How to split up a string on multiple delimiters but only capture some?

Question

I want to split a string on any combination of delimiters I provide. For example, if the string is:

s = 'This, I think,., کباب MAKES , some sense '

And the delimiters are \., ,, and \s. However I want to capture all delimiters except whitespace \s. The output should be:

['This', ',', 'I', 'think', ',.,', 'کباب', 'MAKES', ',', 'some', 'sense']

My solution so far is is using the re module:

pattern = '([\.,\s]+)'  
re.split(pattern, s)

However, this captures whitespace as well. I have tried using other patterns like [(\.)(,)\s]+ but they don't work.

Edit: @PadraicCunningham made an astute observation. For delimiters like Some text ,. , some more text, I'd only want to remove leading and trailing whitespace from ,. , and not whitespace within.

Answer 1

The following approach would be the most simple one, I suppose ...

s = 'This, I think,., کباب MAKES , some sense '
pattern = '([\.,\s]+)'
splitted = [i.strip() for i in re.split(pattern, s) if i.strip()]

The output:

['This', ',', 'I', 'think', ',.,', 'کباب', 'MAKES', ',', 'some', 'sense']

Answer 2

NOTE: According to the new edit on the question, I've improved my old regex. The new one is quite long but trust me, it's work!

I suggest a pattern below as a delimiter of the function re.split():

(?<![,\.\ ])(?=[,\.]+)|(?<=[,\.])(?![,\.\ ])|(?<=[,\.])\ +(?![,\.\ ])|(?<![,\.\ ])\ +(?=[,\.][,\.\ ]+)|(?<![,\.\ ])\ +(?![,\.\ ])

My workaround here doesn't require any pre/post space modification. The thing that make regex work is about how you order the regex expressions with or. My cursory strategy is any patterns that dealing with a space-leading will be evaluated last.

See DEMO

Additional

According to @revo's comment he provided an another shorten version of mine which is

\s+(?=[^.,\s])|\b(?:\s+|(?=[,.]))|(?<=[,.])\b

See DEMO