Python – K list Nested Dictionary Mesh
Given 2 lists, create nested mesh with constant List.
Input : test_list1 = [4, 6], test_list2 = [2, 7], K = [] Output : {4: {2: [], 7: []}, 6: {2: [], 7: []}} Explanation : Nested dictionary initialized with []. Input : test_list1 = [4], test_list2 = [2], K = [1] Output : {4: {2: [1]}} Explanation : Nested dictionary initialized with [1].
Method : Using dictionary comprehension
In this, we use nested dictionary comprehension, inner one for list 2 elements to each element of list 1 as key and outer to assign keys from list 1.
Python3
# Python3 code to demonstrate working of # K list Nested Dictionary Mesh # Using * operator # initializing lists test_list1 = [ 4 , 6 , 8 , 7 ] test_list2 = [ 2 , 7 , 9 , 4 ] # printing original lists print ( "The original list 1 : " + str (test_list1)) print ( "The original list 2 : " + str (test_list2)) # initializing K K = [ None ] # initializing K list mesh res = {idx: {sub2: K for sub2 in test_list2} for idx in test_list1} # printing result print ( "Created Mesh : " + str (res)) |
Output
The original list 1 : [4, 6, 8, 7] The original list 2 : [2, 7, 9, 4] Created Mesh : {4: {2: [None], 7: [None], 9: [None], 4: [None]}, 6: {2: [None], 7: [None], 9: [None], 4: [None]}, 8: {2: [None], 7: [None], 9: [None], 4: [None]}, 7: {2: [None], 7: [None], 9: [None], 4: [None]}}
Time complexity: O(n*n), where n is the length of the dictionary.
Auxiliary Space: O(n), extra space of size n is required