regex split by parenthesis but not all parenthesis

I am trying to split a string containing open and close parenthesis but want to exclude those parenthesis that have a substring right before them.
In the following example:

a = 'abc (xyz pqr) qwe ew (kjlk asd) ue(aad) kljl'

I want to have a list like:

['abc', 'xyz pqr', 'qwe ew', 'kjlk asd', 'ue(aad)', 'kljl']

So I want to keep ue(aad) and do not split by (aad)

I have tried:

y = [x.strip() for x in re.split(r"[^ue()][()]", a) if x.strip()]

Try this:

import re

a = 'abc (xyz pqr) qwe ew (kjlk asd) ue(aad) kljl'
y = [x.strip() for x in re.split(r' (\S*\(.*?\))', a) if x != '']
for i in range(len(y)):
    if y[i][0] == '(' and y[i][-1] == ')':
        y[i] = y[i].strip('()')

print(y)  # => ['abc', 'xyz pqr', 'qwe ew', 'kjlk asd', 'ue(aad)', 'kljl']

The RegEx (\S*\(.*?\)) will match any of the parentheses and any preceding strings, then the loop removes surrounding parentheses from matches that have no preceding strings.

Since the keyword in my case is always known, I was thinking to remove all ue(.*?)s and keep them in a list then split by parenthesis then substitute them.
This way I will be able to split nested parenthesis.
something like:

a = "abc (xyz pqr) qwe ew (kjlk asd) ue(aad) kljl"
ues = re.findall("ue\(.*?\)", a)
j = re.sub("(?<=ue)\(.*?\)", "", a)
y = [x.strip() for x in re.split(r"[()]", j) if x.strip()]
for i in y:
    if "ue" in i:
        print(re.sub("ue", ues.pop(0), i))
    else: 
        print(i)

Update:
The parenthesis that must be ignored will have a substring stuck to it like ue(). So adding a space before will ignore them.

y = [x.strip() for x in re.split(r"[(?<=\s)][()]", a) if x.strip()]  

For your example data, you could use capture groups to keep the result after splitting. In the pattern, capture non whitespace chars except parenthesis before or after the part with parenthesis.

In the list comprehension, first check for x and then you can test again for x.strip()

Note that this does not take any nested/balanced parenthesis into account.

Explanation

• ([^\s()]+\([^()]*\)) Capture group 1, match 1+ non whitespace chars before matching from (...)
• | Or
• (\([^()]*\)[^\s()]+) Capture group 2, match 1+ non whitespace chars after matching from (...)
• | Or
• [()] Match either ( or )

See a Python demo and a regex101 demo.

import re

pattern = r"([^\s()]+\([^()]*\))|(\([^()]*\)[^\s()]+)|[()]"
a = 'abc (xyz pqr) qwe ew (kjlk asd) ue(aad) kljl'

y = [x.strip() for x in re.split(pattern, a) if x and x.strip()]
print(y)

Output

['abc', 'xyz pqr', 'qwe ew', 'kjlk asd', 'ue(aad)', 'kljl']

This is a strange way to accomplish this but it works:

import re

a = "abc (xyz pqr) qwe ew (kjlk asd) ue(aad) kljl"

dissub=re.split("\)\s",a)
newlist=[]
for b in dissub:
    dasplit=re.split("\s\(",b)
    for c in dasplit:

        newlist.append(c)
i=0
while i<len(newlist):
    dacheck=re.search("\(",newlist[i])
    if dacheck:
        newlist[i]+=")"
    i+=1
print(newlist)