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Programming-Idioms

  • Perl
  • C#

Idiom #221 Remove all non-digits characters

Create string t from string s, keeping only digit characters 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

my $t = $s;
$t =~ s/\D+//g;
($t = $s) =~ tr/0-9//cd;

tr/// translates characters, in this case the complement of 0 through 9 as dictated by the c modifier. The d modifier deletes found but unmatched characters, which will be everything except digits. More efficient than using a regex.
using System.Linq;
var t = string.Concat(s.Where(c => char.IsDigit(c)));
(require '[clojure.string :as str])
(let [s "1a22b3c4de5f6"
      t (str/replace s #"[^\d]" "")]
  (println t))

Uses a regex to get only the digit characters.
New implementation...
< > ⌨
programming-idioms.org 
my $t = $s;
$t =~ s/\D+//g;
($t = $s) =~ tr/0-9//cd;
var t = string.Concat(s.Where(c => char.IsDigit(c)));
(let [s "1a22b3c4de5f6"
      t (str/replace s #"[^\d]" "")]
  (println t))
var t = String.fromCharCodes(
	s.codeUnits.where((x) => (x ^ 0x30) <= 9) );
re := regexp.MustCompile("[^\\d]")
t := re.ReplaceAllLiteralString(s, "")
t = filter (`elem` ['0'..'9']) s
let t = s.replaceAll(/\D/g, '')
t = s.replace(/[^\d]/gm,"");
String t = "";
for (char c : s.toCharArray())
    if (isDigit(c)) t = t + c;
String t = s.chars()
    .filter(Character::isDigit)
    .mapToObj(x -> valueOf((char) x))
    .collect(joining());
String t = "";
char c;
int i, n = s.length();
for (i = 0; i < n; ++i)
    if (isDigit(c = s.charAt(i)))
        t = t + c;
StringBuilder tb = new StringBuilder();
for (int i=0; i<s.length(); i++){
    if(Character.isDigit(s.charAt(i)))
        tb.append(s.charAt(i));
}
String t = tb.toString();
String t = s.replaceAll("\\D+", "");
(let ((*t* (remove-if-not #'digit-char-p *s*)))
	(format t "~A~%" *t*))
t := ReplaceRegExpr('[^\d]', s, ''); 
for i := 1 to length(s) do
    if s[i] in ['0'..'9'] then
        t := t + s[i];
t = re.sub(r"\D", "", s)
t = ''.join(x for x in s if x in digits)
t = s.delete("^0-9")
let t: String = s.chars().filter(|c| c.is_digit(10)).collect();