匹配电话号码，正则表达式

Question

I've phone numbers in this format:

 some_text   phone_number some_text
 some_text   (888) 501-7526 some_text

Which is a more pythonic way way to search for the phone numbers

(\(\d\d\d\) \d\d\d-\d\d\d\d)

(\([0-9]+\) [0-9]+-[0-9]+)

or there is a simpler expresion to do this?

Answer 1

I think you are looking for something like this:

(\(\d{3}\) \d{3}-\d{4})

From the Python docs :

{m}

Specifies that exactly m copies of the previous RE should be matched; fewer matches cause the entire RE not to match. For example, a{6} will match exactly six 'a' characters, but not five.

(\\(\\d\\d\\d\\) \\d\\d\\d-\\d\\d\\d\\d) would also work, but, as you said in your question, is rather repetitive. Your other suggested pattern, (\\([0-9]+\\) [0-9]+-[0-9]+) , gives false positives on input such as (1) 2-3 .

Answer 2

Using (\\(\\d{3}\\)\\s*\\d{3}-\\d{4})

>>> import re
>>> s = "some_text   (888) 501-7526 some_text"
>>> pat = re.compile(r'(\(\d{3}\)\s*\d{3}-\d{4})')
>>> pat.search(s).group() 
'(888) 501-7526'

Demo

Explanation:

• (\\(\\d{3}\\)\\s*\\d{3}-\\d{4})/
  • 1st Capturing group (\\(\\d{3}\\)\\s*\\d{3}-\\d{4})
    • \\( matches the character ( literally
    • \\d{3} match a digit [0-9]
      • Quantifier: {3} Exactly 3 times
    • \\) matches the character ) literally
    • \\s* match any white space character [\\r\\n\\t\\f ]
      • Quantifier: * Between zero and unlimited times, as many times as possible, giving back as needed [greedy]
    • \\d{3} match a digit [0-9] Quantifier: {3} Exactly 3 times
    • - matches the character - literally
    • \\d{4} match a digit [0-9] Quantifier: {4} Exactly 4 times

Answer 3

I think the second one would be the more pythonic way. The one above isn't that easy to read, but regular expressions aren't that intuitive at all.

(\\([0-9]+\\) [0-9]+-[0-9]+) will do it, if the lenght of the phone number is not specified. If the length is always the same, you can use (\\([0-9]{3}\\) [0-9]{3}-[0-9]{4}) or (\\(\\d{3}\\) \\d{3}-\\d{4}) .