python - python中列表组合的问题

Question

我有一些 Python 列表，其中包含我想要连接的信息。这些列表类似于这些:

vars1 = ["x1", "x2"]
vars2 = ["y1", "y2"]
main_list = ["a","b","c","d"]

我想要的是获得所有可能的组合(即使我不知道正确的操作名称)以涵盖我在下面列出的所有情况:

[
    ("x1,a,x2", "y1,a,y2"), ("x1,a,x2", "y1,b,y2"),
    ("x1,a,x2", "y1,c,y2"), ("x1,a,x2", "y1,d,y2"),
    ("x1,b,x2", "y1,a,y2"), ("x1,b,x2", "y1,b,y2"), 
    ("x1,b,x2", "y1,c,y2"), ("x1,b,x2", "y1,d,y2")
    ("x1,c,x2", "y1,a,y2"), ("x1,c,x2", "y1,b,y2"),
    ("x1,c,x2", "y1,c,y2"), ("x1,c,x2", "y1,d,y2"),
    ("x1,d,x2", "y1,a,y2"), ("x1,d,x2", "y1,b,y2"),
    ("x1,d,x2", "y1,c,y2"), ("x1,d,x2", "y1,d,y2"),
]

我调查了 itertools.product 函数，但我得不到想要的结果。

如果你能帮助我，我将不胜感激。

Answer 1

你的问题不是很清楚，但这看起来像你想要的(如果我错了请纠正我):

vars1 = ["x1", "x2"]
vars2 = ["y1", "y2"]
main_list = ["a","b","c","d"]
result = []

for a1, a2 in itertools.product(main_list, main_list):
    result.append((','.join((vars1[0], a1, vars1[1])), ','.join((vars2[0], a2, vars2[1]))))

换句话说，形式为 ('x1,<a1>,x2', 'y1,<a2>,y2') 的值对于所有 (<a1>, <a2>)在集合的笛卡尔积中 {'a', 'b', 'c', 'd'}与自己，这确实是itertools.product是为了。

结果:

[('x1,a,x2', 'y1,a,y2'),
 ('x1,a,x2', 'y1,b,y2'),
 ('x1,a,x2', 'y1,c,y2'),
 ('x1,a,x2', 'y1,d,y2'),
 ('x1,b,x2', 'y1,a,y2'),
 ('x1,b,x2', 'y1,b,y2'),
 ('x1,b,x2', 'y1,c,y2'),
 ('x1,b,x2', 'y1,d,y2'),
 ('x1,c,x2', 'y1,a,y2'),
 ('x1,c,x2', 'y1,b,y2'),
 ('x1,c,x2', 'y1,c,y2'),
 ('x1,c,x2', 'y1,d,y2'),
 ('x1,d,x2', 'y1,a,y2'),
 ('x1,d,x2', 'y1,b,y2'),
 ('x1,d,x2', 'y1,c,y2'),
 ('x1,d,x2', 'y1,d,y2')]