Simple Game of Life engine using Numpy

numpy_life.py

#!/usr/bin/env python3

''' Game of Life in Numpy

    Written by PM 2Ring 2016.11.03
'''

import numpy as np
from time import sleep

offsets = (
    (0, 0), (0, 1), (0, 2),
    (1, 0),         (1, 2),
    (2, 0), (2, 1), (2, 2),
)

newcell = np.vectorize(lambda g, k: np.uint8(k == 3 or g and k == 2))

def evolve(grid):
    numrows, numcols = grid.shape
    padded = np.pad(grid, 1, 'constant', constant_values=0)
    counts = empty_grid(numrows, numcols)
    for r, c in offsets:
        counts += padded[r:r+numrows, c:c+numcols]
    return newcell(grid, counts)

def empty_grid(numrows, numcols):
    return np.zeros((numrows, numcols), dtype='uint8')

def show(grid):
    print('\x1bc', '\n'.join([' '.join(['-O'[v] for v in row]) 
        for row in grid]), sep='', end='\n\n')

# - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

glider = '''\
.*.
..*
***
'''

def mirrorX(pattern):
    return '\n'.join([row[::-1] for row in pattern.splitlines()])

def mirrorY(pattern):
    return '\n'.join(reversed(pattern.splitlines()))

def add_pattern(grid, pattern, ox=0, oy=0):
    numrows, numcols = grid.shape
    for r, row in enumerate(pattern.splitlines(), oy):
        if r >= numrows:
            break
        for c, v in enumerate(row, ox):
            if c >= numcols:
                break
            if v not in ' .':
                grid[r][c] = 1

# - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

numrows, numcols = 40, 60

grid = empty_grid(numrows, numcols)

gliderx = mirrorX(glider)
add_pattern(grid, glider, ox=1, oy=3)
add_pattern(grid, gliderx, ox=34, oy=2)

for _ in range(160):
    show(grid)
    grid = evolve(grid)
    sleep(0.1)