Question

我正在尝试构建一个我在dijkstra中使用的图表。我写了这段代码：

def build_graph(self,board):
    G={} # dictionary that use as graph for dijkstra
    len_x=len(board)
    len_y=len(board[0])
    for x in range(0,len(board)):
        for y in range(0,len(board[x])):
            if board[x][y]!=99:
                if (len_y>(y-1)>=0):
                    if board[x][y-1]==90:
                        G[(x,y)]={(x,y-1):6 } # [location ,weight]
                    if board[x][y-1]==10 or board[x][y+1]==12:
                        G[(x,y)]={(x,y-1):1 }  # case 10 or 12 --> clean path
                if (len_y>(y+1)>=0):
                    if board[x][y+1]==90:       
                        G[(x,y)][(x,y+1)]=6
                    if board[x][y+1]==10 or board[x][y+1]==12:     
                        G[(x,y)][(x,y+1)]=1

                if (0<=(x+1)<len_x):
                    if board[x+1][y]==90:
                        G[(x,y)][(x+1,y)]=6
                    if board[x+1][y]==10 or board[x+1][y]==12:     
                         G[(x,y)][(x+1,y)]=1

                if (0<=(x-1)<len_x):
                    if board[x-1][y]==90:
                        G[(x,y)][(x-1,y)]=6
                    if board[x-1][y]==10 or board[x-1][y]==12 :    
                         G[(x,y)][(x-1,y)]=1


    return G

此声明中的键错误（0,0）：

G [（X，Y）] [（X，Y + 1）] = 1 而且我不知道为什么会得到它？

Answer 1

您尝试在G[(0, 0)]创建之前使用字典。尝试在内部循环的顶部为它分配一个空字典，如下所示：

def build_graph(self,board):
    G={} # dictionary that use as graph for dijkstra
    len_x=len(board)
    len_y=len(board[0])
    for x in range(0,len(board)):
        for y in range(0,len(board[x])):
           G[(x,y)] = {}
           ...

python中的键错误（0,0）

1 个答案: