Given a dictionary, write a Python program to convert the given dictionary into list of tuples.
Examples:
Input: { 'Geeks': 10, 'for': 12, 'Geek': 31 }
Output : [ ('Geeks', 10), ('for', 12), ('Geek', 31) ]
Input: { 'dict': 11, 'to': 22, 'list_of_tup': 33}
Output : [ ('dict', 11), ('to', 22), ('list_of_tup', 33) ]
Below are various methods to convert dictionary to list of tuples.
Method #1: Using list comprehension
Python3
# Python code to convert dictionary into list of tuples# Initialization of dictionarydict = {'Geeks': 10, 'for': 12, 'Geek': 31}# Converting into list of tuplelist = [(k, v) for k, v in dict.items()]# Printing list of tupleprint(list) |
[('Geek', 31), ('for', 12), ('Geeks', 10)]
Time Complexity: O(n), where n is the number of key-value pairs.
Auxiliary Space: O(n)
Method #2 : Using items()
Python3
# Python code to convert dictionary into list of tuples# Initialization of dictionarydict = {'Geeks': 10, 'for': 12, 'Geek': 31}# Converting into list of tuplelist = list(dict.items())# Printing list of tupleprint(list) |
[('for', 12), ('Geeks', 10), ('Geek', 31)]
Time complexity: O(n), where n is the number of key-value pairs in the dictionary. This is because the list function takes O(n) time to convert the dictionary into a list of tuples, and the dictionary operations used in the code (e.g., dict.items()) take constant time.
Auxiliary space: O(n), where n is the number of key-value pairs in the dictionary.
Method #3: Using zip
Python3
# Python code to convert dictionary into list of tuples# Initialization of dictionarydict = {'Geeks': 10, 'for': 12, 'Geek': 31}# Using ziplistt = zip(dict.keys(), dict.values())# Converting from zip object to list objectlistt = list(listt)# Printing listprint(listt) |
[('Geek', 31), ('Geeks', 10), ('for', 12)]
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 #4: Using iteration
Python3
# Python code to convert dictionary into list of tuples# Initialization of dictionarydict = {'Geeks': 10, 'for': 12, 'Geek': 31}# Initialization of empty listlist = []# Iterationfor i in dict: k = (i, dict[i]) list.append(k)# Printing listprint(list) |
[('Geeks', 10), ('for', 12), ('Geek', 31)]
Time complexity: The time complexity of the given code is O(n), where n is the number of key-value pairs in the dictionary.
Auxiliary space: The auxiliary space used by the code is also O(n), where n is the number of key-value pairs in the dictionary.
Method #5 : Using collection
Python3
# Python code to convert dictionary into list of tuples# Initialization of dictionaryimport collectionsdict = {'Geeks': 10, 'for': 12, 'Geek': 31}# Importing# Convertinglist_of_tuple = collections.namedtuple('List', 'name value')lists = list(list_of_tuple(*item) for item in dict.items())# Printing listprint(lists) |
[List(name='Geeks', value=10), List(name='for', value=12), List(name='Geek', value=31)]
Time complexity: O(n), where n is the number of key-value pairs in the dictionary.
Auxiliary space: O(n), to store the keys and values in dictionary.
Method #6 : Using map()
Python3
# Python code to convert dictionary into list of tuples# Initialization of dictionarydict = {'Geeks': 10, 'for': 12, 'Geek': 31}k = list(dict.keys())l = list(dict.values())# Converting into list of tuplelist1 = map(lambda a: (k[a], l[a]), range(len(dict)))# Printing list of tupleprint('list is ', [*list1]) |
list is [('Geeks', 10), ('for', 12), ('Geek', 31)]
Method 7: Using a loop and the items() method of the dictionary object.
Approach:
- Initialize an empty list list1.
- Iterate through the items of the dictionary using a for loop and the items() method, which returns a list of key-value pairs.
- For each item in the list, append a tuple of the key and value to the list1.
- Print the list1 of tuples.
Below is the implementation of the above approach:
Python3
# Initialization of dictionarydict1 = {'Geeks': 10, 'for': 12, 'Geek': 31}# Initializing an empty list to store the tupleslist1 = []# Iterating through the dictionary and appending tuples to the listfor key, value in dict1.items(): list1.append((key, value))# Printing the list of tuplesprint('list is', list1) |
list is [('Geeks', 10), ('for', 12), ('Geek', 31)]
Time complexity: O(n), where n is the number of items in the dictionary. We need to iterate through each item to create the tuples.
Auxiliary space: O(n), where n is the number of items in the dictionary. We need to store n tuples in the list.
Method 8: Using a for loop and the sorted() function
Steps:
- Initialize the dictionary.
- Sort the keys of the dictionary using the sorted() function.
- Use a for loop to iterate over the sorted keys and create a list of tuples where the first element of each tuple is a key from the dictionary and the second element of each tuple is the corresponding value from the dictionary.
- Print the list of tuples.
Python3
# Initialization of dictionarydict = { 'Geeks': 10, 'for': 12, 'Geek': 31 }# Sorting the keys of the dictionarysorted_keys = sorted(dict.keys())# Creating a list of tupleslist_of_tuples = [(key, dict[key]) for key in sorted_keys]# Printing the list of tuplesprint('List is', list_of_tuples) |
List is [('Geek', 31), ('Geeks', 10), ('for', 12)]
Time complexity: O(nlogn) for sorting the keys of the dictionary using the sorted() function and O(n) for creating the list of tuples using a for loop. Therefore, the overall time complexity is O(nlogn).
Auxiliary space: O(n) for storing the sorted keys in a list and creating the list of tuples.
