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:
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,但无法获得所需的结果。
如果您能帮助我,我将不胜感激。
最佳答案
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,值的形式为 ('x1,<a1>,x2', 'y1,<a2>,y2')
对于所有人(<a1>, <a2>)
在集合的笛卡尔积 {'a', 'b', 'c', 'd'}
与它本身,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')]
关于python - python 中的列表组合问题,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/53329521/