Given a Dictionary, the task is to write a python program to swap positions of dictionary items. The code given below takes two indices and swap values at those indices.
Input : test_dict = {‘Gfg’ : 4, ‘is’ : 1, ‘best’ : 8, ‘for’ : 10, ‘Lazyroar’ : 9}, i, j = 1, 3
Output : {‘Gfg’: 4, ‘for’: 10, ‘best’: 8, ‘is’: 1, ‘Lazyroar’: 9}
Explanation : (for : 10) and (is : 1) swapped order.
Input : test_dict = {‘Gfg’ : 4, ‘is’ : 1, ‘best’ : 8, ‘for’ : 10, ‘Lazyroar’ : 9}, i, j = 2, 3
Output : {‘Gfg’: 4, ‘is’: 1, ‘for’: 10, ‘best’: 8, ‘Lazyroar’: 9}
Explanation : (for : 10) and (best : 8) swapped order.
Method : Using items() and dict()
This task is achieved in 3 steps:
- First dictionary is converted to equivalent key value pairs in form of tuples,
- Next swap operation is performed in a Pythonic way.
- At last, tuple list is converted back to dictionary, in its required format.
Example:
Python3
# initializing dictionary test_dict = { 'Gfg' : 4 , 'is' : 1 , 'best' : 8 , 'for' : 10 , 'Lazyroar' : 9 } # printing original dictionary print ( "The original dictionary is : " + str (test_dict)) # initializing swap indices i, j = 1 , 3 # conversion to tuples tups = list (test_dict.items()) # swapping by indices tups[i], tups[j] = tups[j], tups[i] # converting back res = dict (tups) # printing result print ( "The swapped dictionary : " + str (res)) |
Output:
The original dictionary is : {‘Gfg’: 4, ‘is’: 1, ‘best’: 8, ‘for’: 10, ‘Lazyroar’: 9}
The swapped dictionary : {‘Gfg’: 4, ‘for’: 10, ‘best’: 8, ‘is’: 1, ‘Lazyroar’: 9}
Time Complexity: O(n)
Auxiliary Space: O(1)
Method 2 ; dictionary manipulation using built-in functions and data types.
Here are the steps:
Initialize a dictionary test_dict with some key-value pairs.
Print the original dictionary.
Initialize two variables i and j with the indices of the key-value pairs that need to be swapped.
Convert the dictionary to a list of tuples using the items() method.
Swap the tuples at indices i and j in the list using tuple unpacking.
Convert the list of tuples back to a dictionary using the dict() method.
Print the swapped dictionary.
Python3
# initializing dictionary test_dict = { 'Gfg' : 4 , 'is' : 1 , 'best' : 8 , 'for' : 10 , 'Lazyroar' : 9 } # printing original dictionary print ( "The original dictionary is : " + str (test_dict)) # initializing swap indices i, j = 1 , 3 # conversion to tuples tups = list (test_dict.items()) # swapping by indices tups[i], tups[j] = tups[j], tups[i] # converting back to dictionary res = dict (tups) # printing result print ( "The swapped dictionary : " + str (res)) |
The original dictionary is : {'Gfg': 4, 'is': 1, 'best': 8, 'for': 10, 'Lazyroar': 9} The swapped dictionary : {'Gfg': 4, 'for': 10, 'best': 8, 'is': 1, 'Lazyroar': 9}
The time complexity of this program is O(n), where n is the number of key-value pairs in the dictionary.
The space complexity of the program is O(n), as we are creating a new list of tuples to store the key-value pairs, and then creating a new dictionary from this list.
METHOD 3:Using loop
APPROACH:
The problem is to swap the positions of key-value pairs in a dictionary in Python using a loop.
ALGORITHM:
1.Create an empty dictionary swapped_dict to store the swapped key-value pairs.
2.Iterate over the key-value pairs of the original dictionary original_dict using a for loop.
3.For each key-value pair, swap the positions of the key and value and add them to the swapped_dict dictionary.
4.Print the swapped_dict dictionary.
Python3
# Original dictionary original_dict = { 'Gfg' : 4 , 'is' : 1 , 'best' : 8 , 'for' : 10 , 'Lazyroar' : 9 } # Create an empty dictionary to store the swapped items swapped_dict = {} # Swap the positions of items using a loop for key, value in original_dict.items(): swapped_dict[value] = key # Print the swapped dictionary print ( "The swapped dictionary :" , swapped_dict) |
The swapped dictionary : {4: 'Gfg', 1: 'is', 8: 'best', 10: 'for', 9: 'Lazyroar'}
Time Complexity:
The time complexity of the algorithm is O(n), where n is the number of key-value pairs in the dictionary. This is because we need to iterate over each key-value pair once.
Space Complexity:
The space complexity of the algorithm is also O(n), where n is the number of key-value pairs in the dictionary. This is because we need to create an empty dictionary to store the swapped key-value pairs. The size of this dictionary will be the same as the original dictionary.