如何将路径的（X，Y）坐标列表分割成单独的X和Y坐标列表？

Question

在下面的函数中，“graph”是由1、0和“two”（分别表示障碍物、开阔区域和目标）组成的二维网格列表，“start”是开始搜索的起点。在

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path的值时，得到的结果如下：

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

这是x和y坐标。在

现在我如何得到坐标，作为一个单独的'x'和'y'坐标列表？在

我的首选输出是

^{pr2}$

当它被打印为“路径”时，显示为“路径”。在

请帮助我，因为我在这一点上陷入了许多搜索，以实现我的首选输出。在

我请求你在这方面给我一些启发。谢谢！！在

Answer 1

你可以用列表理解。在

>>> 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]

Answer 2

您可以转置元组并将其映射到列表，或者使用map和itemgetter。在

from operator import itemgetter
l = [(4, 4), (4, 5), (4, 6), (4, 7), (4, 8), (3, 8), (2, 8), (1, 8), (1, 9)]

a,b = map(itemgetter(0),l), map(itemgetter(1),l)
print(a,b)

a,b = map(list,zip(*l))

print(a,b)
[4, 4, 4, 4, 4, 3, 2, 1, 1] [4, 5, 6, 7, 8, 8, 8, 8, 9]
[4, 4, 4, 4, 4, 3, 2, 1, 1] [4, 5, 6, 7, 8, 8, 8, 8, 9]

您需要在类中添加iter，以便可以迭代对象：

^{pr2}$

对于bfs，使用Queue将是一个有效的解决方案：