Python - Inter Matrix Grouping

Last Updated : 27 Apr, 2023

Given 2 Matrix, with 2 elements in each row, group them on basis on first element.

Input : test_list1 = [[2, 0], [8, 4], [9, 3]], test_list2 = [[8, 1], [9, 7], [2, 10]]
Output : {2: [0, 10], 8: [4, 1], 9: [3, 7]}
Explanation : All values after mapping cross Matrix, 2 mapped to 0 and 10.

Input : test_list1 = [[8, 4], [9, 3]], test_list2 = [[8, 1], [9, 7]]
Output : {8: [4, 1], 9: [3, 7]}
Explanation : All values after mapping cross Matrix.

Method #1 : Using defaultdict() + loop

This is one of the ways in which this task can be performed. In this, we iterate add both the lists and then form groups from similar elements and convert to dictionary with value list.

Python3

# Python3 code to demonstrate working of 
# Inter Matrix Grouping
# Using defaultdict() + loop
from collections import defaultdict

# initializing lists
test_list1 = [[5, 8], [2, 0], [8, 4], [9, 3]]
test_list2 = [[8, 1], [9, 7], [2, 10], [5, 6]]

# printing original list
print("The original list 1 : " + str(test_list1))
print("The original list 2 : " + str(test_list2))

# initializing mapping list 
res = defaultdict(list)

# concatenation matrix using 2 lists
for key, val in test_list1 + test_list2:
    res[key].append(val)

# printing result 
print("The Grouped Matrix : " + str(dict(res)))

Output

The original list 1 : [[5, 8], [2, 0], [8, 4], [9, 3]]
The original list 2 : [[8, 1], [9, 7], [2, 10], [5, 6]]
The Grouped Matrix : {5: [8, 6], 2: [0, 10], 8: [4, 1], 9: [3, 7]}

Time complexity: O(n), where n is the total number of elements in the two input lists.
Auxiliary Space: O(n), where n is the total number of elements in the two input lists, because we are using a defaultdict to store the grouped matrix.

Method #2 : Using dictionary comprehension + dict()

This is yet another way in which this task can be performed. In this, we iterate for all the elements of both matrix by converting each to dictionary and group its values. Its important to have traces of each element's mapping in both the Matrix.

Python3

# Python3 code to demonstrate working of 
# Inter Matrix Grouping
# Using dictionary comprehension + dict()

# initializing lists
test_list1 = [[5, 8], [2, 0], [8, 4], [9, 3]]
test_list2 = [[8, 1], [9, 7], [2, 10], [5, 6]]

# printing original list
print("The original list 1 : " + str(test_list1))
print("The original list 2 : " + str(test_list2))

# mapping dictionaries together, converting each row to dictionary
res = {key: [dict(test_list1)[key], dict(test_list2)[key]] for key in dict(test_list1)}

# printing result 
print("The Grouped Matrix : " + str(dict(res)))

Output

The original list 1 : [[5, 8], [2, 0], [8, 4], [9, 3]]
The original list 2 : [[8, 1], [9, 7], [2, 10], [5, 6]]
The Grouped Matrix : {5: [8, 6], 2: [0, 10], 8: [4, 1], 9: [3, 7]}

Time complexity: O(m*n), because it performs the same number of iterations as the original code.
Auxiliary space: O(m*n) as well, because it creates a dictionary with m * n keys and a list of m * n elements

Method #3: Using extend(),list() and set() methods

Python3

# Python3 code to demonstrate working of
# Inter Matrix Grouping

# initializing lists
test_list1 = [[5, 8], [2, 0], [8, 4], [9, 3]]
test_list2 = [[8, 1], [9, 7], [2, 10], [5, 6], [5, 9]]

# printing original list
print("The original list 1 : " + str(test_list1))
print("The original list 2 : " + str(test_list2))
x = []
y = []
res = dict()
x.extend(test_list1)
x.extend(test_list2)
for i in x:
    y.append(i[0])
y = list(set(y))
for i in y:
    a = []
    for j in x:
        if(j[0] == i):
            a.append(j[1])
    res[i] = a


# printing result
print("The Grouped Matrix : " + str(dict(res)))

Output

The original list 1 : [[5, 8], [2, 0], [8, 4], [9, 3]]
The original list 2 : [[8, 1], [9, 7], [2, 10], [5, 6], [5, 9]]
The Grouped Matrix : {8: [4, 1], 9: [3, 7], 2: [0, 10], 5: [8, 6, 9]}

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