Given a Matrix, map its values with dictionary values.
Input : test_list = [[4, 2, 1], [1, 2, 3]], sub_dict = {1 : “gfg”, 2: “best”, 3 : “CS”, 4 : “Geeks”} Output : [[‘Geeks’, ‘best’, ‘gfg’], [‘gfg’, ‘best’, ‘CS’]] Explanation : Matrix elements are substituted using dictionary. Input : test_list = [[4, 2, 1]], sub_dict = {1 : “gfg”, 2: “best”, 4 : “Geeks”} Output : [[‘Geeks’, ‘best’, ‘gfg’]] Explanation : Matrix elements are substituted using dictionary.
Method #1 : Using loop
This is brute way in which this task can be performed. In this, we iterate for all the elements of Matrix and map from dictionary.
Python3
# Python3 code to demonstrate working of # Mapping Matrix with Dictionary# Using loop# initializing listtest_list = [[4, 2, 1], [1, 2, 3], [4, 3, 1]]# printing original listprint("The original list : " + str(test_list))# initializing dictionarysub_dict = {1 : "gfg", 2: "best", 3 : "CS", 4 : "Geeks"}# Using loop to perform required mappingres = []for sub in test_list: temp = [] for ele in sub: # mapping values from dictionary temp.append(sub_dict[ele]) res.append(temp)# printing result print("Converted Mapped Matrix : " + str(res)) |
The original list : [[4, 2, 1], [1, 2, 3], [4, 3, 1]] Converted Mapped Matrix : [['Geeks', 'best', 'gfg'], ['gfg', 'best', 'CS'], ['Geeks', 'CS', 'gfg']]
Time Complexity: O(n*n) where n is the number of elements in the list “test_list”. The sorted and itemgetter function is used to perform the task and it takes O(nlogn) time.
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the list “test_list”.
Method #2 : Using list comprehension
This is yet another way in which this task can be performed. This is just shorthand to above method.
Python3
# Python3 code to demonstrate working of # Mapping Matrix with Dictionary# Using list comprehension# initializing listtest_list = [[4, 2, 1], [1, 2, 3], [4, 3, 1]]# printing original listprint("The original list : " + str(test_list))# initializing dictionarysub_dict = {1 : "gfg", 2: "best", 3 : "CS", 4 : "Geeks"}# Using list comprehension to perform required mapping# in one line res = [[sub_dict[val] for val in sub] for sub in test_list]# printing result print("Converted Mapped Matrix : " + str(res)) |
The original list : [[4, 2, 1], [1, 2, 3], [4, 3, 1]] Converted Mapped Matrix : [['Geeks', 'best', 'gfg'], ['gfg', 'best', 'CS'], ['Geeks', 'CS', 'gfg']]
Method 3 : using the map() function:
Step-by-step approach:
Initialize the original list test_list with the matrix values.
Print the original list.
Initialize the dictionary sub_dict with the mapping values.
Use the map() function to apply the sub_dict.get method on each element of the sublists of the test_list.
Convert the output of map() to a list using the list() method.
Apply the map() function again to convert each sublist of the test_list to a list of mapped values.
Convert the final output of map() to a list using the list() method.
Print the final result.
Python3
# Python3 code to demonstrate working of # Mapping Matrix with Dictionary# Using map() function# initializing listtest_list = [[4, 2, 1], [1, 2, 3], [4, 3, 1]]# printing original listprint("The original list : " + str(test_list))# initializing dictionarysub_dict = {1 : "gfg", 2: "best", 3 : "CS", 4 : "Geeks"}# Using map() function to perform required mappingres = list(map(lambda x: list(map(sub_dict.get, x)), test_list))# printing result print("Converted Mapped Matrix : " + str(res)) |
The original list : [[4, 2, 1], [1, 2, 3], [4, 3, 1]] Converted Mapped Matrix : [['Geeks', 'best', 'gfg'], ['gfg', 'best', 'CS'], ['Geeks', 'CS', 'gfg']]
Time complexity: The time complexity of the map function is O(n), and the time complexity of the lambda function inside it is also O(n).
Auxiliary space: This approach uses extra memory to store the mapped matrix. The space complexity is O(n^2) as well.
