python - 重新访问 A* 搜索中的已访问节点

Question

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 map 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 SubNode 类来定义函数，时间被定义为到 self 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* 搜索类定义如下。您可以看到我已将条件放在那里以确保不会重新访问节点，当我将打印放在那里时我可以看到多次满足该条件。

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], <强>[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？

Answer 1

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中被添加了两次，所以它被弹出两次，然后被添加到visitedQueue中。

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 将在解压时被丢弃)。

关于您收到的错误:PriorityQueue 不支持自定义排序，而这是您的优先级队列所需要的，因此您必须自定义一个。有一个 example here 。显然，您的 _get_priority 函数需要返回 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 知道何时两个它们中的一部分是相等的，但实际上并非如此，因为您只将值存储在 self.visitedQueue 中。