Extract all numbers (int and floats) after specific word

Question

Assuming I have the following string:

str = """
         HELLO 1 Stop #$**& 5.02‼️ 16.1 
         regex

         5 ,#2.3222
      """

I want to export all numbers , Whether int or float after the word "stop" with no case sensitive . so the expected results will be :

[5.02, 16.1, 5, 2.3222]

The farthest I have come so far is by using PyPi regex from other post here:

regex.compile(r'(?<=stop.*)\d+(?:\.\d+)?', regex.I)

but this expression gives me only [5.02, 16.1]

Answer 1

Yet another one, albeit with the newer regex module:

(?:\G(?!\A)|Stop)\D+\K\d+(?:\.\d+)?

See a demo on regex101.com.

---

In Python, this could be

import regex as re

string = """
         HELLO 1 Stop #$**& 5.02‼️ 16.1 
         regex

         5 ,#2.3222
      """

pattern = re.compile(r'(?:\G(?!\A)|Stop)\D+\K\d+(?:\.\d+)?')

numbers = pattern.findall(string)
print(numbers)

And would yield

['5.02', '16.1', '5', '2.3222']

Don't name your variables after inbuilt-functions, like str, list, dict and the like.

---

If you need to go further and limit your search within some bounds (e.g. all numbers between Stop and end), you could as well use

(?:\G(?!\A)|Stop)(?:(?!end)\D)+\K\d+(?:\.\d+)?
#           ^^^        ^^^

See another demo on regex101.com.