Question

所以我有一个astar算法在迷宫上输出路径。但我只想要在迷宫中实际转弯的节点（表示为元组（行，列））。

实施例。

path = [(10,0),(10,1),(9,1),(8,1),(8,2),(8,3),(7,3)] #given from astar alg  
path = [(10,0),(10,1),(8,1),(8,3),(7,3)] #desired output

以下是我的代码的一部分：

for node in self.path:
            direction1 = (self.path[node][0] - self.path[node+1][0], self.path[node][1] - self.path[node+1][1])
            direction2 = (self.path[node+1][0] - self.path[node+2][0], self.path[node+1][1] - self.path[node+2][1])
            if direction1 == direction2:
                self.path.pop([node+1])
            elif direction1 == None:
                pass
            else:
                pass

Answer 1

迭代self.path使用的索引：

for node in range(len(self.path)):

但是由于你想在结束前停止检查2（以便self.path[node+2]始终有效），只需使用：

for node in range(0,len(self.path)-2):

无论哪种方式，在迭代它时删除路径的元素都可能会导致问题，所以我建议构建新路径然后替换旧路径：

new_path = [self.path[0]] #keep the first element since it will always be needed
for node in range(0,len(self.path)-2):
    direction1 = (self.path[node][0] - self.path[node+1][0], self.path[node][1] - self.path[node+1][1])
    direction2 = (self.path[node+1][0] - self.path[node+2][0], self.path[node+1][1] - self.path[node+2][1])
    if direction1 != direction2:
        new_path.append(self.path[node+1])
new_path.append(self.path[-1]) #add in the last element
self.path = new_path

但这可能更令人困惑，因为你必须处理node+1多node所以你可能想从indice 1开始到len(self.path)-1结束所以你会使用：< / p>

new_path = [self.path[0]] #keep the first element since it will always be needed
for node in range(1,len(self.path)-1):
    direction1 = (self.path[node-1][0] - self.path[node][0], self.path[node-1][1] - self.path[node][1])
    direction2 = (self.path[node][0] - self.path[node+1][0], self.path[node][1] - self.path[node+1][1])
    if direction1 != direction2:
        new_path.append(self.path[node])
new_path.append(self.path[-1]) #add in the last element
self.path = new_path

如何仅输出迷宫路径的转折点

1 个答案: