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。

>>> line = 'hello_toto_is_there'
>>> sep = '_'
>>> [sep + x[1] if x[0] != 0 else x[1] for x in enumerate(line.split(sep))]
['hello', '_toto', '_is', '_there']

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

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

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.

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