Question

这是Conway使用Python制作的Game of Life模拟代码片段。它是非常简单的，目的是检查它是否有效 - 而事实并非如此。无法弄清楚为什么，但据我所知，这是更新的内容。

感谢任何有关它为何更新其原因的输入：

pygame中的Gif（相同代码）： http://imgur.com/6US3Nje 没有正确更新： http://imgur.com/9gubzAF

import pprint,random
#here we make the initial board, 6x6
board = []
for y in range (6):
    row = []
    for x in range (6):
        row.append(random.randint(0,1))
    board.append(row)
#and display it
pprint.pprint(board)

#this function counts the neighbours for each cell
def neighbours(x,y):
    counter = 0
    neighbours = \
    [(x-1,y+1),(x,y+1),(x+1,y+1),\
    (x-1,y),          (x+1,y),\
    (x-1,y-1),(x,y-1),(x+1,y-1)]
    for n in neighbours:
        a, b = n
        try:
            counter += board[a][b]
        except:
            pass
#printed out the counter to check if it counted correctly (it does, as far as I could tell)
    if x == 4 and y == 4:
        print(counter)
    return counter

#function to make a new board based off the old one - basically, the updater. 
#here's where the problem might lie - but for the life of me I cannot tell where and why.
def new(board):
    new_board = []
    for y in range(6):
        new_row = []
        for x in range(6):
            n = neighbours(x,y)
            oldcell = board[x][y]
            #everything is set up to be according to the rules
            #(if dead(0) can only come alive with 3 alive cells
            #(if alive(1) can continue to live with exactly 2 or 3 live neighbours
            if oldcell == 0:
                newcell = 0
                if n == 3:
                    newcell = 1
            elif oldcell == 1:
                newcell = 1
                if n > 3 or n < 2:
                    newcell = 0
            new_row.append(newcell)
        new_board.append(new_row)
    return new_board

#displaying the board for 6 instances
for i in range (6):
    nboard = new(board)
    board = nboard
    pprint.pprint(board)

提前致谢！

Answer 1

您正在循环中混合行和列，有效地将每个新板的整个矩阵转置。循环应该如下所示：

for x in range(6):
    new_row = []
    for y in range(6):
        ...

此外，循环中的大部分if/else逻辑可以简化为单行。是否更容易阅读和理解是由您决定。

def new(board):
    new_board = []
    for x in range(6):
        new_row = []
        for y in range(6):
            n = neighbours(x,y)
            oldcell = board[x][y]
            newcell = int(n in (2, 3)) if oldcell else int(n == 3)
            new_row.append(newcell)
        new_board.append(new_row)
    return new_board

正如所提到的in comments，还有第二个问题（虽然不那么明显）：虽然你的try/except跳过所有的车外索引是一个很好的想法，但它并不能解决问题。 x-1 = x的{{1}}，因为0是Python中的合法索引（索引序列中的最后一个元素），使您的主板“环绕”两个四个边缘。相反，您应该明确检查电路板的边界。如果你愿意，你可以把它变成另一个“有趣”的单行：

[-1]

当然，您也可以保留自己的循环，只需替换def neighbours(x,y): return sum(board[a][b] for a in (x-1,x,x+1) for b in (y-1,y,y+1) if (a,b) != (x,y) and 0 <= a < len(board) and 0 <= b < len(board[a])) try/except的{{1}}。

（已结束）康威的生命游戏未正确更新（Python）

1 个答案: