Question

我有一些Python列表，其中包含我想连接的信息。这些列表类似于以下内容：

git stash push

我想要获得所有可能的组合（甚至我不知道正确的操作名称）以涵盖我在下面提出的所有情况：

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

我对[ ("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"), ]函数进行了调查，但无法获得预期的结果。

如果您能帮助我，我将不胜感激。

Answer 1

您的问题不是很清楚，但这看起来像您想要的（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，集合('x1,<a1>,x2', 'y1,<a2>,y2')与其自身的笛卡尔积中所有(<a1>, <a2>)的形式{'a', 'b', 'c', 'd'}的值，实际上就是itertools.product的含义

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')]

python中的列表组合出现问题

1 个答案: