python实现A*寻路算法
a* 算法简介
a* 算法需要维护两个数据结构:open 集和 closed 集。open 集包含所有已搜索到的待检测节点。初始状态,open集仅包含一个元素:开始节点。closed集包含已检测的节点。初始状态,closed集为空。每个节点还包含一个指向父节点的指针,以确定追踪关系。
a* 算法会给每个搜索到的节点计算一个g+h 的和值f:
- f = g + h
- g:是从开始节点到当前节点的移动量。假设开始节点到相邻节点的移动量为1,该值会随着离开始点越来越远而增大。
- h:是从当前节点到目标节点的移动量估算值。
- 如果允许向4邻域的移动,使用曼哈顿距离。
- 如果允许向8邻域的移动,使用对角线距离。
算法有一个主循环,重复下面步骤直到到达目标节点:
1 每次从open集中取一个最优节点n(即f值最小的节点)来检测。
2 将节点n从open集中移除,然后添加到closed集中。
3 如果n是目标节点,那么算法结束。
4 否则尝试添加节点n的所有邻节点n'。
- 邻节点在closed集中,表示它已被检测过,则无需再添加。
- 邻节点在open集中:
- 如果重新计算的g值比邻节点保存的g值更小,则需要更新这个邻节点的g值和f值,以及父节点;
- 否则不做操作
- 否则将该邻节点加入open集,设置其父节点为n,并设置它的g值和f值。
有一点需要注意,如果开始节点到目标节点实际是不连通的,即无法从开始节点移动到目标节点,那算法在第1步判断获取到的节点n为空,就会退出
关键代码介绍
保存基本信息的地图类
地图类用于随机生成一个供寻路算法工作的基础地图信息
先创建一个map类, 初始化参数设置地图的长度和宽度,并设置保存地图信息的二维数据map的值为0, 值为0表示能移动到该节点。
class map(): def __init__(self, width, height): self.width = width self.height = height self.map = [[0 for x in range(self.width)] for y in range(self.height)]
在map类中添加一个创建不能通过节点的函数,节点值为1表示不能移动到该节点。
def createblock(self, block_num): for i in range(block_num): x, y = (randint(0, self.width-1), randint(0, self.height-1)) self.map[y][x] = 1
在map类中添加一个显示地图的函数,可以看到,这边只是简单的打印出所有节点的值,值为0或1的意思上面已经说明,在后面显示寻路算法结果时,会使用到值2,表示一条从开始节点到目标节点的路径。
def showmap(self): print("+" * (3 * self.width + 2)) for row in self.map: s = '+' for entry in row: s += ' ' + str(entry) + ' ' s += '+' print(s) print("+" * (3 * self.width + 2))
添加一个随机获取可移动节点的函数
def generatepos(self, rangex, rangey): x, y = (randint(rangex[0], rangex[1]), randint(rangey[0], rangey[1])) while self.map[y][x] == 1: x, y = (randint(rangex[0], rangex[1]), randint(rangey[0], rangey[1])) return (x , y)
搜索到的节点类
每一个搜索到将到添加到open集的节点,都会创建一个下面的节点类,保存有entry的位置信息(x,y),计算得到的g值和f值,和该节点的父节点(pre_entry)。
class searchentry(): def __init__(self, x, y, g_cost, f_cost=0, pre_entry=none): self.x = x self.y = y # cost move form start entry to this entry self.g_cost = g_cost self.f_cost = f_cost self.pre_entry = pre_entry def getpos(self): return (self.x, self.y)
算法主函数介绍
下面就是上面算法主循环介绍的代码实现,open集和closed集的数据结构使用了字典,在一般情况下,查找,添加和删除节点的时间复杂度为o(1), 遍历的时间复杂度为o(n), n为字典中对象数目。
def astarsearch(map, source, dest): ... openlist = {} closedlist = {} location = searchentry(source[0], source[1], 0.0) dest = searchentry(dest[0], dest[1], 0.0) openlist[source] = location while true: location = getfastposition(openlist) if location is none: # not found valid path print("can't find valid path") break; if location.x == dest.x and location.y == dest.y: break closedlist[location.getpos()] = location openlist.pop(location.getpos()) addadjacentpositions(map, location, dest, openlist, closedlist) #mark the found path at the map while location is not none: map.map[location.y][location.x] = 2 location = location.pre_entry
我们按照算法主循环的实现来一个个讲解用到的函数。
下面函数就是从open集中获取一个f值最小的节点,如果open集会空,则返回none。
# find a least cost position in openlist, return none if openlist is empty def getfastposition(openlist): fast = none for entry in openlist.values(): if fast is none: fast = entry elif fast.f_cost > entry.f_cost: fast = entry return fast
addadjacentpositions 函数对应算法主函数循环介绍中的尝试添加节点n的所有邻节点n'。
# add available adjacent positions def addadjacentpositions(map, location, dest, openlist, closedlist): poslist = getpositions(map, location) for pos in poslist: # if position is already in closedlist, do nothing if isinlist(closedlist, pos) is none: findentry = isinlist(openlist, pos) h_cost = calheuristic(pos, dest) g_cost = location.g_cost + getmovecost(location, pos) if findentry is none : # if position is not in openlist, add it to openlist openlist[pos] = searchentry(pos[0], pos[1], g_cost, g_cost+h_cost, location) elif findentry.g_cost > g_cost: # if position is in openlist and cost is larger than current one, # then update cost and previous position findentry.g_cost = g_cost findentry.f_cost = g_cost + h_cost findentry.pre_entry = location
getpositions 函数获取到所有能够移动的节点,这里提供了2种移动的方式:
- 允许上,下,左,右 4邻域的移动
- 允许上,下,左,右,左上,右上,左下,右下 8邻域的移动
def getnewposition(map, locatioin, offset): x,y = (location.x + offset[0], location.y + offset[1]) if x < 0 or x >= map.width or y < 0 or y >= map.height or map.map[y][x] == 1: return none return (x, y) def getpositions(map, location): # use four ways or eight ways to move offsets = [(-1,0), (0, -1), (1, 0), (0, 1)] #offsets = [(-1,0), (0, -1), (1, 0), (0, 1), (-1,-1), (1, -1), (-1, 1), (1, 1)] poslist = [] for offset in offsets: pos = getnewposition(map, location, offset) if pos is not none: poslist.append(pos) return poslist
isinlist 函数判断节点是否在open集 或closed集中
# check if the position is in list def isinlist(list, pos): if pos in list: return list[pos] return none
calheuristic 函数简单得使用了曼哈顿距离,这个后续可以进行优化。
getmovecost 函数根据是否是斜向移动来计算消耗(斜向就是2的开根号,约等于1.4)
# imporve the heuristic distance more precisely in future def calheuristic(pos, dest): return abs(dest.x - pos[0]) + abs(dest.y - pos[1]) def getmovecost(location, pos): if location.x != pos[0] and location.y != pos[1]: return 1.4 else: return 1
代码的初始化
可以调整地图的长度,宽度和不可移动节点的数目。
可以调整开始节点和目标节点的取值范围。
width = 10 height = 10 block_num = 15 map = map(width, height) map.createblock(block_num) map.showmap() source = map.generatepos((0,width//3),(0,height//3)) dest = map.generatepos((width//2,width-1),(height//2,height-1)) print("source:", source) print("dest:", dest) astarsearch(map, source, dest) map.showmap()
执行的效果图如下,第一个表示随机生成的地图,值为1的节点表示不能移动到该节点。
第二个图中值为2的节点表示找到的路径。
完整代码
使用python3.7编译
from random import randint class searchentry(): def __init__(self, x, y, g_cost, f_cost=0, pre_entry=none): self.x = x self.y = y # cost move form start entry to this entry self.g_cost = g_cost self.f_cost = f_cost self.pre_entry = pre_entry def getpos(self): return (self.x, self.y) class map(): def __init__(self, width, height): self.width = width self.height = height self.map = [[0 for x in range(self.width)] for y in range(self.height)] def createblock(self, block_num): for i in range(block_num): x, y = (randint(0, self.width-1), randint(0, self.height-1)) self.map[y][x] = 1 def generatepos(self, rangex, rangey): x, y = (randint(rangex[0], rangex[1]), randint(rangey[0], rangey[1])) while self.map[y][x] == 1: x, y = (randint(rangex[0], rangex[1]), randint(rangey[0], rangey[1])) return (x , y) def showmap(self): print("+" * (3 * self.width + 2)) for row in self.map: s = '+' for entry in row: s += ' ' + str(entry) + ' ' s += '+' print(s) print("+" * (3 * self.width + 2)) def astarsearch(map, source, dest): def getnewposition(map, locatioin, offset): x,y = (location.x + offset[0], location.y + offset[1]) if x < 0 or x >= map.width or y < 0 or y >= map.height or map.map[y][x] == 1: return none return (x, y) def getpositions(map, location): # use four ways or eight ways to move offsets = [(-1,0), (0, -1), (1, 0), (0, 1)] #offsets = [(-1,0), (0, -1), (1, 0), (0, 1), (-1,-1), (1, -1), (-1, 1), (1, 1)] poslist = [] for offset in offsets: pos = getnewposition(map, location, offset) if pos is not none: poslist.append(pos) return poslist # imporve the heuristic distance more precisely in future def calheuristic(pos, dest): return abs(dest.x - pos[0]) + abs(dest.y - pos[1]) def getmovecost(location, pos): if location.x != pos[0] and location.y != pos[1]: return 1.4 else: return 1 # check if the position is in list def isinlist(list, pos): if pos in list: return list[pos] return none # add available adjacent positions def addadjacentpositions(map, location, dest, openlist, closedlist): poslist = getpositions(map, location) for pos in poslist: # if position is already in closedlist, do nothing if isinlist(closedlist, pos) is none: findentry = isinlist(openlist, pos) h_cost = calheuristic(pos, dest) g_cost = location.g_cost + getmovecost(location, pos) if findentry is none : # if position is not in openlist, add it to openlist openlist[pos] = searchentry(pos[0], pos[1], g_cost, g_cost+h_cost, location) elif findentry.g_cost > g_cost: # if position is in openlist and cost is larger than current one, # then update cost and previous position findentry.g_cost = g_cost findentry.f_cost = g_cost + h_cost findentry.pre_entry = location # find a least cost position in openlist, return none if openlist is empty def getfastposition(openlist): fast = none for entry in openlist.values(): if fast is none: fast = entry elif fast.f_cost > entry.f_cost: fast = entry return fast openlist = {} closedlist = {} location = searchentry(source[0], source[1], 0.0) dest = searchentry(dest[0], dest[1], 0.0) openlist[source] = location while true: location = getfastposition(openlist) if location is none: # not found valid path print("can't find valid path") break; if location.x == dest.x and location.y == dest.y: break closedlist[location.getpos()] = location openlist.pop(location.getpos()) addadjacentpositions(map, location, dest, openlist, closedlist) #mark the found path at the map while location is not none: map.map[location.y][location.x] = 2 location = location.pre_entry width = 10 height = 10 block_num = 15 map = map(width, height) map.createblock(block_num) map.showmap() source = map.generatepos((0,width//3),(0,height//3)) dest = map.generatepos((width//2,width-1),(height//2,height-1)) print("source:", source) print("dest:", dest) astarsearch(map, source, dest) map.showmap()
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