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python - 为什么这段代码做的是 closed[init[0]][init[1]] 而不是 closed[init[0]][init[0]]?

转载 作者:塔克拉玛干 更新时间:2023-11-03 06:36:35 27 4
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我正在阅读这个 First Search Program - Artificial Intelligence for Robotics 算法,我正在阅读它的 Python 代码。在这里,我们创建了一个封闭数组来检查单元格一旦展开并且不再展开。我们定义了一个名为 closed 的数组,并将其大小定义为我们的网格。作者说它有两个值 0 和 1。0 表示打开,1 表示关闭,但我看到它只是零。

他用1标记起点0,0直到不检查它们,但他把坐标作为0和1放在这条线closed[init[0]][init[1]] = 1。为什么他把0 和 1 而不是 0,0?

python代码在这里:

#grid format
# 0 = navigable space
# 1 = occupied space

grid=[[0,0,1,0,0,0],
[0,0,1,0,0,0],
[0,0,0,0,1,0],
[0,0,1,1,1,0],
[0,0,0,0,1,0]]

init = [0,0]
goal = [len(grid)-1,len(grid[0])-1]


delta=[[-1, 0], #up
[ 0,-1], #left
[ 1, 0], #down
[ 0, 1]] #right

delta_name = ['^','<','V','>'] #The name of above actions
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))]

#We initialize the starting location as checked
closed[init[0]][init[1]] = 1
# we assigned the cordinates and g value
x = init[0]
y = init[1]
g = 0
#our open list will contain our initial value
open = [[g,x,y]]


found = False #flag that is set when search complete
resign= False #Flag set if we can't find expand

#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
resign=True
print('Fail')
print('############# Search terminated without success')
else:
#if there is still elements on our list
#remove node from list
open.sort()
open.reverse() #reverse the list
next = open.pop()
#print('list item')
#print('next')

#Then we assign the three values to x,y and g. Which is our expantion
x = next[1]
y = next[2]
g = next[0]

#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')
else:
#expand winning element and add to new open list
for i in range(len(delta)):
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 x2 and y2 not checked yet and there is not obstacles
if closed[x2][y2] == 0 and grid[x2][y2] == 0:
g2 = g+cost #we increment the cose
open.append([g2,x2,y2])#we add them to our open list
#print('append list item')
#print([g2,x2,y2])
#Then we check them to never expand again
closed[x2][y2] = 1

search()

最佳答案

he put the coordinate as 0 and 1 in this line closed[init[0]][init[1]] = 1

closed[init[0]][init[1]] 并不意味着“将坐标 (0,1) 处的值设置为 1”。它的意思是“使用 init[0] 作为 x 坐标,使用 init[1] 作为 y 坐标,将值设置为 1”。 init[0]为0,init[1]为0,所以closed[init[0]][init[1]] = 1closed[0][0] 设置为 1。

假设起始坐标是 init = [2,5]。将该行更改为 closed[init[2]][init[5]] = 1 是不正确的。这会因 IndexError 而崩溃,因为 init 只有两个元素,所以您只能用 0 或 1 对其进行索引。

关于python - 为什么这段代码做的是 closed[init[0]][init[1]] 而不是 closed[init[0]][init[0]]?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/56808670/

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