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Remove n characters after certain character

I have an string that looks something like this:
*45hello I'm a string *2jwith some *plweird things
I need to remove all the * and the 2 chars that follow those * to get this:
hello I'm a string with some weird things
Is there a practical way to do it without iterating over the string?
Thanks!

Using regular expression:
import re
s = "*45hello I'm a string *2jwith some *plweird things"
s = re.sub(r'\*..', '', s)

You can use regex:
import re
regex = r"\*(.{2})"
test_str = "*45hello I'm a string *2jwith some *plweird things"
# You can manually specify the number of replacements by changing the 4th argument
result = re.sub(regex, '', test_str, 0)

Python Regex: Remove optional characters

I have a regex pattern with optional characters however at the output I want to remove those optional characters. Example:
string = 'a2017a12a'
pattern = re.compile("((20[0-9]{2})(.?)(0[1-9]|1[0-2]))")
result = pattern.search(string)
print(result)
I can have a match like this but what I want as an output is:
desired output = '201712'
Thank you.

You've already captured the intended data in groups and now you can use re.sub to replace the whole match with just contents of group1 and group2.
Try your modified Python code,
import re
string = 'a2017a12a'
pattern = re.compile(".*(20[0-9]{2}).?(0[1-9]|1[0-2]).*")
result = re.sub(pattern, r'\1\2', string)
print(result)
Notice, how I've added .* around the pattern, so any of the extra characters around your data is matched and gets removed. Also, removed extra parenthesis that were not needed. This will also work with strings where you may have other digits surrounding that text like this hello123 a2017a12a some other 99 numbers
Output,
201712
Regex Demo

You can just use re.sub with the pattern \D (=not a number):
>>> import re
>>> string = 'a2017a12a'
>>> re.sub(r'\D', '', string)
'201712'

Try this one:
import re
string = 'a2017a12a'
pattern = re.findall("(\d+)", string) # this regex will capture only digit
print("".join(p for p in pattern)) # combine all digits
Output:
201712

If you want to remove all character from string then you can do this
import re
string = 'a2017a12a'
re.sub('[A-Za-z]+','',string)
Output:
'201712'

You can use re module method to get required output, like:
import re
#method 1
string = 'a2017a12a'
print (re.sub(r'\D', '', string))
#method 2
pattern = re.findall("(\d+)", string)
print("".join(p for p in pattern))
You can also refer below doc for further knowledge.
https://docs.python.org/3/library/re.html

How to add tag for numbers which in brackets using python regex?

The default strings is:
strings123[abc123def456]strings456
Add tag for number:
strings[abc<span>123</span>def<span>456</span>]strings

Search by this:
(\d+)
and replace by :
<span>\1</span>
Regex Demo
Sample Source:
import re
regex = r"(\d+)"
test_str = "strings[abc123def456]strings"
subst = "<span>\\1</span>"
result = re.sub(regex, subst, test_str, 0)
if result:
print (result)

Remove just the alphabet characters from a string

In python,
I have string like
"dsafsadf_afasa_2.2.14_43.33_dsfd"
I need to get just
"2.2.14_43.33"
How do I do it?

Seems like you're trying to remove all alphabets and all the underscores except if the undercore is present inbetween digits,.
>>> s = "dsafsadf_afasa_2.2.14_43.33_dsfd"
>>> re.sub(r'[a-z]|(?<=\D)_(?=\d)|(?<=\d)_(?=\D)|(?<=\D)_(?=\D)|^_+|_+$', '', s)
'2.2.14_43.33'

You can use str.translate if you just want to remove the letters:
s = "dsafsadf_afasa_2.2.14_43.33_dsfd"
from string import ascii_letters
print(s.translate(None,ascii_letters))
which outputs:
__2.2.14_43.33_
For python3:
from string import ascii_letters
print(s.translate({ord(ch):"" for ch in ascii_letters}))
If you really want to remove underscores from the end use strip:
s = "dsafsadf_afasa_2.2.14_43.33_dsfd"
from string import ascii_letters
print(s.translate(None,ascii_letters).strip("_"))
Output:
2.2.14_43.33

You can simply do a re.findall.
import re
p = re.compile(r'\d+(?:[\W_]\d+)*')
test_str = "dsafsadf_afasa_2.2.14_43.33_dsfd"
re.findall(p, test_str)
See demo.
https://regex101.com/r/hF1wE3/2

python regex sub without order

I have following string "3 0ABC, mNone\n" and I want to remove m None and \n. The catch is that 'm', \n and None can be anywhere in the string in any order. I would appreciate any help.
I can do re.sub('[\nm,]','',string) or re.sub('None','',string) but don't know how to combine specially when the order doesn't matter.

If you want to remove m, None and \n you can use them as pattern together in a group. So you can use this regex:
(m|\\n|None)
Working demo
If you use the following code:
import re
p = re.compile(ur'(m|\\n|None)')
test_str = u"3 0ABC, mNone\n"
subst = u""
result = re.sub(p, subst, test_str)
print result
// Will show:
'3 0ABC, '