Python -- Regex match pattern OR end of string

Question

import re
re.findall("(\+?1?[ -.]?\(?\d{3}\)?[ -.]?\d{3}[ -.]?\d{4})(?:[ <$])", "+1.222.222.2222<")

The above code works fine if my string ends with a "<" or space. But if it's the end of the string, it doesn't work. How do I get +1.222.222.2222 to return in this condition:

import re
re.findall("(\+?1?[ -.]?\(?\d{3}\)?[ -.]?\d{3}[ -.]?\d{4})(?:[ <$])", "+1.222.222.2222")

*I removed the "<" and just terminated the string. It returns none in this case. But I'd like it to return the full string -- +1.222.222.2222

POSSIBLE ANSWER:

import re
re.findall("(\+?1?[ -.]?\(?\d{3}\)?[ -.]?\d{3}[ -.]?\d{4})(?:[ <]|$)", "+1.222.222.2222")

If you found an answer you should post it in the Answers section, not in the question itself. — Barmar
– Barmar, Commented Aug 28, 2022 at 3:36

constantstranger · Accepted Answer · 2022-08-28 03:53:00Z

I think you've solved the end-of-string issue, but there are a couple of other potential issues with the pattern in your question:

the - in [ -.] either needs to be escaped as \- or placed in the first or last position within square brackets, e.g. [-. ] or [ .-]; if you search for [] in the docs here you'll find the relevant info:

Ranges of characters can be indicated by giving two characters and separating them 
by a '-', for example [a-z] will match any lowercase ASCII letter, [0-5][0-9] will match
all the two-digits numbers from 00 to 59, and [0-9A-Fa-f] will match any hexadecimal
digit. If - is escaped (e.g. [a\-z]) or if it’s placed as the first or last character
(e.g. [-a] or [a-]), it will match a literal '-'.

you may want to require that either matching parentheses or none are present around the first 3 of 10 digits using (?:$\d{3}$ ?|\d{3}[-. ]?)

Here's a possible tweak incorporating the above

import re
pat = "^((?:\+1[-. ]?|1[-. ]?)?(?:\(\d{3}\) ?|\d{3}[-. ]?)\d{3}[-. ]?\d{4})(?:[ <]|$)"
print( re.findall(pat, "+1.222.222.2222") )
print( re.findall(pat, "+1(222)222.2222") )
print( re.findall(pat, "+1(222.222.2222") )

Output:

['+1.222.222.2222']
['+1(222)222.2222']
[]

ItsLame · Accepted Answer · 2022-08-28 03:42:32Z

0

Maybe try:

import re
re.findall("(\+?1?[ -.]?\(?\d{3}\)?[ -.]?\d{3}[ -.]?\d{4})(?:| |<|$)", "+1.222.222.2222")

null matches any position, +1.222.222.2222
matches space character, +1.222.222.2222
< matches less-than sign character, +1.222.222.2222<
$ end of line, +1.222.222.2222

You can also use regex101 for easier debugging.

answered Aug 28, 2022 at 3:42

ItsLame

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