分离字符串

给定一个字符串,我想生成所有可能的组合.换句话说,将逗号放在字符串中的所有可能方法.

例如:

input:  ["abcd"]
output: ["abcd"]
        ["abc","d"]
        ["ab","cd"]
        ["ab","c","d"]
        ["a","bc","d"]
        ["a","b","cd"]
        ["a","bcd"]
        ["a","b","c","d"]

我有点坚持如何生成所有可能的列表.组合将只给出包含字符串集子集长度的列表,排列将提供所有可能的订购方式.

由于遍历切片,我可以在列表中只使用一个逗号来创建所有情况,但是我不能用两个逗号来表示例如"ab","c","d"和"a","b" ,"光盘"

我的尝试w/slice:

test="abcd"

for x in range(len(test)):
     print test[:x],test[x:]

DSM.. 15

怎么样的:

from itertools import combinations

def all_splits(s):
    for numsplits in range(len(s)):
        for c in combinations(range(1,len(s)), numsplits):
            split = [s[i:j] for i,j in zip((0,)+c, c+(None,))]
            yield split

之后:

>>> for x in all_splits("abcd"):
...     print(x)
...     
['abcd']
['a', 'bcd']
['ab', 'cd']
['abc', 'd']
['a', 'b', 'cd']
['a', 'bc', 'd']
['ab', 'c', 'd']
['a', 'b', 'c', 'd']

Tim Peters.. 15

你当然可以使用itertools它,但我认为直接编写递归生成器更容易:

def gen_commas(s):
    yield s
    for prefix_len in range(1, len(s)):
        prefix = s[:prefix_len]
        for tail in gen_commas(s[prefix_len:]):
            yield prefix + "," + tail

然后

print list(gen_commas("abcd"))

版画

['abcd', 'a,bcd', 'a,b,cd', 'a,b,c,d', 'a,bc,d', 'ab,cd', 'ab,c,d', 'abc,d']

我不确定为什么我觉得这更容易.也许只是因为直接做到这一点很容易;-)