如果有人正在寻找一个反向列表，就像我一样:

stuff = [1, 2, 3, 4]

def reverse(bla, y):
    for subset in itertools.combinations(bla, len(bla)-y):
        print list(subset)
    if y != len(bla):
        y += 1
        reverse(bla, y)

reverse(stuff, 1)

我在这个话题上有点晚了，但我想我可以帮助别人。

你可以使用itertools中的product:

from itertools import product

n = [1, 2, 3]

result = product(n, repeat=3) # You can change the repeat more then n length

print(list(result))

输出:

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

另一个例子，但是改变了repeat参数:

from itertools import product

n = [1, 2, 3]

result = product(n, repeat=4) # Changing repeat to 4
print(list(result))

输出:

(1, 1, 2, 3), (1, 1, 3, 1), (1, 1, 3, 2), (1, 1, 3, 3), (1, 2, 1, 1), 
(1, 2, 1, 2), (1, 2, 1, 3), (1, 2, 2, 1), (1, 2, 2, 2), (1, 2, 2, 3), 
(1, 2, 3, 1), (1, 2, 3, 2), (1, 2, 3, 3), (1, 3, 1, 1), (1, 3, 1, 2), 
(1, 3, 1, 3), (1, 3, 2, 1), (1, 3, 2, 2), (1, 3, 2, 3), (1, 3, 3, 1), 
(1, 3, 3, 2), (1, 3, 3, 3), (2, 1, 1, 1), (2, 1, 1, 2), (2, 1, 1, 3), 
(2, 1, 2, 1), (2, 1, 2, 2), (2, 1, 2, 3), (2, 1, 3, 1), (2, 1, 3, 2),
 (2, 1, 3, 3), (2, 2, 1, 1), (2, 2, 1, 2), (2, 2, 1, 3), (2, 2, 2, 1), 
(2, 2, 2, 2), (2, 2, 2, 3), (2, 2, 3, 1), (2, 2, 3, 2), (2, 2, 3, 3), 
(2, 3, 1, 1), (2, 3, 1, 2), (2, 3, 1, 3), (2, 3, 2, 1), (2, 3, 2, 2), 
(2, 3, 2, 3), (2, 3, 3, 1), (2, 3, 3, 2), (2, 3, 3, 3), (3, 1, 1, 1), 
(3, 1, 1, 2), (3, 1, 1, 3), (3, 1, 2, 1), (3, 1, 2, 2), (3, 1, 2, 3), 
(3, 1, 3, 1), (3, 1, 3, 2), (3, 1, 3, 3), (3, 2, 1, 1), (3, 2, 1, 2), 
(3, 2, 1, 3), (3, 2, 2, 1), (3, 2, 2, 2), (3, 2, 2, 3), (3, 2, 3, 1), 
(3, 2, 3, 2), (3, 2, 3, 3), (3, 3, 1, 1), (3, 3, 1, 2), (3, 3, 1, 3), 
(3, 3, 2, 1), (3, 3, 2, 2), (3, 3, 2, 3), (3, 3, 3, 1), (3, 3, 3, 2), 
(3, 3, 3, 3)]```

看看itertools.combination:

itertools.combinations (iterable, r) 返回元素的r长度子序列 输入迭代对象。 组合是按字典排序顺序发出的。那么，如果 Input iterable已排序，则 组合元组将在 排序顺序。

从2.6开始，电池包括在内!

这是我的实现

def get_combinations(list_of_things):
"""gets every combination of things in a list returned as a list of lists

Should be read : add all combinations of a certain size to the end of a list for every possible size in the
the list_of_things.

"""
list_of_combinations = [list(combinations_of_a_certain_size)
                        for possible_size_of_combinations in range(1,  len(list_of_things))
                        for combinations_of_a_certain_size in itertools.combinations(list_of_things,
                                                                                     possible_size_of_combinations)]
return list_of_combinations

这段代码采用了一个简单的嵌套列表算法…

# FUNCTION getCombos: To generate all combos of an input list, consider the following sets of nested lists...
#
#           [ [ [] ] ]
#           [ [ [] ], [ [A] ] ]
#           [ [ [] ], [ [A],[B] ],         [ [A,B] ] ]
#           [ [ [] ], [ [A],[B],[C] ],     [ [A,B],[A,C],[B,C] ],                   [ [A,B,C] ] ]
#           [ [ [] ], [ [A],[B],[C],[D] ], [ [A,B],[A,C],[B,C],[A,D],[B,D],[C,D] ], [ [A,B,C],[A,B,D],[A,C,D],[B,C,D] ], [ [A,B,C,D] ] ]
#
#  There is a set of lists for each number of items that will occur in a combo (including an empty set).
#  For each additional item, begin at the back of the list by adding an empty list, then taking the set of
#  lists in the previous column (e.g., in the last list, for sets of 3 items you take the existing set of
#  3-item lists and append to it additional lists created by appending the item (4) to the lists in the
#  next smallest item count set. In this case, for the three sets of 2-items in the previous list. Repeat
#  for each set of lists back to the initial list containing just the empty list.
#

def getCombos(listIn = ['A','B','C','D','E','F'] ):
    listCombos = [ [ [] ] ]     # list of lists of combos, seeded with a list containing only the empty list
    listSimple = []             # list to contain the final returned list of items (e.g., characters)

    for item in listIn:
        listCombos.append([])   # append an emtpy list to the end for each new item added
        for index in xrange(len(listCombos)-1, 0, -1):  # set the index range to work through the list
            for listPrev in listCombos[index-1]:        # retrieve the lists from the previous column
                listCur = listPrev[:]                   # create a new temporary list object to update
                listCur.append(item)                    # add the item to the previous list to make it current
                listCombos[index].append(listCur)       # list length and append it to the current list

                itemCombo = ''                          # Create a str to concatenate list items into a str
                for item in listCur:                    # concatenate the members of the lists to create
                    itemCombo += item                   # create a string of items
                listSimple.append(itemCombo)            # add to the final output list

    return [listSimple, listCombos]
# END getCombos()

还可以使用more_itertools包中的powerset函数。

from more_itertools import powerset

l = [1,2,3]
list(powerset(l))

# [(), (1,), (2,), (3,), (1, 2), (1, 3), (2, 3), (1, 2, 3)]

我们也可以验证，它满足OP的要求

from more_itertools import ilen

assert ilen(powerset(range(15))) == 32_768