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Use regex to remove a substring that matches a beginning of a substring through the following comma

I haven't found any helpful Regex tools to help me figure this complicated pattern out.
I have the following string:
Myfirstname Mylastname, Department of Mydepartment, Mytitle, The University of Me; 4-1-1, Hong,Bunk, Tokyo 113-8655, Japan E-mail:my.email#example.jp, Tel:00-00-222-1171, Fax:00-00-225-3386
I am trying to learn enough Regex patterns to remove the substrings one at a time:
E-mail:my.email#example.jp
Tel:00-00-222-1171
Fax:00-00-225-3386
So I think the correct pattern would be to remove a given word (ie., "E-mail", "Tel") all the way through the following comma.
Is type of dynamic pattern possible in Regex?
I am performing the match in Python, however, I don't think that would matter too much.
Also, I know the data string looks comma separated, and it is. However there is no guarantee of preserving the order of those fields. That's why I'm trying to use a Regex match.

How about this regex:
<YOUR_WORD>.*?(?=(,|($)))
Explanation:
It looks for the word specified in <YOUR_WORD> placeholder
It looks for any kind of character afterwards
The search stops when it hits one of the two options:
It finds the character ,
It finds an end of the line
So:
E-mail.*?(?=(,|($)))
Will result in:
E-mail:my.email#example.jp
And
Fax.*?(?=(,|($)))
Will result in:
Fax:00-00-225-3386
If there are edge cases it misses - I would like to know, and whether it affects the performance/ is necessary.

Exact search of a string that has parenthesis using regex

I am new to regexes.
I have the following string : \n(941)\n364\nShackle\n(941)\nRivet\n105\nTop
Out of this string, I want to extract Rivet and I already have (941) as a string in a variable.
My thought process was like this:
Find all the (941)s
filter the results by checking if the string after (941) is followed by \n, followed by a word, and ending with \n
I made a regex for the 2nd part: \n[\w\s\'\d\-\/\.]+$\n.
The problem I am facing is that because of the parenthesis in (941) the regex is taking 941 as a group. In the 3rd step the regex may be wrong, which I can fix later, but 1st I needed help in finding the 2nd (941) so then I can apply the 3rd step on that.
PS.
I know I can use python string methods like find and then loop over the searches, but I wanted to see if this can be done directly using regex only.
I have tried the following regex: (?:...), (941){1} and the make regex literal character \ like this \(941\) with no useful results. Maybe I am using them wrong.
Just wanted to know if it is possible to be done using regex. Though it might be useful for others too or a good share for future viewers.
Thanks!

Assuming:
You want to avoid matching only digits;
Want to match a substring made of word-characters (thus including possible digits);
Try to escape the variable and use it in the regular expression through f-string:
import re
s = '\n(941)\n364\nShackle\n(941)\nRivet\n105\nTop'
var1 = '(941)'
var2 = re.escape(var1)
m = re.findall(fr'{var2}\n(?!\d+\n)(\w+)', s)[0]
print(m)
Prints:
Rivet

If you have text in a variable that should be matched exactly, use re.escape() to escape it when substituting into the regexp.
s = '\n(941)\n364\nShackle\n(941)\nRivet\n105\nTop'
num = '(941)'
re.findall(rf'(?<=\n{re.escape(num)}\n)[\w\s\'\d\-\/\.]+(?=\n)', s)
This puts (941)\n in a lookbehind, so it's not included in the match. This avoids a problem with the \n at the end of one match overlapping with the \n at the beginning of the next.

Python regex match all sentences include either wordA or wordB [duplicate]

I'm creating a javascript regex to match queries in a search engine string. I am having a problem with alternation. I have the following regex:
.*baidu.com.*[/?].*wd{1}=
I want to be able to match strings that have the string 'word' or 'qw' in addition to 'wd', but everything I try is unsuccessful. I thought I would be able to do something like the following:
.*baidu.com.*[/?].*[wd|word|qw]{1}=
but it does not seem to work.

replace [wd|word|qw] with (wd|word|qw) or (?:wd|word|qw).
[] denotes character sets, () denotes logical groupings.

Your expression:
.*baidu.com.*[/?].*[wd|word|qw]{1}=
does need a few changes, including [wd|word|qw] to (wd|word|qw) and getting rid of the redundant {1}, like so:
.*baidu.com.*[/?].*(wd|word|qw)=
But you also need to understand that the first part of your expression (.*baidu.com.*[/?].*) will match baidu.com hello what spelling/handle????????? or hbaidu-com/ or even something like lkas----jhdf lkja$##!3hdsfbaidugcomlaksjhdf.[($?lakshf, because the dot (.) matches any character except newlines... to match a literal dot, you have to escape it with a backslash (like \.)
There are several approaches you could take to match things in a URL, but we could help you more if you tell us what you are trying to do or accomplish - perhaps regex is not the best solution or (EDIT) only part of the best solution?

Regex match only if word count between 1-50

So I have this code:
(r'\[quote\](.+?)\[/quote\]')
What I want to do is to change the regex so it only matches if the text within [quote] [/quote] is between 1-50 words.
Is there any easy way to do this?
Edit: Removed confusing html code in the regex example. I am NOT trying to match HTML.

Sure there is, depending on how you define a "word."
I would do so separately from regex, but if you want to use regex, you could probably do:
r"\[quote\](.+?\s){1,49}[/quote\]"
That will match between 2 and 50 words (since it demands a trailing \s, it can't match ONE)
Crud, that also won't match the LAST word, so let's do this instead:
r"\[quote\](.+?(?:\s.+?){1,49})\[/quote\]"

This is a definite misuse of regexes for a lot of reasons, not the least of which is the problem matching [X]HTML as #Hyperboreus noted, but if you really insist you could do something along the lines of ([a-zA-Z0-9]\s){1}{49}.
For the record, I don't recommend this.

string consists of punctuation

I want to check if string contains punctuation or not so a continuous sequence of exclamation, question & both.
By continuous, it means more than 2 times. Just like below,
#If sentence contains !!!
exc = re.compile(r"(.)\!{2}")
word["cont_exclamation"] = if exc.search(sent[i]) else not(found)
#If sentence contains ???
reg = re.compile(r"(.)\?{2}")
word["cont_question"] = if reg.search(sent[i]) else not(found)
But now I want to find both, exclamation and question marks so for example, hello??! or hey!! or dude!?!
Also, what if I want ? and ! both but more than 2 any of them.
I dont know regex properly so any help would be great.

Use the regex '[?!]{3,}' which means match the ? or ! characters 3 or more times (if continous = more than two times). Quoting is not needed inside character class.
Add more punctuation characters to the char class as needed

try re.compile(r"(.)[\?\!]{2}")

regex = re.compile(r"(.)(\?|\!){2}")
edit: Typing "regex tutorial" into google gives more info than you possibly need. This tutorial looks particularly well-balanced between conciseness and completeness.
Particularly (i.m.o.) useful tricks that are often not mentioned:
use +? and *? to switch from greedy to lazy match. I.e. match as few characters as possible instead of as much as possible. Example text: #ab# #de# --> #.*?# matches #ab# only (not #ab# #de#)
parentheses create a capture group by default. If you don't want this, you can use (?:...).
Most importantly, comment each regexp with a human-readable explanation. Future-you will be grateful. :-)