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how to extract the front and back of a designated special token using regex?

How to extract the front and back of a designated special token(in this case, -, not #)?
And if those that are connected by - are more than two, I want to extract those too. (In the example, Bill-Gates-Foundation)
e.g)
from 'Meinda#Bill-Gates-Foundation#drug-delivery' -> ['Bill-Gates-Foundation', 'drug-delivery']
I tried p = re.compile('#(\D+)\*(\D+)')
but that was not what I wanted.

You can exclude matchting the # char and repeat 1 or more times the -
#([^\s#-]+(?:-[^\s#-]+)+)
Explanation
# Match literally
( Capture group 1 (returned by re.findall)
[^\s#-]+ Match 1+ non whitespace chars except - and #
(?:-[^\s#-]+)+ Repeat 1+ times matching - and again 1+ non whitespace chars except - and #
) Close group 1
Regex demo
import re
pattern = r"#([^\s#-]+(?:-[^\s#-]+)+)"
s = r"Meinda#Bill-Gates-Foundation#drug-delivery"
print(re.findall(pattern, s))
Output
['Bill-Gates-Foundation', 'drug-delivery']

#ahmet-buğra-buĞa gave an answer with regex.
If you don't have to use regex, then it is easier way is to just use split.
test_str = "Meinda#Bill-Gates-Foundation#drug-delivery"
test_str.split("#")[1:]
This outputs
['Bill-Gates-Foundation', 'drug-delivery']
You can make it a function like so
def get_list_of_strings_after_first(original_str, token_to_split_on):
return original_str.split("#")[1:]
get_list_of_strings_after_first("Meinda#Bill-Gates-Foundation#drug-delivery", "#")
This give the same output
['Bill-Gates-Foundation', 'drug-delivery']

How to create regex to match a string that contains only hexadecimal numbers and arrows?

I am using a string that uses the following characters:
0-9
a-f
A-F
-
>
The mixture of the greater than and hyphen must be:
->
-->
Here is the regex that I have so far:
[0-9a-fA-F\-\>]+
I tried these others using exclusion with ^ but they didn't work:
[^g-zG-Z][0-9a-fA-F\-\>]+
^g-zG-Z[0-9a-fA-F\-\>]+
[0-9a-fA-F\-\>]^g-zG-Z+
[0-9a-fA-F\-\>]+^g-zG-Z
[0-9a-fA-F\-\>]+[^g-zG-Z]
Here are some samples:
"0912adbd->12d1829-->218990d"
"ab2c8d-->82a921->193acd7"

Firstly, you don't need to escape - and >
Here's the regex that worked for me:
^([0-9a-fA-F]*(->)*(-->)*)*$
Here's an alternative regex:
^([0-9a-fA-F]*(-+>)*)*$
What does the regex do?
^ matches the beginning of the string and $ matches the ending.
* matches 0 or more instances of the preceding token
Created a big () capturing group to match any token.
[0-9a-fA-F] matches any character that is in the range.
(->) and (-->) match only those given instances.
Putting it into a code:
import re
regex = "^([0-9a-fA-F]*(->)*(-->)*)*$"
re.match(re.compile(regex),"0912adbd->12d1829-->218990d")
re.match(re.compile(regex),"ab2c8d-->82a921->193acd7")
re.match(re.compile(regex),"this-failed->so-->bad")
You can also convert it into a boolean:
print(bool(re.match(re.compile(regex),"0912adbd->12d1829-->218990d")))
print(bool(re.match(re.compile(regex),"ab2c8d-->82a921->193acd7")))
print(bool(re.match(re.compile(regex),"this-failed->so-->bad")))
Output:
True
True
False
I recommend using regexr.com to check your regex.

If there must be an arrow present, and not at the start or end of the string using a case insensitive pattern:
^[a-f\d]+(?:-{1,2}>[a-f\d]+)+$
Explanation
^ Start of string
[a-f\d]+ Match 1+ chars a-f or digits
(?: Non capture group to repeat as a whole
-{1,2}>[a-f\d]+ Match - or -- and > followed by 1+ chars a-f or digits
)+ Close the non capture group and repeat 1+ times
$ End of string
See a regex demo and a Python demo.
import re
pattern = r"^[a-f\d]+(?:-{1,2}>[a-f\d]+)+$"
s = ("0912adbd->12d1829-->218990d\n"
"ab2c8d-->82a921->193acd7\n"
"test")
print(re.findall(pattern, s, re.I | re.M))
Output
[
'0912adbd->12d1829-->218990d',
'ab2c8d-->82a921->193acd7'
]

You can construct the regex by steps. If I understand your requirements, you want a sequence of hexadecimal numbers (like a01d or 11efeb23, separated by arrows with one or two hyphens (-> or -->).
The hex part's regex is [0-9a-fA-F]+ (assuming it cannot be empty).
The arrow's regex can be -{1,2}> or (->|-->).
The arrow is only needed before each hex number but the first, so you'll build the final regex in two parts: the first number, then the repetition of arrow and number.
So the general structure will be:
NUMBER(ARROW NUMBER)*
Which gives the following regex:
[0-9a-fA-F]+(-{1,2}>[0-9a-fA-F]+)*

Pattern to extract, expand and form a sentence based on a certain delimiter

I was trying out to solve a problem on regex:
There is an input sentence which is of one of these forms: Number1,2,3 or Number1/2/3 or Number1-2-3 these are the 3 delimiters: , / -
The expected output is: Number1,Number2,Number3
Pattern I've tried so far:
(?\<=,)\[^,\]+(?=,)
but this misses out on the edge cases i.e. 1st element and last element. I am also not able to generate for '/'.

You could separate out the key from values, then use a list comprehension to build the output you want.
inp = "Number1,2,3"
matches = re.search(r'(\D+)(.*)', inp)
output = [matches[1] + x for x in re.split(r'[,/]', matches[2])]
print(output) # ['Number1', 'Number2', 'Number3']

You can do it in several steps: 1) validate the string to match your pattern, and once validated 2) add the first non-digit chunk to the numbers while replacing - and / separator chars with commas:
import re
texts = ['Number1,2,3', 'Number1/2/3', 'Number1-2-3']
for text in texts:
m = re.search(r'^(\D+)(\d+(?=([,/-]))(?:\3\d+)*)$', text)
if m:
print( re.sub(r'(?<=,)(?=\d)', m.group(1).replace('\\', '\\\\'), text.replace('/',',').replace('-',',')) )
else:
print(f"NO MATCH in '{text}'")
See this Python demo.
Output:
Number1,Number2,Number3
Number1,Number2,Number3
Number1,Number2,Number3
The ^(\D+)(\d+(?=([,/-]))(?:\3\d+)*)$ regex validates your three types of input:
^ - start of string
(\D+) - Group 1: one or more non-digits
(\d+(?=([,/-]))(?:\3\d+)*) - Group 2: one or more digits, and then zero or more repetitions of ,, / or - and one or more digits (and the separator chars should be consistent due to the capture used in the positive lookahead and the \3 backreference to that value used in the non-capturing group)
$ - end of string.
The re.sub pattern, (?<=,)(?=\d), matches a location between a comma and a digit, the Group 1 value is placed there (note the .replace('\\', '\\\\') is necessary since the replacement is dynamic).

import re
for text in ("Number1,2,3", "Number1-2-3", "Number1/2/3"):
print(re.sub(r"(\D+)(\d+)[/,-](\d+)[/,-](\d+)", r"\1\2,\1\3,\1\4", text))
\D+ matches "Number" or any other non-number text
\d+ matches a number (or more than one)
[/,-] matches any of /, ,, -
The rest is copy paste 3 times.
The substitution consists of backreferences to the matched "Number" string (\1) and then each group of the (\d+)s.
This works if you're sure that it's always three numbers divided by that separator. This does not ensure that it's the same separator between each number. But it's short.
Output:
Number1,Number2,Number3
Number1,Number2,Number3
Number1,Number2,Number3

If you can make use of the pypi regex module you can use the captures collection with a named capture group.
([^\d\s,/]+)(?<num>\d+)([,/-])(?<num>\d+)(?:\3(?<num>\d+))*(?!\S)
([^\d\s,/]+) Capture group 1, match 1+ chars other than the listed
(?<num>\d+) Named capture group num matching 1+ digits
([,/-]) Capture either , / - in group 3
(?<num>\d+) Named capture group num matching 1+ digits
(?:\3(?<num>\d+))* Optionally repeat a backreference to group 3 to keep the separators the same and match 1+ digits in group num
(?!\S) Assert a whitspace boundary to the right to prevent a partial match
Regex demo | Python demo
import regex as re
pattern = r"([^\d\s,/]+)(?<num>\d+)([,/-])(?<num>\d+)(?:\3(?<num>\d+))*(?!\S)"
s = "Number1,2,3 or Number4/5/6 but not Number7/8,9"
matches = re.finditer(pattern, s)
for _, m in enumerate(matches, start=1):
print(','.join([m.group(1) + c for c in m.captures("num")]))
Output
Number1,Number2,Number3
Number4,Number5,Number6

Python Regex: Capture overlapping parts

Given a string s = "<foo>abcaaa<bar>a<foo>cbacba<foo>c" I'm trying to write a regular expression which will extract portions of: angle brackets with the text inside and the surrounding text. Like this:
<foo>abcaaa
abcaaa<bar>a
a<foo>cbacba
cbacba<foo>c
So expected output should look like this:
["<foo>abcaaa", "abcaaa<bar>a", "a<foo>cbacba", "cbacba<foo>c"]
I found this question How to find overlapping matches with a regexp? which brought me little bit closer to the desired result but still my regex doesn't work.
regex = r"(?=([a-c]*)\<(\w+)\>([a-c]*))"
Any ideas how to solve this problem?

You can match overlapping content with standard regex syntax by using capturing groups inside lookaround assertions, since those may match parts of the string without consuming the matched substring and hence precluding it from further matches. In this specific example, we match either the beginning of the string or a > as anchor for the lookahead assertion which captures our actual targets:
(?:\A|>)(?=([a-c]*<\w+>[a-c]*))
See regex demo.
In python we then use the property of re.findall() to only return matches captured in groups when capturing groups are present in the expression:
text = '<foo>abcaaa<bar>a<foo>cbacba<foo>c'
expr = r'(?:\A|>)(?=([a-c]*<\w+>[a-c]*))'
captures = re.findall(expr, text)
print(captures)
Output:
['<foo>abcaaa', 'abcaaa<bar>a', 'a<foo>cbacba', 'cbacba<foo>c']

You need to set the left- and right-hand boundaries to < or > chars or start/end of string.
Use
import re
text = "<foo>abcaaa<bar>a<foo>cbacba<foo>c"
print( re.findall(r'(?=(?<![^<>])([a-c]*<\w+>[a-c]*)(?![^<>]))', text) )
# => ['<foo>abcaaa', 'abcaaa<bar>a', 'a<foo>cbacba', 'cbacba<foo>c']
See the Python demo online and the regex demo.
Pattern details
(?= - start of a positive lookahead to enable overlapping matches
(?<![^<>]) - start of string, < or >
([a-c]*<\w+>[a-c]*) - Group 1 (the value extracted): 0+ a, b or c chars, then <, 1+ word chars, > and again 0+ a, b or c chars
(?![^<>]) - end of string, < or > must follow immediately
) - end of the lookahead.

You may use this regex code in python:
>>> s = '<foo>abcaaa<bar>a<foo>cbacba<foo>c'
>>> reg = r'([^<>]*<[^>]*>)(?=([^<>]*))'
>>> print ( [''.join(i) for i in re.findall(reg, s)] )
['<foo>abcaaa', 'abcaaa<bar>a', 'a<foo>cbacba', 'cbacba<foo>c']
RegEx Demo
RegEx Details:
([^<>]*<[^>]*>): Capture group #1 to match 0 or more characters that are not < and > followed by <...> string.
(?=([^<>]*)): Lookahead to assert that we have 0 or more non-<> characters ahead of current position. We have capture group #2 inside this lookahead.

Remove duplicate words in a string using regex

I'm working on my regex skills and i find one of my strings having duplicate words at the starting. I would like to remove the duplicate and just have one word of it -
server_server_dev1_check_1233.zzz
server_server_qa1_run_1233.xyz
server_server_dev2_1233.qqa
server_dev1_1233.zzz
data_data_dev9_check_660.log
I used the below regex but i get both server_server in my output,
((.*?))_(?!\D)
How can i have my output just to one server_ if there are two or more and if its only one server_, then take as is?
The output doesn't have to contain the digits and also the part after . i.e. .zzz, .xyz etc
Expected output -
server_dev1_check
server_qa1_run
server_dev2
server_dev1
data_dev9_check

you could back reference the word in your search expression:
>>> s = "server_server_dev1_check_1233.zzz"
>>> re.sub(r"(.*_)\1",r"\1",s)
'server_dev1_check_1233.zzz'
and use the "many times" suffix so if there are more than 2 occurrences it still works:
'server_server_server_dev1_check_1233.zzz'
>>> re.sub(r"(.*_)\1{1,}",r"\1",s)
'server_dev1_check_1233.zzz'
getting rid of the suffix is not the hardest part, just capture the rest and discard the end:
>>> re.sub(r"(.*_)\1{1,}(.*)(_\d+\..*)",r"\1\2",s)
'server_dev1_check'

You may use a single re.sub call to match and remove what you do not need and match and capture what you need:
re.sub(r'^([^_]+)(?:_\1)*(.*)_\d+\.\w+$', r'\1\2', s)
See the regex demo
Details
^ - start of string
([^_]+) - Capturing group 1: any 1+ chars other than _
(?:_\1)* - zero or more repetitions of _ followed with the same substring as in Group 1 (thanks to the inline backreference \1 that retrieves the text from Group 1)
(.*) - Group 2: any 0+ chars, as many as possible
_ - an underscore
\d+ - 1+ digits
\. - a dot
\w+ - 1+ word chars ([^.]+ will also do, 1 or more chars other than .)
$ - end of string.
The replacement pattern is \1\2, i.e. the contents of Group 1 and 2 are concatenated and make up the resulting value.
Python demo:
import re
rx = r'^([^_]+)(?:_\1)*(.*)_\d+\.\w+$'
strs = ["server_server_dev1_check_1233.zzz", "server_server_qa1_run_1233.xyz", "server_server_dev2_1233.qqa", "server_dev1_1233.zzz", "data_data_dev9_check_660.log"]
for s in strs:
print(re.sub(rx, r'\1\2', s))
Output:
server_dev1_check
server_qa1_run
server_dev2
server_dev1
data_dev9_check