How do I create binary matrix with specific pattern of 1s and 0s efficiently?-CodePudding

How can I create a binary table of this alternating pattern efficiently in python? The 1s repeat for 4 elements then 0s for another 4 elements and so on and so forth as shown below:

CodePudding user response：

Consider using numpy.tile

import numpy as np

one_zero = np.tile([1, 0], 12).reshape(4, 6)
"""
array([[1, 0, 1, 0, 1, 0],
       [1, 0, 1, 0, 1, 0],
       [1, 0, 1, 0, 1, 0],
       [1, 0, 1, 0, 1, 0]])
"""
zero_one = np.tile([0, 1], 12).reshape(4, 6)
"""
array([[0, 1, 0, 1, 0, 1],
       [0, 1, 0, 1, 0, 1],
       [0, 1, 0, 1, 0, 1],
       [0, 1, 0, 1, 0, 1]])
"""
ar = np.tile([[1, 0], [0, 1]], 12).reshape(8, 6)
"""
array([[1, 0, 1, 0, 1, 0],
       [1, 0, 1, 0, 1, 0],
       [1, 0, 1, 0, 1, 0],
       [1, 0, 1, 0, 1, 0],
       [0, 1, 0, 1, 0, 1],
       [0, 1, 0, 1, 0, 1],
       [0, 1, 0, 1, 0, 1],
       [0, 1, 0, 1, 0, 1]])
"""

CodePudding user response：

Using list comprehension:

table = [('101010', '010101')[x//4%2 == 1] for x in range(100)]

You may prefer use 0b101010 and 0b010101 instead of string, but the result printed will be 42 and 21.