Validating Telephone Numbers

December 13, 2011

When I was a kid, telephones had rotary dials, not push buttons, and exchanges had names; my grandmother was in the Underhill 8 exchange. If you were calling someone in the same exchange as you were, you only had to dial the last four digits of the number. Long distance calling generally involved a human operator.

Modern American telephone numbers have ten digits, segmented as a three-digit area code, a three-digit exchange code, and a four-digit number. Within an area code, you need only dial (the verb hasn’t changed, even though telephones no longer have a dial) the seven-digit exchange code and number; otherwise, you must dial the complete ten-digit number, often with a prefix.

Our exercise today asks you to validate a telephone number, as if written on an input form. Telephone numbers can be written as ten digits, or with dashes, spaces, or dots between the three segments, or with the area code parenthesized; both the area code and any white space between segments are optional. Thus, all of the following are valid telephone numbers: 1234567890, 123-456-7890, 123.456.7890, (123)456-7890, (123) 456-7890 (note the white space following the area code), and 456-7890. The following are not valid telephone numbers: 123-45-6789, 123:4567890, and 123/456-7890.

Your task is to write a phone number validator that follows the rules given above; your function should either return a valid telephone number or an indication that the input is invalid. Be sure to write a proper test suite. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

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17 Responses to “Validating Telephone Numbers”

  1. Ajay said

    Your posts are extremely useful!
    But… why LISP

  2. programmingpraxis said

    @geirskjootskift: The task requires you to return the number if it is valid, not just a true or false.

    @Ajay: Thank you. It’s Scheme, not Lisp. And because I want to, and because Scheme gives me options not (easily) available in C or Java or Python; Scheme gives me things like garbage collection and big integers for free, and in what other language could I add list comprehensions and pattern matching to the standard library, and write generators and streams and other little goodies?

  3. kawas44 said

    Clojure solution using regexp like geirskjootskift,
    – use two regexp, deliberately permissive on numbers of seps
    – use regexp groups to extract a normalized phone number.

    (defn is-phone [phone]
      (let [phn-re [#"(?:(\d{3})[-. ]*)?(\d{3})[-. ]*(\d{4})"
                    #"\((\d{3})\)\s*(\d{3})[-. ]*(\d{4})"]
            mtchs (map #(re-matches %1 phone) phn-re)]
        (when-let [m (first (remove nil? mtchs))]
          (apply str (apply concat (next m))))))
    

    Some tests with valid and invalid entries

    (defn test-phone []
      (let [phones ["0123456789" "012 345 6789" "012.345.6789" "012-345-6789"
                    "3456789" "345 6789" "345.6789" "345-6789"
                    "(012)3456789" "(012) 3456789"
                    "(012)345 6789" "(012) 345 6789"
                    "(012)345.6789" "(012) 345.6789"
                    "(012)345-6789" "(012) 345-6789"
                    ;; invalid phones
                    "12-345-6789" "123-45-6789" "123-456-789"
                    "123-45-67890" "123:4567890" "123/456-7890" "123456"
                    "12345678" "123456789" "12345678901" "123-456-7890 x123"]]
        (doseq [phn phones]
          (if-let [m (is-phone phn)]
            (println phn "... ok \t" m)
            (println phn "... KO !!")))))
    
    (test-phone)
    
  4. tomjleo said

    Here is my solution (also using Regex…)

    import re
    import itertools

    def isTelephone(n):
    regex=re.compile(“([0-9]{10})|([0-9]{3}[-]{1}[0-9]{3}[-]{1}[0-9]{4})|([0-9]{3}[.]{1}[0-9]{3}[.]{1}[0-9]{4})|([\(]{1}[0-9]{3}[\)][ ]{1}[0-9]{3}[-]{1}[0-9]{4})|([0-9]{3}[-]{1}[0-9]{4})”)
    chain = itertools.chain(*regex.findall(n))
    if n in list(chain):
    return n
    else:
    return False

    def test():
    test_numbers=[‘1234567890′,
    ‘123-456-7890′,
    ‘123.456.7890’,
    ‘(123) 456-7890′,
    ‘456-7890′,
    ‘123-45-6789′,
    ‘123 4567890′,
    ‘123/456-7890′]
    for num in test_numbers:
    print(isTelephone(num))

    test()

  5. Tom said

    Here is my solution using python and regex

    import re
    import itertools
    
    def isTelephone(n):
        regex=re.compile("([0-9]{10})|([0-9]{3}[-]{1}[0-9]{3}[-]{1}[0-9]{4})|([0-9]{3}[.]{1}[0-9]{3}[.]{1}[0-9]{4})|([\(]{1}[0-9]{3}[\)][ ]{1}[0-9]{3}[-]{1}[0-9]{4})|([0-9]{3}[-]{1}[0-9]{4})")
        chain = itertools.chain(*regex.findall(n)) 
        if n in list(chain):
            return n
        else:
            return False
        
    def test():
        test_numbers=['1234567890',
                      '123-456-7890',
                      '123.456.7890',
                      '(123) 456-7890',
                      '456-7890',
                      '123-45-6789',
                      '123 4567890',
                      '123/456-7890']
        for num in test_numbers:
            print(isTelephone(num))
            
    test()
    
    
  6. Ok.. returned on normalized form. Still don’t see no reason to implement a parser “by hand” :-) I also prefer not putting all patterns in as few regexps as possible. Readability for the win.

    VALID_PHONE_PATTERNS = [
    "^(\d{10})$",
    "^(\d{3})\-(\d{3})\-(\d{4})$",
    "^(\d{3})\.(\d{3})\.(\d{4})$",
    "^\((\d{3})\)\s?(\d{3})\-(\d{4})$",
    "^(\d{3})\-(\d{4})$"]

    def validate_phone_number(phn):

    for pattern in VALID_PHONE_PATTERNS:
    res = compile(pattern).findall(phn)
    if res:
    return "".join(res[0])
    return None

  7. markmain said

    Your rules state that 123/456-7890 is not valid; while it is the least common use, it is frequently used by people. If I were writing code I would use it.

  8. Mike said

    I would normally use a list of regex’s like some of the previous solutions. I used one big regular expression; however, because having two problems is better than having just one problem (see http://regex.info/blog/2006-09-15/247).

    normalize() returns a tuple with the areacode, exchange, and number, or raises a ValueError if the input is not a valid phone number. The areacode is None if one was not provided.

    import re
    
    pn_re = re.compile(r"""(?x)		# verbose mode
    \A					# start of string
    \s*
    (?:
       (?P<lparen>\()?			# optional left paren
       \s*
       (?P<areacode>\d{3})			# areacode is 3 digits
       \s*
       (?P<rparen>(?(lparen)		# if there was a lparen
                     \)			# then match an rparen
                    |(?P<sep1>[-.]?))       # else match an optional separator
       )
       \s*
    )? 					# areacode section is optional
    (?P<exchange>\d{3})			# exchange is three digits
    \s*
    (?P<sep2>(?(sep1)(?P=sep1)|[-.]?))	# if there was a first separator, the
    					# second one must match it
    \s*
    (?P<number>\d{4})			# number is 4 digits
    \s*
    \Z 					# end of string
    """)
    
    def normalize(phone_number):
        mo = pn_re.match(phone_number)
        if mo:
            return mo.group('areacode','exchange','number')
        
        else:
            raise ValueError('Improperly formatted phone number: ' + phone_number )
    
    
    def test():
        valid = ["1234567890", "123-456-7890", "123.456.7890",
                 "(123)456-7890", "(123) 456-7890", "987-6543",
                 "  123 - 456 - 7890  "]
        
        invalid = ["12-345-6789", "123-45-6789", "123-456-789",
              "123-45-67890", "123:4567890", "123/456-7890", "123456",
              "12345678", "123456789", "12345678901", "123-456-7890 x123",
              "123--456--7890", "(123)-456-7890", "12 3 -456- 789 0",
              "123 . 456 - 7890"]
    
        for number in valid + invalid:
            try:
                normal = normalize(number)
    
                assert number in valid, "{} not in valid set".format(number)
                if normal[0] is not None:
                    assert normal == ('123','456','7890'), (number, normal)
                else:
                    assert normal == (None, '987', '6543'), (number, normal)
    
            except ValueError:
                assert number in invalid
    
    
    

    @praxis, I don’t know Scheme, but was wondering how your code would handle the last 4 invalid test cases I added?

  9. Yuushi said
    import re
    
    no_bracket = re.compile('^([0-9]{3})?[-\. ]*[0-9]{3}[-\.]*[0-9]{4}$')
    bracket = re.compile('^\([0-9]{3}\)[-\. ]*[0-9]{3}[-\.]*[0-9]{4}$')
    
    def tel_no_match(number):
        if number[0] == '(':
            z = bracket.match(number)
        else: z = no_bracket.match(number)
    
        if z and z.start() == 0 and z.end() == len(number):
            return number
        return None
    

    Figured it was easier to break it into 2 options, as flagging the case (123-456-7890 is a bit annoying. Also assumes you can mix and match – and ., so 123-456.7890 is ok.

  10. […] is a solution to a programmingpraxis exercise -validating telephone […]

  11. OK, initially missed a couple of requirements, so this is iteration 3 of my solution – over time and over budget ;-) But at least you get the unit tests.

    import unittest
    import re
    
    tele_re = re.compile(r"((\([0-9]{3}\))|([0-9]{3}))?[- \.]?[0-9]{3}[- \.]?[0-9]{4}")
    
    def is_tele(s):
        return tele_re.match(s)
    
    def tele(s):
        if is_tele(s):
            return "".join([c for c in s if '0' <= c <= '9' ])
    
    class TestTelephoneNumbers(unittest.TestCase):
        def setUp(self):
            self.good = [ "1234567890", "123-456-7890", "123.456.7890", "(123)456-7890", "(123) 456-7890", "456-7890" ]
            self.bad = [ "123-45-6789", "123:4567890", "123/456-7890", "(123 456 7890", "123)456 7890", "(456)7890" ]
        def test_good(self):
            for tele_num in self.good:
                self.assertTrue(is_tele(tele_num), tele_num)
        def test_bad(self):
            for tele_num in self.bad:
                self.assertFalse(is_tele(tele_num), tele_num)
        def test_extract(self):
            for tele_num in self.good[0:-1]:
                self.assertEqual(tele(tele_num), self.good[0], msg="%s extract failed" % tele_num)
            self.assertEqual(tele(self.good[-1]), "4567890", msg="%s extract failed" % self.good[-1])
           
    
    if __name__ == '__main__':
        unittest.main()
    
    
  12. One more thing – I’ve just read the suggested solution and liked the trick with working backwards.

  13. Mike said

    My regex-based version above was clearly a tounge-in-cheek solution. Here is a more useful solution.

    It uses a regular expression to group the input string into digits and non-digits. The non-digits are normalized to form a punctuation string, which is compared with a set of valid punctuation strings. Other valid punctuation, e.g., 123/456-7890, can easily be added by adding the punctuation string, e.g., ” /-“, to validpunctuation.

    import re
    
    pattern = re.compile(r"""(?x)     # verbose mode
        \A                            # match beginning of string 
        (\D*)                         #     leading whitespace and punctuation
        (\d{3})?                      # optional areacode
        (\D*)                         #     whitespace and punctuation
        (\d{3})                       # 3-digit exchange
        (\D*)                         #     whitespace and punctuation
        (\d{4})                       # 4-digit number
        \s*                           #     trailing whitespace
        \Z                            # match end of string
        """
        )
    
    validpunctuation = {
        "   ", "()-", "().", " --", " ..",  # with areacode
        "  ", " -", " ."                    # and without
        }
    
    def validate(phonenumber):
        mo = pattern.match(phonenumber)
        
        if mo:
            punctuation = mo.group(1,3,5)
            number = mo.group(2,4,6)
    
            if not mo.group(2):
                punctuation = punctuation[1:]
                number = number[1:]
    
            if ''.join(s.strip() or ' ' for s in punctuation) in validpunctuation:
                return '.'.join(number)
    
        return None
    
    
  14. Diego Giuliani said

    Here is my solution using python:
    Note that I’m still a n00b in Phython so I know there are some semantic bugs, e.g. phones like “(123-456 7890″ or “123)456 7890″ will match

    import re
    
    def match(str):
        pattern = """
            ^           #Begin of the string
            [\(]?       # Possible open parenthesis
            (\d{3})?  # First 3 digits - Optional
            [\)]?       # Possible open parenthesis
            [-|.| ]?    # Possible separators
            (\d{3})    # Second group of 3 digits
            [-|.| ]?    # Possible separators
            (\d{4})    # Last group of digits
            $
        """
        return re.search(pattern,str,re.VERBOSE)
    
    def test():
        phones = [
        "1234567890",
        "123-456-7890",
        "123.456.7890",
        "(123)456-7890",
        "(123) 456-7890",
        "456-7890",
        "123-45-6789",
        "123:4567890",
        "123/456-7890"
        ]
        for phone in phones:
            print match(phone) != None
    
    
  15. Tom said

    @Diego Giuliani you should avoid using str as a variable because str is actually a built-in function see here -> http://docs.python.org/library/functions.html#str

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