使用re.split，并且你的正则表达式来自变量，并且你有多个分隔符，你可以像下面这样使用:

# BashSpecialParamList is the special param in bash,
# such as your separator is the bash special param
BashSpecialParamList = ["$*", "$@", "$#", "$?", "$-", "$$", "$!", "$0"]
# aStr is the the string to be splited
aStr = "$a Klkjfd$0 $? $#%$*Sdfdf"

reStr = "|".join([re.escape(sepStr) for sepStr in BashSpecialParamList])

re.split(f'({reStr})', aStr)

# Then You can get the result:
# ['$a Klkjfd', '$0', ' ', '$?', ' ', '$#', '%', '$*', 'Sdfdf']

参考:GNU Bash特殊参数

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.

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

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']

将所有分隔符:(\W)替换为分隔符+ new_分隔符:(\W;) 由new_separator分隔符拆分:(;)

def split_and_keep(seperator, s):
  return re.split(';', re.sub(seperator, lambda match: match.group() + ';', s))

print('\W', 'foo/bar spam\neggs')

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 = ["'''", '+=', '+', "/", "*", "\\'", '\\"', "-=", "-", " ", '"""', "(", ")"]))