python for dynamo路径规划代码
时间: 2023-11-20 19:55:24 浏览: 108
以下是一个使用Python编写的Dynamo路径规划代码的示例:
```python
import numpy as np
import matplotlib.pyplot as plt
# 定义起点和终点
start = (0, 0)
goal = (10, 10)
# 定义障碍物
obstacles = [(5, 5), (6, 6), (7, 7), (8, 8)]
# 定义地图大小
map_size = 11
# 定义代价地图
cost_map = np.ones((map_size, map_size))
for obstacle in obstacles:
cost_map[obstacle] = np.inf
# 定义动态规划函数
def dynamic_programming(start, goal, cost_map):
value_map = np.zeros_like(cost_map)
policy_map = np.zeros_like(cost_map, dtype=int)
directions = np.array([[0, 1], [1, 0], [0, -1], [-1, 0]])
def is_valid_state(state):
x, y = state
if x < 0 or x >= map_size or y < 0 or y >= map_size:
return False
if np.isinf(cost_map[state]):
return False
return True
def get_valid_neighbors(state):
neighbors = []
for direction in directions:
neighbor = tuple(np.array(state) + direction)
if is_valid_state(neighbor):
neighbors.append(neighbor)
return neighbors
def get_value(state):
if state == goal:
return 0
if np.isinf(cost_map[state]):
return np.inf
neighbors = get_valid_neighbors(state)
values = []
for neighbor in neighbors:
value = cost_map[neighbor] + value_map[neighbor]
values.append(value)
return np.min(values)
def get_policy(state):
if state == goal:
return -1
if np.isinf(cost_map[state]):
return -1
neighbors = get_valid_neighbors(state)
values = []
for neighbor in neighbors:
value = cost_map[neighbor] + value_map[neighbor]
values.append(value)
return np.argmin(values)
for i in range(map_size):
for j in range(map_size):
state = (i, j)
value_map[state] = get_value(state)
policy_map[state] = get_policy(state)
return value_map, policy_map
# 运行动态规划函数
value_map, policy_map = dynamic_programming(start, goal, cost_map)
# 可视化结果
plt.imshow(value_map, cmap='jet')
plt.colorbar()
for obstacle in obstacles:
plt.plot(obstacle[1], obstacle[0], 'sk')
plt.plot(start[1], start[0], 'bo')
plt.plot(goal[1], goal[0], 'ro')
plt.show()
```
该代码使用动态规划算法计算从起点到终点的最短路径,并考虑了障碍物的影响。在代码中,我们首先定义了起点、终点和障碍物的位置,然后定义了地图大小和代价地图。接着,我们实现了一个动态规划函数,该函数计算了每个状态的最小代价和最优策略。最后,我们使用Matplotlib库可视化了结果。
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