Question

我和我在一起很长

s = 'abcdabcdabcdabcdabcdefghi'

我想将其拆分为K个子字符串，其中每个子字符串的长度至少应为1，即非空。我想要所有这些可能的组合。

如果K为3，我期望的输出必须类似于以下内容

[['abcda', 'bcdabcdabcda', 'bcdefghi'], [.....], [....], ... ]

我想通过列表理解来做到这一点，但是我被困住了。有可能实现吗？还有其他更快的选择吗？？

Answer 1

使用itertools.combinations，可以获得分离索引对：

>>> s = 'abcdef'
>>> k = 3
>>> list(combinations(range(1, len(s)), k-1))
[(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)]

使用该索引对获取字符串切片

(1, 2)->（s[:1]，s[1:2]，s[2:]）
(1, 3)->（s[:1]，s[1:3]，s[3:]）
...
(4, 5)->（s[:4]，s[4:5]，s[5:]）

>>> from itertools import combinations
>>> s = 'abcdef'

>>> k = 3
>>> [[s[i:j] for i, j in zip((None,) + idxs, idxs + (None,))]
...  for idxs in combinations(range(1, len(s)), k-1)]
[['a', 'b', 'cdef'], ['a', 'bc', 'def'], ..., ['abcd', 'e', 'f']]


>>> k = 4
>>> [[s[i:j] for i, j in zip((None,) + idxs, idxs + (None,))]
...  for idxs in combinations(range(1, len(s)), k-1)]
[['a', 'b', 'c', 'def'], ['a', 'b', 'cd', 'ef'], ..., ['abc', 'd', 'e', 'f']]

s[:1] == s[0:1] == s[None:1]
s[2:] == s[2:len(s)] == s[2:None]

Answer 2

您可以找到列表中的所有切片，以便没有任何扩展库，像这样的切片部分都不会为空：

s = 'abcd'
substrings = []

# find slice of first part - from a|bcd to ab|cd
for first_slice in range(len(s)-2):
    # find slice of second and last part, for bcd - from b|cd to bc|d
    #                                     for cd -  just c|d
    for second_slice in range(first_slice+1, len(s)-1):
        substrings.append([s[:first_slice+1], s[first_slice+1: second_slice+1], s[second_slice+1:]])

print(substrings) # -> [['a', 'b', 'cd'], ['a', 'bc', 'd'], ['ab', 'c', 'd']]

s = 'abcdabcdabcdabcdabcdefghi'
print(len(substrings)) # -> 276

使用列表理解将字符串拆分为K个子字符串

2 个答案: