TypeError：'float' 对象在 3d 数组中不可下标

Question

伙计们，我正在尝试编写一个函数来获取 3-D 数组并检查其中有多少单元格是空的。 但我会收到以下错误

in checkpoint
if m[i][j][0] == 0:
TypeError: 'float' object is not subscriptable

我的功能如下

def checkpoint(m, i, j):
    c = 0
    if m[i][j][0] == 0:
        c += 1.0
    if m[i][j][1] == 0:
        c += 1.0
    if m[i][j][2] == 0:
        c += 1.0
    if m[i][j][3] == 0:
        c += 1.0
return c

它来自我正在使用的一个大型模块 这是使用它的功能

def check(m, size):
flag = 2
for i in range(size):
    for j in range(size):
        print(i, j, "/n")
        c = checkpoint(m, i, j)
        s = summ(m, i, j)
        if c == 2:
            if s == 2 or -2:
                flag = 1.0
                if m[i][j][0] == 0:
                    if m[i][j][1] == 0:
                        m[i][j][0] = m[i][j][1] = (-s/2)
                        fix(m, i, j, 0, size)
                        fix(m, i, j, 1, size)
                    elif m[i][j][2] == 0:
                        m[i][j][0] = m[i][j][2] = (-s/2)
                        fix(m, i, j, 0, size)
                        fix(m, i, j, 2, size)
                    else:
                        m[i][j][0] = m[i][j][3] = (-s/2)
                        fix(m, i, j, 0, size)
                        fix(m, i, j, 3, size)
                elif m[i][j][1] == 0:
                    if m[i][j][2] == 0:
                        m[i][j][1] = m[i][j][2] = (-s/2)
                        fix(m, i, j, 1, size)
                        fix(m, i, j, 2, size)
                    elif m[i][j][3] == 0:
                        m[i][j][1] = m[i][j][3] = (-s/2)
                        fix(m, i, j, 1, size)
                        fix(m, i, j, 3, size)
                else:
                    m[i][j][2] = m[i][j][3] = (-s/2)
        if c == 3:
            flag = 1.0
            if m[i][j][0] == 0:
                m[i][j][0] = -s
            elif m[i][j][1] == 0:
                m[i][j][1] = -s
            elif m[i][j][2] == 0:
                m[i][j][2] = -s
            else:
                m[i][j][3] = -s
return m, flag

任何评论都将不胜感激

更新：

我拼命地在模块内运行该函数，我发现检查函数中 i 和 j 的第一次迭代和第二次迭代没有任何问题。但之后将面临错误。

这是我的输出：the output of the code that I'm trying to run

如您所见，在 i in check 函数的第一次迭代中没有任何问题。
这是我的修复功能。它相对于刚刚更改的单元格的箭头更改了其他一些单元格。

def fix(m, i, j, k, size):
ip = i - 1
jp = j - 1
iz = i + 1
jz = j + 1
if ip < 0:
    ip = size - 1
if jp < 0:
    jp = size - 1
if iz > size - 1:
    iz = 0
if jz > size - 1:
    jz = 0
kp = (k+2) % 4
if k == 0:
    m[i][jz][kp] = -1 * m[i][j][k]
if k == 1:
    m[iz][j][kp] = -1 * m[i][j][k]
if k == 2:
    m[i][jp][kp] = -1 * m[i][j][k]
if k == 3:
    m[ip][j][kp] = -1 * m[i][j][k]
return m

在这里您可以找到整个包裹： my code

Answer 1

这意味着 m 实际上不是 3d 数组，代码试图在 float 上执行 [i] 或 [i][j] 或 [i][j][0]。

“容器”是“可编写脚本的”，如 here 所述