Regex: Remove the letters with length 1-3 which are before the dot

If I have an input something like this

input="AB. Hello word."
the output should be 
output="Hello word."

Another example is

input="AB′. Hello word"
output = Hello Word

I want to produce a code which is generalized for any group of letter in any language. This is my code

text="A. Hello word."
text = re.sub(r'A. w{1,2}.*', '', text)
text

output = llo word.

So I can change ‘A’ with any other letter, but for some reason isn’t working well.

I tried also this one

text="Ab. Hello word."
text = re.sub(r'A+. w{1,2}.*', '', text)
text
output = Ab. Hello word.

but isn’t working as well.

Don’t use a regex for this, just .split() on it, you can just split once and take the last half [-1]

>>> "Ab. Hello world.".split(".", 1)[-1].strip()
'Hello world.'
>>> "Hello world".split(".", 1)[-1]
'Hello world'

You may use this regex for matching:

b[A-Za-z]{1,3}′?.

Replace it with "".

RegEx Demo

RegEx Details:

• b: Word boundary
• [A-Za-z]{1,3}: Match 1 to 3 letters
• ′?: Match an optional ′
• .: Match a dot

This solution may be useful:

a = "AB. Hello word."
print(a[a.find(".")+1:])

Try this:

import re

regex = r"^[^.]{1,3}.s*"

test_str = ("AB. Hello word.n"
    "AB′. Hello word.n"
    "A. Hello word.n"
    "Ab. Hello word.n")

subst = ""

# You can manually specify the number of replacements by changing the 4th argument
result = re.sub(regex, subst, test_str, 0, re.MULTILINE)

if result:
    print (result)

Output:

Hello word.
Hello word.
Hello word.
Hello word.

regex101

Rextester

Use this generic regex pattern:

“^.{0,}.”

Expalination:

^ Finds a match as the beginning of a string.
.{0,} Matches any string that contains a sequence of zero or more characters.
. ending with a . (dot)