Question

我有一个n个单词的数组作为字符串，如：

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

我需要做的是创建由空格分隔的这些单词的所有可能组合以及与原始字符串的组合。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，但无法让它做我需要的东西。我现在所拥有的：

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

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

以下几乎按要求运作：

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：

澄清这应该如何适用于长度为4的数组：输入：['an'，'even'，'greater'，'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函数为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']

Answer 2

你可以尝试这个（兼容python 2.x和python 3.x）：

s = ["this", "is", "just", "a", "simple", "test"] # the input
sepCount = len(s) - 1 # separator count of the input
output = [] # output

for i in range(0, 2 ** sepCount): # iterate through all possible combinations
    t = s # modified string
    j = i # for converting to binary
    for k in reversed(range(sepCount)):
        if j % 2 == 0:
            t = t[ : k] + [" ".join(t[k : k + 2])] + t [k + 2 :] # replace separator to " "
        j = j // 2
    output.append(t)

print(output)

输出：

[['this is just a simple test'],
['this is just a simple', 'test'],
['this is just a', 'simple test'],
['this is just a', 'simple', 'test'],
['this is just', 'a simple test'],
['this is just', 'a simple', 'test'],
['this is just', 'a', 'simple test'],
['this is just', 'a', 'simple', 'test'],
['this is', 'just a simple test'],
['this is', 'just a simple', 'test'],
['this is', 'just a', 'simple test'],
['this is', 'just a', 'simple', 'test'],
['this is', 'just', 'a simple test'],
['this is', 'just', 'a simple', 'test'],
['this is', 'just', 'a', 'simple test'],
['this is', 'just', 'a', 'simple', 'test'],
['this', 'is just a simple test'],
['this', 'is just a simple', 'test'],
['this', 'is just a', 'simple test'],
['this', 'is just a', 'simple', 'test'],
['this', 'is just', 'a simple test'],
['this', 'is just', 'a simple', 'test'],
['this', 'is just', 'a', 'simple test'],
['this', 'is just', 'a', 'simple', 'test'],
['this', 'is', 'just a simple test'],
['this', 'is', 'just a simple', 'test'],
['this', 'is', 'just a', 'simple test'],
['this', 'is', 'just a', 'simple', 'test'],
['this', 'is', 'just', 'a simple test'],
['this', 'is', 'just', 'a simple', 'test'],
['this', 'is', 'just', 'a', 'simple test'],
['this', 'is', 'just', 'a', 'simple', 'test']]

动机：对于长度为n的列表，有n-1个分隔符（,）。有2 ^（n-1）种用空格替换, s的方法。通过迭代所有这些2 ^（n-1）种可能的方式，您可以生成由空格分隔的这些单词的所有可能组合。

Python：使用空格生成多个字符串组合

2 个答案: