python - python字符串按分隔符拆分所有可能的排列

Question

这可能与Python 3.3: Split string and create all combination等类似问题密切相关，但我无法从中推断出 Pythonic 解决方案。

问题是：

假设有一个 str 'hi|guys|whats|app'，我需要用分隔符拆分该 str 的所有排列。例子：

#splitting only once
['hi','guys|whats|app']
['hi|guys','whats|app']
['hi|guys|whats','app']
#splitting only twice
['hi','guys','whats|app']
['hi','guys|whats','app']
#splitting only three times
...
etc

我可以编写一个回溯算法，但是 python（例如 itertools）是否提供了一个简化该算法的库？

提前致谢！！

标签： pythonstringsplitpermutation

Answer 1

一种方法，一旦你拆分了字符串，就是用来itertools.combinations定义列表中的拆分点，其他位置应该再次融合。

def lst_merge(lst, positions, sep='|'):
    '''merges a list on points other than positions'''
    '''A, B, C, D and 0, 1 -> A, B, C|D'''
    a = -1
    out = []
    for b in list(positions)+[len(lst)-1]:
        out.append('|'.join(lst[a+1:b+1]))
        a = b
    return out

def split_comb(s, split=1, sep='|'):
    from itertools import combinations
    l = s.split(sep)
    return [lst_merge(l, pos, sep=sep)
            for pos in combinations(range(len(l)-1), split)]

例子

>>> split_comb('hi|guys|whats|app', 0)
[['hi|guys|whats|app']]

>>> split_comb('hi|guys|whats|app', 1)
[['hi', 'guys|whats|app'],
 ['hi|guys', 'whats|app'],
 ['hi|guys|whats', 'app']]

>>> split_comb('hi|guys|whats|app', 2)
[['hi', 'guys', 'whats|app'],
 ['hi', 'guys|whats', 'app'],
 ['hi|guys', 'whats', 'app']]

>>> split_comb('hi|guys|whats|app', 3)
[['hi', 'guys', 'whats', 'app']]

>>> split_comb('hi|guys|whats|app', 4)
[] ## impossible

理由

ABCD -> A B C D
         0 1 2

combinations of split points: 0/1 or 0/2 or 1/2

0/1 -> merge on 2 -> A B CD
0/2 -> merge on 1 -> A BC D
1/2 -> merge on 0 -> AB C D

泛型函数

这是一个通用版本，像上面一样工作，但也-1作为参数 for split，在这种情况下，它将输出所有组合

def lst_merge(lst, positions, sep='|'):
    a = -1
    out = []
    for b in list(positions)+[len(lst)-1]:
        out.append('|'.join(lst[a+1:b+1]))
        a = b
    return out

def split_comb(s, split=1, sep='|'):
    from itertools import combinations, chain
    
    l = s.split(sep)
    
    if split == -1:
        pos = chain.from_iterable(combinations(range(len(l)-1), r)
                                  for r in range(len(l)+1))
    else:
        pos = combinations(range(len(l)-1), split)
        
    return [lst_merge(l, pos, sep=sep)
            for pos in pos]

例子：

>>> split_comb('hi|guys|whats|app', -1)
[['hi|guys|whats|app'],
 ['hi', 'guys|whats|app'],
 ['hi|guys', 'whats|app'],
 ['hi|guys|whats', 'app'],
 ['hi', 'guys', 'whats|app'],
 ['hi', 'guys|whats', 'app'],
 ['hi|guys', 'whats', 'app'],
 ['hi', 'guys', 'whats', 'app']]