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)]```

我来晚了，但我想分享我找到的解决这个问题的方法: 具体来说，我想要做顺序组合，所以对于“STAR”，我想要“STAR”，“TA”，“AR”，但不是“SR”。

lst = [S, T, A, R]
lstCombos = []
for Length in range(0,len(lst)+1):
    for i in lst:
        lstCombos.append(lst[lst.index(i):lst.index(i)+Length])

可以通过在最后一行之前添加额外的if来过滤重复:

lst = [S, T, A, R]
lstCombos = []
for Length in range(0,len(lst)+1):
    for i in lst:
         if not lst[lst.index(i):lst.index(i)+Length]) in lstCombos:
             lstCombos.append(lst[lst.index(i):lst.index(i)+Length])

如果由于某种原因，这将在输出中返回空白列表，这发生在我身上，我添加:

for subList in lstCombos:
    if subList = '':
         lstCombos.remove(subList)

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!

这里是另一个解决方案(一行程序)，涉及到使用itertools.combination函数，但这里我们使用了双链表理解式(而不是for循环或sum):

def combs(x):
    return [c for i in range(len(x)+1) for c in combinations(x,i)]

演示:

>>> combs([1,2,3,4])
[(), 
 (1,), (2,), (3,), (4,), 
 (1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4), 
 (1, 2, 3), (1, 2, 4), (1, 3, 4), (2, 3, 4), 
 (1, 2, 3, 4)]

如果你不想使用组合库，这里是解决方案:

nums = [1,2,3]
p = [[]]
fnl = [[],nums]

for i in range(len(nums)):
    for j in range(i+1,len(nums)):
        p[-1].append([i,j])

for i in range(len(nums)-3):
    p.append([])
    for m in p[-2]:
        p[-1].append(m+[m[-1]+1])

for i in p:
    for j in i:
        n = []
        for m in j:
            if m < len(nums):
                n.append(nums[m])
        if n not in fnl:
            fnl.append(n)

for i in nums:
    if [i] not in fnl:
        fnl.append([i])

print(fnl)

输出:

[[], [1, 2, 3], [1, 2], [1, 3], [2, 3], [1], [2], [3]]

这一行代码给出了所有的组合(如果原始列表/set包含n个不同的元素，则在0到n个元素之间)，并使用本机方法itertools.combination:

Python 2

from itertools import combinations

input = ['a', 'b', 'c', 'd']

output = sum([map(list, combinations(input, i)) for i in range(len(input) + 1)], [])

Python 3

from itertools import combinations

input = ['a', 'b', 'c', 'd']

output = sum([list(map(list, combinations(input, i))) for i in range(len(input) + 1)], [])

输出将是:

[[],
 ['a'],
 ['b'],
 ['c'],
 ['d'],
 ['a', 'b'],
 ['a', 'c'],
 ['a', 'd'],
 ['b', 'c'],
 ['b', 'd'],
 ['c', 'd'],
 ['a', 'b', 'c'],
 ['a', 'b', 'd'],
 ['a', 'c', 'd'],
 ['b', 'c', 'd'],
 ['a', 'b', 'c', 'd']]

在网上试试吧:

http://ideone.com/COghfX