Split a string with multiple delimiters, and keep *some* of the delimiters, but not all

user_1907765 0
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I have a string that can look something like this:

1. "foo bar"
2. "foo bar foo:bar"
3. "foo bar "
4. "foo bar      "
5. "foo bar foo:bar:baz"

I want to split this string so that it would end up with the following results:

1. ['foo', 'bar']
2. ['foo', 'bar', 'foo', ':', 'bar']
3. / 4. ['foo', 'bar', '']
5. ['foo', 'bar', 'foo', ':', 'bar', ':', 'baz']

In other words, following these rules:

  1. Split the string on every occurrence of a space.

    a. If there are one or more spaces at the end of a string, add one empty string to the split list

    b. Any spaces before the last non-space character in a string should be consumed, and not add to the split list.

  2. Split the string on every occurrence of a colon, and do not consume the colon.

The XY problem is this, in case it's relevant:

I want to mimic Bash tab-completion behaviour. When you type a command into a Bash interpreter, it will split the command into an array COMP_WORDS, and it will follow the above rules - splitting the words based on spaces and colons, with colons placed into their own array element, and spaces ignored unless they're at the end of a string. I want to recreate this behaviour in Python, given a string that looks like a command that a user would type.

I've seen this question about splitting a string and keeping the separators using re.split. And this question about splitting using multiple delimiters. But my use case is more complicated, and neither question seems to cover it. I tried the following to at least split on spaces and colons:

print(re.split('(:)|(?: )', splitstr))

But even that doesn't work. When splitstr is "foo bar foo:bar" returns this:

['foo', None, 'bar', None, 'foo', ':', 'bar']

Any idea how this could be done in Python?

EDIT: My requirements weren't clear - I would want "foo bar " (with any number of spaces at the end) to return the list ["foo", "bar", ""] (with just one empty string at the end of the list.)

python-3.x python regex
3 Answers

There is no need to use regular expressions for this task. String methods work just as well, and might be more readable.

def split_comp(s: str) -> 'list[str]':
    trailing = s.endswith(' ')
    s = s.replace(':', ' : ')  # insert split marks before/after every colon
    parts = s.split()
    return parts if not trailing else [*parts, ' ']

This technique can be used for any delimiters – pick one delimiter to split on, then replace/pad those to remove/keep with it.

user_53499160

You can use a re.findall approach here with:

[^:\s]+|:|(?<=\S)(?=\s+$)

See the regex demo. Details:

  • [^:\s]+ - one or more chars other than whitespace and :
  • | - or
  • : - a colon
  • | - or
  • (?<=\S)(?=\s+$) - any empty string that is located between a non-whitespace and one or more whitespaces at the end of string.

See the Python demo.

import re
l = ['foo bar', 'foo bar foo:bar', 'foo bar ', 'foo     bar     ']
rx = re.compile(r'[^:\s]+|:|(?<=\S)(?=\s+$)')
for s in l:
    if s.rstrip() != s:
        s = s.rstrip() + " "
    print(f"'{s}'", '=>', rx.findall(s))

Output:

'foo bar' => ['foo', 'bar']
'foo bar foo:bar' => ['foo', 'bar', 'foo', ':', 'bar']
'foo bar ' => ['foo', 'bar', '']
'foo     bar ' => ['foo', 'bar', '']
user_38329700

Maybe there are shorter ways, but here is my suggestion:

def func(s):
    if s[-1]==' ':
        l=s.split()+['']
    else:
        l=s.split()
    def f(l):
        m=l.copy()
        res=[]
        for i in m:
            if i!=':' and ':' in i:
                temp=[i[:i.find(':')]]+[':']+[i[i.find(':')+1:]]
                res.extend(temp)
            else:
                res.append(i)
        return res
    while any(i!=':' and ':' in i for i in l):
        l=f(l)
    return l

Examples:

>>> func("foo bar")
['foo', 'bar']

>>> func("foo bar foo:bar")
['foo', 'bar', 'foo', ':', 'bar']

>>> func("foo bar ")
['foo', 'bar', '']

>>> func("foo bar      ")
['foo', 'bar', '']
user_82285580
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