Here is am going to use modulo operator on the index of A which will loop back on the array A and will help to get the pattern. you can use (i%n) & (i%m) in A(n*m) to get this pattern.
Code :
import numpy as np
A = np.array([[1,2],[3,4]]) // converting the list to arrray
B = np.array([[0,0,0,0],[0,0,0,0],[0,0,0,0],[0,0,0,0]]) //
converting the list to arrray
for i in range(4):
for j in range(4):
B[i][j] = A[i%2][j%2] // assign the values of B
print(B)
Sample Run :
[[1 2 1 2]
[3 4 3 4]
[1 2 1 2]
[3 4 3 4]]
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