Iterate over an sqaure matrix, from the inside out

in2out.py

'''
Created this for a school project. It works for both, matrices with even or odd orders.
'''

def iterate(mat):
    order = len(mat)

    cycles = order // 2

    # odd
    if order % 2 == 1:
        yield mat[cycles][cycles]
        calc_n = lambda i: i * 2 + 2
        calc_c = lambda i : cycles + 1 + i
    else:
        calc_c = lambda i : cycles + i
        calc_n = lambda i: i * 2 + 1


    for i in range(cycles):
        x = calc_c(i)
        y = calc_c(i)

        n = calc_n(i)

        # left
        for _ in range(n):
            yield mat[y][x]
            x -= 1

        # down
        for _ in range(n):
            yield mat[y][x]
            y -= 1

        # right
        for _ in range(n):
            yield mat[y][x]
            x += 1

        # up
        for _ in range(n):
            yield mat[y][x]
            y += 1

from pprint import pprint

# Create a simple matrix, for test
q = 4
m = [ [ str(j * q + i).rjust(2) for i in range(q) ] for j in range(q) ]
pprint(m[::-1])
# Out:
# [['12', '13', '14', '15'],
#  [' 8', ' 9', '10', '11'],
#  [' 4', ' 5', ' 6', ' 7'],
#  [' 0', ' 1', ' 2', ' 3']]

print(list(iterate(m)))
# Out:
# ['10', ' 9', ' 5', ' 6', '15', '14', '13', '12', ' 8', ' 4', ' 0', ' 1', ' 2', ' 3', ' 7', '11']