我想在有限的点网格上找到可能的路径。比如说，起点是（x，y）。然后路径中的下一个点（m，n）由条件
给出因此，生成这些＆＃39;路径的所有可能组合。可能的。
以下是我的尝试。
def lisy(x,y):
return [(i,j) for i in range(4,0,-1) for j in range(4,0,-1) if(i!=x and j<y)]
def recurse(x,y):
if (not lisy(x,y)):
return (x,y)
else:
return [(x,y), [recurse(i,j) for i,j in lisy(x,y)]]
输出：
In [89]: recurse(1,4)
Out[89]:
[(1, 4),
[[(4, 3),
[[(3, 2), [(4, 1), (2, 1), (1, 1)]],
(3, 1),
[(2, 2), [(4, 1), (3, 1), (1, 1)]],
(2, 1),
[(1, 2), [(4, 1), (3, 1), (2, 1)]],
(1, 1)]],
[(4, 2), [(3, 1), (2, 1), (1, 1)]],
(4, 1),
[(3, 3),
[[(4, 2), [(3, 1), (2, 1), (1, 1)]],
(4, 1),
[(2, 2), [(4, 1), (3, 1), (1, 1)]],
(2, 1),
[(1, 2), [(4, 1), (3, 1), (2, 1)]],
(1, 1)]],
[(3, 2), [(4, 1), (2, 1), (1, 1)]],
(3, 1),
[(2, 3),
[[(4, 2), [(3, 1), (2, 1), (1, 1)]],
(4, 1),
[(3, 2), [(4, 1), (2, 1), (1, 1)]],
(3, 1),
[(1, 2), [(4, 1), (3, 1), (2, 1)]],
(1, 1)]],
[(2, 2), [(4, 1), (3, 1), (1, 1)]],
(2, 1)]]
这给了我一个从每个点可能的新点的嵌套列表。
有谁能告诉我如何处理从recurse(1,4)获得的列表？
edit1：
有效地，我从一个给定的起始点跳跃（在一个4x4网格[有限]中），满足上述三个条件，直到满足停止标准，即。 m,n > 0
答案 0 :(得分：1)
我在我的生成器gridpaths()的docstring中阐明了我正在使用的要求。请注意，我将网格的水平大小作为全局变量，并且网格的垂直大小无关紧要，路径点的x坐标可以达到但不超过该全局值，并且x坐标为非连续的路径点可以相等（尽管连续的路径点必须具有不同的x坐标）。我改变了例程的名称，但保留了你拥有的参数。这个版本的代码增加了路径上最后一个点的y坐标必须为1的要求，并且接受参数也更安全。
这是一个列表生成器，所以我的测试代码显示了生成器的大小，然后打印所有列表。
def gridpaths(x, y):
"""Generate all paths starting at (x,y) [x and y must be positive
integers] where, if (m,n) is the next point in the path after
(x,y), then m and n are positive integers, m <= xsize [xsize is a
global variable], m != x, and n < y, and so on for all consecutive
path points. The final point in the path must have a y-coordinate
of 1. Paths are yielded in lexicographic order."""
def allgridpaths(x, y, pathsofar):
"""Generate all such paths continuing from pathssofar without
the y == 1 requirement for the final path point."""
newpath = pathsofar + [(x, y)]
yield newpath
for m in range(1, xsize+1):
if m != x:
for n in range(1, y):
for path in allgridpaths(m, n, newpath):
yield path
x, y = max(int(x), 1), max(int(y), 1) # force positive integers
for path in allgridpaths(x, y, []):
# Only yield paths that end at y == 1
if path[-1][1] == 1:
yield path
# global variable: horizontal size of grid
xsize = 4
print(sum(1 for p in gridpaths(1, 4)), 'paths total.')
for p in gridpaths(1, 4):
print(p)
打印输出显示4x4网格中的点（1,4）产生48条路径。事实上，gridpaths(x, y)将返回(xsize - 1) * xsize ** (y - 2)个路径，这些路径可以非常快速地增长。这就是我编写列表生成器而不是列表列表的原因。如果您的要求与我的要求不同，请告诉我。上述代码的打印输出为：
48 paths total.
[(1, 4), (2, 1)]
[(1, 4), (2, 2), (1, 1)]
[(1, 4), (2, 2), (3, 1)]
[(1, 4), (2, 2), (4, 1)]
[(1, 4), (2, 3), (1, 1)]
[(1, 4), (2, 3), (1, 2), (2, 1)]
[(1, 4), (2, 3), (1, 2), (3, 1)]
[(1, 4), (2, 3), (1, 2), (4, 1)]
[(1, 4), (2, 3), (3, 1)]
[(1, 4), (2, 3), (3, 2), (1, 1)]
[(1, 4), (2, 3), (3, 2), (2, 1)]
[(1, 4), (2, 3), (3, 2), (4, 1)]
[(1, 4), (2, 3), (4, 1)]
[(1, 4), (2, 3), (4, 2), (1, 1)]
[(1, 4), (2, 3), (4, 2), (2, 1)]
[(1, 4), (2, 3), (4, 2), (3, 1)]
[(1, 4), (3, 1)]
[(1, 4), (3, 2), (1, 1)]
[(1, 4), (3, 2), (2, 1)]
[(1, 4), (3, 2), (4, 1)]
[(1, 4), (3, 3), (1, 1)]
[(1, 4), (3, 3), (1, 2), (2, 1)]
[(1, 4), (3, 3), (1, 2), (3, 1)]
[(1, 4), (3, 3), (1, 2), (4, 1)]
[(1, 4), (3, 3), (2, 1)]
[(1, 4), (3, 3), (2, 2), (1, 1)]
[(1, 4), (3, 3), (2, 2), (3, 1)]
[(1, 4), (3, 3), (2, 2), (4, 1)]
[(1, 4), (3, 3), (4, 1)]
[(1, 4), (3, 3), (4, 2), (1, 1)]
[(1, 4), (3, 3), (4, 2), (2, 1)]
[(1, 4), (3, 3), (4, 2), (3, 1)]
[(1, 4), (4, 1)]
[(1, 4), (4, 2), (1, 1)]
[(1, 4), (4, 2), (2, 1)]
[(1, 4), (4, 2), (3, 1)]
[(1, 4), (4, 3), (1, 1)]
[(1, 4), (4, 3), (1, 2), (2, 1)]
[(1, 4), (4, 3), (1, 2), (3, 1)]
[(1, 4), (4, 3), (1, 2), (4, 1)]
[(1, 4), (4, 3), (2, 1)]
[(1, 4), (4, 3), (2, 2), (1, 1)]
[(1, 4), (4, 3), (2, 2), (3, 1)]
[(1, 4), (4, 3), (2, 2), (4, 1)]
[(1, 4), (4, 3), (3, 1)]
[(1, 4), (4, 3), (3, 2), (1, 1)]
[(1, 4), (4, 3), (3, 2), (2, 1)]
[(1, 4), (4, 3), (3, 2), (4, 1)]