如果你只有一个分隔符，你可以使用列表推导式:

text = 'foo,bar,baz,qux'  
sep = ','

附加/将分隔符:

result = [x+sep for x in text.split(sep)]
#['foo,', 'bar,', 'baz,', 'qux,']
# to get rid of trailing
result[-1] = result[-1].strip(sep)
#['foo,', 'bar,', 'baz,', 'qux']

result = [sep+x for x in text.split(sep)]
#[',foo', ',bar', ',baz', ',qux']
# to get rid of trailing
result[0] = result[0].strip(sep)
#['foo', ',bar', ',baz', ',qux']

分隔符作为它自己的元素:

result = [u for x in text.split(sep) for u in (x, sep)]
#['foo', ',', 'bar', ',', 'baz', ',', 'qux', ',']
results = result[:-1]   # to get rid of trailing

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

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, '<'))

另一个例子，在非字母数字上进行分割，并保留分隔符

import re
a = "foo,bar@candy*ice%cream"
re.split('([^a-zA-Z0-9])',a)

输出:

['foo', ',', 'bar', '@', 'candy', '*', 'ice', '%', 'cream']

解释

re.split('([^a-zA-Z0-9])',a)

() <- keep the separators
[] <- match everything in between
^a-zA-Z0-9 <-except alphabets, upper/lower and numbers.

安装wrs时“不拆卸分离器”

pip install wrs

(由Rao Hamza开发)

import wrs
text  = "Now inbox “how to make spam ad” Invest in hard email marketing."
splitor = 'email | spam | inbox'
list = wrs.wr_split(splitor, text)
print(list)

结果: ['现在'，'收件箱'如何制作'，'垃圾广告'努力投资'，'电子邮件营销'。]

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