Java Regex String#replaceAll Alternative

Question

I\'ve been trying to devise a method of replacing multiple String#replaceAll calls with a Pattern/Matcher instance in the hopes that it would be faster than my current metho

Answer 1

This is a more dynamic version of previous answer to another similar question.

Here is a helper method for searching for any @keyword@ you want. They don't have to be 3 characters long.

private static String replace(String input, Map replacement) {
    StringJoiner regex = new StringJoiner("|", "@(", ")@");
    for (String keyword : replacement.keySet())
        regex.add(Pattern.quote(keyword));
    StringBuffer output = new StringBuffer();
    Matcher m = Pattern.compile(regex.toString()).matcher(input);
    while (m.find())
        m.appendReplacement(output, Matcher.quoteReplacement(replacement.get(m.group(1))));
    return m.appendTail(output).toString();
}

The above runs on Java 8+. In Java 9+, this can be done with lambda expression. The following also fixes the potential issue of a short keyword being a substring of a longer one, by sorting the keywords descending by length.

private static String replace(String input, Map replacement) {
    String regex = replacement.keySet().stream()
            .sorted(Comparator.comparingInt(String::length).reversed())
            .map(Pattern::quote).collect(Collectors.joining("|", "@(", ")@"));
    return Pattern.compile(regex).matcher(input)
            .replaceAll(m -> Matcher.quoteReplacement(replacement.get(m.group(1))));
}

Test

Map replacement = new HashMap<>();
replacement.put("bla", "hello,");
replacement.put("red", "world!");
replacement.put("Hold", "wait");
replacement.put("Better", "more");
replacement.put("a?b*c", "special regex characters");
replacement.put("foo @ bar", "with spaces and the @ boundary character work");

System.out.println(replace("@bla@This is a @red@line @bla@of text", replacement));
System.out.println(replace("But @Hold@, this can do @Better@!", replacement));
System.out.println(replace("It can even handle @a?b*c@ without dying", replacement));
System.out.println(replace("Keyword @foo @ bar@ too", replacement));

Output

hello,This is a world!line hello,of text
But wait, this can do more!
It can even handle special regex characters without dying
Keyword with spaces and the @ boundary character work too