如何在Python中获得1到n个列表列表的所有组合

Question

我有一个列表列表，例如： [[a,b,c],[1,2,3],[x,y]]

我想产生[a],[b],...,[a,1],[a,2],...,[a,1,x],[a,1,y]

查看解决方案，我已经看到itertools.combinations如何产生单个列表的所有组合， itertools.product如何产生最高级别的组合，即上面的示例中的3个元素

我不确定如何遍历1到n个列表的所有组合，而不分解列表结构的列表，并使用itertools.combinations进行一些布尔检查以确保不合并同一列表中的元素。

Answer 1

我认为这是您要寻找的：

>>> import itertools
>>> x = [['a', 'b', 'c'], [1, 2, 3], ['x', 'y']] # your list of lists
>>> [tup for sublist in [itertools.product(*x[:n+1]) for n in range(len(x))] 
         for tup in sublist]
[('a',), ('b',), ('c',), 
 ('a', 1), ('a', 2), ('a', 3), 
 ('b', 1), ('b', 2), ('b', 3), 
 ('c', 1), ('c', 2), ('c', 3), 
 ('a', 1, 'x'), ('a', 1, 'y'), ('a', 2, 'x'), ('a', 2, 'y'), ('a', 3, 'x'), ('a', 3, 'y'), 
 ('b', 1, 'x'), ('b', 1, 'y'), ('b', 2, 'x'), ('b', 2, 'y'), ('b', 3, 'x'), ('b', 3, 'y'), 
 ('c', 1, 'x'), ('c', 1, 'y'), ('c', 2, 'x'), ('c', 2, 'y'), ('c', 3, 'x'), ('c', 3, 'y')]

您只需要将itertools.product放在列表列表的所有前缀上（即x[:1] ， x[:2] ，...）。 外部列表推导仅用于展平内部列表推导生成的列表列表。

Answer 2

先前的帖子提供了涉及嵌套理解的简洁解决方案，但是缺少可能的子列表集（例如('a', 'x')几种产品。 我将尝试以更具可读性的方式将其分解：

lst = [['a', 'b', 'c'], [1, 2, 3], ['x', 'y']]
result = []  # collect your products

# n sublists: iterate over all 'sub_lengthes'
for length in xrange(1, len(lst)+1):
    # iterate over all possible combinations of sublists
    for c in itertools.combinations(lst, length):
        # iterate over all products for each combination
        for res in itertools.product(*c):
            result.append(res)

print(result)

>>> result
# 3 + 3 + 2 = 8 singletons 
[('a',), ('b',), ('c',), (1,), (2,), (3,), ('x',), ('y',), 
# 3*3 + 3*2 + 3*2 = 21 pairs
('a', 1), ('a', 2), ('a', 3), ('b', 1), ('b', 2), ('b', 3), ('c', 1), ('c', 2), ('c', 3), 
('a', 'x'), ('a', 'y'), ('b', 'x'), ('b', 'y'), ('c', 'x'), ('c', 'y'),
(1, 'x'), (1, 'y'), (2, 'x'), (2, 'y'), (3, 'x'), (3, 'y'), 
# 3*3*2 = 18 triplets
('a', 1, 'x'), ('a', 1, 'y'), ('a', 2, 'x'), ('a', 2, 'y'), ('a', 3, 'x'), ('a', 3, 'y'), ('b', 1, 'x'), ('b', 1, 'y'), ('b', 2, 'x'), ('b', 2, 'y'), ('b', 3, 'x'), ('b', 3, 'y'), ('c', 1, 'x'), ('c', 1, 'y'), ('c', 2, 'x'), ('c', 2, 'y'), ('c', 3, 'x'), ('c', 3, 'y')]

Answer 3

您可以在列表推导中使用itertools.combinations和itertools.product()来计算所有单，对和三元组的乘积：

>>> from itertools import product
>>> lst = [['a','b','c'],[1,2,3],['x','y']]
>>> [[list(product(*t)) for t in combinations(lst,i)] for i in range(1,len(lst)+1)]
[[[('a',), ('b',), ('c',)], [(1,), (2,), (3,)], [('x',), ('y',)]], 
 [[('a', 1), ('a', 2), ('a', 3), ('b', 1), ('b', 2), ('b', 3), ('c', 1), ('c', 2), ('c', 3)], [('a', 'x'), ('a', 'y'), ('b', 'x'), ('b', 'y'), ('c', 'x'), ('c', 'y')], [(1, 'x'), (1, 'y'), (2, 'x'), (2, 'y'), (3, 'x'), (3, 'y')]],
[[('a', 1, 'x'), ('a', 1, 'y'), ('a', 2, 'x'), ('a', 2, 'y'), ('a', 3, 'x'), ('a', 3, 'y'), ('b', 1, 'x'), ('b', 1, 'y'), ('b', 2, 'x'), ('b', 2, 'y'), ('b', 3, 'x'), ('b', 3, 'y'), ('c', 1, 'x'), ('c', 1, 'y'), ('c', 2, 'x'), ('c', 2, 'y'), ('c', 3, 'x'), ('c', 3, 'y')]]]