我发现这种基于生成器的方法更令人满意:

def split_keep(string, sep):
    """Usage:
    >>> list(split_keep("a.b.c.d", "."))
    ['a.', 'b.', 'c.', 'd']
    """
    start = 0
    while True:
        end = string.find(sep, start) + 1
        if end == 0:
            break
        yield string[start:end]
        start = end
    yield string[start:]

它避免了需要找出正确的正则表达式，而在理论上应该相当便宜。它不创建新的字符串对象，并将大部分迭代工作委托给高效的find方法。

．.． 在Python 3.8中，它可以短到:

def split_keep(string, sep):
    start = 0
    while (end := string.find(sep, start) + 1) > 0:
        yield string[start:end]
        start = end
    yield string[start:]

这里有一个简单的.split解决方案，不需要regex。

这是一个没有删除分隔符的Python split()的答案，所以不完全是最初的帖子所要求的，但另一个问题被关闭为这个问题的副本。

def splitkeep(s, delimiter):
    split = s.split(delimiter)
    return [substr + delimiter for substr in split[:-1]] + [split[-1]]

随机测试:

import random

CHARS = [".", "a", "b", "c"]
assert splitkeep("", "X") == [""]  # 0 length test
for delimiter in ('.', '..'):
    for _ in range(100000):
        length = random.randint(1, 50)
        s = "".join(random.choice(CHARS) for _ in range(length))
        assert "".join(splitkeep(s, delimiter)) == s

如果你想拆分字符串，同时通过regex保留分隔符，而不捕获组:

def finditer_with_separators(regex, s):
    matches = []
    prev_end = 0
    for match in regex.finditer(s):
        match_start = match.start()
        if (prev_end != 0 or match_start > 0) and match_start != prev_end:
            matches.append(s[prev_end:match.start()])
        matches.append(match.group())
        prev_end = match.end()
    if prev_end < len(s):
        matches.append(s[prev_end:])
    return matches

regex = re.compile(r"[\(\)]")
matches = finditer_with_separators(regex, s)

如果假设regex被封装到捕获组中:

def split_with_separators(regex, s):
    matches = list(filter(None, regex.split(s)))
    return matches

regex = re.compile(r"([\(\)])")
matches = split_with_separators(regex, s)

这两种方法也将删除空组，在大多数情况下是无用和恼人的。

你也可以用字符串数组而不是正则表达式分割字符串，就像这样:

def tokenizeString(aString, separators):
    #separators is an array of strings that are being used to split the string.
    #sort separators in order of descending length
    separators.sort(key=len)
    listToReturn = []
    i = 0
    while i < len(aString):
        theSeparator = ""
        for current in separators:
            if current == aString[i:i+len(current)]:
                theSeparator = current
        if theSeparator != "":
            listToReturn += [theSeparator]
            i = i + len(theSeparator)
        else:
            if listToReturn == []:
                listToReturn = [""]
            if(listToReturn[-1] in separators):
                listToReturn += [""]
            listToReturn[-1] += aString[i]
            i += 1
    return listToReturn
    

print(tokenizeString(aString = "\"\"\"hi\"\"\" hello + world += (1*2+3/5) '''hi'''", separators = ["'''", '+=', '+', "/", "*", "\\'", '\\"', "-=", "-", " ", '"""', "(", ")"]))

另一个在Python 3上工作良好的非正则表达式解决方案

# Split strings and keep separator
test_strings = ['<Hello>', 'Hi', '<Hi> <Planet>', '<', '']

def split_and_keep(s, sep):
   if not s: return [''] # consistent with string.split()

   # Find replacement character that is not used in string
   # i.e. just use the highest available character plus one
   # Note: This fails if ord(max(s)) = 0x10FFFF (ValueError)
   p=chr(ord(max(s))+1) 

   return s.replace(sep, sep+p).split(p)

for s in test_strings:
   print(split_and_keep(s, '<'))


# If the unicode limit is reached it will fail explicitly
unicode_max_char = chr(1114111)
ridiculous_string = '<Hello>'+unicode_max_char+'<World>'
print(split_and_keep(ridiculous_string, '<'))

在下面的代码中，对这个问题有一个简单、高效且经过测试的答案。代码中有解释其中所有内容的注释。

我保证它并不像看起来那么可怕——它实际上只有13行代码!其余的都是注释、文档和断言

def split_including_delimiters(input: str, delimiter: str):
    """
    Splits an input string, while including the delimiters in the output
    
    Unlike str.split, we can use an empty string as a delimiter
    Unlike str.split, the output will not have any extra empty strings
    Conequently, len(''.split(delimiter))== 0 for all delimiters,
       whereas len(input.split(delimiter))>0 for all inputs and delimiters
    
    INPUTS:
        input: Can be any string
        delimiter: Can be any string

    EXAMPLES:
         >>> split_and_keep_delimiter('Hello World  ! ',' ')
        ans = ['Hello ', 'World ', ' ', '! ', ' ']
         >>> split_and_keep_delimiter("Hello**World**!***", "**")
        ans = ['Hello', '**', 'World', '**', '!', '**', '*']
    EXAMPLES:
        assert split_and_keep_delimiter('-xx-xx-','xx') == ['-', 'xx', '-', 'xx', '-'] # length 5
        assert split_and_keep_delimiter('xx-xx-' ,'xx') == ['xx', '-', 'xx', '-']      # length 4
        assert split_and_keep_delimiter('-xx-xx' ,'xx') == ['-', 'xx', '-', 'xx']      # length 4
        assert split_and_keep_delimiter('xx-xx'  ,'xx') == ['xx', '-', 'xx']           # length 3
        assert split_and_keep_delimiter('xxxx'   ,'xx') == ['xx', 'xx']                # length 2
        assert split_and_keep_delimiter('xxx'    ,'xx') == ['xx', 'x']                 # length 2
        assert split_and_keep_delimiter('x'      ,'xx') == ['x']                       # length 1
        assert split_and_keep_delimiter(''       ,'xx') == []                          # length 0
        assert split_and_keep_delimiter('aaa'    ,'xx') == ['aaa']                     # length 1
        assert split_and_keep_delimiter('aa'     ,'xx') == ['aa']                      # length 1
        assert split_and_keep_delimiter('a'      ,'xx') == ['a']                       # length 1
        assert split_and_keep_delimiter(''       ,''  ) == []                          # length 0
        assert split_and_keep_delimiter('a'      ,''  ) == ['a']                       # length 1
        assert split_and_keep_delimiter('aa'     ,''  ) == ['a', '', 'a']              # length 3
        assert split_and_keep_delimiter('aaa'    ,''  ) == ['a', '', 'a', '', 'a']     # length 5
    """

    # Input assertions
    assert isinstance(input,str), "input must be a string"
    assert isinstance(delimiter,str), "delimiter must be a string"

    if delimiter:
        # These tokens do not include the delimiter, but are computed quickly
        tokens = input.split(delimiter)
    else:
        # Edge case: if the delimiter is the empty string, split between the characters
        tokens = list(input)
        
    # The following assertions are always true for any string input and delimiter
    # For speed's sake, we disable this assertion
    # assert delimiter.join(tokens) == input

    output = tokens[:1]

    for token in tokens[1:]:
        output.append(delimiter)
        if token:
            output.append(token)
    
    # Don't let the first element be an empty string
    if output[:1]==['']:
        del output[0]
        
    # The only case where we should have an empty string in the output is if it is our delimiter
    # For speed's sake, we disable this assertion
    # assert delimiter=='' or '' not in output
        
    # The resulting strings should be combinable back into the original string
    # For speed's sake, we disable this assertion
    # assert ''.join(output) == input

    return output