Java8: Create HashMap with character count of a String

Question

Wondering is there more simple way than computing the character count of a given string as below?

String word = "AAABBB";
    Map<String, Integer> charCount = new HashMap();
    for(String charr: word.split("")){
        Integer added = charCount.putIfAbsent(charr, 1);
        if(added != null)
            charCount.computeIfPresent(charr,(k,v) -> v+1);
    }

    System.out.println(charCount);

Solution 1

Simplest way to count occurrence of each character in a string, with full Unicode support (Java 11+)¹:

String word = "AAABBB";
Map<String, Long> charCount = word.codePoints().mapToObj(Character::toString)
        .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));
System.out.println(charCount);

^{1) Java 8 version with full Unicode support is at the end of the answer.}

Output

{A=3, B=3}

---

UPDATE: For Java 8+ (doesn't support characters from supplemental planes, e.g. emoji):

Map<String, Long> charCount = IntStream.range(0, word.length())
        .mapToObj(i -> word.substring(i, i + 1))
        .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

---

UPDATE 2: Also for Java 8+.

I was mistaken, thinking that codePoints() wasn't added until Java 9. It was added in Java 8 to the CharSequence interface, so it doesn't show in javadoc for String in Java 8, and shows as added in Java 9 for later versions of the javadoc.

However, the Character.toString​(int codePoint) method wasn't added until Java 11, so to use the Character.toString​(char c) method, we can use chars() in Java 8:

Map<String, Long> charCount = word.chars().mapToObj(c -> Character.toString((char) c))
        .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

Or for full Unicode support, incl. supplemental planes, we can use codePoints() and the String(int[] codePoints, int offset, int count) constructor, in Java 8:

Map<String, Long> charCount = word.codePoints()
        .mapToObj(cp -> new String(new int[] { cp }, 0, 1))
        .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

Solution 2

     String str = "Hello Manash";
    Map<Character,Long> hm = str.chars().mapToObj(c-> 
    (char)c).collect(Collectors.groupingBy(c->c,Collectors.counting()));
    System.out.println(hm);

Solution 3

Try the below approaches:

Approach 1:

    String str = "abcaadcbcb";
    
    Map<Character, Integer> charCount = str.chars()
            .boxed()
            .collect(toMap(
                    k -> (char) k.intValue(),
                    v -> 1,         // 1 occurence
                    Integer::sum));
    System.out.println("Char Counts:\n" + charCount);

Approach 2:

    String str = "abcaadcbcb";
    Map<Character, Integer> charCount = new HashMap<>();
    for (char c : str.toCharArray()) {
        charCount.merge(c,          // key = char
                1,                  // value to merge
                Integer::sum);      // counting
    }
    System.out.println("Char Counts:\n" + charCount);

Output:

    Char Counts:
    {a=3, b=3, c=3, d=1}

Solution 4

String str = "abcaadcbcb";

Map<String, Long> charCount  = 
Arrays.asList(str.split("")).stream().collect(Collectors.groupingBy(Function.identity(),Collectors.counting()));
    

View more solutions

Comments

[nice_dev]

For ANSI characters, you can just have an array of size 256 and compute it.

[Andreas]

@vivek_23 Which ANSI character set would that be? Or did you mean ASCII and 128?

[Holger]

@vivek_23 that is the windows code page 1252, not ANSI. The Unicode standard matches the iso-latin-1 character set for the first 256 codepoints. Referring to the windows code page 1252 is an unnecessary complication, as that code page does not match in the 128-159 range.

[nice_dev]

@Holger Ahh! Thanks for the correction. Deleted my previous comment to avoid confusion.

[OTUser]

Am sorry, is there a simple way for Java 8?

[Andreas]

chars() requires Java 9, and better solution using codePoints() instead of chars() already posted 13 minutes earlier.

[mm6]

@Andreas agree withcodePoints()solution, butchars()introduce in java 8 String.chars()

[Andreas]

That would be CharSequence.chars(), not String.chars(), but I accept your correction. Javadoc for Java 11 show method as added to String in Java 9, which is what lead me astray.

[Holger]

charCount.put(charr,charCount.getOrDefault(charr,0)+1); can be simplified to charCount.merge(charr, 1, Integer::sum); By the way, you should use new HashMap<>()…

[Holger]

Speaking of “full Unicode support” and Emojis, it’s worth pointing out that even using codepoints is not necessarily providing the intended semantics. E.g. "ā̧👩‍🇮🇩" has 10 chars, 7 codepoints, but only three characters; the first one demonstrates that this is not only an Emoji issue. The only solution, I currently know of, is to process grapheme clusters, e.g. with Java 9+: Pattern.compile("\\X").matcher(example).results() .collect(Collectors.groupingBy(MatchResult::group, Collectors.counting())).

[user2901351]

MIght you format your code snippet as coded, to allow for greater readability? Thanks.

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