How to split String for a digit and letters in java

Question

Test data are e.g.

1a, 12a, 1ab, 12ab, 123a, 123abc

so if as input we have:

String input = "1a";

The output will be

String number = "1";
String letter = "a";

Like you can notice in this String there are sometimes 1-3digits(0-9) and sometimes 1-3letters(A-Z).

My first attempt:

I tried to use .substring()

But it will work only if there would've been for example always the same amount of digits or letters

My second attempt was:

.split(" ");

But it will work only if there will be a space or any other sign between.

PS. Thanks for a response in answers. I checked most of your answers and they all work. The question now which one is the best?

Answer 1

A simple solution without regular expressions: Find the index of the first Letter and split the string at this position.

private String[] splitString(String s) {
  // returns an OptionalInt with the value of the index of the first Letter
  OptionalInt firstLetterIndex = IntStream.range(0, s.length())
    .filter(i -> Character.isLetter(s.charAt(i)))
    .findFirst();

  // Default if there is no letter, only numbers
  String numbers = s;
  String letters = "";
  // if there are letters, split the string at the first letter
  if(firstLetterIndex.isPresent()) {
    numbers = s.substring(0, firstLetterIndex.getAsInt());
    letters = s.substring(firstLetterIndex.getAsInt());
  }

  return new String[] {numbers, letters};
}

Gives you:

splitString("123abc") 
returns ["123", "abc"]

splitString("123") 
returns ["123", ""]

splitString("abc") 
returns ["", "abc"]

Answer 2

If your string sequence starts with digits and ends with letters, then the below code will work.


int asciRepresentation, startCharIndex = -1;
    for(int i = 0; i < str.length(); i++) {
        asciRepresentation = (int) str.charAt(i);
        if (asciRepresentation > 47 && asciRepresentation < 58)
            strB.append(str.charAt(i));
        else {
            startCharIndex = i;
            break;
        }
    }
    System.out.println(strB.toString());
    if (startCharIndex != -1)
        System.out.println(str.substring(startCharIndex, str.length()));

Answer 3

You can use regex:

String str = "1a, 12a, 1ab, 12ab, 123a, 123abc";
Pattern p = Pattern.compile("(?<digit>\\d{1,3})(?<letter>[a-z]{1,3})");
Matcher m = p.matcher(str);

while (m.find()){
    System.out.println(m.group("digit")+"/"+m.group("letter"));
}
// Ouput:
// 1/a
// 12/a
// 1/ab...

Answer 4

Below you have my proposal. Works correctly for mentioned test data

(1a, 12a, 1ab, 12ab, 123a, 123abc)

Solution:

public ArrayList<String> split(String text) {

Pattern pattern = Pattern.compile("(\\d+)([a-zA-Z]+)");
Matcher matcher = pattern.matcher(text);
ArrayList<String> result = new ArrayList<>();

if (matcher.find() && matcher.groupCount() == 2) {
  result.add(matcher.group(1));
  result.add(matcher.group(2));
}
return result;
}

Answer 5

Solution:

(also take a look at the edit that I made at the end of my answer)

"\\b(\\d{1,3})([a-z]{1,3})(?=,*|\\b)"

Example:

String s = "1a, 12a, 1ab, 12ab, 123a, 123abc";
Pattern p = Pattern.compile("\\b(\\d{1,3})([a-z]{1,3})(?=,*|\\b)");
Matcher m = p.matcher(s);
while(m.find()) {
    System.out.println("Group: "+ m.group() + ", letters: " + m.group(1) + ", digits: " + m.group(2));
}

Output that you get:

Group: 1a, letters: 1, digits: a
Group: 12a, letters: 12, digits: a
Group: 1ab, letters: 1, digits: ab
Group: 12ab, letters: 12, digits: ab
Group: 123a, letters: 123, digits: a
Group: 123abc, letters: 123, digits: abc

Explanation:

\\b(\\d{1,3})([a-z]{1,3})(?=,*|\\b) whole regex

\\b - word boundary

\\d{1,3} - digit, from one to three times

[a-z]{1,3} - characters from a to z from one to three times

(?=,*|\\b) - this is positive lookahead, you say that after these letters you want to be present , or word boundary, but you don't want them to be present in the matching group (called with m.group())

() - matching groups are in parenthesis - in my regex I used two matching groups: #1: (\\d{1,3}) #2: ([a-z]{1,3}) (they are printed with m.group(1) and m.group(2))

If you're not very familiar to regular expressions syntax yet, you might want to have a look at Java API Documentation of class Pattern. There is a list of available uses of regular expressions. It's worth giving regular expressions a try, as it might save a lot of your time when working with Strings in the future.

---

Edit:

Actually this regex can be changed to:

(?<=\\b)(\\d{1,3})([a-z]{1,3})(?=\\b)

There is a positive lookbehind (?<=\\b) - it means that you want digits to preceded by word boundary (including commas in the lookahead and lookbehind was redundant so I deleted it).