I am trying to write a Python regex pattern that will allow me to capture words in a given text that have letters separated by the same symbol or space.
For example, in the text "This is s u p e r and s.u.p.e.r and s👌u👌p👌e👌r and s!u.p!e.r", my goal is to extract the words "s u p e r", "s.u.p.e.r", and s👌u👌p👌e👌r. However, I want to exclude "s!u.p!e.r" because it does not have the same consistent separating symbol within the word.
I'm currently using the following:
x="This is s u p e r and s.u.p.e.r and s👌u👌p👌e👌r and s!u.p!e.r"


pattern = r"(?:\b\w[^\w\d]){2,}"

re.findall(pattern, x)


['s u p e r ', 's.u.p.e.r ', 's👌u👌p👌e👌r ', 's!u.p!e.']

I'm just curious if it's possible to exclude the cases that do not have the same symbol.

How to capture words with letters separated by a consistent symbol in Python regex?
You may consider using
pattern = r"(?<!\S)\w(?=(\W))(?:\1\w)+(?!\S)"
results = [m.group() for m in re.finditer(pattern, x)]

See the Python demo and the regex demo.
import re
x="This is s u p e r and s.u.p.e.r and s👌u👌p👌e👌r and s!u.p!e.r"
pattern = r"(?<!\S)\w(?=(\W))(?:\1\w)+(?!\S)"
print([m.group() for m in re.finditer(pattern, x)])
# => ['s u p e r', 's.u.p.e.r', 's👌u👌p👌e👌r']

Pattern details

(?<!\S) - left-hand whitespace boundary
\w - a word char
(?=(\W)) - a positive lookahead that requires the next char to e a non-word char capturing it into Group 1 (\1)
(?:\1\w)+ - one or more repetitions of the same char as captured in Group 1 and then a single word char
(?!\S) - right-hand whitespace boundary

