Check that both `lookbehind` conditions are satisfied in `RegEx` - python-3.x

Question

I'm trying to check if a username is preceded either by RT # or by RT# by using lookbehind mechanism paired with conditionals, as explained in this tutorial.
The regex and the example are shown in Example 1:
Example 1
import re
text = 'RT #u1, #u2, u3, #u4, rt #u5:, #u3.#u1^, rt#u3'
mt_regex = r'(?i)(?<!RT )&(?<!RT)#(\w+)'
mt_pat = re.compile(mt_regex)
re.findall(mt_pat, text)
which outputs [] (empty list), while the desired output should be:
['u2', 'u4', 'u3', 'u1']
What am I missing? Thanks in advance.

Answer 1

The other answer is obviously correct and rightfully accepted, but here is what I think might work for you without negative lookbehinds. The benefit is that you are not limited to the single space character using \s*:
(?i)(?:^|[,.])\s*#(\w+)
See the online demo
(?i) - Turn of case sensitivity. Note you can also use re.IGNORECASE.
(?:^|[,.]) - Non-capture group to match start of string or literal comma/dot.
\s* - Zero or more spaces.
# - Literally match "#".
(\w+) - Open capture group and match word characters, short for [A-Za-z0-9_].
This print ['u2', 'u4', 'u3', 'u1']

Answer 2

If we break down your regex:
r"(?i)(?<!RT )&(?<!RT)#(\w+)"
(?i) match the remainder of the pattern, case insensitive match
(?<!RT ) negative lookbehind
asserts that 'RT ' does not match
& matches the character '&' literally
(?<!RT) negative lookbehind
asserts that 'RT' does not match
# matches the character '#' literally
(\w+) Capturing Group
matches [a-zA-Z0-9_] between one and unlimited times
You have the & character that is preventing your regex matching:
import re
text = "RT #u1, #u2, u3, #u4, rt #u5:, #u3.#u1^, rt#u3"
mt_regex = r"(?i)(?<!RT )(?<!RT)#(\w+)"
mt_pat = re.compile(mt_regex)
print(re.findall(mt_pat, text))
# ['u2', 'u4', 'u3', 'u1']
See this regex here