Split string into array of character strings

Question

I need to split a String into an array of single character Strings.

Eg, splitting "cat" would give the array "c", "a", "t"

Answer 1

"cat".split("(?!^)")

This will produce

array ["c", "a", "t"]

Answer 2

"cat".toCharArray()

But if you need strings

"cat".split("")

Edit: which will return an empty first value.

Answer 3

String str = "cat";
char[] cArray = str.toCharArray();

Answer 4

If characters beyond Basic Multilingual Plane are expected on input (some CJK characters, new emoji...), approaches such as "ab".split("(?!^)") cannot be used, because they break such characters (results into array ["a", "?", "?", "b"]) and something safer has to be used:

"ab".codePoints()
    .mapToObj(cp -> new String(Character.toChars(cp)))
    .toArray(size -> new String[size]);

Answer 5

split("(?!^)") does not work correctly if the string contains surrogate pairs. You should use split("(?<=.)").

String[] splitted = "花ab".split("(?<=.)");
System.out.println(Arrays.toString(splitted));

output:

[花, a, b, , , ]

Answer 6

To sum up the other answers...

This works on all Java versions:

"cat".split("(?!^)")

This only works on Java 8 and up:

"cat".split("")

Answer 7

An efficient way of turning a String into an array of one-character Strings would be to do this:

String[] res = new String[str.length()];
for (int i = 0; i < str.length(); i++) {
    res[i] = Character.toString(str.charAt(i));
}

However, this does not take account of the fact that a char in a String could actually represent half of a Unicode code-point. (If the code-point is not in the BMP.) To deal with that you need to iterate through the code points ... which is more complicated.

This approach will be faster than using String.split(/* clever regex*/), and it will probably be faster than using Java 8+ streams. It is probable faster than this:

String[] res = new String[str.length()];
int 0 = 0;
for (char ch: str.toCharArray[]) {
    res[i++] = Character.toString(ch);
}  

because toCharArray has to copy the characters to a new array.

Answer 8

for(int i=0;i<str.length();i++)
{
System.out.println(str.charAt(i));
}

Answer 9

If the original string contains supplementary Unicode characters, then split() would not work, as it splits these characters into surrogate pairs. To correctly handle these special characters, a code like this works:

String[] chars = new String[stringToSplit.codePointCount(0, stringToSplit.length())];
for (int i = 0, j = 0; i < stringToSplit.length(); j++) {
    int cp = stringToSplit.codePointAt(i);
    char c[] = Character.toChars(cp);
    chars[j] = new String(c);
    i += Character.charCount(cp);
}

Answer 10

Maybe you can use a for loop that goes through the String content and extract characters by characters using the charAt method.

Combined with an ArrayList<String> for example you can get your array of individual characters.

Answer 11

In my previous answer I mixed up with JavaScript. Here goes an analysis of performance in Java.

I agree with the need for attention on the Unicode Surrogate Pairs in Java String. This breaks the meaning of methods like String.length() or even the functional meaning of Character because it's ultimately a technical object which may not represent one character in human language.

I implemented 4 methods that split a string into list of character-representing strings (Strings corresponding to human meaning of characters). And here's the result of comparison:

A line is a String consisting of 1000 arbitrary chosen emojis and 1000 ASCII characters (1000 times <emoji><ascii>, total 2000 "characters" in human meaning).

(discarding 256 and 512 measures)

Implementations:

• codePoints (java 11 and above)

    public static List<String> toCharacterStringListWithCodePoints(String str) {
        if (str == null) {
            return Collections.emptyList();
        }
        return str.codePoints()
            .mapToObj(Character::toString)
            .collect(Collectors.toList());
    }

• classic

    public static List<String> toCharacterStringListWithIfBlock(String str) {
        if (str == null) {
            return Collections.emptyList();
        }
        List<String> strings = new ArrayList<>();
        char[] charArray = str.toCharArray();
        int delta = 1;
        for (int i = 0; i < charArray.length; i += delta) {
            delta = 1;
            if (i < charArray.length - 1 && Character.isSurrogatePair(charArray[i], charArray[i + 1])) {
                delta = 2;
                strings.add(String.valueOf(new char[]{ charArray[i], charArray[i + 1] }));
            } else {
                strings.add(Character.toString(charArray[i]));
            }
        }
        return strings;
    }

• regex

    static final Pattern p = Pattern.compile("(?<=.)");
    public static List<String> toCharacterStringListWithRegex(String str) {
        if (str == null) {
            return Collections.emptyList();
        }
        return Arrays.asList(p.split(str));
    }

Annex (RAW DATA):

codePoints;classic;regex;lines
45;44;84;256
14;20;98;512
29;42;91;1024
52;56;99;2048
87;121;174;4096
175;221;375;8192
345;411;839;16384
667;826;1285;32768
1277;1536;2440;65536
2426;2938;4238;131072

Answer 12

We can do this simply by

const string = 'hello';
console.log([...string]); // -> ['h','e','l','l','o']

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Spread_syntax says

Spread syntax (...) allows an iterable such as an array expression or string to be expanded...

So, strings can be quite simply spread into arrays of characters.