使用列表理解将字符串拆分为k个子字符串

使用 itertools.combinations ,您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) -gt; s[:1] , s[1:2] , s[2:] )
• (1, 3) -gt; S[(1)] , s[1:3] , s[3:] )
• …
• (4, 5) -gt; 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]

您可以找到列表中的所有切片,这样如果没有这样的扩展库,任何切片零件都不会是空的:

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