Given a list and a number N, the task here is to write a python program to convert it to matrix where each row has N elements more than previous row elements from list.
Input : test_list = [4, 6, 8, 1, 2, 9, 0, 10, 12, 3, 9, 1], N = 3 Output : [[4, 6, 8], [4, 6, 8, 1, 2, 9], [4, 6, 8, 1, 2, 9, 0, 10, 12], [4, 6, 8, 1, 2, 9, 0, 10, 12, 3, 9, 1]] Explanation : Each row has 3 elements more than previous row.
Input : test_list = [4, 6, 8, 1, 2, 9, 0, 10, 12, 3, 9, 1], N = 4 Output : [[4, 6, 8, 1], [4, 6, 8, 1, 2, 9, 0, 10], [4, 6, 8, 1, 2, 9, 0, 10, 12, 3, 9, 1]] Explanation : Each row has 4 elements more than previous row.
Method 1: Using loop and slicing
In this, we perform the task of getting chunks using slicing, and the conversion of the list into a Matrix is done using a loop.
Program:
Python3
# Initializing listtest_list = [4, 6, 8, 1, 2, 9, 0, 10, 12, 3, 9, 1]# Printing original listprint("The original list is : " + str(test_list))# initializing NN = 3# Empty listres = []for idx in range(0, len(test_list) // N): # Getting incremented chunks res.append(test_list[0: (idx + 1) * N])# Printing resultprint("Constructed Chunk Matrix : " + str(res)) |
The original list is : [4, 6, 8, 1, 2, 9, 0, 10, 12, 3, 9, 1] Constructed Chunk Matrix : [[4, 6, 8], [4, 6, 8, 1, 2, 9], [4, 6, 8, 1, 2, 9, 0, 10, 12], [4, 6, 8, 1, 2, 9, 0, 10, 12, 3, 9, 1]]
Time Complexity: O(m*n), where m is the row and n is the column of the matrix
Auxiliary Space: O(k), where k is the size of the matrix
Method 2: Using list comprehension and list slicing
In this, we perform the task of setting values using list comprehension as a shorthand. Rest all the operations are done similarly to the above method.
Program:
Python3
# initializing listtest_list = [4, 6, 8, 1, 2, 9, 0, 10, 12, 3, 9, 1]# printing original listprint("The original list is : " + str(test_list))# initializing NN = 3# getting incremented chunks# using list comprehension as shorthandres = [test_list[0: (idx + 1) * N] for idx in range(0, len(test_list) // N)]# printing resultprint("Constructed Chunk Matrix : " + str(res)) |
The original list is : [4, 6, 8, 1, 2, 9, 0, 10, 12, 3, 9, 1] Constructed Chunk Matrix : [[4, 6, 8], [4, 6, 8, 1, 2, 9], [4, 6, 8, 1, 2, 9, 0, 10, 12], [4, 6, 8, 1, 2, 9, 0, 10, 12, 3, 9, 1]]
Time Complexity: O(m*n)
Auxiliary Space: O(k)
Method 3: Using the numpy library’s reshape() method.
Here’s the step-by-step approach:
- Import the numpy library.
- Initialize the input list and the chunk size.
- Convert the input list to a numpy array.
- Reshape the array to have N columns and as many rows as necessary.
- Convert the chunk matrix back to a Python list.
- Print the result.
Python3
import numpy as np# initializing listtest_list = [4, 6, 8, 1, 2, 9, 0, 10, 12, 3, 9, 1]# printing original listprint("The original list is : " + str(test_list))# initializing NN = 3# convert list to numpy arrayarr = np.array(test_list)# reshape array into chunk matrixchunk_matrix = arr.reshape(-1, N)# convert chunk matrix back to listres = chunk_matrix.tolist()# printing resultprint("Constructed Chunk Matrix : " + str(res)) |
OUTPUT:
The original list is : [4, 6, 8, 1, 2, 9, 0, 10, 12, 3, 9, 1] Constructed Chunk Matrix : [[4, 6, 8], [1, 2, 9], [0, 10, 12], [3, 9, 1]]
Time complexity: O(n), where n is the size of the input list.
Auxiliary space: O(n) for the numpy array, but it is converted back to a Python list, so the overall space complexity is also O(n).
Method 4: Using a recursive function
- Create a list called test_list containing 12 integers.
- Initialize a variable called N to 3, which will be used as the chunk size.
- Define a function called chunk_list
- In the chunk_list function, check if the length of the list is less than or equal to the chunk size. If so, return a list containing the original list.
- If the length of the list is greater than the chunk size, slice the list from the beginning to the chunk size and append it to a new list using the + operator. Then recursively call the chunk_list function with the remaining part of the list until the sublist is less than or equal to the chunk size.
- Call the chunk_list function with test_list and N as arguments and assign the result to a variable called res.
- Print the constructed chunk matrix using the print() function, the string “Constructed Chunk Matrix: “, and the str() function to convert the result to a string.
Python3
# Initializing listtest_list = [4, 6, 8, 1, 2, 9, 0, 10, 12, 3, 9, 1]# Printing original listprint("The original list is : " + str(test_list))# Initializing NN = 3def chunk_list(lst, n): if len(lst) <= n: return [lst] else: return [lst[:n]] + chunk_list(lst[n:], n)res = chunk_list(test_list, N)# Printing resultprint("Constructed Chunk Matrix : " + str(res)) |
The original list is : [4, 6, 8, 1, 2, 9, 0, 10, 12, 3, 9, 1] Constructed Chunk Matrix : [[4, 6, 8], [1, 2, 9], [0, 10, 12], [3, 9, 1]]
Time complexity: O(N * (L/N)) = O(L), where L is the length of the original list and N is the chunk size.
Auxiliary space: O(L), as it creates a new list for each chunk of the original list.
Method 5: Using the itertools module
can use the zip_longest function from the itertools module to chunk the list into sublists of size N. This method fills the last sublist with a specified value (in this case, None) if its length is less than N.
Python3
from itertools import zip_longest# Initializing listtest_list = [4, 6, 8, 1, 2, 9, 0, 10, 12, 3, 9, 1]# printing original listprint("The original list is: " + str(test_list))# Initializing NN = 3def chunk_list(lst, n): return [list(filter(None, sublist)) for sublist in zip_longest(*[iter(lst)] * n)]res = chunk_list(test_list, N)# Printing resultprint("Constructed Chunk Matrix: " + str(res)) |
The original list is: [4, 6, 8, 1, 2, 9, 0, 10, 12, 3, 9, 1] Constructed Chunk Matrix: [[4, 6, 8], [1, 2, 9], [10, 12], [3, 9, 1]]
Time Complexity: O(N * M), where N is the length of the list and M is the chunk size.
Auxiliary Space: O(N), where N is the length of the list, as the result is stored in a new list.
