I want to efficiently print a matrix in Python that follows a repeating specific pattern in the columns of 3 1s in a column, the rest of the columns 0s then the column 1s switch and so on for 1000 rows as shown below:

100000
100000
100000
010000
010000
010000
001000
001000
001000
000100
000100
000100
000010
000010
000010
000001
000001
000001
100000
100000
100000
010000
010000
010000
 ...

CodePudding user response：

First, you can create a diagonal matrix of size (6, 6) with only ones on the diagonal:

>>> arr = np.diag(np.ones(6))

Then, you can repeat each rows of that matrix 3times:

>>> arr = np.repeat(arr, repeats=3, axis=0)
>>> arr
[[1. 0. 0. 0. 0. 0.]
 [1. 0. 0. 0. 0. 0.]
 [1. 0. 0. 0. 0. 0.]
 [0. 1. 0. 0. 0. 0.]
 [0. 1. 0. 0. 0. 0.]
 [0. 1. 0. 0. 0. 0.]
 [0. 0. 1. 0. 0. 0.]
 [0. 0. 1. 0. 0. 0.]
 [0. 0. 1. 0. 0. 0.]
 [0. 0. 0. 1. 0. 0.]
 [0. 0. 0. 1. 0. 0.]
 [0. 0. 0. 1. 0. 0.]
 [0. 0. 0. 0. 1. 0.]
 [0. 0. 0. 0. 1. 0.]
 [0. 0. 0. 0. 1. 0.]
 [0. 0. 0. 0. 0. 1.]
 [0. 0. 0. 0. 0. 1.]
 [0. 0. 0. 0. 0. 1.]]

Finally, use np.tile to tile this matrix the number of times you want. In your case, as you want 1000 rows, you can repeat the array 1000 // 18 1 = 56 times, and only keep the first 1000 rows.

>>> arr = np.tile(arr, (56, 1))[:1000]

CodePudding user response：

Build an identity matrix, and then take out the matrix you need by generating the row indices (elegant but inefficient):

>>> np.eye(6, dtype=int)[np.arange(1000) // 3 % 6]
array([[1, 0, 0, 0, 0, 0],
       [1, 0, 0, 0, 0, 0],
       [1, 0, 0, 0, 0, 0],
       ...,
       [0, 0, 1, 0, 0, 0],
       [0, 0, 1, 0, 0, 0],
       [0, 0, 0, 1, 0, 0]])