CODEWITHSUNDEEP
javaregexsplit
Matt
Matt

Reputation: 11317

Split string into array of character strings

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

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

Upvotes: 152

Views: 459331

Answers (13)

chaotic3quilibrium
chaotic3quilibrium

Reputation: 5924

I found this question because I am attempting to validate the contents of a fixed length String identifier.

Rather than use a Regex, I wanted to figure out how to use the Java Collections Library, i.e. Set<String>, instead.

Here's what I created to validate a String fixed length (6 characters) identifier alpha characters only prefixed with "X":

public static final Set<String> CHARACTERS_ALPHA_INVALID = Set.of("B", "D", "Q", "U");

public static final Set<String> CHARACTERS_VALID = IntStream.range(0, 26)
    .mapToObj(index -> String.valueOf((char) (index + 65)))
    .filter(charAsString -> !CHARACTERS_ALPHA_INVALID.contains(charAsString))
    .collect(Collectors.toUnmodifiableSet());

public static boolean isCharactersValid(String candidate) {
  return Optional.ofNullable(candidate)
      .filter(candidateNonNull -> !candidateNonNull.isEmpty())
      .map(candidateNonEmpty ->
          Arrays.stream(candidateNonEmpty.split(""))
              .allMatch(CHARACTERS_VALID::contains))
      .orElseThrow(() -> new IllegalArgumentException("candidate must not be null or empty"));
}

public static boolean isInstanceOfNewXPrefixedSize6UserCode(String candidate) {
  return Optional.ofNullable(candidate)
      .map(candidateNonNull ->
          candidateNonNull.startsWith("X") && (candidateNonNull.length() == 6)
              && isCharactersValid(candidateNonNull))
      .orElse(false);
}

Upvotes: 0

Masaru Tomimitsu
Masaru Tomimitsu

Reputation: 39

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).

Comparison of different splitting methods

(discarding 256 and 512 measures) enter image description here

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

Upvotes: 1

Masaru Tomimitsu
Masaru Tomimitsu

Reputation: 39

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.

Upvotes: 0

user4910279
user4910279

Reputation:

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, 🌹, 🌺, 🌷]

Upvotes: 6

Daniel
Daniel

Reputation: 1676

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);
}

Upvotes: 1

Lezorte
Lezorte

Reputation: 503

To sum up the other answers...

This works on all Java versions:

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

This only works on Java 8 and up:

"cat".split("")

Upvotes: 5

Jan Moln&#225;r
Jan Moln&#225;r

Reputation: 420

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

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

Upvotes: 12

Stephen C
Stephen C

Reputation: 718708

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.

Upvotes: 3

JV More
JV More

Reputation: 45

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

Upvotes: 2

coberty
coberty

Reputation: 1499

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

This will produce

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

Upvotes: 142

Yuriy Faktorovich
Yuriy Faktorovich

Reputation: 68667

"cat".toCharArray()

But if you need strings

"cat".split("")

Edit: which will return an empty first value.

Upvotes: 135

reef
reef

Reputation: 1833

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.

Upvotes: 1

Raman
Raman

Reputation: 1517

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

Upvotes: 48

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