我仍然对Python还是陌生的，这是我关于stackoverflow的第一个问题，我在实施A *算法方面遇到了一周的麻烦。
我得到的代码找到了一个具有直墙的目标，但是正如您将看到的，只要我将墙扩展到起点以下，并且它必须向后或围绕它，它就会永远循环。
我一直在用力撞墙，试图修复它，以及如何实现升高代码，以使其停止循环。任何帮助将不胜感激。
我的代码：
class Node:
"""A node class for A* Pathfinding"""
def __init__(self, parent=None, position=None):
self.parent = parent
self.position = position
self.g = 0
self.h = 0
self.f = 0
def __eq__(self, other):
return self.position == other.position
def astar(maze, start, end):
"""Returns a list of tuples as a path from the given start to the given end in the given maze"""
# Create start and end node
start_node = Node(None, start)
start_node.g = start_node.h = start_node.f = 0
end_node = Node(None, end)
end_node.g = end_node.h = end_node.f = 0
# Initialize both open and closed list
open_list = []
closed_list = []
# Add the start node
open_list.append(start_node)
# Loop until you find the end
while len(open_list) > 0:
# Get the current node
current_node = open_list[0]
current_index = 0
for index, item in enumerate(open_list):
if item.f < current_node.f:
current_node = item
current_index = index
# Pop current off open list, add to closed list
open_list.pop(current_index)
closed_list.append(current_node)
# Found the goal
if current_node == end_node:
path = []
current = current_node
while current is not None:
path.append(current.position)
current = current.parent
return path[::-1] # Return reversed path
# Generate children
children = []
for new_position in [(0, -1), (0, 1), (-1, 0), (1, 0), (-1, -1), (-1, 1), (1, -1), (1, 1)]: # Adjacent squares
# Get node position
node_position = (current_node.position[0] + new_position[0], current_node.position[1] + new_position[1])
# Make sure within range
if node_position[0] > (len(maze) - 1) or node_position[0] < 0 or node_position[1] > (
len(maze[len(maze) - 1]) - 1) or node_position[1] < 0:
continue
# Make sure walkable terrain
if maze[node_position[0]][node_position[1]] != 0:
print("node -", node_position[0],node_position[1], "is blocked")
continue
# Create new node
new_node = Node(current_node, node_position)
# Append
children.append(new_node)
# Loop through children
for child in children:
# Child is on the closed list
for closed_child in closed_list:
if child == closed_child:
continue
# Create the f, g, and h values
child.g = current_node.g + 1
child.h = ((child.position[0] - end_node.position[0]) ** 2) + (
(child.position[1] - end_node.position[1]) ** 2)
child.f = child.g + child.h
# Child is already in the open list
for open_node in open_list:
if child == open_node and child.g > open_node.g:
continue
# Add the child to the open list
open_list.append(child)
def main():
maze = [[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 1, 1, 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, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]
start = (0, 0)
end = (1, 8)
path = astar(maze, start, end)
print(path)
main()
您具有这两件作品：
# Child is on the closed list
for closed_child in closed_list:
if child == closed_child:
continue
# Child is already in the open list
for open_node in open_list:
if child == open_node and child.g > open_node.g:
continue
continue将继续内部for循环。因此，它没有任何作用。您正在寻找类似这样的东西：
# Child is on the closed list
is_in_closed = False
for closed_child in closed_list:
if child == closed_child:
is_in_closed = True
break
if is_in_closed:
continue
# Child is already in the open list
is_in_open = False
for open_node in open_list:
if child == open_node and child.g > open_node.g:
is_in_open = True
break
if is_in_open:
continue
另外两个评论：
您的距离度量仅计算步数。因此，对角线与水平/垂直步骤一样昂贵。您可能需要更改它，并获得实际长度或近似值，才能找到真正最短的路径。
您的启发式是到目标的平方距离。这不是允许的启发式方法，因为它高估了目标的实际成本。结果，您可能找不到正确的最短路径。对于您的成本函数（步数），更好的和可接受的启发式方法是max(child.position[0] - end_node.position[0], child.position[1] - end_node.position[1])（您至少需要此步数才能达到目标）。