the implementation of the above approach:

Python3

# Using a dictionary comprehension
res = {i: None for i in range(10)}

# printing result
print("The dictionary with None values : " + str(res))

Output

The dictionary with None values : {0: None, 1: None, 2: None, 3: None, 4: None, 5: None, 6: None, 7: None, 8: None, 9: None}

Time complexity: O(n), where n is the number of elements in the range.
Auxiliary space: O(n), where n is the number of elements in the range. This is because the dictionary stores n key-value pairs.