Python program to Swap Keys and Values in Dictionary
Last Updated :
17 Apr, 2023
Dictionary is quite a useful data structure in programming that is usually used to hash a particular key with value so that they can be retrieved efficiently. Let’s discuss various ways of swapping the keys and values in Python Dictionary.
Method#1 (Does not work when there are multiple same values): One naive solution maybe something like just swapping the key and values respectively.
Example:
Input : {'A': 67, 'B': 23, 'C': 45, 'D': 56, 'E': 12, 'F': 69, 'G': 67, 'H': 23}
Output: {67: 'G', 69: 'F', 23: 'H', 56: 'D', 12: 'E', 45: 'C'}
Python3
old_dict = {'A': 67, 'B': 23, 'C': 45, 'D': 56, 'E': 12, 'F': 69, 'G': 67, 'H': 23}
new_dict = dict([(value, key) for key, value in old_dict.items()])
print ("Original dictionary is : ")
print(old_dict)
print()
print ("Dictionary after swapping is : ")
print("keys: values")
for i in new_dict:
print(i, " : ", new_dict[i])
|
Output
Original dictionary is :
{'D': 56, 'E': 12, 'C': 45, 'A': 67, 'F': 69, 'H': 23, 'B': 23, 'G': 67}
Dictionary after swapping is :
keys: values
67 : G
69 : F
23 : B
56 : D
12 : E
45 : C
But there is a problem with this approach. In our example, we have multiple keys with the same values i.e. (‘A’, 67) and (‘G’, 67) and the other keys having the same values are (‘B’, 23) and (‘H’, 23). But in the result, we obtained only one key from each. i.e we obtained only (‘G’, 67) and (‘B’, 23). So, here’s another approach to deal with this problem:
Method#2 (Handles multiple same values): In this approach, we will check if the value is already present or not. If present then just append it to the list.
Example:
Input : {'A': 67, 'B': 23, 'C': 45, 'E': 12, 'F': 69, 'G': 67, 'H': 23}
Output: {45: ['C'], 67: ['A', 'G'], 12: ['E'], 69: ['F'], 23: ['B', 'H']}
Python3
old_dict = {'A': 67, 'B': 23, 'C': 45, 'E': 12, 'F': 69, 'G': 67, 'H': 23}
print ("Original dictionary is : ")
print(old_dict)
print()
new_dict = {}
for key, value in old_dict.items():
if value in new_dict:
new_dict[value].append(key)
else:
new_dict[value]=[key]
print ("Dictionary after swapping is : ")
print("keys: values")
for i in new_dict:
print(i, " :", new_dict[i])
|
Output
Original dictionary is :
{'F': 69, 'G': 67, 'H': 23, 'A': 67, 'C': 45, 'B': 23, 'E': 12}
Dictionary after swapping is :
keys: values
45 : ['C']
67 : ['G', 'A']
12 : ['E']
69 : ['F']
23 : ['H', 'B']
Method 3: Using List Comprehension and Set
Python3
dict1 = {"a": 1, "b": 2, "c": 3, "d": 2}
dict2 = {}
dict2 = dict([(v, [k for k, v1 in dict1.items() if v1 == v])
for v in set(dict1.values())])
print(dict2)
|
Output
{1: ['a'], 2: ['b', 'd'], 3: ['c']}
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 4: Using get() Method
Python3
dict1 = {"a": 1, "b": 2, "c": 3, "d": 2}
dict2 = {}
for k, v in dict1.items():
dict2[v] = dict2.get(v, []) + [k]
print(dict2)
|
Output
{1: ['a'], 2: ['b', 'd'], 3: ['c']}
Time complexity: O(n), where n is the number of key-value pairs in the dictionary.
Auxiliary space: O(n), where n is the number of key-value pairs in the dictionary.
Method 5: Using Dictionary SetDefault() Method
Python3
dict1 = {"a": 1, "b": 2, "c": 3, "d": 2}
dict2 = {}
for i in dict1:
dict2.setdefault(dict1[i], []).append(i)
print(dict2)
|
Output
{1: ['a'], 2: ['b', 'd'], 3: ['c']}
Method 6: Using DefaultDict
Python3
from collections import defaultdict
dict1 = {"a": 1, "b": 2, "c": 3, "d": 2}
dict2 = defaultdict(list)
{dict2[v].append(k) for k, v in dict1.items()}
print(dict(dict2))
|
Output
{1: ['a'], 2: ['b', 'd'], 3: ['c']}
Time complexity: The time complexity of the given program is O(n), where n is the number of items in the dictionary. .
Auxiliary space: The auxiliary space used by the program is O(n), where n is the number of items in the dictionary.
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