Python | Initialize common value to keys

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Sometimes, while working with Python, we can be confronted with an issue in which we need to assign each key of dictionary with a common value. This type of problem is not occasional but can occur many times while programming. Let’s discuss certain ways in which this task can be performed.

Method #1 : Using defaultdict() + lambda The defaultdict can be initialized using a function which by default assigns each new key with the common key. This is most recommended way to perform this task.

Python3

# Python3 code to demonstrate working of
# Initialize common value to keys
# Using defaultdict()
from collections import defaultdict
 
# Initialize dictionary
test_dict = dict()
 
# printing original dictionary
print("The original dictionary is : " + str(test_dict))
 
# Initialize common value to keys
# Using defaultdict()
res = defaultdict(lambda: 4, test_dict)
res_demo = res['Geeks']
 
# printing result
print("The value of key is :  " + str(res_demo))

Output :

The original dictionary is : {}
The value of key is :  4

Time Complexity: O(1)
Space Complexity: O(n) -> where n is the number of elements in the dictionary

Method #2 : Using get() + default value This method is just a display hack to perform this task. It doesn’t create the actual list, but just prints the default value passed to get function and hence the result.

Python3

# Python3 code to demonstrate working of
# Initialize common value to keys
# Using get() + default value
 
# Initialize dictionary
test_dict = dict()
 
# printing original dictionary
print("The original dictionary is : " + str(test_dict))
 
# Initialize common value to keys
# Using get() + default value
res_demo = test_dict.get('Geeks', 4)
 
# printing result
print("The value of key is :  " + str(res_demo))

Output :

The original dictionary is : {}
The value of key is :  4

Time Complexity: O(1)
Space Complexity: O(n) -> where n is the number of elements in the dictionary

Method #3: Using a loop to assign the common value to keys

Step-by-step approach to understanding the code you provided:

Initializes an empty dictionary named ‘test_dict‘.
Print the original dictionary using the ‘print()’ function and converts the dictionary to a string using the ‘str()’ function.
Initializes a variable named ‘common_value’ to 4.
Iterate a for loop that iterates over the list [‘Geeks’, ‘for’, ‘Geeks’]. Inside the for loop, the current key in the iteration is used to assign the ‘common_value’ to the key in the ‘test_dict’ dictionary using the key-value assignment syntax.
After the for loop finishes, the dictionary ‘test_dict’ will have three key-value pairs, where each key is ‘Geeks’ and the value is 4.
Now, accesses the key ‘Geeks’ in the dictionary using the indexing operator, and assigns the resulting value to the variable ‘res_demo’.
Print the value of ‘res_demo’ using the ‘print()’ function and converts the value to a string using the ‘str()’ function.

Below is the implementation of the above approach:

Python3

# Initialize dictionary
test_dict = {}
 
# printing original dictionary
print("The original dictionary is: " + str(test_dict))
 
# Initialize common value to keys using a loop
common_value = 4
for key in ['Geeks', 'for', 'Geeks']:
    test_dict[key] = common_value
 
# Access a key that we set a common value for
res_demo = test_dict['Geeks']
 
# printing result
print("The value of key is: " + str(res_demo))

Output

The original dictionary is: {}
The value of key is: 4

This method will have a time complexity of O(n), where n is the number of keys in the dictionar= because we need to iterate over each key to assign the common value.

The space complexity is also O(n), because we are storing n key-value pairs in the dictionary.