重新访问 A* 搜索中的已访问节点
Revisiting visited nodes in A* Search
我正在尝试在定向运动中应用 A* 搜索以获得最佳路线。输入是两个文件 - 一个解释地形的图像文件和一个定义高程的文本文件。我根据预设值计算地形难度,以定义可以穿越地形的速度。我还根据坡度定义了海拔难度(下坡速度更快,反之亦然)。
地形和海拔数据存储在矩阵(列表的列表)中。因此,输入是与地图上的点相同的索引。提供了两个输入 - 例如:
start = [230,327]
end = [241,347]
问题是我的代码不断重新访问已存在于已访问队列中的节点。节点定义如下:
class Node:
def __init__(self,value,parent,start=[],goal=[]):
self.children = []
self.parent = parent
self.value = value
self.timeToGoal = 0.0000
self.timeTravelled = 0.0000
if parent:
timeToParent = self.parent.timeTravelled
[parentX, parentY] = parent.value
[currentX, currentY] = self.value
xDiff = abs(currentX - parentX)
yDiff = abs(currentX - parentX)
distance = 12.7627
if xDiff == 0 and yDiff != 0:
distance = 10.29
elif xDiff != 0 and yDiff == 0:
distance = 7.55
# distanceFromParent = math.sqrt(((currentX - parentX) ** 2) - (currentY - parentY) ** 2)
speedFromParent = 1.388 * calculateTerrainDifficulty( terrainMap[currentX][currentY]) * calculateElevationDifficulty(elevationMap[parentX][parentY], elevationMap[currentX][currentY], distance)
timeTravelledFromParent = 0
if speedFromParent != 0:
timeTravelledFromParent = distance / speedFromParent
else:
"Error: Speed from Parent Cannot Be Zero"
self.timeTravelled = timeToParent + timeTravelledFromParent
self.path = parent.path[:]
self.path.append(value)
self.start = parent.start
self.goal = parent.goal
else:
self.path = [value]
self.start = start
self.goal = goal
def GetTime(self):
pass
def CreateChildren(self):
pass
我还使用了一个子节点 class 来定义函数,时间被定义为到自我的时间 + 到目标的毕达哥拉斯斜边距离:
class SubNode(Node):
def __init__(self, value, parent, start=[], goal=[]):
super(SubNode, self).__init__(value, parent, start, goal)
self.timeToGoal = self.GetTime()
def GetTime(self):
if self.value == self.goal:
return 0
[currentX, currentY] = self.value
[targetX, targetY] = self.goal
parentTime = 0
if self.parent:
parentTime = self.timeTravelled
heuristicTime = 99999.99
# Pythagorean Hypotenuse - Straight-line Distance
distance = math.sqrt(((int(currentX) - int(targetX)) ** 2) + (int(currentY)- int(targetY)) ** 2)
speed = 1.38 * calculateTerrainDifficulty(terrainMap[currentX][currentY])
if speed != 0:
heuristicTime = distance / speed
heuristicTime=heuristicTime+parentTime
return heuristicTime
def CreateChildren(self):
if not self.children:
dirs = [-1, 0, 1]
[xVal, yVal] = self.value
for xDir in dirs:
newXVal = xVal + xDir
if newXVal < 0 or newXVal > 394: continue
for yDir in dirs:
newYVal = yVal + yDir
if ((xVal == newXVal) and (yVal == newYVal)) or (newYVal < 0 or newYVal > 499) or (
calculateTerrainDifficulty(terrainMap[newXVal][newYVal]) == 0):
continue
child = SubNode([newXVal, newYVal], self)
self.children.append(child)
A* 搜索 class 定义如下。你可以看到我已经把条件放在那里以确保节点不会被重新访问,当我把打印放在那里时我可以看到多次满足条件。
class AStarSearch:
def __init__(self, start, goal):
self.path = []
self.visitedQueue = []
self.priorityQueue = PriorityQueue()
self.start = start
self.goal = goal
def Search(self):
startNode = SubNode(self.start, 0, self.start, self.goal)
count = 0
self.priorityQueue.put((0, count, startNode))
while (not self.path and self.priorityQueue.qsize()):
closestChild = self.priorityQueue.get()[2]e
closestChild.CreateChildren()
self.visitedQueue.append(closestChild.value)
for child in closestChild.children:
if child.value not in self.visitedQueue:
count += 1
if not child.timeToGoal:
self.path = child.path
break
self.priorityQueue.put((child.timeToGoal, child.value, child))
if not self.path:
print("Not possible to reach goal")
return self.path
由于某些原因,我的程序不断地重新访问某些节点(正如我在打印访问队列时从输出中看到的那样。我该如何避免这种情况?
[[230, 327], [231, 326], [229, 326], [231, 325], [231, 328], [229, 328], [231, 327], [229, 327], [231, 327], [229, 327], [229, 325], [231, 324], [230 , 323], [231, 329], [229, 329], [231, 327], [229, 327], [229, 324], [231, 330], [231, 323], [229, 330 ], [229, 331]]
我面临的另一个问题是:
TypeError: unorderable types: SubNode() < SubNode()
有没有办法在不改变 python 优先级队列的使用的情况下克服这个问题?
您需要在 closestChild 而不是其子项上添加测试:
closestChild = self.priorityQueue.get()[2]e
if closesChild.value not in self.visitedQueue:
closestChild.CreateChildren()
self.visitedQueue.append(closestChild.value)
否则,您可以说您访问了 n1
,然后访问了 n2
,两者都链接到节点 n3
。 n3
在priorityqueue
中添加了两次,所以弹出两次,然后在visitedQueue
中添加了两次。
条件 if child.value not in self.visitedQueue:
可以用来加快速度(通过保持较小的优先级队列),但不是必需的(因为 priorityQueue
中不需要的对象在展开时将被丢弃)。
关于您收到的错误:PriorityQueue
不支持自定义排序,这是您的优先队列所需要的,因此您必须自定义一个。有一个example here。显然你的 _get_priority
函数需要 return timeTravelled
而不是 item[1]
编辑 3:我们(tobias_k 和我)首先说您需要为 SubNode
实现 __eq__
函数,以便 python 知道它们中的两个何时相等,但实际上并非如此,因为您只存储 self.visitedQueue.
中的值
我正在尝试在定向运动中应用 A* 搜索以获得最佳路线。输入是两个文件 - 一个解释地形的图像文件和一个定义高程的文本文件。我根据预设值计算地形难度,以定义可以穿越地形的速度。我还根据坡度定义了海拔难度(下坡速度更快,反之亦然)。
地形和海拔数据存储在矩阵(列表的列表)中。因此,输入是与地图上的点相同的索引。提供了两个输入 - 例如:
start = [230,327]
end = [241,347]
问题是我的代码不断重新访问已存在于已访问队列中的节点。节点定义如下:
class Node:
def __init__(self,value,parent,start=[],goal=[]):
self.children = []
self.parent = parent
self.value = value
self.timeToGoal = 0.0000
self.timeTravelled = 0.0000
if parent:
timeToParent = self.parent.timeTravelled
[parentX, parentY] = parent.value
[currentX, currentY] = self.value
xDiff = abs(currentX - parentX)
yDiff = abs(currentX - parentX)
distance = 12.7627
if xDiff == 0 and yDiff != 0:
distance = 10.29
elif xDiff != 0 and yDiff == 0:
distance = 7.55
# distanceFromParent = math.sqrt(((currentX - parentX) ** 2) - (currentY - parentY) ** 2)
speedFromParent = 1.388 * calculateTerrainDifficulty( terrainMap[currentX][currentY]) * calculateElevationDifficulty(elevationMap[parentX][parentY], elevationMap[currentX][currentY], distance)
timeTravelledFromParent = 0
if speedFromParent != 0:
timeTravelledFromParent = distance / speedFromParent
else:
"Error: Speed from Parent Cannot Be Zero"
self.timeTravelled = timeToParent + timeTravelledFromParent
self.path = parent.path[:]
self.path.append(value)
self.start = parent.start
self.goal = parent.goal
else:
self.path = [value]
self.start = start
self.goal = goal
def GetTime(self):
pass
def CreateChildren(self):
pass
我还使用了一个子节点 class 来定义函数,时间被定义为到自我的时间 + 到目标的毕达哥拉斯斜边距离:
class SubNode(Node):
def __init__(self, value, parent, start=[], goal=[]):
super(SubNode, self).__init__(value, parent, start, goal)
self.timeToGoal = self.GetTime()
def GetTime(self):
if self.value == self.goal:
return 0
[currentX, currentY] = self.value
[targetX, targetY] = self.goal
parentTime = 0
if self.parent:
parentTime = self.timeTravelled
heuristicTime = 99999.99
# Pythagorean Hypotenuse - Straight-line Distance
distance = math.sqrt(((int(currentX) - int(targetX)) ** 2) + (int(currentY)- int(targetY)) ** 2)
speed = 1.38 * calculateTerrainDifficulty(terrainMap[currentX][currentY])
if speed != 0:
heuristicTime = distance / speed
heuristicTime=heuristicTime+parentTime
return heuristicTime
def CreateChildren(self):
if not self.children:
dirs = [-1, 0, 1]
[xVal, yVal] = self.value
for xDir in dirs:
newXVal = xVal + xDir
if newXVal < 0 or newXVal > 394: continue
for yDir in dirs:
newYVal = yVal + yDir
if ((xVal == newXVal) and (yVal == newYVal)) or (newYVal < 0 or newYVal > 499) or (
calculateTerrainDifficulty(terrainMap[newXVal][newYVal]) == 0):
continue
child = SubNode([newXVal, newYVal], self)
self.children.append(child)
A* 搜索 class 定义如下。你可以看到我已经把条件放在那里以确保节点不会被重新访问,当我把打印放在那里时我可以看到多次满足条件。
class AStarSearch:
def __init__(self, start, goal):
self.path = []
self.visitedQueue = []
self.priorityQueue = PriorityQueue()
self.start = start
self.goal = goal
def Search(self):
startNode = SubNode(self.start, 0, self.start, self.goal)
count = 0
self.priorityQueue.put((0, count, startNode))
while (not self.path and self.priorityQueue.qsize()):
closestChild = self.priorityQueue.get()[2]e
closestChild.CreateChildren()
self.visitedQueue.append(closestChild.value)
for child in closestChild.children:
if child.value not in self.visitedQueue:
count += 1
if not child.timeToGoal:
self.path = child.path
break
self.priorityQueue.put((child.timeToGoal, child.value, child))
if not self.path:
print("Not possible to reach goal")
return self.path
由于某些原因,我的程序不断地重新访问某些节点(正如我在打印访问队列时从输出中看到的那样。我该如何避免这种情况?
[[230, 327], [231, 326], [229, 326], [231, 325], [231, 328], [229, 328], [231, 327], [229, 327], [231, 327], [229, 327], [229, 325], [231, 324], [230 , 323], [231, 329], [229, 329], [231, 327], [229, 327], [229, 324], [231, 330], [231, 323], [229, 330 ], [229, 331]]
我面临的另一个问题是:
TypeError: unorderable types: SubNode() < SubNode()
有没有办法在不改变 python 优先级队列的使用的情况下克服这个问题?
您需要在 closestChild 而不是其子项上添加测试:
closestChild = self.priorityQueue.get()[2]e
if closesChild.value not in self.visitedQueue:
closestChild.CreateChildren()
self.visitedQueue.append(closestChild.value)
否则,您可以说您访问了 n1
,然后访问了 n2
,两者都链接到节点 n3
。 n3
在priorityqueue
中添加了两次,所以弹出两次,然后在visitedQueue
中添加了两次。
条件 if child.value not in self.visitedQueue:
可以用来加快速度(通过保持较小的优先级队列),但不是必需的(因为 priorityQueue
中不需要的对象在展开时将被丢弃)。
关于您收到的错误:PriorityQueue
不支持自定义排序,这是您的优先队列所需要的,因此您必须自定义一个。有一个example here。显然你的 _get_priority
函数需要 return timeTravelled
而不是 item[1]
编辑 3:我们(tobias_k 和我)首先说您需要为 SubNode
实现 __eq__
函数,以便 python 知道它们中的两个何时相等,但实际上并非如此,因为您只存储 self.visitedQueue.