Question

扫雷，仍然是。我找到了一种方法，但我知道必须有一种简化的方法来做到这一点。我必须在矩阵中放置一个数字来表示围绕它的炸弹数量（“b”）。这就是我所拥有的，我知道必须采用更短的方式。

def check(y,x):
    if ((y < 0) or (y >= len(mat1)) or (x < 0) or (x >= len(mat1))):
        return (False)
    else:
        return mat1[y][x]

def addscores():
    for x in range(len(mat1)):
        for y in range(len(mat1)):
            if mat1[y][x] != "b":
                if check(y-1,x-1) == "b" or check(y,x-1) == "b" or check(y+1,x-1) == "b" or check(y+1,x) == "b" or check(y+1,x+1) == "b" or check(y,x+1) == "b" or check(y-1,x+1) == "b" or check(y-1,x) =="b":
                    mat1[y][x] = 1
                if check(y-1,x-1) == "b":
                    if check(y,x-1) == "b" or check(y+1,x-1) == "b" or check(y+1,x) == "b" or check(y+1,x+1) == "b" or check(y,x+1) == "b" or check(y-1,x+1) == "b" or check(y-1,x) == "b":
                        mat1[y][x] = 2
                if check(y,x-1) == "b":
                    if check(y+1,x-1) == "b" or check(y+1,x) == "b" or check(y+1,x+1) == "b" or check(y,x+1) == "b" or check(y-1,x+1) == "b" or check(y-1,x) == "b":
                        mat1[y][x] = 2
                if check(y+1,x-1) == "b":
                    if check(y+1,x) == "b" or check(y+1,x+1) == "b" or check(y,x+1) == "b" or check(y-1,x+1) == "b" or check(y-1,x) == "b":
                         mat1[y][x] = 2
                if check(y+1,x) == "b":
                    if check(y+1,x+1) == "b" or check(y,x+1) == "b" or check(y-1,x+1) == "b" or check(y-1,x) == "b":
                        mat1[y][x] = 2
                if check(y+1,x+1) == "b":
                    if check(y,x+1) == "b" or check(y-1,x+1) == "b" or check(y-1,x) == "b":
                        mat1[y][x] = 2
                if check(y,x+1) == "b":
                    if check(y-1,x+1) == "b" or check(y-1,x) == "b":
                        mat1[y][x] = 2
                if check(y-1,x+1) == "b":
                    if check(y-1,x) == "b":
                        mat1[y][x] = 2
                if check(y-1,x-1) == "b":
                    if check(y,x-1) == "b":
                        if check(y+1,x-1) == "b" or check(y+1,x) == "b" or check(y+1,x+1) == "b" or check(y,x+1) == "b" or check(y-1,x+1) == "b" or check(y-1,x) == "b":
                            mat1[y][x] = 3
                    if check(y+1,x-1) == "b":
                        if check(y+1,x) == "b" or check(y+1,x+1) == "b" or check(y,x+1) == "b" or check(y-1,x+1) == "b" or check(y-1,x) == "b":
                            mat1[y][x] = 3
                    if check(y+1,x) == "b":
                        if check(y+1,x+1) == "b" or check(y,x+1) == "b" or check(y-1,x+1) == "b" or check(y-1,x) == "b":
                            mat1[y][x] = 3
                    if check(y+1,x+1) == "b":
                        if check(y,x+1) == "b" or check(y-1,x+1) == "b" or check(y-1,x) == "b":
                            mat1[y][x] = 3
                    if check(y,x+1) == "b":
                        if check(y-1,x+1) == "b" or check(y-1,x) == "b":
                            mat1[y][x] = 3
                if check(y-1,x-1) == "b":
                    if check(y,x-1) == "b":
                        if check(y+1,x-1) == "b":
                            if check(y+1,x) == "b" or check(y+1,x+1) == "b" or check(y,x+1) == "b" or check(y-1,x+1) == "b" or check(y-1,x) == "b":
                                mat1[y][x] = 4
                    if check(y+1,x-1) == "b":
                        if check(y+1,x) == "b":
                            if check(y+1,x+1) == "b" or check(y,x+1) == "b" or check(y-1,x+1) == "b" or check(y-1,x) == "b":
                                mat1[y][x] = 4
                    if check(y+1,x) == "b":
                        if check(y+1,x+1) == "b":
                            if check(y,x+1) == "b" or check(y-1,x+1) == "b" or check(y-1,x) == "b":
                                mat1[y][x] = 4
                    #ETC

Answer 1

def check(y,x):
    if ((y < 0) or (x < 0) or (y >= len(mat1)) or (x >= len(mat1))):
        return False
    else:
        if mat1[y][x] == 'b':
            return 1
        else:
            return 0

def check_all(y,x):
    if mat1[y][x] != 'b':        
        return sum([check(y + yy, x + xx) for xx in range(-1,2) for yy in range(-1,2)])
    else:
        return 'b'

def addscores():
    for x in range(len(mat1)):
        for y in range(len(mat1)):
            mat1[y][x] = check_all(y,x)

如果我已经明白你正在做什么，这段代码 - 尤其是check_all函数 - 应该解决你的问题。你是对的，肯定有一个更短的方法，使用循环（或列表推导，在这种情况下），而不是必须单独写出每个检查。

我尽可能保留了您的代码，因为我没有足够的上下文来知道更改是否会破坏任何内容。

Answer 2

您可以使用以下内容：

def add_scores():
    def bounded_range(i):
        return range(max(0, i - 1), min(len(mat1) - 1, x + 1))

    for x in range(len(mat1)):
        for y in range(len(mat1)):
            Xs = bounded_range(x)
            Ys = bounded_range(y)
            mat1[x][y] = len([0 for _x in Xs for _y in Ys
                    if not (x == _x and y == _x) and mat1[_x][_y] == 'b'])

Answer 3

如果你想弄清楚如何编写扫雷程序，你可以查看Python Cookbook上的MineSweep，它显示了开发游戏GUI版本的12个步骤。

检查矩阵中“b”周围的值

3 个答案: