【已解决】Python Bresenham 3D算法

放一段使用Python实现Bresenham 3D 算法的代码,并通过Matplot可视化

python 复制代码
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from numba import njit

@njit
def bresenham_safe(grid, x0, y0, z0, x1, y1, z1, value_to_fill):
    start_point = [int(x0), int(y0), int(z0)]
    end_point = [int(x1), int(y1), int(z1)]

    steep_xy = (abs(end_point[1] - start_point[1]) > abs(end_point[0] - start_point[0]))
    if steep_xy:
        start_point[0], start_point[1] = start_point[1], start_point[0]
        end_point[0], end_point[1] = end_point[1], end_point[0]

    steep_xz = (abs(end_point[2] - start_point[2]) > abs(end_point[0] - start_point[0]))
    if steep_xz:
        start_point[0], start_point[2] = start_point[2], start_point[0]
        end_point[0], end_point[2] = end_point[2], end_point[0]

    delta = [abs(end_point[0] - start_point[0]), abs(end_point[1] - start_point[1]), abs(end_point[2] - start_point[2])]

    error_xy = delta[0] / 2
    error_xz = delta[0] / 2

    step = [
        -1 if start_point[0] > end_point[0] else 1,
        -1 if start_point[1] > end_point[1] else 1,
        -1 if start_point[2] > end_point[2] else 1
    ]

    y = start_point[1]
    z = start_point[2]

    for x in range(start_point[0], end_point[0], step[0]):
        point = [x, y, z]

        if steep_xz:
            point[0], point[2] = point[2], point[0]
        if steep_xy:
            point[0], point[1] = point[1], point[0]

        if 0 <= point[0] < grid.shape[0] and 0 <= point[1] < grid.shape[1] and 0 <= point[2] < grid.shape[2]:
            grid[point[0], point[1], point[2]] = value_to_fill

        error_xy -= delta[1]
        error_xz -= delta[2]

        if error_xy < 0:
            y += step[1]
            error_xy += delta[0]

        if error_xz < 0:
            z += step[2]
            error_xz += delta[0]

@njit
def get_free_area(obstacle, x, y, z):
    free = np.zeros_like(obstacle)
    obstacle = obstacle > 0
    xs, ys, zs = np.where(obstacle)
    for ox, oy, oz in zip(xs, ys, zs):
        bresenham_safe(free, ox, oy, oz, x, y, z, 1)
    free -= obstacle
    return free

# 创建三维网格和障碍物示例
grid = np.zeros((65, 65, 12))
obstacle = np.zeros_like(grid)
start = np.zeros_like(grid)
obstacle[20:30, 4:60, 2:9] = 1

# 获取自由区域
x = 8
y = 8
z = 11
import time
start_time = time.time()
free_area = get_free_area(obstacle, x, y, z)
end_time = time.time()
execution_time = end_time - start_time
print("□□□□□□□□□□程序执行时间为:", execution_time, "秒") 
start[8,8,11]=1

# 翻转颜色映射
cmap = plt.cm.gray
cmap_inverted = cmap.reversed()
# # 可视化
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')

# 设置坐标轴范围和比例,让显示比例正常
x_dim, y_dim, z_dim = obstacle.shape
max_dim = max(x_dim, y_dim, z_dim)
ax.set_xlim(0, max_dim)
ax.set_ylim(0, max_dim)
ax.set_zlim(0, max_dim)


# ------------------------------------------------------------------------------
# x_indices, y_indices, z_indices = np.where(free_area)
ax.voxels(free_area, facecolors='green',)
ax.voxels(obstacle, facecolors='red',)
ax.voxels(start, facecolors='blue')
# print(grid)
ax.set_xlabel('X')
ax.set_ylabel('Y')
ax.set_zlabel('Z')
plt.show()
相关推荐
happylifetree7 小时前
Python09:核心语法-数据存储与运算-字面量
python
longlongzihan7 小时前
LeetCode 17电话号码的字母组合:回溯算法(DFS)详解
c++·算法·leetcode·深度优先
心之语歌7 小时前
Tkinter 画布基本梳理
运维·服务器·python
封印师请假去地球钓鱼7 小时前
边解边变的问题:从“决策依赖“一词出发
人工智能·算法
Wx-bishekaifayuan7 小时前
django个性化旅游路线推荐平台49005-计算机课程设计、毕业设计
spring boot·后端·python·django·课程设计·express·旅游
今晚打老虎8 小时前
c++之提高A(前缀和)(第三课)
数据结构·c++·算法
小白快快跑哦8 小时前
python-字符串全解(六):正则表达式-量词
python·正则表达式·字符串
禹凕9 小时前
机器学习之Selenium(Machina Learning about Selenium)
爬虫·python·selenium·测试工具·机器学习
不会就选b9 小时前
算法日常・每日刷题--<贪心>26
数据结构·算法
ebiobiz11 小时前
基于 GD32 Embedded Builder (GEB) 与 Nimmake 的 MCU 工程搭建指南
c++·python·单片机·嵌入式硬件·mcu