我正在研究A-star algorithm，我已经实现了我的代码，只需在four directions中移动，如下所示，结果fixed heuristic：
[11, 11, 0, 4, 5]
############## Search is success
[0, -1, -1, -1, -1, -1]
[1, -1, -1, -1, -1, -1]
[2, -1, -1, -1, -1, -1]
[3, -1, 8, 9, 10, 11]
[4, 5, 6, 7, -1, 12]
['V', ' ', ' ', ' ', ' ', ' ']
['V', ' ', ' ', ' ', ' ', ' ']
['V', ' ', ' ', ' ', ' ', ' ']
['V', ' ', ' ', '>', '>', 'V']
['>', '>', '>', '^', ' ', '*']
我尝试使用Euclid distance计算heuristic，如下所示：
h = math.sqrt((x - goal[0])**2 + (y - goal[1])**2)
我添加了eight动作，如下所示，delta和delta name：
delta = [[1, 0, 1],
[0, 1, 1],
[-1, 0, 1],
[0, -1, 1],
[-1, -1, math.sqrt(2)],
[-1, 1, math.sqrt(2)],
[1, -1, math.sqrt(2)],
[1, 1, math.sqrt(2)]]
delta_name = ['^','','/','<','V','>','','/']
它给了我一些错误，如下所示：
1-
File "<ipython-input-24-224819b0ad4c>", line 55
delta_name = ['^','','/','<','V','>','','/']
^
SyntaxError: invalid syntax
阿拉伯数字-
IndexError Traceback (most recent call last)
<ipython-input-25-bc33334a69ba> in <module>
174
175
--> 176 search()
<ipython-input-25-bc33334a69ba> in search()
140 x2=x-delta[action[x][y]][0]
141 y2=y-delta[action[x][y]][1]
--> 142 policy[x2][y2]= delta_name[action[x][y]]
143 x=x2
144 y=y2
IndexError: list index out of range
请问我该如何修复它们？ 你能找到我的三角洲的运动吗，哪一个是向上，对，向下....等？
这是我的代码：
import random
Import math
grid = [[0, 1, 0, 0, 0, 0],
[0, 1, 0, 0, 0, 0],
[0, 1, 0, 0, 0, 0],
[0, 1, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0]]
heuristic = [[9, 8, 7, 6, 5, 4],
[8, 7, 6, 5, 4, 3],
[7, 6, 5, 4, 3, 2],
[6, 5, 4, 3, 2, 1],
[5, 4, 3, 2, 1, 0]]
init = [0,0]
goal = [len(grid)-1,len(grid[0])-1]
#Below the four potential actions to the single field
'''
delta = [[-1 , 0], #up
[ 0 ,-1], #left
[ 1 , 0], #down
[ 0 , 1]] #right
'''
delta = [[1, 0, 1],
[0, 1, 1],
[-1, 0, 1],
[0, -1, 1],
[-1, -1, math.sqrt(2)],
[-1, 1, math.sqrt(2)],
[1, -1, math.sqrt(2)],
[1, 1, math.sqrt(2)]]
#delta_name = ['^','<','V','>'] #The name of above actions
delta_name = ['^','','/','<','V','>','','/']
cost = 1
def search():
#open list elements are of the type [g,x,y]
closed = [[0 for row in range(len(grid[0]))] for col in range(len(grid))]
action = [[-1 for row in range(len(grid[0]))] for col in range(len(grid))]
#We initialize the starting location as checked
closed[init[0]][init[1]] = 1
expand=[[-1 for row in range(len(grid[0]))] for col in range(len(grid))]
# we assigned the cordinates and g value
x = init[0]
y = init[1]
g = 0
#h = heuristic[x][y]
h = math.sqrt((x - goal[0])**2 + (y - goal[1])**2)
f = g + h
#our open list will contain our initial value
open = [[f, g, h, x, y]]
found = False #flag that is set when search complete
resign = False #Flag set if we can't find expand
count = 0
#print('initial open list:')
#for i in range(len(open)):
#print(' ', open[i])
#print('----')
while found is False and resign is False:
#Check if we still have elements in the open list
if len(open) == 0: #If our open list is empty, there is nothing to expand.
resign = True
print('Fail')
print('############# Search terminated without success')
print()
else:
#if there is still elements on our list
#remove node from list
open.sort() #sort elements in an increasing order from the smallest g value up
open.reverse() #reverse the list
next = open.pop() #remove the element with the smallest g value from the list
#print('list item')
#print('next')
#Then we assign the three values to x,y and g. Which is our expantion.
x = next[3]
y = next[4]
g = next[1]
expand[x][y] = count
count+=1
#Check if we are done
if x == goal[0] and y == goal[1]:
found = True
print(next) #The three elements above this "if".
print('############## Search is success')
print()
else:
#expand winning element and add to new open list
for i in range(len(delta)): #going through all our actions the four actions
#We apply the actions to x and y with additional delta to construct x2 and y2
x2 = x + delta[i][0]
y2 = y + delta[i][1]
#if x2 and y2 falls into the grid
if x2 >= 0 and x2 < len(grid) and y2 >=0 and y2 <= len(grid[0])-1:
if closed[x2][y2] == 0 and grid[x2][y2] == 0:
g2 = g + cost
#h2 = heuristic[x2][y2]
h2 = math.sqrt((x2 - goal[0])**2 + (y2 - goal[1])**2)
f2 = g2 + h2
open.append([f2,g2,h2,x2,y2])
#print('append list item')
#print([g2,x2,y2])
#Then we check them to never expand again
closed[x2][y2] = 1
action[x2][y2] = i
for i in range(len(expand)):
print(expand[i])
print()
policy=[[' ' for row in range(len(grid[0]))] for col in range(len(grid))]
x=goal[0]
y=goal[1]
policy[x][y]='*'
while x !=init[0] or y !=init[1]:
x2=x-delta[action[x][y]][0]
y2=y-delta[action[x][y]][1]
policy[x2][y2]= delta_name[action[x][y]]
x=x2
y=y2
for i in range(len(policy)):
print(policy[i])
search()
使用语法突出显示
在第 delta_name = ['^','','/','<','V','>','','/'] 行中，您没有"字符"列表，因为在第二个字符串中，字符转义了右引号，因此它不会在您想要的位置关闭。大多数编辑器中的语法突出显示会告诉你这一点。
如果你想要一个包含单个字符的字符串，那么你必须把它写成'\'。
有关详细信息，请参阅 https://docs.python.org/3/reference/lexical_analysis.html#string-and-bytes-literals -"反斜杠 (( 字符用于转义具有特殊含义的字符，例如换行符、反斜杠本身或引号字符。
你在这里写一个字符串，''在这里意味着另一件事。 例如'n'，它表示换行符，'t'，它表示水龙头或空格。等。 如果要将其用作方向，可以在字符串内部使用双斜杠来取消反斜杠的效果并得到您想要的。 '\' .正如其他评论员上面提到的。