如何将路径的（X，Y）坐标列表拆分为单独的 X 和 Y 坐标列表？

Question

在下面的函数中，'graph' 是一个由 1、0 和 '2'（分别是障碍物、开放区域和目标）组成的 2d 网格列表，'start' 是开始搜索的点。

def bfs（图形，开始）：

fringe = [[start]]
# Special case: start == goal
if start.val == 'g':
    return [start]
start.visited = True
# Calculate width and height dynamically. We assume that "graph" is dense.
width = len(graph[0])
height = len(graph)
# List of possible moves: up, down, left, right.

moves = [(-1, 0), (1, 0), (0, -1), (0, 1)]
while fringe:
    # Get first path from fringe and extend it by possible moves.

    path = fringe.pop(0)
    #print path
    node = path[-1]
    pos = node.pos
    # Using moves list (without all those if's with +1, -1 etc.) has huge benefit:
    # moving logic is not duplicated. It will save you from many silly errors.

    for move in moves:
        # Check out of bounds. Note that it's the ONLY place where we check it.
        if not (0 <= pos[0] + move[0] < height and 0 <= pos[1] + move[1] < width):
            continue
        neighbor = graph[pos[0] + move[0]][pos[1] + move[1]]
        if neighbor.val == 'g':
            return path + [neighbor]
        elif neighbor.val == 'o' and not neighbor.visited:
            neighbor.visited = True
            fringe.append(path + [neighbor])  # creates copy of list
raise Exception('Path not found!')

TRANSLATE = {0: 'o', 1: 'x', 2: 'g'}
graph = [[Node(TRANSLATE[x], (i, j)) for j, x in enumerate(row)] for i, row in enumerate(graph)]
# Find path
path = bfs(graph, graph[4][4])

当我打印路径的值时，我得到以下内容：

路径 [(4, 4), (4, 5), (4, 6), (4, 7), (4, 8), (3, 8), (2, 8), (1, 8), (1, 9)]

这些是 x 和 y 坐标。

现在如何将坐标作为单独的“x”和“y”坐标列表？

我的首选输出是

X_list=[4,4,4,4,4,3,2,1,1]
Y_list=[4,5,6,7,8,8,8,8,9]

PS：当我检查打印路径的“类型”时，它将其显示为“实例”。

请帮助我，因为经过大量搜索以实现我的首选输出后，我陷入了困境。

我请求你在这方面启发我。非常感谢！！

Answer 1

您可以使用列表理解。

>>> l = [(4, 4), (4, 5), (4, 6), (4, 7), (4, 8), (3, 8), (2, 8), (1, 8), (1, 9)]
>>> la = [x for x,y in l]
>>> lb = [y for x,y in l]
>>> la
[4, 4, 4, 4, 4, 3, 2, 1, 1]
>>> lb
[4, 5, 6, 7, 8, 8, 8, 8, 9]