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 combinations 类似的问题密切相关，但我无法从中推断出 Pythonic 解决方案。

问题是：

假设有一个str，例如'hi|guys|whats|app'，我需要用分隔符分割那个str的所有排列。示例：

#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

我可以编写回溯算法，但是 python（例如 itertools）是否提供了简化该算法的库？

提前致谢！！

Answer 1

如果你想要所有个分区，试试partitions from more-itertools:

from more_itertools import partitions

s = 'hi|guys|whats|app'

for p in partitions(s.split('|')):
    print(list(map('|'.join, p)))

输出：

['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']

如果您只想要一定数量的拆分，那么不要在 all 分隔符处拆分然后重新连接部分，您可以获取分隔符索引的组合并相应地获取子字符串：

from itertools import combinations

s = 'hi|guys|whats|app'
splits = 2

indexes = [i for i, c in enumerate(s) if c == '|']
for I in combinations(indexes, splits):
    print([s[i+1:j] for i, j in zip([-1, *I], [*I, None])])

输出：

['hi', 'guys', 'whats|app']
['hi', 'guys|whats', 'app']
['hi|guys', 'whats', 'app']

Answer 2

一种方法，分割字符串后，使用itertools.combinations定义列表中的分割点，其他位置应再次融合。

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

通用函数

这是一个通用版本，与上面一样工作，但也将-1 作为split 的参数，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]

示例：

>>> 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']]

Answer 3

您可以找到index all '|' 然后在所有组合中将'|' 替换为',' 然后拆分基','，如下所示：

>>> from itertools import combinations
>>> st = 'hi|guys|whats|app'
>>> idxs_rep = [idx for idx, s in enumerate(st) if s=='|']

>>> def combs(x):
...    return [c for i in range(len(x)+1) for c in combinations(x,i)]

>>> for idxs in combs(idxs_rep):        
...    lst_st = list(st)
...    for idx in idxs:
...        lst_st[idx] = ','
...    st2 = ''.join(lst_st)
...    print(st2.split(','))

['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']

Answer 4

使用combinations 和chain 的一种方法

from itertools import combinations, chain


def partition(alist, indices):
    # https://stackoverflow.com/a/1198876/4001592
    pairs = zip(chain([0], indices), chain(indices, [None]))
    return (alist[i:j] for i, j in pairs)


s = 'hi|guys|whats|app'
delimiter_count = s.count("|")
splits = s.split("|")


for i in range(1, delimiter_count + 1):
    print("split", i)
    for combination in combinations(range(1, delimiter_count + 1), i):
        res = ["|".join(part) for part in partition(splits, combination)]
        print(res)

输出

split 1
['hi', 'guys|whats|app']
['hi|guys', 'whats|app']
['hi|guys|whats', 'app']
split 2
['hi', 'guys', 'whats|app']
['hi', 'guys|whats', 'app']
['hi|guys', 'whats', 'app']
split 3
['hi', 'guys', 'whats', 'app']

这个想法是生成所有方法来选择（或删除）分隔符 1、2、3 次并从那里生成分区。

Answer 5

这是我想出的递归函数：

def splitperms(string, i=0):
    if len(string) == i:
        return [[string]]
    elif string[i] == "|":
        return [*[[string[:i]] + split for split in splitperms(string[i + 1:])], *splitperms(string, i + 1)]
    else:
        return splitperms(string, i + 1)

输出：

>>> splitperms('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'], ['hi|guys|whats|app']]
>>>