Python:生成多个带空格的字符串组合

Question

我有一个包含 n 个单词的字符串数组，例如:

input: ["just", "a", "test"]

我需要做的是创建这些单词的所有可能组合，这些单词由空格分隔并与原始字符串组合。例如，上面应该创建:

output: [["just", "a", "test"], ["just a", "test"], ["just a test"], ["just", "a test"]]

我一直在使用 itertools，但无法让它执行我需要的操作。我现在拥有的:

iterable = ['just', 'a', 'test']

for n in chain.from_iterable(combinations(iterable, n) for n in range(len(iterable)+1)):
    print(n)

以下几乎可以按要求工作:

iterable = ['just', 'a', 'test']
L = [''.join(reversed(x)).rstrip()
     for x in product(*[(c, c+' ') for c in reversed(iterable)])]
print(L)

谢谢。

编辑:

为了阐明这对于长度为 4 的数组应该如何工作: 输入:['an', 'even', 'bigger', 'test']`

output: 
['an', 'even', 'bigger', 'test']
['an even', 'bigger', 'test']
['an even bigger', 'test']
['an even bigger test']

['an', 'even bigger', 'test']
['an even', 'bigger test']
['an', 'even bigger test']
['an', 'even', 'bigger test']

Answer 1

这是一个解决方案。 partitions 函数是 courtesy of @Kiwi .

from itertools import combinations

iterable = ['just', 'a', 'test', 'and', 'another']

n = len(iterable)

def partitions(items, k):

    def split(indices):
        i=0
        for j in indices:
            yield items[i:j]
            i = j
        yield items[i:]

    for indices in combinations(range(1, len(items)), k-1):
        yield list(split(indices))

for i in range(1, n+1):
    for x in partitions(iterable, i):
        print([' '.join(y) for y in x])

['just a test and another']
['just', 'a test and another']
['just a', 'test and another']
['just a test', 'and another']
['just a test and', 'another']
['just', 'a', 'test and another']
['just', 'a test', 'and another']
['just', 'a test and', 'another']
['just a', 'test', 'and another']
['just a', 'test and', 'another']
['just a test', 'and', 'another']
['just', 'a', 'test', 'and another']
['just', 'a', 'test and', 'another']
['just', 'a test', 'and', 'another']
['just a', 'test', 'and', 'another']
['just', 'a', 'test', 'and', 'another']