元组列表的所有可能组合

Question

我正在 python 中创建“Boggle”，并且我有一个表示游戏板上坐标的元组列表：

all_coordinates = [(0, 0), (1, 0), (2, 0), (0, 1), (1, 1), (2, 1), (0, 2), (1, 2), (2, 2), (0, 3), (1, 3), (2, 3)]

我正在尝试创建一个新的元组列表，它将代表板上所有可能的路径。

它看起来像这样：

[[(0,0),(1,0)], ... , [(0,0),(1,0),(2,0),(2,1)] , ... , [(2, 1), (2, 2), (2, 3)], ...]

我尝试使用itertools.combinations和itertools.permutations但它似乎没有完成这项工作，例如以下路径：

[(2,1),(1,1),(1,0),(2,0)]

没有出现在列表中。

这个特定的 function 不一定必须是 output 的“有效”路径（有效 = 每次水平、垂直或对角移动一步），只是代表棋盘的元组中的所有可能组合。 我有一个 function 来检查某个路径是否返回有效单词。 我正在尝试打印出在板上返回有效单词的所有可能路径。

Answer 1

itertools.permutations确实产生了所有的排列，包括你所说的[(2,1),(1,1),(1,0),(2,0)]一个。 请注意，对permutations的每次调用都会为您提供特定长度的所有排列：

>>> all_coordinates = [(0, 0), (1, 0), (2, 0), (0, 1), (1, 1), (2, 1), (0, 2), (1, 2), (2, 2), (0, 3), (1, 3), (2, 3)]
>>> from itertools import permutations
>>> ((2,1),(1,1),(1,0),(2,0)) in permutations(all_coordinates, 4)
True

如果您想查看从长度 2 到长度 4 的所有排列，请尝试：

for k in range(2, 5):
    for p in permutations(all_coordinates, k):
        print(p)

结果序列很长； 正如其他人指出的那样，您可能想提出另一种方法来生成仅包含相邻坐标的路径（例如广度优先搜索）。

（编辑）只是为了好玩，这是一种非常快速的 DFS 方法，通过仅查看相邻坐标来构建长度为 4 的所有路径：

>>> def print_paths(path):
...     print(path)
...     if len(path) >= 4:
...         return
...     x, y = path[-1]
...     for dx in range(-1, 2):
...         for dy in range(-1, 2):
...             c = x + dx, y + dy
...             if c not in path and c in all_coordinates:
...                 print_paths(path + [c])
...
>>> print_paths([(2, 1)])
[(2, 1)]
[(2, 1), (1, 0)]
[(2, 1), (1, 0), (0, 0)]
[(2, 1), (1, 0), (0, 0), (0, 1)]
[(2, 1), (1, 0), (0, 0), (1, 1)]
[(2, 1), (1, 0), (0, 1)]
[(2, 1), (1, 0), (0, 1), (0, 0)]
[(2, 1), (1, 0), (0, 1), (0, 2)]
[(2, 1), (1, 0), (0, 1), (1, 1)]
[(2, 1), (1, 0), (0, 1), (1, 2)]
[(2, 1), (1, 0), (1, 1)]
[(2, 1), (1, 0), (1, 1), (0, 0)]
[(2, 1), (1, 0), (1, 1), (0, 1)]
[(2, 1), (1, 0), (1, 1), (0, 2)]
[(2, 1), (1, 0), (1, 1), (1, 2)]
[(2, 1), (1, 0), (1, 1), (2, 0)]
[(2, 1), (1, 0), (1, 1), (2, 2)]
[(2, 1), (1, 0), (2, 0)]
[(2, 1), (1, 0), (2, 0), (1, 1)]
[(2, 1), (1, 1)]
[(2, 1), (1, 1), (0, 0)]
[(2, 1), (1, 1), (0, 0), (0, 1)]
[(2, 1), (1, 1), (0, 0), (1, 0)]
[(2, 1), (1, 1), (0, 1)]
[(2, 1), (1, 1), (0, 1), (0, 0)]
[(2, 1), (1, 1), (0, 1), (0, 2)]
[(2, 1), (1, 1), (0, 1), (1, 0)]
[(2, 1), (1, 1), (0, 1), (1, 2)]
[(2, 1), (1, 1), (0, 2)]
[(2, 1), (1, 1), (0, 2), (0, 1)]
[(2, 1), (1, 1), (0, 2), (0, 3)]
[(2, 1), (1, 1), (0, 2), (1, 2)]
[(2, 1), (1, 1), (0, 2), (1, 3)]
[(2, 1), (1, 1), (1, 0)]
[(2, 1), (1, 1), (1, 0), (0, 0)]
[(2, 1), (1, 1), (1, 0), (0, 1)]
[(2, 1), (1, 1), (1, 0), (2, 0)]
(...)

Answer 2

itertools.product()完全符合您的要求。

>>> import itertools
>>> tuple1 = ("A", "B", "C")
>>> tuple2 = ("D", "E", "F")
>>> list(itertools.product(tuple1, tuple2))
[('A', 'D'), ('A', 'E'), ('A', 'F'), ('B', 'D'), ('B', 'E'), ('B', 'F'), ('C', 'D'), ('C', 'E'), ('C', 'F')]
>>> [''.join(x) for x in itertools.product(tuple1, tuple2)]
['AD', 'AE', 'AF', 'BD', 'BE', 'BF', 'CD', 'CE', 'CF']