Question

我已经找过其他类似的案例，但没有一个适用于我......或者如果他们这样做我不明白怎么做。

我要做的是过滤2D数组（这个）：

blocks = [
[  [0], [0], [0], [0], [0], [0], [0], [0], [0], [0]  ],

[  [0], [0], [0], [0], [0], [0], [0], [0], [0], [0]  ],

[  [0], [0], [0], [0], [0], [0], [0], [0], [0], [0]  ],

[  [0], [0], [0], [0], [0], [0], [0], [0], [0], [0]  ],

[  [0], [0], [0], [0], [0], [0], [0], [0], [0], [0]  ],

[  [0], [0], [0], [0], [0], [0], [0], [0], [0], [0]  ],

[  [0], [0], [0], [0], [0], [0], [0], [0], [0], [0]  ],

[  [0], [0], [0], [0], [0], [0], [0], [0], [0], [0]  ],

[  [0], [0], [0], [0], [0], [0], [0], [0], [0], [0]  ],

[  [0], [0], [0], [0], [0], [0], [0], [0], [0], [0]  ]

并找出当前部分是什么。对不起，如果我的代码有点草率，我是python的新手。这是我的代码的一部分，

def drawBlock():
for blockL1 in blocks:
    for blokL2 in blocks:
        if blocks[blockL1[blockL2[1]]] == 0:
            pygame.draw.rect(screen, (0, 255, 255), Rect((blockL1 * 32, blockL2 *32), (32, 32)))
            pygame.draw.rect(screen, color, Rect(i * 32, ))
        global blockL2
        blockL2 += 1
    global blockL1
    blockL1 += 1

Answer 1

你在这里遇到了几个问题。

主要问题是您使用for blank in blank_holder。

写作时

for blockL1 in blocks:

与写作几乎相同

for ii in range(len(blocks)):
    blockL1 = blocks[ii]
    ...

意味着blockL1不是整数，也不是块的索引..它是块的元素！

其值如下：[ [0], [0], [0], [0], [0], [0], [0], [0], [0], [0] ]

哪个列表是列表...其中blocks是包含列表的列表列表...是的

当你写

for blockL1 in blocks:
    for blockL2 in blocks:

blockL2的值将是单个列表[0]

假设您要查看blockL2中包含的一个值，您可以更改：

if blocks[blockL1[blockL2[1]]] == 0:

为：

if blockL2[0] == 0: # the 1 index would go out of bounds for a list length 1

但是，此代码看起来需要blockL1和blockL2的索引值，因此我建议您循环浏览其索引列表（意为ints）...而不是它们的元素

def drawBlock():
    for blockL1 in range(len(blocks)):
        for blockL2 in range(len(blockL1)):
            if blocks[blockL1[blockL2[0]]] == 0:
                pygame.draw.rect(screen, (0, 255, 255), Rect((blockL1 * 32, blockL2 *32), (32, 32)))

希望这有助于一些人！

TypeError：＆＃39; int＆＃39;对象不可订阅

1 个答案: