Python – Resize Keys in dictionary

Last Updated : 03 May, 2023

Given Dictionary, resize keys to K by getting starting k elements from keys.

Input : test_dict = {“geeksforgeeks” :3, “best” :3, “coding” :4, “practice” :3}, K = 3
Output : {‘gee’: 3, ‘bes’: 3, ‘cod’: 4, ‘pra’: 3}
Explanation : Keys resized to have 3 elements.

Input : test_dict = {“geeksforgeeks” :3, “best” :3, “coding” :4, “practice” :3}, K = 4
Output : {‘geek’: 3, ‘best’: 3, ‘codi’: 4, ‘prac’: 3}
Explanation : Keys resized to have 4 elements.

Method #1 : Using slicing + loop

In this, resizing is done using slicing of dictionary keys, loop is used to iterate for all the keys of dictionary.

Python3

# Python3 code to demonstrate working of
# Resize Keys in dictionary
# Using slicing + loop
 
# initializing dictionary
test_dict = {"geeksforgeeks": 3, "best": 3, "coding": 4, "practice": 3}
 
# printing original dictionary
print("The original dictionary is : " + str(test_dict))
 
# initializing K
K = 2
 
# reforming dictionary
res = dict()
for key in test_dict:
 
    # resizing to K prefix keys
    res[key[:K]] = test_dict[key]
 
# printing result
print("The required result : " + str(res))

Output:

The original dictionary is : {‘geeksforgeeks’: 3, ‘best’: 3, ‘coding’: 4, ‘practice’: 3} The required result : {‘ge’: 3, ‘be’: 3, ‘co’: 4, ‘pr’: 3}

Time Complexity: O(N*K), where N is the number of elements in the dictionary and K is the length of the each dictionary.

Auxiliary Space: O(N)

Method #2 : Using dictionary comprehension + slicing

In this, we perform task of reforming dictionary in one liner using dictionary comprehension.

Python3

# Python3 code to demonstrate working of
# Resize Keys in dictionary
# Using dictionary comprehension + slicing
 
# initializing dictionary
test_dict = {"geeksforgeeks": 3, "best": 3, "coding": 4, "practice": 3}
 
# printing original dictionary
print("The original dictionary is : " + str(test_dict))
 
# initializing K
K = 2
 
# reforming dictionary
res = {key[:K]: test_dict[key] for key in test_dict}
 
# printing result
print("The required result : " + str(res))

Output:

The original dictionary is : {‘geeksforgeeks’: 3, ‘best’: 3, ‘coding’: 4, ‘practice’: 3} The required result : {‘ge’: 3, ‘be’: 3, ‘co’: 4, ‘pr’: 3}

Method #3: Using map() function with lambda function

Initialize a lambda function that takes in a key-value pair from the original dictionary and slices the first K characters of the key to create a new key.
Use the map() function to apply the lambda function to each key-value pair in the original dictionary.
Convert the resulting map object to a dictionary using the dict() constructor to create the new dictionary with resized keys and corresponding values.

Python3

# Python3 code to demonstrate working of
# Resize Keys in dictionary
# Using map() + lambda
 
# initializing dictionary
test_dict = {"geeksforgeeks": 3, "best": 3, "coding": 4, "practice": 3}
 
# printing original dictionary
print("The original dictionary is : " + str(test_dict))
 
# initializing K
K = 2
 
# reforming dictionary
res = dict(map(lambda x: (x[0][:K], x[1]), test_dict.items()))
 
# printing result
print("The required result : " + str(res))

Output

The original dictionary is : {'geeksforgeeks': 3, 'best': 3, 'coding': 4, 'practice': 3}
The required result : {'ge': 3, 'be': 3, 'co': 4, 'pr': 3}

Time complexity: O(N), where N is the number of key-value pairs in the dictionary.
Auxiliary space: O(N), to store the new dictionary with resized keys and corresponding values.