Python实现三维数组中目标物体的定位与重标记

现有一个10×10×10的三维数组，背景值为0，物体区域标记为1，包含两个物体：

• 一个位于坐标范围(0,0,0)-(2,2,2)
• 另一个位于坐标范围(3,5,5)-(5,6,8)

需要定位坐标(1,1,1)所属的物体，并将该物体的所有1重标记为5，如何实现？

原三维数组如下：

a = [
[[1,1,1,0,0,0,0,0,0,0],
 [1,1,1,0,0,0,0,0,0,0],
 [1,1,1,0,0,0,0,0,0,0],
 [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],
 [0,0,0,0,0,0,0,0,0,0]],
[
    [1,1,1,0,0,0,0,0,0,0],
    [1,1,1,0,0,0,0,0,0,0],
    [1,1,1,0,0,0,0,0,0,0],
    [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],
    [0,0,0,0,0,0,0,0,0,0]],
[
    [1,1,1,0,0,0,0,0,0,0],
    [1,1,1,0,0,0,0,0,0,0],
    [1,1,1,0,0,0,0,0,0,0],
    [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],
    [0,0,0,0,0,0,0,0,0,0]],
[
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0]],
[
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0]],
[
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,1,1,1,0,0,0,0],
    [0,0,0,1,1,1,0,0,0,0],
    [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],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,1,1,1,0,0,0,0],
    [0,0,0,1,1,1,0,0,0,0],
    [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],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,1,1,1,0,0,0,0],
    [0,0,0,1,1,1,0,0,0,0],
    [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],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,1,1,1,0,0,0,0],
    [0,0,0,1,1,1,0,0,0,0],
    [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],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0]]
]

可以采用**广度优先搜索（BFS）**遍历目标坐标所属的连通区域，将所有1替换为5。这种方法适用于任意形状的连通物体，通用性强。

步骤说明 #

1. 确认目标坐标(1,1,1)的值为1，属于待标记的物体区域
2. 定义三维空间的六个遍历方向（上下、左右、前后）
3. 使用队列存储待访问的坐标，用集合记录已访问的坐标避免重复处理
4. 遍历队列中的每个坐标，将其值设为5，并检查六个方向的相邻坐标，符合条件则加入队列继续处理

# 三维方向向量：上下、左右、前后六个方向
directions = [(-1,0,0), (1,0,0),
              (0,-1,0), (0,1,0),
              (0,0,-1), (0,0,1)]

# 目标坐标
target_x, target_y, target_z = 1, 1, 1

# 检查目标坐标是否属于物体区域
if a[target_x][target_y][target_z] != 1:
    print("目标坐标不属于物体区域")
else:
    # 初始化队列和已访问集合
    queue = [(target_x, target_y, target_z)]
    visited = set()
    visited.add((target_x, target_y, target_z))
    
    # 广度优先遍历
    while queue:
        x, y, z = queue.pop(0)
        # 将当前位置标记为5
        a[x][y][z] = 5
        
        # 遍历所有相邻方向
        for dx, dy, dz in directions:
            nx, ny, nz = x + dx, y + dy, z + dz
            # 检查相邻坐标是否在数组范围内、值为1且未被访问
            if 0 <= nx < 10 and 0 <= ny < 10 and 0 <= nz < 10:
                if a[nx][ny][nz] == 1 and (nx, ny, nz) not in visited:
                    visited.add((nx, ny, nz))
                    queue.append((nx, ny, nz))

# 验证结果（打印前3层的前3行前3列，应为全5）
for layer in range(3):
    print(f"第{layer}层：")
    for row in range(3):
        print(a[layer][row][:3])

简化方案（仅针对规则立方体） #

如果已知目标物体是规则的立方体，也可以直接遍历坐标范围完成标记，代码更简洁：

# 遍历目标物体的坐标范围(0,0,0)-(2,2,2)
for x in range(3):
    for y in range(3):
        for z in range(3):
            if a[x][y][z] == 1:
                a[x][y][z] = 5

内容的提问来源于stack exchange，提问作者Donnerhorn