检查矩阵中其他值的对角线值的问题。 （Python）

Question

所以我和我的朋友试图在 python 中重新创建 Conway 的生活游戏，但在尝试检查单元矩阵中对角相邻的值时遇到了问题。我们的代码查找与所讨论的值对角线的值，但由于某种原因，它似乎无法找到它们。例如，具有 3 个相邻单元（2 个相邻单元和 1 个对角线）的单元将作为 2 个相邻单元返回。 为了调试，我们让它列出了所有活细胞的coridinates及其邻居计数。 这是我们的代码：

initial_frame = [
     [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, 0, 0, 0, 0, 0, 0, 0],
     [0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0],
     [0, 0, 0, 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, 0, 0, 0, 0]]

next_frame = [
     [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, 0, 0, 0, 0, 0, 0, 0],
     [0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0],
     [0, 0, 0, 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, 0, 0, 0, 0]]

row = []

neighborcount = 0

next_frame = initial_frame

while True:

    for e in range(1, 10):

        for a in range(1, 10):

            row.append(initial_frame[e][a])

        print(row)

        row = []

    print("\n\n\n\n")

    input()

    for i in range(1, 10):

        for o in range(1, 10):

            neighborcount = 0

            #Down 1

            if initial_frame[(o + 1)][i] == 1:

                neighborcount += 1



            #Up 1

            if initial_frame[(o - 1)][i] == 1:

                neighborcount += 1



            #Right 1

            if initial_frame[o][(i + 1)] == 1:

                neighborcount += 1



            #Left 1

            if initial_frame[o][(i - 1)] == 1:

                neighborcount += 1



            #Down 1, Right 1

            if initial_frame[(o + 1)][(i + 1)] == 1:

                neighborcount += 1



            #Down 1, Left 1

            if initial_frame[(o + 1)][(i - 1)] == 1:

                neighborcount += 1



            #Up 1, Left 1

            if initial_frame[(o - 1)][(i - 1)] == 1:

                neighborcount += 1



            #Up 1, Right 1

            if initial_frame[(o - 1)][(i + 1)] == 1:

                neighborcount += 1



            #If dead cell has exactly 3 neighbors, set it to be born

            if initial_frame[o][i] == 0 and neighborcount == 3:

                next_frame[o][i] = 1



            #If living cell:

            if initial_frame[o][i] == 1:

                #does not have either 2 or 3 neighbors, set it to die

                if neighborcount != 2 and neighborcount != 3:

                    next_frame[o][i] = 0

                print(str(o) + ", " + str(i) + ": " + str(neighborcount))

            #reset neighbors

            neighborcount = 0

    #Project set values onto real board

    initial_frame = next_frame

Answer 1

问题是，在这一行

next_frame = initial_frame

Python 实际上并不复制整个数组。 next_frame 刚刚开始引用 initial_frame 所引用的任何内容，因此 next_frame is initial_frame 将返回 true。

这可以通过在生成计算结束时交换数组来解决。像这样：

@@ -30,8 +30,6 @@ row = []

neighborcount = 0

-next_frame = initial_frame
-
while True:

    for e in range(1, 10):
@@ -120,6 +118,8 @@ while True:

            #If dead cell has exactly 3 neighbors, set it to be born

+            next_frame[o][i] = initial_frame[o][i]
+
            if initial_frame[o][i] == 0 and neighborcount == 3:

                next_frame[o][i] = 1
@@ -143,5 +143,6 @@ while True:
            neighborcount = 0

    #Project set values onto real board
-
+    garbage_arr = initial_frame
    initial_frame = next_frame
+    next_frame = garbage_arr

结果代码：

initial_frame = [
     [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, 0, 0, 0, 0, 0, 0, 0],
     [0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0],
     [0, 0, 0, 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, 0, 0, 0, 0]]

next_frame = [
     [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, 0, 0, 0, 0, 0, 0, 0],
     [0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0],
     [0, 0, 0, 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, 0, 0, 0, 0]]

row = []

neighborcount = 0

while True:

    for e in range(1, 10):

        for a in range(1, 10):

            row.append(initial_frame[e][a])

        print(row)

        row = []

    print("\n\n\n\n")

    input()

    for i in range(1, 10):

        for o in range(1, 10):

            neighborcount = 0

            #Down 1

            if initial_frame[(o + 1)][i] == 1:

                neighborcount += 1



            #Up 1

            if initial_frame[(o - 1)][i] == 1:

                neighborcount += 1



            #Right 1

            if initial_frame[o][(i + 1)] == 1:

                neighborcount += 1



            #Left 1

            if initial_frame[o][(i - 1)] == 1:

                neighborcount += 1



            #Down 1, Right 1

            if initial_frame[(o + 1)][(i + 1)] == 1:

                neighborcount += 1



            #Down 1, Left 1

            if initial_frame[(o + 1)][(i - 1)] == 1:

                neighborcount += 1



            #Up 1, Left 1

            if initial_frame[(o - 1)][(i - 1)] == 1:

                neighborcount += 1



            #Up 1, Right 1

            if initial_frame[(o - 1)][(i + 1)] == 1:

                neighborcount += 1



            #If dead cell has exactly 3 neighbors, set it to be born

            next_frame[o][i] = initial_frame[o][i]

            if initial_frame[o][i] == 0 and neighborcount == 3:

                next_frame[o][i] = 1



            #If living cell:

            if initial_frame[o][i] == 1:

                #does not have either 2 or 3 neighbors, set it to die

                if neighborcount != 2 and neighborcount != 3:

                    next_frame[o][i] = 0

                print(str(o) + ", " + str(i) + ": " + str(neighborcount))

            #reset neighbors

            neighborcount = 0

    #Project set values onto real board
    garbage_arr = initial_frame
    initial_frame = next_frame
    next_frame = garbage_arr

Answer 2

避免相邻坐标到达边缘问题的一种方法是在矩阵的每一侧填充一行和一列零。这使您可以使用偏移列表来访问 8 个相邻值，而无需进行任何边界检查。

[编辑] 我刚刚注意到您已经填充了矩阵。 （我相应地调整了我的例子）。似乎其中一个问题是您使用的 range(1,10) 只会通过 1..9 但您需要通过 1..10 所以它们应该是 range(1,11)

例如：

next_Frame  = [ line.copy() for line in initial_frame ]
offsets     = [ (-1,-1),(-1,0),(-1,1),(0,-1),(0,1),(1,-1),(1,0),(1,1) ]
for row in range(1,11):
    for col in range(1,11):
        neighbors = sum([ initial_frame[row+r][col+c] for r,c in offsets ])
        if next_Frame[row][col] == 0 and neighbors == 3:
            next_Frame[row][col] = 1 
        elif next_Frame[row][col] == 1 and neighbors not in [2,3] 
            next_Frame[row][col] = 0