Regular expression to extract monetary values from an invoice text

Question

I have a regular expression: \b[-+]?(?:\d{1,3}\.)(?:\d{3}\.)*(?:\d*) (Python) that matches numeric values in strings like this:

Amount: 12.234.55222 EUR
Some Text 123.222.22 maybe more text
1.245.455.2
22.34565 Could be at the beginning
It could be at the end of a string 21.1
221. It could be a number like this too (for US invoices, I saw a lot of different stuff)

Which is what I want. But it also matches the first 2 parts of a date like this: 08.05.2023

I know this is happening because of the first and the last group, but I don't know how to prevent that. I only want to match values that stand by themselves.
Can somebody point me in the right direction?

Edit: I forgot to mention that I've tried it with a negative look behind, but that didn't work:

\b([-+]?(?:\d{1,3}\.)(?:\d{3}\.)*(?:\d*))(?!(?:\.d{4}))

Maybe I'm doing the look behind wrong?

Answer 1

With the idea of @user24714692 I've done it this way:

def float_check(text: str) -> bool:
    try:
        float(text)
        return True
    except ValueError:
            return False
        
def match_amount(amount: float, text: str) -> bool:
    pattern = r"^([-+]?(?:\d{1,3}\.)(?:\d{3}\.)*(?:\d*))$"
    # find all numbers, that match a pattern like i.e.: (123./,)*23

    # replace all comma, because the US is doing the numbers like this 291,234.23
    text = text.replace(",", ".")

    numbers = [
        pot_number for pot_number in text.split() if re.findall(pattern, pot_number)
    ]

    for number in numbers:
        # now replace all dots but the last one
        number = re.sub(r"\.(?=.*\.)", "", number)
        # and convert the numbers to float, and check if the number equals the given amount
        if float_check(number) and float(number) == amount:
            return True
    return False
    
print(match_amount(23.45, "some text with numbers here"))

Works pretty fine now, thank you very much.