Given a List, convert it to dictionary, with separate keys for index and values.
Input : test_list = [3, 5, 7, 8, 2, 4, 9], idx, val = “1”, “2”
Output : {‘1’: [0, 1, 2, 3, 4, 5, 6], ‘2’: [3, 5, 7, 8, 2, 4, 9]}
Explanation : Index and values mapped at similar index in diff. keys., as “1” and “2”.Input : test_list = [3, 5, 7], idx, val = “1”, “2”
Output : {‘1’: [0, 1, 2], ‘2’: [3, 5, 7]}
Explanation : Index and values mapped at similar index in diff. keys., as “1” and “2”.
Method : Using loop + enumerate()
In this, we iterate for list elements using enumerate() to get index along with values, and append the values and indices accordingly in separate dictionaries accordingly.
Python3
# Python3 code to demonstrate working of # Convert List to Index and Value dictionary# Using loop + enumerate()# initializing listtest_list = [3, 5, 7, 8, 2, 4, 9]# printing original listprint("The original list is : " + str(test_list))# initializing keys for index and vals idx, val = "indx", "vals"# initializing empty meshres = {idx : [], val : []}for id, vl in enumerate(test_list): res[idx].append(id) res[val].append(vl) # printing resultsprint("Constructed dictionary : " + str(res)) |
The original list is : [3, 5, 7, 8, 2, 4, 9]
Constructed dictionary : {'indx': [0, 1, 2, 3, 4, 5, 6], 'vals': [3, 5, 7, 8, 2, 4, 9]}
Time Complexity: O(n), where n is the length of the dictionary
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the dictionary
Method 2: Using zip() and dictionary comprehension:
First uses zip() to create a list of tuples where each tuple contains an index-value pair. Then, it uses dictionary comprehension to convert the list of tuples to a dictionary with keys “indx” and “vals“, where each key’s value is a list of the corresponding indices or values. Finally, it prints the resulting dictionary.
Python3
test_list = [3, 5, 7, 8, 2, 4, 9]# using zip() to create a list of tuples, where each tuple contains an index-value pairpairs = list(zip(range(len(test_list)), test_list))# using dictionary comprehension to convert the list of tuples to a dictionary with keys "indx" and "vals"res = { "indx": [i for i, v in pairs], "vals": [v for i, v in pairs] }# print the resulting dictionaryprint(res) |
{'indx': [0, 1, 2, 3, 4, 5, 6], 'vals': [3, 5, 7, 8, 2, 4, 9]}
Time Complexity: O(N), where N is the length of the input list.
Auxiliary Space: O(N)
Method 3: Use the built-in dict() constructor with a generator expression
Step-by-step approach:
- Creates a dictionary named result with two keys:
‘indx’: a list of the indices of the values in test_list, created using the range() function and the len() function to generate a list of integers from 0 to 6 (the length of test_list minus 1).
‘vals’: a reference to the original test_list. - Finally, the code prints the original list and the constructed dictionary
Python3
test_list = [3, 5, 7, 8, 2, 4, 9]result = {'indx': list(range(len(test_list))), 'vals': test_list}print("The original list is:", test_list)print("Constructed dictionary:", result) |
The original list is: [3, 5, 7, 8, 2, 4, 9]
Constructed dictionary: {'indx': [0, 1, 2, 3, 4, 5, 6], 'vals': [3, 5, 7, 8, 2, 4, 9]}
Time complexity: O(n), where n is the length of the list, because it only requires iterating over the list once to create the key-value pairs.
Auxiliary space: O(n), because the resulting dictionary will contain n key-value pairs.
