Ignore empty lines [\t\s] spaces or tabs

So I have this regex:

And I am basically, trying to ignore all the line numbers followed by space or nothing. My current regex: ^[\d\s].+(?:[A-Z\s]*)*$

The line numbers followed by nothing are actually not ignored.

Solution

You might use a negative lookahead to assert that what follows is not 1+ digits followed by 0+ times a whitespace character:

^(?!\d+\s*$)\d+.+$

^ Start of the string
(?!\d+\s*$) Negative lookahead to assert what is on the right is not 1+ digits followed by 0+ times a whitespace character and the end of the string
\d+.+ Match 1+ times a digit and 1+ times any character
$ End of the string

See the regex demo | Python demo

Example using findall:

import re
regex = r"^(?!\d+\s*$)\d+.+$"
test_str = ("Here goes some text. {tag} A wonderful day. It's soon cristmas.\n"
    "2 Happy 2019, soon. {Some useful tag!} Something else goes here.\n"
    "3 Happy ending. Yeppe! See you.\n"
    "4\n"
    "5 Happy KKK!\n"
    "6 Happy B-Day!\n"
    "7\n"
    "8 Universe is cool!\n"
    "9\n"
    "10 {Tagish}.\n"
    "11\n"
    "12 {Slugish}. Here goes another line. {Slugish} since this is a new sentence.\n"
    "13\n"
    "14 endline.")
print(re.findall(regex, test_str, re.MULTILINE));

When there is a dot after the digit, you could use:

^(?!\d+\.\s*$)\d+.+$