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, '<'))

之前发布的一些答案，会重复分隔符，或者有一些我在自己的情况下遇到的其他错误。你可以使用这个函数:

def split_and_keep_delimiter(input, delimiter):
    result      = list()
    idx         = 0
    while delimiter in input:
        idx     = input.index(delimiter);
        result.append(input[0:idx+len(delimiter)])
        input = input[idx+len(delimiter):]
    result.append(input)
    return result

re.split的文档中提到:

根据出现的模式拆分字符串。如果捕获 括号是在模式中使用的，然后是文本中的所有组 模式也作为结果列表的一部分返回。

所以你只需要用一个捕获组来包装分隔符:

>>> re.split('(\W)', 'foo/bar spam\neggs')
['foo', '/', 'bar', ' ', 'spam', '\n', 'eggs']

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:]

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