Given list values and keys list, convert these values to key value pairs in form of list of dictionaries.
Input : test_list = [“Gfg”, 3, “is”, 8], key_list = [“name”, “id”]
Output : [{‘name’: ‘Gfg’, ‘id’: 3}, {‘name’: ‘is’, ‘id’: 8}]
Explanation : Values mapped by custom key, “name” -> “Gfg”, “id” -> 3.Input : test_list = [“Gfg”, 10], key_list = [“name”, “id”]
Output : [{‘name’: ‘Gfg’, ‘id’: 10}]
Explanation : Conversion of lists to list of records by keys mapping.
Method #1 : Using loop + dictionary comprehension
This is one of the ways in which this task can be performed. In this, we perform mapping values using dictionary comprehension. The iteration is performed using loop.
Python3
# Python3 code to demonstrate working of # Convert List to List of dictionaries# Using dictionary comprehension + loop# initializing liststest_list = ["Gfg", 3, "is", 8, "Best", 10, "for", 18, "Geeks", 33]# printing original listprint("The original list : " + str(test_list))# initializing key list key_list = ["name", "number"]# loop to iterate through elements# using dictionary comprehension# for dictionary constructionn = len(test_list)res = []for idx in range(0, n, 2): res.append({key_list[0]: test_list[idx], key_list[1] : test_list[idx + 1]})# printing result print("The constructed dictionary list : " + str(res)) |
The original list : [‘Gfg’, 3, ‘is’, 8, ‘Best’, 10, ‘for’, 18, ‘Geeks’, 33] The constructed dictionary list : [{‘name’: ‘Gfg’, ‘number’: 3}, {‘name’: ‘is’, ‘number’: 8}, {‘name’: ‘Best’, ‘number’: 10}, {‘name’: ‘for’, ‘number’: 18}, {‘name’: ‘Geeks’, ‘number’: 33}]
Time Complexity: O(n*n) where n is the number of elements in the dictionary
Auxiliary Space: O(n), where n is the number of elements in the dictionary
Method #2 : Using dictionary comprehension + list comprehension
The combination of above functions is used to solve this problem. In this, we perform a similar task as above method. But difference is that its performed as shorthand.
Python3
# Python3 code to demonstrate working of # Convert List to List of dictionaries# Using zip() + list comprehension# initializing liststest_list = ["Gfg", 3, "is", 8, "Best", 10, "for", 18, "Geeks", 33]# printing original listprint("The original list : " + str(test_list))# initializing key list key_list = ["name", "number"]# using list comprehension to perform as shorthandn = len(test_list)res = [{key_list[0]: test_list[idx], key_list[1]: test_list[idx + 1]} for idx in range(0, n, 2)]# printing result print("The constructed dictionary list : " + str(res)) |
The original list : [‘Gfg’, 3, ‘is’, 8, ‘Best’, 10, ‘for’, 18, ‘Geeks’, 33] The constructed dictionary list : [{‘name’: ‘Gfg’, ‘number’: 3}, {‘name’: ‘is’, ‘number’: 8}, {‘name’: ‘Best’, ‘number’: 10}, {‘name’: ‘for’, ‘number’: 18}, {‘name’: ‘Geeks’, ‘number’: 33}]
Method #3: Using zip function and dictionary comprehension
The zip() function creates pairs of corresponding elements from the two lists, and the enumerate() function is used to iterate over the key_list to assign the keys to the dictionary. Finally, the dictionary comprehension is used to construct the dictionaries.
Python3
# initializing liststest_list = ["Gfg", 3, "is", 8, "Best", 10, "for", 18, "Geeks", 33]# initializing key list key_list = ["name", "number"]# using zip() function and dictionary comprehensionres = [{key_list[i]: val for i, val in enumerate(pair)} for pair in zip(test_list[::2], test_list[1::2])]# printing result print("The constructed dictionary list : " + str(res)) |
The constructed dictionary list : [{'name': 'Gfg', 'number': 3}, {'name': 'is', 'number': 8}, {'name': 'Best', 'number': 10}, {'name': 'for', 'number': 18}, {'name': 'Geeks', 'number': 33}]
Time complexity: O(n), where n is the length of the input list test_list.
Auxiliary space: O(n), where n is the length of the input list test_list.
Method 4: Use the groupby() function from the itertools module.
Step-by-step approach:
- Import the itertools module.
- Initialize an empty list res.
- Use the groupby() function to group the test_list into pairs based on the remainder when the index is divided by 2.
- For each pair of elements, create a dictionary with keys “name” and “number“.
- Append each dictionary to the res list.
- Convert the res list to a string and print the result.
Below is the implementation of the above approach:
Python3
# Python3 code to demonstrate working of # Convert List to List of dictionaries# Using groupby() function from itertools module# import itertools moduleimport itertools# initializing liststest_list = ["Gfg", 3, "is", 8, "Best", 10, "for", 18, "Geeks", 33]# printing original listprint("The original list : " + str(test_list))# initializing key listkey_list = ["name", "number"]# using groupby() function to group elements into pairsres = []for pair in zip(test_list[::2], test_list[1::2]): res.append({key_list[0]: pair[0], key_list[1]: pair[1]})# printing resultprint("The constructed dictionary list : " + str(res)) |
The original list : ['Gfg', 3, 'is', 8, 'Best', 10, 'for', 18, 'Geeks', 33]
The constructed dictionary list : [{'name': 'Gfg', 'number': 3}, {'name': 'is', 'number': 8}, {'name': 'Best', 'number': 10}, {'name': 'for', 'number': 18}, {'name': 'Geeks', 'number': 33}]
Time complexity: O(N), where N is the length of the input list.
Auxiliary space: O(N), since we create a new list of tuples using the zip() function.
