Replace number with repeated characters [closed] - regex

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Say I have a string: 8, and I wanted to replace numbers [0-9] with a . that would repeat as many times as the number. What regex string would I use?
e.g.: input string
8
output
........

Answer 1

You can't do this using only regular expressions so you'll have to rely on a language or library feature which allows substitution of matched strings with the result of calling a function with the match as an argument.
In Ruby:
"8".gsub(/[0-9]/) { |x| '.' * x.to_i } # => "........"
"812".gsub(/[0-9]/) { |x| '.' * x.to_i } # => "..........."
"a1b2".gsub(/[0-9]/) { |x| '.' * x.to_i } # => "a.b.."
In JavaScript:
function replaceNumbersWithDots(str) {
return (''+str).replace(/[0-9]/g, function(m) {
var s='', num=parseInt(m, 10);
for (i=0; i<num; i++) { s+= '.'; }
return s;
});
}
replaceNumbersWithDots('8'); // => "........"
replaceNumbersWithDots('812'); // => ".........."
replaceNumbersWithDots('a1b2'); // => "a.b.."
In Java:
public static void main(String args[]) throws Exception {
System.out.println(replaceNumbersWithDots("8")); // => "........"
System.out.println(replaceNumbersWithDots("812")); // => "..........."
System.out.println(replaceNumbersWithDots("a1b2")); // => "a.b.."
}
public static String replaceNumbersWithDots(String s) {
Pattern pattern = Pattern.compile("[0-9]");
Matcher matcher = pattern.matcher(s);
StringBuffer buf = new StringBuffer();
while (matcher.find()) {
int x = Integer.parseInt(matcher.group());
matcher.appendReplacement(buf, stringOfDots(x));
}
return buf.toString();
}
public static String stringOfDots(int x) {
String s = "";
for (int i=0; i<x; i++) { s += "."; }
return s;
}

Answer 2

This can't be done with standard regular expressions alone. As #m.buettner points out, you can in certain languages specify a function that processes the replacement. For example, with Python
>>> import re
>>> s = '8'
>>> re.sub(r'\d', lambda m: '.'*int(m.group()), s)
'........'
But maybe you don't even need a regex? Since you're only looking for single-character matches (namely \d), you can perhaps do something like this:
Initialize some sort of string buffer to hold the resultant string
Loop over the characters of your input string
If the character is a digit, parse it as an integer and append that many .s to your buffer.
Otherwise, append the character itself to your buffer.
Here's an implementation of this in Java:
String s = "8";
StringBuilder sb = new StringBuilder();
for (char c : s.toCharArray()) {
if (Character.isDigit(c)) {
int n = c - '0';
for (int i = 0; i < n; i++)
sb.append('.');
} else {
sb.append(c);
}
}
System.out.println(sb);
........

Answer 3

AS you didn't provide a lang, i made an exemple of solving it in php
$tr = array();
foreach(range(0, 9) as $nr)
{
$tr[$nr] = str_repeat('.', $nr);
}
echo strtr("Hello 8", $tr);
// Gives: "Hello ........"