Python – Assigning Subsequent Rows to Matrix first row elements

Given a (N + 1) * N Matrix, assign each column of 1st row of matrix, the subsequent row of Matrix.

Input : test_list = [[5, 8, 10], [2, 0, 9], [5, 4, 2], [2, 3, 9]] Output : {5: [2, 0, 9], 8: [5, 4, 2], 10: [2, 3, 9]} 
Explanation : 5 paired with 2nd row, 8 with 3rd and 10 with 4th 
Input : test_list = [[5, 8], [2, 0], [5, 4]] Output : {5: [2, 0], 8: [5, 4]} 
Explanation : 5 paired with 2nd row, 8 with 3rd.

Method #1: Using dictionary comprehension

This is one of the ways in which this task can be performed. In this, we iterate for rows and corresponding columns using loop and assign value list accordingly in a one-liner way using dictionary comprehension.

Python3

# Python3 code to demonstrate working of
# Assigning Subsequent Rows to Matrix first row elements
# Using dictionary comprehension

# initializing list
test_list = [[5, 8, 9], [2, 0, 9], [5, 4, 2], [2, 3, 9]]

# printing original list
print("The original list : " + str(test_list))

# pairing each 1st col with next rows in Matrix
res = {test_list[0][ele] :  test_list[ele + 1] for ele in range(len(test_list) - 1)}

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

Output

The original list : [[5, 8, 9], [2, 0, 9], [5, 4, 2], [2, 3, 9]]
The Assigned Matrix : {5: [2, 0, 9], 8: [5, 4, 2], 9: [2, 3, 9]}

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 #2 : Using zip() + list slicing + dict()

This is yet another way in which this task can be performed. In this, we slice elements to be first row and subsequent rows using list slicing and zip() performs the task of required grouping. Returned 

Python3

# Python3 code to demonstrate working of
# Assigning Subsequent Rows to Matrix first row elements
# Using zip() + list slicing + dict()

# initializing list
test_list = [[5, 8, 9], [2, 0, 9], [5, 4, 2], [2, 3, 9]]

# printing original list
print("The original list : " + str(test_list))

# dict() to convert back to dict type
# slicing and pairing using zip() and list slicing
res = dict(zip(test_list[0], test_list[1:]))

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

Output

The original list : [[5, 8, 9], [2, 0, 9], [5, 4, 2], [2, 3, 9]]
The Assigned Matrix : {5: [2, 0, 9], 8: [5, 4, 2], 9: [2, 3, 9]}

Method #3 : Using for loop and slicing

Python3

# Python3 code to demonstrate working of
# Assigning Subsequent Rows to Matrix first row elements

# initializing list
test_list = [[5, 8, 9], [2, 0, 9], [5, 4, 2], [2, 3, 9]]

# printing original list
print("The original list : " + str(test_list))

# pairing each 1st col with next rows in Matrix
a=test_list[0]
b=test_list[1:len(test_list)]
d=dict()
for i in range(0,len(a)):
    d[a[i]]=b[i]


# printing result
print("The Assigned Matrix : " + str(d))

Output

The original list : [[5, 8, 9], [2, 0, 9], [5, 4, 2], [2, 3, 9]]
The Assigned Matrix : {5: [2, 0, 9], 8: [5, 4, 2], 9: [2, 3, 9]}