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);
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()));