Many times while working with numbers in data science we come across the problem in which we need to work with data science we need to transform a number to a matrix of consecutive numbers and hence this problem has a good application. Let’s discuss certain ways in which this problem can be solved.
Method #1: Using list comprehension List comprehension can be used to accomplish this particular task by using the range function for each list that needs to be constructed consecutively.
Python3
# Python3 code to demonstrate# matrix creation of n * n# using list comprehension# initializing NN = 4# printing dimensionprint("The dimension : " + str(N))# using list comprehension# matrix creation of n * nres = [list(range(1 + N * i, 1 + N * (i + 1))) for i in range(N)]# print resultprint("The created matrix of N * N: " + str(res)) |
The dimension : 4 The created matrix of N * N: [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]
Time Complexity: O(n*n)
Auxiliary Space: O(n*n)
Method #2 : Using next() + itertools.count() The count function can be used start the count of numbers and next function does the task of creation of sublist consecutively. List comprehension handles the processing.
Python3
# Python3 code to demonstrate# matrix creation of n * n# using next() + itertools.count()import itertools# initializing NN = 4# printing dimensionprint("The dimension : " + str(N))# using next() + itertools.count()# matrix creation of n * ntemp = itertools.count(1)res = [[next(temp) for i in range(N)] for i in range(N)]# print resultprint("The created matrix of N * N: " + str(res)) |
The dimension : 4 The created matrix of N * N: [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]
Method #3 Using a list comprehension and the enumerate() function:
In this code, the list comprehension iterates over the rows of the matrix (using the enumerate() function to get the index of each row), and generates the elements of the row using a nested list comprehension. The nested list comprehension iterates over the columns of the matrix and generates the elements of the column using the formula i * n + j, where i is the index of the row and j is the index of the column.
Python3
# Initialize the size of the matrixn = 4# Create the matrix using a list comprehension and the enumerate() functionmatrix = [[i * n + j for j in range(1, n+1)] for i, _ in enumerate(range(n))]# Print the matrixprint("The created matrix of N * N: "+ str(matrix))#This code is contributed by Edula Vinay Kumar Reddy |
The created matrix of N * N: [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]
Time complexity: O(N^2)
Auxiliary Space: O(N^2)
Method #4 : Using for loop, while loop and slicing
Python3
# Python3 code to demonstrate# matrix creation of n * n# initializing NN = 4# printing dimensionprint("The dimension : " + str(N))# using list comprehension# matrix creation of n * nx=N*Ny=[]for i in range(1,x+1): y.append(i)res=[]i=0while(i<len(y)): a=y[i:i+N] res.append(a) i+=N# print resultprint("The created matrix of N * N: " + str(res)) |
The dimension : 4 The created matrix of N * N: [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]
Time Complexity : O(N*N)
Auxiliary Space : O(N)
Method #5 : Using nested for loop:
1.Initialize the value of N.
2.Create an empty list called ‘res’ to store the matrix.
3.Using nested for loops, create a matrix with dimensions N x N, and populate it with values from 1 to N x N, in row-major order.
a. For each row i, create an empty list called ‘row’.
b. For each column j in row i, append the value 1 + N * i + j to the list ‘row’.
c. Append the list ‘row’ to the list ‘res’.
4.Print the created matrix using the print() function.
Python3
# Python3 code to demonstrate# matrix creation of n * n# using nested for loops# initializing NN = 4# printing dimensionprint("The dimension : " + str(N))# initializing empty matrixres = []for i in range(N): row = [] for j in range(N): row.append(1 + N * i + j) res.append(row)# print resultprint("The created matrix of N * N: " + str(res))#This code is contributed by Jyothi pinjala. |
The dimension : 4 The created matrix of N * N: [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]
Time Complexity:
The time complexity of the given code is O(N^2), since it iterates over N rows and N columns in the nested for loops.
Space Complexity:
The space complexity of the given code is O(N^2), since it stores N^2 elements in the ‘res’ list.
