I have the output of a command in tabular form. I’m parsing this output from a result file and storing it in a string. Each element in one row is separated by one or more whitespace characters, thus I’m using regular expressions to match 1 or more spaces and split it. However, a space is being inserted between every element:

>>> str1="a    b     c      d" # spaces are irregular
>>> str1
'a    b     c      d'
>>> str2=re.split("( )+", str1)
>>> str2
['a', ' ', 'b', ' ', 'c', ' ', 'd'] # 1 space element between!!!

Is there a better way to do this?

After each split str2 is appended to a list.

By using (,), you are capturing the group, if you simply remove them you will not have this problem.

>>> str1 = "a    b     c      d"
>>> re.split(" +", str1)
['a', 'b', 'c', 'd']

However there is no need for regex, str.split without any delimiter specified will split this by whitespace for you. This would be the best way in this case.

>>> str1.split()
['a', 'b', 'c', 'd']

If you really wanted regex you can use this ('\s' represents whitespace and it’s clearer):

>>> re.split("\s+", str1)
['a', 'b', 'c', 'd']

or you can find all non-whitespace characters

>>> re.findall(r'\S+',str1)
['a', 'b', 'c', 'd']

The str.split method will automatically remove all white space between items:

>>> str1 = "a    b     c      d"
>>> str1.split()
['a', 'b', 'c', 'd']

Docs are here: http://docs.python.org/library/stdtypes.html#str.split

When you use re.split and the split pattern contains capturing groups, the groups are retained in the output. If you don’t want this, use a non-capturing group instead.

Its very simple actually. Try this:

str1="a    b     c      d"
splitStr1 = str1.split()
print splitStr1

Here’s the simplest way to explain this. Here’s what I’m using:

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

Here’s what I want:

someMethod('\W', 'foo/bar spam\neggs')
-> ['foo', '/', 'bar', ' ', 'spam', '\n', 'eggs']

The reason is that I want to split a string into tokens, manipulate it, then put it back together again.

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

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

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

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

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 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.

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']

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]')

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.

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 spam\neggs')

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

I want to use input from a user as a regex pattern for a search over some text. It works, but how I can handle cases where user puts characters that have meaning in regex?

For example, the user wants to search for Word (s): regex engine will take the (s) as a group. I want it to treat it like a string "(s)" . I can run replace on user input and replace the ( with \( and the ) with \) but the problem is I will need to do replace for every possible regex symbol.

Do you know some better way ?

Use the re.escape() function for this:

4.2.3 re Module Contents

escape(string)

Return string with all non-alphanumerics backslashed; this is useful if you want to match an arbitrary literal string that may have regular expression metacharacters in it.

A simplistic example, search any occurence of the provided string optionally followed by ‘s’, and return the match object.

def simplistic_plural(word, text):
    word_or_plural = re.escape(word) + 's?'
    return re.match(word_or_plural, text)

You can use re.escape():

re.escape(string) Return string with all non-alphanumerics backslashed; this is useful if you want to match an arbitrary literal string that may have regular expression metacharacters in it.

>>> import re
>>> re.escape('^a.*$')
'\\^a\\.\\*\\$'

If you are using a Python version < 3.7, this will escape non-alphanumerics that are not part of regular expression syntax as well.

If you are using a Python version < 3.7 but >= 3.3, this will escape non-alphanumerics that are not part of regular expression syntax, except for specifically underscore (_).

Unfortunately, re.escape() is not suited for the replacement string:

>>> re.sub('a', re.escape('_'), 'aa')
'\\_\\_'

A solution is to put the replacement in a lambda:

>>> re.sub('a', lambda _: '_', 'aa')
'__'

because the return value of the lambda is treated by re.sub() as a literal string.

Please give a try:

\Q and \E as anchors

Put an Or condition to match either a full word or regex.

Ref Link : How to match a whole word that includes special characters in regex

I have a string which is like this:

this is "a test"

I’m trying to write something in Python to split it up by space while ignoring spaces within quotes. The result I’m looking for is:

['this','is','a test']

PS. I know you are going to ask “what happens if there are quotes within the quotes, well, in my application, that will never happen.

You want split, from the built-in shlex module.

>>> import shlex
>>> shlex.split('this is "a test"')
['this', 'is', 'a test']

This should do exactly what you want.

Have a look at the shlex module, particularly shlex.split.

>>> import shlex
>>> shlex.split('This is "a test"')
['This', 'is', 'a test']

I see regex approaches here that look complex and/or wrong. This surprises me, because regex syntax can easily describe “whitespace or thing-surrounded-by-quotes”, and most regex engines (including Python’s) can split on a regex. So if you’re going to use regexes, why not just say exactly what you mean?:

test = 'this is "a test"'  # or "this is 'a test'"
# pieces = [p for p in re.split("( |[\\\"'].*[\\\"'])", test) if p.strip()]
# From comments, use this:
pieces = [p for p in re.split("( |\\\".*?\\\"|'.*?')", test) if p.strip()]

Explanation:

[\\\"'] = double-quote or single-quote
.* = anything
( |X) = space or X
.strip() = remove space and empty-string separators

shlex probably provides more features, though.

Depending on your use case, you may also want to check out the csv module:

import csv
lines = ['this is "a string"', 'and more "stuff"']
for row in csv.reader(lines, delimiter=" "):
    print(row)

Output:

['this', 'is', 'a string']
['and', 'more', 'stuff']

I use shlex.split to process 70,000,000 lines of squid log, it’s so slow. So I switched to re.

Please try this, if you have performance problem with shlex.

import re

def line_split(line):
    return re.findall(r'[^"\s]\S*|".+?"', line)

Since this question is tagged with regex, I decided to try a regex approach. I first replace all the spaces in the quotes parts with \x00, then split by spaces, then replace the \x00 back to spaces in each part.

Both versions do the same thing, but splitter is a bit more readable then splitter2.

import re

s = 'this is "a test" some text "another test"'

def splitter(s):
    def replacer(m):
        return m.group(0).replace(" ", "\x00")
    parts = re.sub('".+?"', replacer, s).split()
    parts = [p.replace("\x00", " ") for p in parts]
    return parts

def splitter2(s):
    return [p.replace("\x00", " ") for p in re.sub('".+?"', lambda m: m.group(0).replace(" ", "\x00"), s).split()]

print splitter2(s)

It seems that for performance reasons re is faster. Here is my solution using a least greedy operator that preserves the outer quotes:

re.findall("(?:\".*?\"|\S)+", s)

Result:

['this', 'is', '"a test"']

It leaves constructs like aaa"bla blub"bbb together as these tokens are not separated by spaces. If the string contains escaped characters, you can match like that:

>>> a = "She said \"He said, \\\"My name is Mark.\\\"\""
>>> a
'She said "He said, \\"My name is Mark.\\""'
>>> for i in re.findall("(?:\".*?[^\\\\]\"|\S)+", a): print(i)
...
She
said
"He said, \"My name is Mark.\""

Please note that this also matches the empty string "" by means of the \S part of the pattern.

The main problem with the accepted shlex approach is that it does not ignore escape characters outside quoted substrings, and gives slightly unexpected results in some corner cases.

I have the following use case, where I need a split function that splits input strings such that either single-quoted or double-quoted substrings are preserved, with the ability to escape quotes within such a substring. Quotes within an unquoted string should not be treated differently from any other character. Some example test cases with the expected output:

 input string        | expected output
===============================================
 'abc def'           | ['abc', 'def']
 "abc \\s def"       | ['abc', '\\s', 'def']
 '"abc def" ghi'     | ['abc def', 'ghi']
 "'abc def' ghi"     | ['abc def', 'ghi']
 '"abc \\" def" ghi' | ['abc " def', 'ghi']
 "'abc \\' def' ghi" | ["abc ' def", 'ghi']
 "'abc \\s def' ghi" | ['abc \\s def', 'ghi']
 '"abc \\s def" ghi' | ['abc \\s def', 'ghi']
 '"" test'           | ['', 'test']
 "'' test"           | ['', 'test']
 "abc'def"           | ["abc'def"]
 "abc'def'"          | ["abc'def'"]
 "abc'def' ghi"      | ["abc'def'", 'ghi']
 "abc'def'ghi"       | ["abc'def'ghi"]
 'abc"def'           | ['abc"def']
 'abc"def"'          | ['abc"def"']
 'abc"def" ghi'      | ['abc"def"', 'ghi']
 'abc"def"ghi'       | ['abc"def"ghi']
 "r'AA' r'.*_xyz$'"  | ["r'AA'", "r'.*_xyz$'"]

I ended up with the following function to split a string such that the expected output results for all input strings:

import re

def quoted_split(s):
    def strip_quotes(s):
        if s and (s[0] == '"' or s[0] == "'") and s[0] == s[-1]:
            return s[1:-1]
        return s
    return [strip_quotes(p).replace('\\"', '"').replace("\\'", "'") \
            for p in re.findall(r'"(?:\\.|[^"])*"|\'(?:\\.|[^\'])*\'|[^\s]+', s)]

The following test application checks the results of other approaches (shlex and csv for now) and the custom split implementation:

#!/bin/python2.7

import csv
import re
import shlex

from timeit import timeit

def test_case(fn, s, expected):
    try:
        if fn(s) == expected:
            print '[ OK ] %s -> %s' % (s, fn(s))
        else:
            print '[FAIL] %s -> %s' % (s, fn(s))
    except Exception as e:
        print '[FAIL] %s -> exception: %s' % (s, e)

def test_case_no_output(fn, s, expected):
    try:
        fn(s)
    except:
        pass

def test_split(fn, test_case_fn=test_case):
    test_case_fn(fn, 'abc def', ['abc', 'def'])
    test_case_fn(fn, "abc \\s def", ['abc', '\\s', 'def'])
    test_case_fn(fn, '"abc def" ghi', ['abc def', 'ghi'])
    test_case_fn(fn, "'abc def' ghi", ['abc def', 'ghi'])
    test_case_fn(fn, '"abc \\" def" ghi', ['abc " def', 'ghi'])
    test_case_fn(fn, "'abc \\' def' ghi", ["abc ' def", 'ghi'])
    test_case_fn(fn, "'abc \\s def' ghi", ['abc \\s def', 'ghi'])
    test_case_fn(fn, '"abc \\s def" ghi', ['abc \\s def', 'ghi'])
    test_case_fn(fn, '"" test', ['', 'test'])
    test_case_fn(fn, "'' test", ['', 'test'])
    test_case_fn(fn, "abc'def", ["abc'def"])
    test_case_fn(fn, "abc'def'", ["abc'def'"])
    test_case_fn(fn, "abc'def' ghi", ["abc'def'", 'ghi'])
    test_case_fn(fn, "abc'def'ghi", ["abc'def'ghi"])
    test_case_fn(fn, 'abc"def', ['abc"def'])
    test_case_fn(fn, 'abc"def"', ['abc"def"'])
    test_case_fn(fn, 'abc"def" ghi', ['abc"def"', 'ghi'])
    test_case_fn(fn, 'abc"def"ghi', ['abc"def"ghi'])
    test_case_fn(fn, "r'AA' r'.*_xyz$'", ["r'AA'", "r'.*_xyz$'"])

def csv_split(s):
    return list(csv.reader([s], delimiter=' '))[0]

def re_split(s):
    def strip_quotes(s):
        if s and (s[0] == '"' or s[0] == "'") and s[0] == s[-1]:
            return s[1:-1]
        return s
    return [strip_quotes(p).replace('\\"', '"').replace("\\'", "'") for p in re.findall(r'"(?:\\.|[^"])*"|\'(?:\\.|[^\'])*\'|[^\s]+', s)]

if __name__ == '__main__':
    print 'shlex\n'
    test_split(shlex.split)
    print

    print 'csv\n'
    test_split(csv_split)
    print

    print 're\n'
    test_split(re_split)
    print

    iterations = 100
    setup = 'from __main__ import test_split, test_case_no_output, csv_split, re_split\nimport shlex, re'
    def benchmark(method, code):
        print '%s: %.3fms per iteration' % (method, (1000 * timeit(code, setup=setup, number=iterations) / iterations))
    benchmark('shlex', 'test_split(shlex.split, test_case_no_output)')
    benchmark('csv', 'test_split(csv_split, test_case_no_output)')
    benchmark('re', 'test_split(re_split, test_case_no_output)')

Output:

shlex

[ OK ] abc def -> ['abc', 'def']
[FAIL] abc \s def -> ['abc', 's', 'def']
[ OK ] "abc def" ghi -> ['abc def', 'ghi']
[ OK ] 'abc def' ghi -> ['abc def', 'ghi']
[ OK ] "abc \" def" ghi -> ['abc " def', 'ghi']
[FAIL] 'abc \' def' ghi -> exception: No closing quotation
[ OK ] 'abc \s def' ghi -> ['abc \\s def', 'ghi']
[ OK ] "abc \s def" ghi -> ['abc \\s def', 'ghi']
[ OK ] "" test -> ['', 'test']
[ OK ] '' test -> ['', 'test']
[FAIL] abc'def -> exception: No closing quotation
[FAIL] abc'def' -> ['abcdef']
[FAIL] abc'def' ghi -> ['abcdef', 'ghi']
[FAIL] abc'def'ghi -> ['abcdefghi']
[FAIL] abc"def -> exception: No closing quotation
[FAIL] abc"def" -> ['abcdef']
[FAIL] abc"def" ghi -> ['abcdef', 'ghi']
[FAIL] abc"def"ghi -> ['abcdefghi']
[FAIL] r'AA' r'.*_xyz$' -> ['rAA', 'r.*_xyz$']

csv

[ OK ] abc def -> ['abc', 'def']
[ OK ] abc \s def -> ['abc', '\\s', 'def']
[ OK ] "abc def" ghi -> ['abc def', 'ghi']
[FAIL] 'abc def' ghi -> ["'abc", "def'", 'ghi']
[FAIL] "abc \" def" ghi -> ['abc \\', 'def"', 'ghi']
[FAIL] 'abc \' def' ghi -> ["'abc", "\\'", "def'", 'ghi']
[FAIL] 'abc \s def' ghi -> ["'abc", '\\s', "def'", 'ghi']
[ OK ] "abc \s def" ghi -> ['abc \\s def', 'ghi']
[ OK ] "" test -> ['', 'test']
[FAIL] '' test -> ["''", 'test']
[ OK ] abc'def -> ["abc'def"]
[ OK ] abc'def' -> ["abc'def'"]
[ OK ] abc'def' ghi -> ["abc'def'", 'ghi']
[ OK ] abc'def'ghi -> ["abc'def'ghi"]
[ OK ] abc"def -> ['abc"def']
[ OK ] abc"def" -> ['abc"def"']
[ OK ] abc"def" ghi -> ['abc"def"', 'ghi']
[ OK ] abc"def"ghi -> ['abc"def"ghi']
[ OK ] r'AA' r'.*_xyz$' -> ["r'AA'", "r'.*_xyz$'"]

re

[ OK ] abc def -> ['abc', 'def']
[ OK ] abc \s def -> ['abc', '\\s', 'def']
[ OK ] "abc def" ghi -> ['abc def', 'ghi']
[ OK ] 'abc def' ghi -> ['abc def', 'ghi']
[ OK ] "abc \" def" ghi -> ['abc " def', 'ghi']
[ OK ] 'abc \' def' ghi -> ["abc ' def", 'ghi']
[ OK ] 'abc \s def' ghi -> ['abc \\s def', 'ghi']
[ OK ] "abc \s def" ghi -> ['abc \\s def', 'ghi']
[ OK ] "" test -> ['', 'test']
[ OK ] '' test -> ['', 'test']
[ OK ] abc'def -> ["abc'def"]
[ OK ] abc'def' -> ["abc'def'"]
[ OK ] abc'def' ghi -> ["abc'def'", 'ghi']
[ OK ] abc'def'ghi -> ["abc'def'ghi"]
[ OK ] abc"def -> ['abc"def']
[ OK ] abc"def" -> ['abc"def"']
[ OK ] abc"def" ghi -> ['abc"def"', 'ghi']
[ OK ] abc"def"ghi -> ['abc"def"ghi']
[ OK ] r'AA' r'.*_xyz$' -> ["r'AA'", "r'.*_xyz$'"]

shlex: 0.281ms per iteration
csv: 0.030ms per iteration
re: 0.049ms per iteration

So performance is much better than shlex, and can be improved further by precompiling the regular expression, in which case it will outperform the csv approach.

To preserve quotes use this function:

def getArgs(s):
    args = []
    cur = ''
    inQuotes = 0
    for char in s.strip():
        if char == ' ' and not inQuotes:
            args.append(cur)
            cur = ''
        elif char == '"' and not inQuotes:
            inQuotes = 1
            cur += char
        elif char == '"' and inQuotes:
            inQuotes = 0
            cur += char
        else:
            cur += char
    args.append(cur)
    return args

Speed test of different answers:

import re
import shlex
import csv

line = 'this is "a test"'

%timeit [p for p in re.split("( |\\\".*?\\\"|'.*?')", line) if p.strip()]
100000 loops, best of 3: 5.17 µs per loop

%timeit re.findall(r'[^"\s]\S*|".+?"', line)
100000 loops, best of 3: 2.88 µs per loop

%timeit list(csv.reader([line], delimiter=" "))
The slowest run took 9.62 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 2.4 µs per loop

%timeit shlex.split(line)
10000 loops, best of 3: 50.2 µs per loop

Hmm, can’t seem to find the “Reply” button… anyway, this answer is based on the approach by Kate, but correctly splits strings with substrings containing escaped quotes and also removes the start and end quotes of the substrings:

  [i.strip('"').strip("'") for i in re.split(r'(\s+|(?<!\\)".*?(?<!\\)"|(?<!\\)\'.*?(?<!\\)\')', string) if i.strip()]

This works on strings like 'This is " a \\\"test\\\"\\\'s substring"' (the insane markup is unfortunately necessary to keep Python from removing the escapes).

If the resulting escapes in the strings in the returned list are not wanted, you can use this slightly altered version of the function:

[i.strip('"').strip("'").decode('string_escape') for i in re.split(r'(\s+|(?<!\\)".*?(?<!\\)"|(?<!\\)\'.*?(?<!\\)\')', string) if i.strip()]

To get around the unicode issues in some Python 2 versions, I suggest:

from shlex import split as _split
split = lambda a: [b.decode('utf-8') for b in _split(a.encode('utf-8'))]

As an option try tssplit:

In [1]: from tssplit import tssplit
In [2]: tssplit('this is "a test"', quote='"', delimiter='')
Out[2]: ['this', 'is', 'a test']

I suggest:

test string:

s = 'abc "ad" \'fg\' "kk\'rdt\'" zzz"34"zzz "" \'\''

to capture also “” and ”:

import re
re.findall(r'"[^"]*"|\'[^\']*\'|[^"\'\s]+',s)

result:

['abc', '"ad"', "'fg'", '"kk\'rdt\'"', 'zzz', '"34"', 'zzz', '""', "''"]

to ignore empty “” and ”:

import re
re.findall(r'"[^"]+"|\'[^\']+\'|[^"\'\s]+',s)

result:

['abc', '"ad"', "'fg'", '"kk\'rdt\'"', 'zzz', '"34"', 'zzz']

If you don’t care about sub strings than a simple

>>> 'a short sized string with spaces '.split()

Performance:

>>> s = " ('a short sized string with spaces '*100).split() "
>>> t = timeit.Timer(stmt=s)
>>> print "%.2f usec/pass" % (1000000 * t.timeit(number=100000)/100000)
171.39 usec/pass

Or string module

>>> from string import split as stringsplit; 
>>> stringsplit('a short sized string with spaces '*100)

Performance: String module seems to perform better than string methods

>>> s = "stringsplit('a short sized string with spaces '*100)"
>>> t = timeit.Timer(s, "from string import split as stringsplit")
>>> print "%.2f usec/pass" % (1000000 * t.timeit(number=100000)/100000)
154.88 usec/pass

Or you can use RE engine

>>> from re import split as resplit
>>> regex = '\s+'
>>> medstring = 'a short sized string with spaces '*100
>>> resplit(regex, medstring)

Performance

>>> s = "resplit(regex, medstring)"
>>> t = timeit.Timer(s, "from re import split as resplit; regex='\s+'; medstring='a short sized string with spaces '*100")
>>> print "%.2f usec/pass" % (1000000 * t.timeit(number=100000)/100000)
540.21 usec/pass

For very long strings you should not load the entire string into memory and instead either split the lines or use an iterative loop

Try this:

  def adamsplit(s):
    result = []
    inquotes = False
    for substring in s.split('"'):
      if not inquotes:
        result.extend(substring.split())
      else:
        result.append(substring)
      inquotes = not inquotes
    return result

Some test strings:

'This is "a test"' -> ['This', 'is', 'a test']
'"This is \'a test\'"' -> ["This is 'a test'"]