这是解释此问题的最简单方法。这是我正在使用的：
re.split('\W', 'foo/bar spam\neggs')
-> ['foo', 'bar', 'spam', 'eggs']这就是我想要的：
someMethod('\W', 'foo/bar spam\neggs')
-> ['foo', '/', 'bar', ' ', 'spam', '\n', 'eggs']原因是我想将字符串拆分为标记，对其进行操作，然后将其重新放回原处。
Answers:
>>> re.split('(\W)', 'foo/bar spam\neggs')
['foo', '/', 'bar', ' ', 'spam', '\n', 'eggs']['foo', '/bar', ' spam', '\neggs']吗？
re.split('% ', re.sub('% ', '%% ', '5.000% Additional Whatnot'))->['5.000%', 'Additional Whatnot']
如果要在换行符上拆分，请使用splitlines(True)。
>>> 'line 1\nline 2\nline without newline'.splitlines(True)
['line 1\n', 'line 2\n', 'line without newline']（这不是一个通用的解决方案，但是请在此处添加此功能，以防万一有人来这里而意识到此方法不存在。）
另一个在Python 3上运行良好的无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, '<'))如果只有1个分隔符，则可以使用列表推导：
text = 'foo,bar,baz,qux'
sep = ','追加/前置分隔符：
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']分隔符是它自己的元素：
result = [u for x in text.split(sep) for u in (x, sep)]
#['foo', ',', 'bar', ',', 'baz', ',', 'qux', ',']
results = result[:-1] # to get rid of trailingif x以确保产生的块split具有一定的内容，即result = [x + sep for x in text.split(sep) if x]
result = [sep+x for x in data.split(sep)] result[0] = result[0][len(sep):]
另一个示例，拆分非字母数字并保留分隔符
import re
a = "foo,bar@candy*ice%cream"
re.split('([^a-zA-Z0-9])',a)输出：
['foo', ',', 'bar', '@', 'candy', '*', 'ice', '%', 'cream']说明
re.split('([^a-zA-Z0-9])',a)
() <- keep the separators
[] <- match everything in between
^a-zA-Z0-9 <-except alphabets, upper/lower and numbers.您还可以使用字符串数组而不是正则表达式来拆分字符串，如下所示：
def tokenizeString(aString, separators):
#separators is an array of strings that are being used to split the 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 = ["'''", '+=', '+', "/", "*", "\\'", '\\"', "-=", "-", " ", '"""', "(", ")"]))# This keeps all separators in result
##########################################################################
import re
st="%%(c+dd+e+f-1523)%%7"
sh=re.compile('[\+\-//\*\<\>\%\(\)]')
def splitStringFull(sh, st):
ls=sh.split(st)
lo=[]
start=0
for l in ls:
if not l : continue
k=st.find(l)
llen=len(l)
if k> start:
tmp= st[start:k]
lo.append(tmp)
lo.append(l)
start = k + llen
else:
lo.append(l)
start =llen
return lo
#############################
li= splitStringFull(sh , st)
['%%(', 'c', '+', 'dd', '+', 'e', '+', 'f', '-', '1523', ')%%', '7']一种懒惰和简单的解决方案
假设您的正则表达式模式是 split_pattern = r'(!|\?)'
首先，您添加与新分隔符相同的字符，例如“ [cut]”
new_string = re.sub(split_pattern, '\\1[cut]', your_string)
然后拆分新的分隔符， new_string.split('[cut]')
[cut]某个位置时，它将失败。
如果要拆分字符串同时用正则表达式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)如果假设正则表达式包含在捕获组中：
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)两种方式都将删除在大多数情况下无用且烦人的空组。
我在尝试拆分文件路径时遇到了类似的问题，并且很难找到一个简单的答案。这对我有用，并且不需要将分隔符替换回拆分文本中：
my_path = 'folder1/folder2/folder3/file1'
import re
re.findall('[^/]+/|[^/]+', my_path)
返回：
['folder1/', 'folder2/', 'folder3/', 'file1']
re.findall('[^/]+/?', my_path)（如使斜线可选的使用?，而不是提供两种选择用|。
我发现这种基于生成器的方法更加令人满意：
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:]它在理论上应该相当便宜，而无需找出正确的正则表达式。它不会创建新的字符串对象，而是将大部分迭代工作委托给有效的find方法。
...并且在python 3.8中可以很短：
def split_keep(string, sep):
start = 0
while (end := string.find(sep, start) + 1) > 0:
yield string[start:end]
start = end
yield string[start:]这是一个.split无需正则表达式的简单解决方案。
这是对Python split（）的答案，没有删除定界符，因此与原始帖子所要求的不完全相同，但另一个问题已作为与此问题的重复而关闭。
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 idx in range(100000):
length = random.randint(1, 50)
s = "".join(random.choice(CHARS) for _ in range(length))
assert "".join(splitkeep(s, delimiter)) == s
\W？我在Google上失败了。