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']
最佳答案
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']
关于Python:生成多个带空格的字符串组合,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/50005856/