为什么这段代码会执行 closed[init[0]][init[1]] 而不是 closed[init[0]][init[0]]？

Question

我正在阅读第一个搜索程序 - 用于机器人算法的人工智能，并且正在阅读它的 python 代码。在这里，我们创建了一个封闭的数组来检查单元格一旦扩展并且不再扩展它们。我们定义了一个名为 closed 的数组，它的大小作为我们的网格。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。

他将起点0,0标记为1直到不检查它们，但他将坐标为0和1在这一行中关闭[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()

Answer 1

他把坐标为0和1在这一行close[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”。这意味着“使用init[0]作为x坐标，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]] = 1 将 closed[0][0] 设置为 1。

假设起始坐标是init = [2,5]。将行更改为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，所以您只能使用 0 或 1 对其进行索引。