我找到了这种算法，但是似乎创建者没有测试是否存在没有路径的情况。如果没有路径，而且我不知道解决方案，似乎open_list的长度会越来越大。这是我的第一篇文章，非常抱歉我犯了任何错误，非常感谢您的帮助。
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:
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, 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, 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, 1, 0, 0, 0, 0, 0]]
start = (0, 0)
end = (7, 6)
path = astar(maze, start, end)
return path
print(main())
答案 0 :(得分：0)
closed_list应该是set，而不是list，以检查是否已经访问过Node。这样可以删除检查节点是否已访问的内部循环，并非常有效地执行此操作；
但这不仅是一种优化：它使continue可以在外循环结束时恢复应有的执行。这是代码中的主要错误：continue仅使您进入了内部循环的最后，实际上使访问的检查无用：您不断重复添加相同的节点，无论它们是否已经添加是否被访问过。
为了在Node中包含set，Node必须是可哈希的。在这里，我返回元组position
修改后的代码将返回一个路径（如果存在）或“无”。
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 __hash__(self): #<-- added a hash method
return hash(self.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 = set() # <-- closed_list must be a set
# 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.add(current_node) # <-- change append to add
# 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:
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
if child in closed_list: # <-- remove inner loop so continue takes you to the end of the outer loop
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, 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, 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, 1, 0, 0, 0, 0, 0]]
start = (0, 0)
end = (7, 6)
path = astar(maze, start, end)
return path
print(main())