In Python, how do I split a string and keep the separators?

You can also split a string with an array of strings instead of a regular expression, like this:

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

# 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']

Another no-regex solution that works well on Python 3

# 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, '<'))

If you are splitting on newline, use splitlines(True).

>>> 'line 1nline 2nline without newline'.splitlines(True)
['line 1n', 'line 2n', 'line without newline']

(Not a general solution, but adding this here in case someone comes here not realizing this method existed.)

If one wants to split string while keeping separators by regex without capturing group:

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)

If one assumes that regex is wrapped up into capturing group:

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)

Both ways also will remove empty groups which are useless and annoying in most of the cases.

If you have only 1 separator, you can employ list comprehensions:

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

Appending/prepending separator:

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']

Separator as it’s own element:

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

another example, split on non alpha-numeric and keep the separators

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

output:

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

explanation

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

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

One Lazy and Simple Solution

Assume your regex pattern is split_pattern = r'(!|?)'

First, you add some same character as the new separator, like ‘[cut]’

new_string = re.sub(split_pattern, '\1[cut]', your_string)

Then you split the new separator, new_string.split('[cut]')

I had a similar issue trying to split a file path and struggled to find a simple answer.
This worked for me and didn’t involve having to substitute delimiters back into the split text:

my_path = 'folder1/folder2/folder3/file1'

import re

re.findall('[^/]+/|[^/]+', my_path)

returns:

['folder1/', 'folder2/', 'folder3/', 'file1']

I found this generator based approach more satisfying:

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:]

It avoids the need to figure out the correct regex, while in theory should be fairly cheap. It doesn’t create new string objects and, delegates most of the iteration work to the efficient find method.

… and in Python 3.8 it can be as short as:

def split_keep(string, sep):
    start = 0
    while (end := string.find(sep, start) + 1) > 0:
        yield string[start:end]
        start = end
    yield string[start:]

1. replace all seperator: (W) with seperator + new_seperator: (W;)

2. split by the new_seperator: (;)

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

print('W', 'foo/bar spamneggs')

Here is a simple .split solution that works without regex.

This is an answer for Python split() without removing the delimiter, so not exactly what the original post asks but the other question was closed as a duplicate for this one.

def splitkeep(s, delimiter):
    split = s.split(delimiter)
    return [substr + delimiter for substr in split[:-1]] + [split[-1]]

Random tests:

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

May I just leave it here

s = 'foo/bar spamneggs'
print(s.replace('/', '+++/+++').replace(' ', '+++ +++').replace('n', '+++n+++').split('+++'))

['foo', '/', 'bar', ' ', 'spam', 'n', 'eggs']

Use re.split and also your regular expression comes from variable and also you have multi separator ,you can use as the following:

# 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']

reference: GNU Bash Special Parameters

install wrs "WITHOUT REMOVING SPLITOR" BY DOING

pip install wrs

(developed by Rao Hamza)

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)

result:
[‘now ‘, ‘inbox “how to make ‘, ‘spam ad” invest in hard ‘, ’email marketing.’]

Some of those answers posted before, will repeat delimiter, or have some other bugs which I faced in my case. You can use this function, instead:

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

In the below code, there is a simple, very efficient and well tested answer to this question. The code has comments explaining everything in it.

I promise it’s not as scary as it looks – it’s actually only 13 lines of code! The rest are all comments, docs and assertions

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

>>> 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']