Conway's game of life neighbor count
3
def neighbors(matrix, r, c):
live_neighbors = 0
if matrix[r][c-1] != 0:
live_neighbors += 1
if matrix[r-1][c] != 0:
live_neighbors += 1
if matrix[r-1][c+1] != 0:
live_neighbors += 1
if matrix[r][c-1] != 0:
live_neighbors += 1
if matrix[r][c+1] != 0:
live_neighbors += 1
if matrix[r+1][c-1] != 0:
live_neighbors += 1
if matrix[r+1][c] != 0:
live_neighbors += 1
if matrix[r+1][c+1] != 0:
live_neighbors += 1
return live_neighbors
This is the code I have written so far. How do I count the neighbors of the border cells? If I use this code I would get an index out of range error.
python python-3.x matrix conways-game-of-life cellular-automata
edited Nov 13 '18 at 23:17
blhsing
29.5k41336
asked Nov 13 '18 at 22:58
BrowserMBrowserM
214
add a comment |
3
def neighbors(matrix, r, c):
live_neighbors = 0
if matrix[r][c-1] != 0:
live_neighbors += 1
if matrix[r-1][c] != 0:
live_neighbors += 1
if matrix[r-1][c+1] != 0:
live_neighbors += 1
if matrix[r][c-1] != 0:
live_neighbors += 1
if matrix[r][c+1] != 0:
live_neighbors += 1
if matrix[r+1][c-1] != 0:
live_neighbors += 1
if matrix[r+1][c] != 0:
live_neighbors += 1
if matrix[r+1][c+1] != 0:
live_neighbors += 1
return live_neighbors
This is the code I have written so far. How do I count the neighbors of the border cells? If I use this code I would get an index out of range error.
python python-3.x matrix conways-game-of-life cellular-automata
edited Nov 13 '18 at 23:17
blhsing
29.5k41336
asked Nov 13 '18 at 22:58
BrowserMBrowserM
214
One major consideration is borders: does the value in position [0, 0] have 3 neighbors? Or does it "roll over" and still have 8?
– Brad Solomon
Nov 14 '18 at 3:16
Either way, you may also findscipy.signal.convolveuseful
– Brad Solomon
Nov 14 '18 at 3:17
add a comment |
3
3
3
def neighbors(matrix, r, c):
live_neighbors = 0
if matrix[r][c-1] != 0:
live_neighbors += 1
if matrix[r-1][c] != 0:
live_neighbors += 1
if matrix[r-1][c+1] != 0:
live_neighbors += 1
if matrix[r][c-1] != 0:
live_neighbors += 1
if matrix[r][c+1] != 0:
live_neighbors += 1
if matrix[r+1][c-1] != 0:
live_neighbors += 1
if matrix[r+1][c] != 0:
live_neighbors += 1
if matrix[r+1][c+1] != 0:
live_neighbors += 1
return live_neighbors
This is the code I have written so far. How do I count the neighbors of the border cells? If I use this code I would get an index out of range error.
python python-3.x matrix conways-game-of-life cellular-automata
edited Nov 13 '18 at 23:17
blhsing
29.5k41336
asked Nov 13 '18 at 22:58
BrowserMBrowserM
214
def neighbors(matrix, r, c):
live_neighbors = 0
if matrix[r][c-1] != 0:
live_neighbors += 1
if matrix[r-1][c] != 0:
live_neighbors += 1
if matrix[r-1][c+1] != 0:
live_neighbors += 1
if matrix[r][c-1] != 0:
live_neighbors += 1
if matrix[r][c+1] != 0:
live_neighbors += 1
if matrix[r+1][c-1] != 0:
live_neighbors += 1
if matrix[r+1][c] != 0:
live_neighbors += 1
if matrix[r+1][c+1] != 0:
live_neighbors += 1
return live_neighbors
This is the code I have written so far. How do I count the neighbors of the border cells? If I use this code I would get an index out of range error.
python python-3.x matrix conways-game-of-life cellular-automata
python python-3.x matrix conways-game-of-life cellular-automata
edited Nov 13 '18 at 23:17
blhsing
29.5k41336
asked Nov 13 '18 at 22:58
BrowserMBrowserM
214
edited Nov 13 '18 at 23:17
blhsing
29.5k41336
asked Nov 13 '18 at 22:58
BrowserMBrowserM
214
edited Nov 13 '18 at 23:17
blhsing
29.5k41336
edited Nov 13 '18 at 23:17
blhsing
29.5k41336
edited Nov 13 '18 at 23:17
blhsing
29.5k41336
29.5k41336
asked Nov 13 '18 at 22:58
BrowserMBrowserM
214
asked Nov 13 '18 at 22:58
BrowserMBrowserM
214
asked Nov 13 '18 at 22:58
BrowserMBrowserM
214
214
One major consideration is borders: does the value in position [0, 0] have 3 neighbors? Or does it "roll over" and still have 8?
– Brad Solomon
Nov 14 '18 at 3:16
Either way, you may also findscipy.signal.convolveuseful
– Brad Solomon
Nov 14 '18 at 3:17
add a comment |
One major consideration is borders: does the value in position [0, 0] have 3 neighbors? Or does it "roll over" and still have 8?
– Brad Solomon
Nov 14 '18 at 3:16
Either way, you may also findscipy.signal.convolveuseful
– Brad Solomon
Nov 14 '18 at 3:17
One major consideration is borders: does the value in position [0, 0] have 3 neighbors? Or does it "roll over" and still have 8?
– Brad Solomon
Nov 14 '18 at 3:16
One major consideration is borders: does the value in position [0, 0] have 3 neighbors? Or does it "roll over" and still have 8?
– Brad Solomon
Nov 14 '18 at 3:16
Either way, you may also find
scipy.signal.convolve useful– Brad Solomon
Nov 14 '18 at 3:17
Either way, you may also find
scipy.signal.convolve useful– Brad Solomon
Nov 14 '18 at 3:17
add a comment |
3 Answers
3
active
oldest
votes
1
A possible solution without all those if statements:
def neighbors(matrix, r, c):
def get(r, c):
if 0 <= r < len(matrix) and 0 <= c < len(matrix[r]):
return matrix[r][c]
else:
return 0
neighbors_list = [get(r-1, c-1), get(r-1, c), get(r-1, c+1),
get(r , c-1), get(r , c+1),
get(r+1, c-1), get(r+1, c), get(r+1, c+1)]
return sum(map(bool, neighbors_list))
matrix = [ [0, 0, 0, 0, 0],
[0, 0, 0, 0, 1],
[0, 0, 0, 1, 1],
[0, 0, 0, 1, 1],
[0, 0, 1, 1, 1] ]
print(neighbors(matrix, 0, 0)) # 0
print(neighbors(matrix, 1, 2)) # 1
print(neighbors(matrix, 3, 2)) # 4
print(neighbors(matrix, 4, 4)) # 3
In case the cells would only have values either 0 or 1, the neighbors function would simply return sum(neighbors_list).
answered Nov 15 '18 at 10:33
myrmicamyrmica
41418
add a comment |
1
You can use a helper function to check for boundaries:
def neighbors(matrix, r, c):
def get(r, c):
return 0 <= r < len(matrix) and 0 <= c < len(matrix[r]) and matrix[r][c]
live_neighbors = 0
if get(r, c-1) != 0:
live_neighbors += 1
if get(r-1, c) != 0:
live_neighbors += 1
if get(r-1, c+1) != 0:
live_neighbors += 1
if get(r, c-1) != 0:
live_neighbors += 1
if get(r, c+1) != 0:
live_neighbors += 1
if get(r+1, c-1) != 0:
live_neighbors += 1
if get(r+1, c) != 0:
live_neighbors += 1
if get(r+1, c+1) != 0:
live_neighbors += 1
return live_neighbors
You can also use itertools.product in a generator expression for sum instead of if statements to count all the live neighbors:
from itertools import product
def neighbors(matrix, r, c):
def get(r, c):
return 0 <= r < len(matrix) and 0 <= c < len(matrix[r]) and matrix[r][c]
return sum(get(r + i, c + j) for i, j in product(range(-1, 2), 2) if i or j)
answered Nov 13 '18 at 23:17
blhsingblhsing
29.5k41336
why did you write len(matrix(r))?
– BrowserM
Nov 14 '18 at 9:57
Oops that was a typo indeed. Fixed. Thanks.
– blhsing
Nov 14 '18 at 13:15
Maybe inverse theif/elseso you can use comparison chaining?if 0 <= r < len(matrix) and 0 <= c < len(matrix[r]): return matrix[r][c] else: return 0. In fact, you could just doreturn 0 <= r < len(matrix) and 0 <= c < len(matrix[r]) and matrix[r][c]and then drop the!= 0in theifclauses.
– tobias_k
Nov 15 '18 at 10:39
@tobias_k True thanks. I also took the opportunity to further optimize the rest of the code.
– blhsing
Nov 15 '18 at 14:08
add a comment |
0
Closest to what you have would be:
def neighbors(matrix, r, c):
live_neighbors = 0
if c and matrix[r][c-1] != 0:
live_neighbors += 1
if r and matrix[r-1][c] != 0:
live_neighbors += 1
if r and matrix[r-1][c+1] != 0:
live_neighbors += 1
if c and matrix[r][c-1] != 0:
live_neighbors += 1
if c < len(matrix[r])-1 and matrix[r][c+1] != 0:
live_neighbors += 1
if r < len(matrix)-1 and matrix[r+1][c-1] != 0:
live_neighbors += 1
if r < len(matrix)-1 and matrix[r+1][c] != 0:
live_neighbors += 1
if r < len(matrix)-1 and matrix[r+1][c+1] != 0:
live_neighbors += 1
return live_neighbors
and so forth.
answered Nov 15 '18 at 10:41
kabanuskabanus
11.4k31339
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
A possible solution without all those if statements:
def neighbors(matrix, r, c):
def get(r, c):
if 0 <= r < len(matrix) and 0 <= c < len(matrix[r]):
return matrix[r][c]
else:
return 0
neighbors_list = [get(r-1, c-1), get(r-1, c), get(r-1, c+1),
get(r , c-1), get(r , c+1),
get(r+1, c-1), get(r+1, c), get(r+1, c+1)]
return sum(map(bool, neighbors_list))
matrix = [ [0, 0, 0, 0, 0],
[0, 0, 0, 0, 1],
[0, 0, 0, 1, 1],
[0, 0, 0, 1, 1],
[0, 0, 1, 1, 1] ]
print(neighbors(matrix, 0, 0)) # 0
print(neighbors(matrix, 1, 2)) # 1
print(neighbors(matrix, 3, 2)) # 4
print(neighbors(matrix, 4, 4)) # 3
In case the cells would only have values either 0 or 1, the neighbors function would simply return sum(neighbors_list).
answered Nov 15 '18 at 10:33
myrmicamyrmica
41418
add a comment |
1
A possible solution without all those if statements:
def neighbors(matrix, r, c):
def get(r, c):
if 0 <= r < len(matrix) and 0 <= c < len(matrix[r]):
return matrix[r][c]
else:
return 0
neighbors_list = [get(r-1, c-1), get(r-1, c), get(r-1, c+1),
get(r , c-1), get(r , c+1),
get(r+1, c-1), get(r+1, c), get(r+1, c+1)]
return sum(map(bool, neighbors_list))
matrix = [ [0, 0, 0, 0, 0],
[0, 0, 0, 0, 1],
[0, 0, 0, 1, 1],
[0, 0, 0, 1, 1],
[0, 0, 1, 1, 1] ]
print(neighbors(matrix, 0, 0)) # 0
print(neighbors(matrix, 1, 2)) # 1
print(neighbors(matrix, 3, 2)) # 4
print(neighbors(matrix, 4, 4)) # 3
In case the cells would only have values either 0 or 1, the neighbors function would simply return sum(neighbors_list).
answered Nov 15 '18 at 10:33
myrmicamyrmica
41418
add a comment |
1
1
1
A possible solution without all those if statements:
def neighbors(matrix, r, c):
def get(r, c):
if 0 <= r < len(matrix) and 0 <= c < len(matrix[r]):
return matrix[r][c]
else:
return 0
neighbors_list = [get(r-1, c-1), get(r-1, c), get(r-1, c+1),
get(r , c-1), get(r , c+1),
get(r+1, c-1), get(r+1, c), get(r+1, c+1)]
return sum(map(bool, neighbors_list))
matrix = [ [0, 0, 0, 0, 0],
[0, 0, 0, 0, 1],
[0, 0, 0, 1, 1],
[0, 0, 0, 1, 1],
[0, 0, 1, 1, 1] ]
print(neighbors(matrix, 0, 0)) # 0
print(neighbors(matrix, 1, 2)) # 1
print(neighbors(matrix, 3, 2)) # 4
print(neighbors(matrix, 4, 4)) # 3
In case the cells would only have values either 0 or 1, the neighbors function would simply return sum(neighbors_list).
answered Nov 15 '18 at 10:33
myrmicamyrmica
41418
A possible solution without all those if statements:
def neighbors(matrix, r, c):
def get(r, c):
if 0 <= r < len(matrix) and 0 <= c < len(matrix[r]):
return matrix[r][c]
else:
return 0
neighbors_list = [get(r-1, c-1), get(r-1, c), get(r-1, c+1),
get(r , c-1), get(r , c+1),
get(r+1, c-1), get(r+1, c), get(r+1, c+1)]
return sum(map(bool, neighbors_list))
matrix = [ [0, 0, 0, 0, 0],
[0, 0, 0, 0, 1],
[0, 0, 0, 1, 1],
[0, 0, 0, 1, 1],
[0, 0, 1, 1, 1] ]
print(neighbors(matrix, 0, 0)) # 0
print(neighbors(matrix, 1, 2)) # 1
print(neighbors(matrix, 3, 2)) # 4
print(neighbors(matrix, 4, 4)) # 3
In case the cells would only have values either 0 or 1, the neighbors function would simply return sum(neighbors_list).
answered Nov 15 '18 at 10:33
myrmicamyrmica
41418
answered Nov 15 '18 at 10:33
myrmicamyrmica
41418
answered Nov 15 '18 at 10:33
myrmicamyrmica
41418
answered Nov 15 '18 at 10:33
myrmicamyrmica
41418
41418
add a comment |
add a comment |
1
You can use a helper function to check for boundaries:
def neighbors(matrix, r, c):
def get(r, c):
return 0 <= r < len(matrix) and 0 <= c < len(matrix[r]) and matrix[r][c]
live_neighbors = 0
if get(r, c-1) != 0:
live_neighbors += 1
if get(r-1, c) != 0:
live_neighbors += 1
if get(r-1, c+1) != 0:
live_neighbors += 1
if get(r, c-1) != 0:
live_neighbors += 1
if get(r, c+1) != 0:
live_neighbors += 1
if get(r+1, c-1) != 0:
live_neighbors += 1
if get(r+1, c) != 0:
live_neighbors += 1
if get(r+1, c+1) != 0:
live_neighbors += 1
return live_neighbors
You can also use itertools.product in a generator expression for sum instead of if statements to count all the live neighbors:
from itertools import product
def neighbors(matrix, r, c):
def get(r, c):
return 0 <= r < len(matrix) and 0 <= c < len(matrix[r]) and matrix[r][c]
return sum(get(r + i, c + j) for i, j in product(range(-1, 2), 2) if i or j)
answered Nov 13 '18 at 23:17
blhsingblhsing
29.5k41336
why did you write len(matrix(r))?
– BrowserM
Nov 14 '18 at 9:57
Oops that was a typo indeed. Fixed. Thanks.
– blhsing
Nov 14 '18 at 13:15
Maybe inverse theif/elseso you can use comparison chaining?if 0 <= r < len(matrix) and 0 <= c < len(matrix[r]): return matrix[r][c] else: return 0. In fact, you could just doreturn 0 <= r < len(matrix) and 0 <= c < len(matrix[r]) and matrix[r][c]and then drop the!= 0in theifclauses.
– tobias_k
Nov 15 '18 at 10:39
@tobias_k True thanks. I also took the opportunity to further optimize the rest of the code.
– blhsing
Nov 15 '18 at 14:08
add a comment |
1
You can use a helper function to check for boundaries:
def neighbors(matrix, r, c):
def get(r, c):
return 0 <= r < len(matrix) and 0 <= c < len(matrix[r]) and matrix[r][c]
live_neighbors = 0
if get(r, c-1) != 0:
live_neighbors += 1
if get(r-1, c) != 0:
live_neighbors += 1
if get(r-1, c+1) != 0:
live_neighbors += 1
if get(r, c-1) != 0:
live_neighbors += 1
if get(r, c+1) != 0:
live_neighbors += 1
if get(r+1, c-1) != 0:
live_neighbors += 1
if get(r+1, c) != 0:
live_neighbors += 1
if get(r+1, c+1) != 0:
live_neighbors += 1
return live_neighbors
You can also use itertools.product in a generator expression for sum instead of if statements to count all the live neighbors:
from itertools import product
def neighbors(matrix, r, c):
def get(r, c):
return 0 <= r < len(matrix) and 0 <= c < len(matrix[r]) and matrix[r][c]
return sum(get(r + i, c + j) for i, j in product(range(-1, 2), 2) if i or j)
answered Nov 13 '18 at 23:17
blhsingblhsing
29.5k41336
why did you write len(matrix(r))?
– BrowserM
Nov 14 '18 at 9:57
Oops that was a typo indeed. Fixed. Thanks.
– blhsing
Nov 14 '18 at 13:15
Maybe inverse theif/elseso you can use comparison chaining?if 0 <= r < len(matrix) and 0 <= c < len(matrix[r]): return matrix[r][c] else: return 0. In fact, you could just doreturn 0 <= r < len(matrix) and 0 <= c < len(matrix[r]) and matrix[r][c]and then drop the!= 0in theifclauses.
– tobias_k
Nov 15 '18 at 10:39
@tobias_k True thanks. I also took the opportunity to further optimize the rest of the code.
– blhsing
Nov 15 '18 at 14:08
add a comment |
1
1
1
You can use a helper function to check for boundaries:
def neighbors(matrix, r, c):
def get(r, c):
return 0 <= r < len(matrix) and 0 <= c < len(matrix[r]) and matrix[r][c]
live_neighbors = 0
if get(r, c-1) != 0:
live_neighbors += 1
if get(r-1, c) != 0:
live_neighbors += 1
if get(r-1, c+1) != 0:
live_neighbors += 1
if get(r, c-1) != 0:
live_neighbors += 1
if get(r, c+1) != 0:
live_neighbors += 1
if get(r+1, c-1) != 0:
live_neighbors += 1
if get(r+1, c) != 0:
live_neighbors += 1
if get(r+1, c+1) != 0:
live_neighbors += 1
return live_neighbors
You can also use itertools.product in a generator expression for sum instead of if statements to count all the live neighbors:
from itertools import product
def neighbors(matrix, r, c):
def get(r, c):
return 0 <= r < len(matrix) and 0 <= c < len(matrix[r]) and matrix[r][c]
return sum(get(r + i, c + j) for i, j in product(range(-1, 2), 2) if i or j)
answered Nov 13 '18 at 23:17
blhsingblhsing
29.5k41336
You can use a helper function to check for boundaries:
def neighbors(matrix, r, c):
def get(r, c):
return 0 <= r < len(matrix) and 0 <= c < len(matrix[r]) and matrix[r][c]
live_neighbors = 0
if get(r, c-1) != 0:
live_neighbors += 1
if get(r-1, c) != 0:
live_neighbors += 1
if get(r-1, c+1) != 0:
live_neighbors += 1
if get(r, c-1) != 0:
live_neighbors += 1
if get(r, c+1) != 0:
live_neighbors += 1
if get(r+1, c-1) != 0:
live_neighbors += 1
if get(r+1, c) != 0:
live_neighbors += 1
if get(r+1, c+1) != 0:
live_neighbors += 1
return live_neighbors
You can also use itertools.product in a generator expression for sum instead of if statements to count all the live neighbors:
from itertools import product
def neighbors(matrix, r, c):
def get(r, c):
return 0 <= r < len(matrix) and 0 <= c < len(matrix[r]) and matrix[r][c]
return sum(get(r + i, c + j) for i, j in product(range(-1, 2), 2) if i or j)
answered Nov 13 '18 at 23:17
blhsingblhsing
29.5k41336
edited Nov 15 '18 at 14:07
answered Nov 13 '18 at 23:17
blhsingblhsing
29.5k41336
answered Nov 13 '18 at 23:17
blhsingblhsing
29.5k41336
answered Nov 13 '18 at 23:17
blhsingblhsing
29.5k41336
29.5k41336
why did you write len(matrix(r))?
– BrowserM
Nov 14 '18 at 9:57
Oops that was a typo indeed. Fixed. Thanks.
– blhsing
Nov 14 '18 at 13:15
Maybe inverse theif/elseso you can use comparison chaining?if 0 <= r < len(matrix) and 0 <= c < len(matrix[r]): return matrix[r][c] else: return 0. In fact, you could just doreturn 0 <= r < len(matrix) and 0 <= c < len(matrix[r]) and matrix[r][c]and then drop the!= 0in theifclauses.
– tobias_k
Nov 15 '18 at 10:39
@tobias_k True thanks. I also took the opportunity to further optimize the rest of the code.
– blhsing
Nov 15 '18 at 14:08
add a comment |
why did you write len(matrix(r))?
– BrowserM
Nov 14 '18 at 9:57
Oops that was a typo indeed. Fixed. Thanks.
– blhsing
Nov 14 '18 at 13:15
Maybe inverse theif/elseso you can use comparison chaining?if 0 <= r < len(matrix) and 0 <= c < len(matrix[r]): return matrix[r][c] else: return 0. In fact, you could just doreturn 0 <= r < len(matrix) and 0 <= c < len(matrix[r]) and matrix[r][c]and then drop the!= 0in theifclauses.
– tobias_k
Nov 15 '18 at 10:39
@tobias_k True thanks. I also took the opportunity to further optimize the rest of the code.
– blhsing
Nov 15 '18 at 14:08
why did you write len(matrix(r))?
– BrowserM
Nov 14 '18 at 9:57
why did you write len(matrix(r))?
– BrowserM
Nov 14 '18 at 9:57
Oops that was a typo indeed. Fixed. Thanks.
– blhsing
Nov 14 '18 at 13:15
Oops that was a typo indeed. Fixed. Thanks.
– blhsing
Nov 14 '18 at 13:15
Maybe inverse the
if/else so you can use comparison chaining? if 0 <= r < len(matrix) and 0 <= c < len(matrix[r]): return matrix[r][c] else: return 0. In fact, you could just do return 0 <= r < len(matrix) and 0 <= c < len(matrix[r]) and matrix[r][c] and then drop the != 0 in the if clauses.– tobias_k
Nov 15 '18 at 10:39
Maybe inverse the
if/else so you can use comparison chaining? if 0 <= r < len(matrix) and 0 <= c < len(matrix[r]): return matrix[r][c] else: return 0. In fact, you could just do return 0 <= r < len(matrix) and 0 <= c < len(matrix[r]) and matrix[r][c] and then drop the != 0 in the if clauses.– tobias_k
Nov 15 '18 at 10:39
@tobias_k True thanks. I also took the opportunity to further optimize the rest of the code.
– blhsing
Nov 15 '18 at 14:08
@tobias_k True thanks. I also took the opportunity to further optimize the rest of the code.
– blhsing
Nov 15 '18 at 14:08
add a comment |
0
Closest to what you have would be:
def neighbors(matrix, r, c):
live_neighbors = 0
if c and matrix[r][c-1] != 0:
live_neighbors += 1
if r and matrix[r-1][c] != 0:
live_neighbors += 1
if r and matrix[r-1][c+1] != 0:
live_neighbors += 1
if c and matrix[r][c-1] != 0:
live_neighbors += 1
if c < len(matrix[r])-1 and matrix[r][c+1] != 0:
live_neighbors += 1
if r < len(matrix)-1 and matrix[r+1][c-1] != 0:
live_neighbors += 1
if r < len(matrix)-1 and matrix[r+1][c] != 0:
live_neighbors += 1
if r < len(matrix)-1 and matrix[r+1][c+1] != 0:
live_neighbors += 1
return live_neighbors
and so forth.
answered Nov 15 '18 at 10:41
kabanuskabanus
11.4k31339
add a comment |
0
Closest to what you have would be:
def neighbors(matrix, r, c):
live_neighbors = 0
if c and matrix[r][c-1] != 0:
live_neighbors += 1
if r and matrix[r-1][c] != 0:
live_neighbors += 1
if r and matrix[r-1][c+1] != 0:
live_neighbors += 1
if c and matrix[r][c-1] != 0:
live_neighbors += 1
if c < len(matrix[r])-1 and matrix[r][c+1] != 0:
live_neighbors += 1
if r < len(matrix)-1 and matrix[r+1][c-1] != 0:
live_neighbors += 1
if r < len(matrix)-1 and matrix[r+1][c] != 0:
live_neighbors += 1
if r < len(matrix)-1 and matrix[r+1][c+1] != 0:
live_neighbors += 1
return live_neighbors
and so forth.
answered Nov 15 '18 at 10:41
kabanuskabanus
11.4k31339
add a comment |
0
0
0
Closest to what you have would be:
def neighbors(matrix, r, c):
live_neighbors = 0
if c and matrix[r][c-1] != 0:
live_neighbors += 1
if r and matrix[r-1][c] != 0:
live_neighbors += 1
if r and matrix[r-1][c+1] != 0:
live_neighbors += 1
if c and matrix[r][c-1] != 0:
live_neighbors += 1
if c < len(matrix[r])-1 and matrix[r][c+1] != 0:
live_neighbors += 1
if r < len(matrix)-1 and matrix[r+1][c-1] != 0:
live_neighbors += 1
if r < len(matrix)-1 and matrix[r+1][c] != 0:
live_neighbors += 1
if r < len(matrix)-1 and matrix[r+1][c+1] != 0:
live_neighbors += 1
return live_neighbors
and so forth.
answered Nov 15 '18 at 10:41
kabanuskabanus
11.4k31339
Closest to what you have would be:
def neighbors(matrix, r, c):
live_neighbors = 0
if c and matrix[r][c-1] != 0:
live_neighbors += 1
if r and matrix[r-1][c] != 0:
live_neighbors += 1
if r and matrix[r-1][c+1] != 0:
live_neighbors += 1
if c and matrix[r][c-1] != 0:
live_neighbors += 1
if c < len(matrix[r])-1 and matrix[r][c+1] != 0:
live_neighbors += 1
if r < len(matrix)-1 and matrix[r+1][c-1] != 0:
live_neighbors += 1
if r < len(matrix)-1 and matrix[r+1][c] != 0:
live_neighbors += 1
if r < len(matrix)-1 and matrix[r+1][c+1] != 0:
live_neighbors += 1
return live_neighbors
and so forth.
answered Nov 15 '18 at 10:41
kabanuskabanus
11.4k31339
answered Nov 15 '18 at 10:41
kabanuskabanus
11.4k31339
answered Nov 15 '18 at 10:41
kabanuskabanus
11.4k31339
answered Nov 15 '18 at 10:41
kabanuskabanus
11.4k31339
11.4k31339
add a comment |
add a comment |
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One major consideration is borders: does the value in position [0, 0] have 3 neighbors? Or does it "roll over" and still have 8?
– Brad Solomon
Nov 14 '18 at 3:16
Either way, you may also find
scipy.signal.convolveuseful– Brad Solomon
Nov 14 '18 at 3:17