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")

Asked By: Gary

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Source

Answer 1

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']
[]

Answered By: constantstranger

Answer 2

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 By: ItsLame

Python — Regex match pattern OR end of string

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