How can I make a square with a specified circumference and add margin?-CodePudding

I am trying to make a square path of a specified length:

I made a function - and if I put 20 then I get a 6x6 matrix.

How can I add a margin of 0's of eg. 3 fields thickness?

like this

0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 1 1 1 1 1 1 0 0 0
0 0 0 1 1 1 1 1 1 0 0 0
0 0 0 1 1 1 1 1 1 0 0 0
0 0 0 1 1 1 1 1 1 0 0 0
0 0 0 1 1 1 1 1 1 0 0 0
0 0 0 1 1 1 1 1 1 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0

def square(length): return [
    [1 for _ in range(length//4 1)]
    for _ in range(length//4 1)
]

for x in square(24):
    print(x)

CodePudding user response：

You can prepare a line pattern of 0s and 1s then build a 2D matrix by intersecting them.

def square(size,margin=3):
    p = [0]*margin   [1]*(size-2*margin)   [0]*margin
    return [[r*c for r in p] for c in p]

for row in square(20):print(*row)

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0
0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0
0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0
0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0
0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0
0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0
0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0
0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0
0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0
0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0
0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0
0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0
0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0
0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

CodePudding user response：

Here's one way. One caution here is that, because of the way I duplicated the zero rows, those are all the same list. If you modify one of the zero rows, it will modify all of them.

def square(length): 
    zeros = [0]*(length//4 7)
    sq = [zeros] * 3
    sq.extend( [
        ([0,0,0]   [1 for _ in range(length//4 1)]   [0,0,0] )
        for _ in range(length//4 1)
    ])
    sq.extend( [zeros]*3 )
    return sq

for x in square(24):
    print(x)

Here's a numpy method.

import numpy as np
def square(length): 
    c = length//4 1
    sq = np.zeros((c 6,c 6)).astype(int)
    sq[3:c 3,3:c 3] = np.ones((c,c))
    return sq

print( square(24) )

CodePudding user response：

One way to do this is to build it as a flat string, then use textwrap to style the output into the right number of lines:

import textwrap

# The number of 1's in a row/column
count = 6
# The number of 0's to pad with
margin = 3
# The total 'size' of a row/column
size = margin   count   margin

pad_rows = "0" * size * margin
core = (("0" * margin)   ("1" * count)   ("0" * margin)) * count
print('\n'.join(textwrap.wrap(pad_rows   core   pad_rows, size)))