Question

Java’s string split(regex) function splits at all instances of the regex. Python’s partition function only splits at the first instance of the given separator, and returns a tuple of {left,separator,right}.

How do I achieve what partition does in Java?

e.g.

"foo bar hello world".partition(" ")

should become

"foo", " ", "bar hello world"

• Is there an external library which
  provides this utility already?

• how would I achieve it without
  an external library?

• And can it be achieved without an external library and without Regex?

NB. I’m not looking for split(” “,2) as it doesn’t return the separator character.

Answer 1

The String.split(String regex, int limit) is close to what you want. From the documentation:

The limit parameter controls the number of times the pattern is applied and therefore affects the length of the resulting array.

• If the limit n is greater than zero then the pattern will be applied at most n - 1 times, the array’s length will be no greater than n, and the array’s last entry will contain all input beyond the last matched delimiter.
• If n is non-positive then the pattern will be applied as many times as possible and the array can have any length.
  • If n is zero then the pattern will be applied as many times as possible, the array can have any length, and trailing empty strings will be discarded.

Here’s an example to show these differences (as seen on ideone.com):

static void dump(String[] ss) {
    for (String s: ss) {
        System.out.print("[" + s + "]");
    }
    System.out.println();
}
public static void main(String[] args) {
    String text = "a-b-c-d---";

    dump(text.split("-"));
    // prints "[a][b][c][d]"

    dump(text.split("-", 2));
    // prints "[a][b-c-d---]"

    dump(text.split("-", -1));
    // [a][b][c][d][][][]
    
}

---

A partition that keeps the delimiter

If you need a similar functionality to the partition, and you also want to get the delimiter string that was matched by an arbitrary pattern, you can use Matcher, then taking substring at appropriate indices.

Here’s an example (as seen on ideone.com):

static String[] partition(String s, String regex) {
    Matcher m = Pattern.compile(regex).matcher(s);
    if (m.find()) {
        return new String[] {
            s.substring(0, m.start()),
            m.group(),
            s.substring(m.end()),
        };
    } else {
        throw new NoSuchElementException("Can't partition!");
    }
}
public static void main(String[] args) {
    dump(partition("james007bond111", "\\d+"));
    // prints "[james][007][bond111]"
}

The regex \d+ of course is any digit character (\d) repeated one-or-more times (+).