# 所有节点的g值并没有初始化为无穷大
# 当两个子节点的f值一样时，程序选择最先搜索到的一个作为父节点加入closed
# 对相同数值的不同对待，导致不同版本的A*算法找到等长的不同路径
# 最后closed表中的节点很多，如何找出最优的一条路径
# 撞墙之后产生较多的节点会加入closed表，此时开始删除closed表中不合理的节点，1.1版本的思路
# 1.2版本思路，建立每一个节点的方向指针，指向f值最小的上个节点
# 参考《无人驾驶概论》、《基于A*算法的移动机器人路径规划》王淼驰，《人工智能及应用》鲁斌import numpy
from pylab import *
import copy# 定义一个含有障碍物的20×20的栅格地图
# 10表示可通行点
# 0表示障碍物
# 7表示起点
# 5表示终点
map_grid = numpy.full((20, 20), int(10), dtype=numpy.int8)
map_grid[3, 3:8] = 0
map_grid[3:10, 7] = 0
map_grid[10, 3:8] = 0
map_grid[17, 13:17] = 0
map_grid[10:17, 13] = 0
map_grid[10, 13:17] = 0
map_grid[5, 2] = 7
map_grid[15, 15] = 5class AStar(object):
"""
创建一个A*算法类
""" def __init__(self):
"""
初始化
"""
# self.g = 0 # g初始化为0
self.start = numpy.array([5, 2]) # 起点坐标
self.goal = numpy.array([15, 15]) # 终点坐标
self.open = numpy.array([[], [], [], [], [], []]) # 先创建一个空的open表, 记录坐标，方向，g值，f值
self.closed = numpy.array([[], [], [], [], [], []]) # 先创建一个空的closed表
self.best_path_array = numpy.array([[], []]) # 回溯路径表 def h_value_tem(self, son_p):
"""
计算拓展节点和终点的h值
:param son_p:子搜索节点坐标
:return:
"""
h = (son_p[0] - self.goal[0]) ** 2 + (son_p[1] - self.goal[1]) ** 2
h = numpy.sqrt(h) # 计算h
return h # def g_value_tem(self, son_p, father_p):
# """
# 计算拓展节点和父节点的g值
# 其实也可以直接用1或者1.414代替
# :param son_p:子节点坐标
# :param father_p:父节点坐标，也就是self.current_point
# :return:返回子节点到父节点的g值，但不是全局g值
# """
# g1 = father_p[0] - son_p[0]
# g2 = father_p[1] - son_p[1]
# g = g1 ** 2 + g2 ** 2
# g = numpy.sqrt(g)
# return g def g_accumulation(self, son_point, father_point):
"""
累计的g值
:return:
"""
g1 = father_point[0] - son_point[0]
g2 = father_point[1] - son_point[1]
g = g1 ** 2 + g2 ** 2
g = numpy.sqrt(g) + father_point[4] # 加上累计的g值
return g def f_value_tem(self, son_p, father_p):
"""
求出的是临时g值和h值加上累计g值得到全局f值
:param father_p: 父节点坐标
:param son_p: 子节点坐标
:return:f
"""
f = self.g_accumulation(son_p, father_p) + self.h_value_tem(son_p)
return f def child_point(self, x):
"""
拓展的子节点坐标
:param x: 父节点坐标
:return: 子节点存入open表，返回值是每一次拓展出的子节点数目，用于撞墙判断
当搜索的节点撞墙后，如果不加处理，会陷入死循环
"""
# 开始遍历周围8个节点
for j in range(-1, 2, 1):
for q in range(-1, 2, 1): if j == 0 and q == 0: # 搜索到父节点去掉
continue
m = [x[0] + j, x[1] + q]
print(m)
if m[0] < 0 or m[0] > 19 or m[1] < 0 or m[1] > 19: # 搜索点出了边界去掉
continue if map_grid[int(m[0]), int(m[1])] == 0: # 搜索到障碍物去掉
continue record_g = self.g_accumulation(m, x)
record_f = self.f_value_tem(m, x) # 计算每一个节点的f值 x_direction, y_direction = self.direction(x, m) # 每产生一个子节点，记录一次方向 para = [m[0], m[1], x_direction, y_direction, record_g, record_f] # 将参数汇总一下
print(para) # 在open表中，则去掉搜索点，但是需要更新方向指针和self.g值
# 而且只需要计算并更新self.g即可，此时建立一个比较g值的函数
a, index = self.judge_location(m, self.open)
if a == 1:
# 说明open中已经存在这个点 if record_f <= self.open[5][index]:
self.open[5][index] = record_f
self.open[4][index] = record_g
self.open[3][index] = y_direction
self.open[2][index] = x_direction continue # 在closed表中,则去掉搜索点
b, index2 = self.judge_location(m, self.closed)
if b == 1: if record_f <= self.closed[5][index2]:
self.closed[5][index2] = record_f
self.closed[4][index2] = record_g
self.closed[3][index2] = y_direction
self.closed[2][index2] = x_direction
self.closed = numpy.delete(self.closed, index2, axis=1)
self.open = numpy.c_[self.open, para]
continue self.open = numpy.c_[self.open, para] # 参数添加到open中
print(self.open) def judge_location(self, m, list_co):
"""
判断拓展点是否在open表或者closed表中
:return:返回判断是否存在，和如果存在，那么存在的位置索引
"""
jud = 0
index = 0
for i in range(list_co.shape[1]): if m[0] == list_co[0, i] and m[1] == list_co[1, i]: jud = jud + 1 index = i
break
else:
jud = jud
# if a != 0:
# continue
return jud, index def direction(self, father_point, son_point):
"""
建立每一个节点的方向，便于在closed表中选出最佳路径
非常重要的一步，不然画出的图像参考1.1版本
x记录子节点和父节点的x轴变化
y记录子节点和父节点的y轴变化
如（0，1）表示子节点在父节点的方向上变化0和1
:return:
"""
x = son_point[0] - father_point[0]
y = son_point[1] - father_point[1]
return x, y def path_backtrace(self):
"""
回溯closed表中的最短路径
:return:
"""
best_path = [15, 15] # 回溯路径的初始化
self.best_path_array = numpy.array([[15], [15]])
j = 0
while j <= self.closed.shape[1]:
for i in range(self.closed.shape[1]):
if best_path[0] == self.closed[0][i] and best_path[1] == self.closed[1][i]:
x = self.closed[0][i]-self.closed[2][i]
y = self.closed[1][i]-self.closed[3][i]
best_path = [x, y]
self.best_path_array = numpy.c_[self.best_path_array, best_path]
break # 如果已经找到，退出本轮循环，减少耗时
else:
continue
j = j+1
# return best_path_array def main(self):
"""
main函数
:return:
"""
best = self.start # 起点放入当前点，作为父节点
h0 = self.h_value_tem(best)
init_open = [best[0], best[1], 0, 0, 0, h0] # 将方向初始化为（0，0），g_init=0,f值初始化h0
self.open = numpy.column_stack((self.open, init_open)) # 起点放入open,open初始化 ite = 1 # 设置迭代次数小于200，防止程序出错无限循环
while ite <= 1000: # open列表为空，退出
if self.open.shape[1] == 0:
print('没有搜索到路径！')
return self.open = self.open.T[numpy.lexsort(self.open)].T # open表中最后一行排序(联合排序） # 选取open表中最小f值的节点作为best，放入closed表 best = self.open[:, 0]
print('检验第%s次当前点坐标*******************' % ite)
print(best)
self.closed = numpy.c_[self.closed, best] if best[0] == 15 and best[1] == 15: # 如果best是目标点，退出
print('搜索成功！')
return self.child_point(best) # 生成子节点并判断数目
print(self.open)
self.open = numpy.delete(self.open, 0, axis=1) # 删除open中最优点 # print(self.open) ite = ite+1class MAP(object):
"""
画出地图
"""
def draw_init_map(self):
"""
画出起点终点图
:return:
"""
plt.imshow(map_grid, cmap=plt.cm.hot, interpolation='nearest', vmin=0, vmax=10)
# plt.colorbar()
xlim(-1, 20) # 设置x轴范围
ylim(-1, 20) # 设置y轴范围
my_x_ticks = numpy.arange(0, 20, 1)
my_y_ticks = numpy.arange(0, 20, 1)
plt.xticks(my_x_ticks)
plt.yticks(my_y_ticks)
plt.grid(True)
# plt.show() def draw_path_open(self, a):
"""
画出open表中的坐标点图
:return:
"""
map_open = copy.deepcopy(map_grid)
for i in range(a.closed.shape[1]):
x = a.closed[:, i] map_open[int(x[0]), int(x[1])] = 1 plt.imshow(map_open, cmap=plt.cm.hot, interpolation='nearest', vmin=0, vmax=10)
# plt.colorbar()
xlim(-1, 20) # 设置x轴范围
ylim(-1, 20) # 设置y轴范围
my_x_ticks = numpy.arange(0, 20, 1)
my_y_ticks = numpy.arange(0, 20, 1)
plt.xticks(my_x_ticks)
plt.yticks(my_y_ticks)
plt.grid(True)
# plt.show() def draw_path_closed(self, a):
"""
画出closed表中的坐标点图
:return:
"""
print('打印closed长度：')
print(a.closed.shape[1])
map_closed = copy.deepcopy(map_grid)
for i in range(a.closed.shape[1]):
x = a.closed[:, i] map_closed[int(x[0]), int(x[1])] = 5 plt.imshow(map_closed, cmap=plt.cm.hot, interpolation='nearest', vmin=0, vmax=10)
# plt.colorbar()
xlim(-1, 20) # 设置x轴范围
ylim(-1, 20) # 设置y轴范围
my_x_ticks = numpy.arange(0, 20, 1)
my_y_ticks = numpy.arange(0, 20, 1)
plt.xticks(my_x_ticks)
plt.yticks(my_y_ticks)
plt.grid(True)
# plt.show() def draw_direction_point(self, a):
"""
从终点开始，根据记录的方向信息，画出搜索的路径图
:return:
"""
print('打印direction长度：')
print(a.best_path_array.shape[1])
map_direction = copy.deepcopy(map_grid)
for i in range(a.best_path_array.shape[1]):
x = a.best_path_array[:, i] map_direction[int(x[0]), int(x[1])] = 6 plt.imshow(map_direction, cmap=plt.cm.hot, interpolation='nearest', vmin=0, vmax=10)
# plt.colorbar()
xlim(-1, 20) # 设置x轴范围
ylim(-1, 20) # 设置y轴范围
my_x_ticks = numpy.arange(0, 20, 1)
my_y_ticks = numpy.arange(0, 20, 1)
plt.xticks(my_x_ticks)
plt.yticks(my_y_ticks)
plt.grid(True) def draw_three_axes(self, a):
"""
将三张图画在一个figure中
:return:
"""
plt.figure()
ax1 = plt.subplot(221) ax2 = plt.subplot(222)
ax3 = plt.subplot(223)
ax4 = plt.subplot(224)
plt.sca(ax1)
self.draw_init_map()
plt.sca(ax2)
self.draw_path_open(a)
plt.sca(ax3)
self.draw_path_closed(a)
plt.sca(ax4)
self.draw_direction_point(a) plt.show()if __name__ == '__main__': a1 = AStar()
a1.main()
a1.path_backtrace()
m1 = MAP()
m1.draw_three_axes(a1)
A*算法基于栅格地图的全局路径规划
![[python] A*算法基于栅格地图的全局路径规划](https://img.zhankr.net/jgu5mdlpbfd266545.png)
