Splitting a string at every n-th character

In JavaScript this is how we can spl开发者_开发知识库it a string at every 3-rd character

"foobarspam".match(/.{1,3}/g)

I am trying to figure out how to do this in Java. Any pointers?

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You could do it like this:

String s = "1234567890";
System.out.println(java.util.Arrays.toString(s.split("(?<=\\G...)")));

which produces:

[123, 456, 789, 0]

The regex (?<=\G...) matches an empty string that has the last match (\G) followed by three characters (...) before it ((?<= ))

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Java does not provide very full-featured splitting utilities, so the Guava libraries do:

Iterable<String> pieces = Splitter.fixedLength(3).split(string);

Check out the Javadoc for Splitter; it's very powerful.

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import java.util.ArrayList;
import java.util.List;

public class Test {
    public static void main(String[] args) {
        for (String part : getParts("foobarspam", 3)) {
            System.out.println(part);
        }
    }
    private static List<String> getParts(String string, int partitionSize) {
        List<String> parts = new ArrayList<String>();
        int len = string.length();
        for (int i=0; i<len; i+=partitionSize)
        {
            parts.add(string.substring(i, Math.min(len, i + partitionSize)));
        }
        return parts;
    }
}

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As an addition to Bart Kiers answer I want to add that it is possible instead of using the three dots ... in the regex expression which are representing three characters you can write .{3} which has the same meaning.

Then the code would look like the following:

String bitstream = "00101010001001010100101010100101010101001010100001010101010010101";
System.out.println(java.util.Arrays.toString(bitstream.split("(?<=\\G.{3})")));

With this it would be easier to modify the string length and the creation of a function is now reasonable with a variable input string length. This could be done look like the following:

public static String[] splitAfterNChars(String input, int splitLen){
    return input.split(String.format("(?<=\\G.{%1$d})", splitLen));
}

An example in IdeOne: http://ideone.com/rNlTj5

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Late Entry.

Following is a succinct implementation using Java8 streams, a one liner:

String foobarspam = "foobarspam";
AtomicInteger splitCounter = new AtomicInteger(0);
Collection<String> splittedStrings = foobarspam
                                    .chars()
                                    .mapToObj(_char -> String.valueOf((char)_char))
                                    .collect(Collectors.groupingBy(stringChar -> splitCounter.getAndIncrement() / 3
                                                                ,Collectors.joining()))
                                    .values();

Output:

[foo, bar, spa, m]

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This a late answer, but I am putting it out there anyway for any new programmers to see:

If you do not want to use regular expressions, and do not wish to rely on a third party library, you can use this method instead, which takes between 89920 and 100113 nanoseconds in a 2.80 GHz CPU (less than a millisecond). It's not as pretty as Simon Nickerson's example, but it works:

   /**
     * Divides the given string into substrings each consisting of the provided
     * length(s).
     * 
     * @param string
     *            the string to split.
     * @param defaultLength
     *            the default length used for any extra substrings. If set to
     *            <code>0</code>, the last substring will start at the sum of
     *            <code>lengths</code> and end at the end of <code>string</code>.
     * @param lengths
     *            the lengths of each substring in order. If any substring is not
     *            provided a length, it will use <code>defaultLength</code>.
     * @return the array of strings computed by splitting this string into the given
     *         substring lengths.
     */
    public static String[] divideString(String string, int defaultLength, int... lengths) {
        java.util.ArrayList<String> parts = new java.util.ArrayList<String>();

        if (lengths.length == 0) {
            parts.add(string.substring(0, defaultLength));
            string = string.substring(defaultLength);
            while (string.length() > 0) {
                if (string.length() < defaultLength) {
                    parts.add(string);
                    break;
                }
                parts.add(string.substring(0, defaultLength));
                string = string.substring(defaultLength);
            }
        } else {
            for (int i = 0, temp; i < lengths.length; i++) {
                temp = lengths[i];
                if (string.length() < temp) {
                    parts.add(string);
                    break;
                }
                parts.add(string.substring(0, temp));
                string = string.substring(temp);
            }
            while (string.length() > 0) {
                if (string.length() < defaultLength || defaultLength <= 0) {
                    parts.add(string);
                    break;
                }
                parts.add(string.substring(0, defaultLength));
                string = string.substring(defaultLength);
            }
        }

        return parts.toArray(new String[parts.size()]);
    }

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Using plain java:

    String s = "1234567890";
    List<String> list = new Scanner(s).findAll("...").map(MatchResult::group).collect(Collectors.toList());
    System.out.printf("%s%n", list);

Produces the output:

[123, 456, 789]

Note that this discards leftover characters (0 in this case).

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You can also split a string at every n-th character and put them each, in each index of a List :

Here I made a list of Strings named Sequence :

List < String > Sequence

Then I'm basically splitting the String "KILOSO" by every 2 words. So 'KI' 'LO' 'SO' would be incorporate in separate index of the List called Sequence.

String S = KILOSO

Sequence = Arrays.asList(S.split("(?<=\G..)"));

So when I'm doing :

System.out.print(Sequence)

It should print :

[KI, LO, SO]

to verify I can write :

System.out.print(Sequence.get(1))

it will print :

LO

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I recently encountered this issue, and here is the solution I came up with

final int LENGTH = 10;
String test = "Here is a very long description, it is going to be past 10";

Map<Integer,StringBuilder> stringBuilderMap = new HashMap<>();
for ( int i = 0; i < test.length(); i++ ) {
    int position = i / LENGTH; // i<10 then 0, 10<=i<19 then 1, 20<=i<30 then 2, etc.

    StringBuilder currentSb = stringBuilderMap.computeIfAbsent( position, pos -> new StringBuilder() ); // find sb, or create one if not present
    currentSb.append( test.charAt( i ) ); // add the current char to our sb
}

List<String> comments = stringBuilderMap.entrySet().stream()
        .sorted( Comparator.comparing( Map.Entry::getKey ) )
        .map( entrySet -> entrySet.getValue().toString() )
        .collect( Collectors.toList() );
//done



// here you can see the data
comments.forEach( cmt -> System.out.println( String.format( "'%s' ... length= %d", cmt, cmt.length() ) ) );
// PRINTS:
// 'Here is a ' ... length= 10
// 'very long ' ... length= 10
// 'descriptio' ... length= 10
// 'n, it is g' ... length= 10
// 'oing to be' ... length= 10
// ' past 10' ... length= 8

// make sure they are equal
String joinedString = String.join( "", comments );
System.out.println( "\nOriginal strings are equal " + joinedString.equals( test ) );
// PRINTS: Original strings are equal true

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I'd start with something like

public List<String> split(String str, int interval) {
    if (str.length() <= interval) {
        return List.of(str);
    }
    var subStrings = new ArrayList<String>();
    int pointer = 0;
    while (str.length() > pointer) {
        String substring = str.substring(pointer, pointer + interval);
        subStrings.add(substring);
        pointer += interval;
    }
    return subStrings;
}