使用 java stream 查找字符串中每个字符的计数

Question

we have this string: String input1 = "abbccd";

expected output: ab2c2d (note: if count=1, it shouldn't show in output).

the following code outputs a1,b2 c2 d2 on separate lines. Any suggestion to fix and improve?

input1.chars()
      .mapToObj(s -> Character.toLowerCase(Character.valueOf((char) s)))
      .collect(Collectors.groupingBy(Function.identity(), LinkedHashMap::new, Collectors.counting()))
      .entrySet().stream()
      .forEach(n -> {System.out.println(n.getKey()+""+n.getValue());});

Answer 1

Make the last forEach a map instead.

Instead of n.getValue() only add that part if n.getValue is not 1.

Then collect by joining.

At that point you will have a string you can print.

So, assuming we don't want to change your first part:

"abbccd".chars()
        .mapToObj(s -> Character.toLowerCase((char)s)) // notice here Character.valueOf was redundant, we're already dealing with a char
        .collect(Collectors.groupingBy(Function.identity(), LinkedHashMap::new, Collectors.counting()))
        .entrySet().stream()
        .map(n -> n.getKey()+""+(n.getValue() == 1 ? "" : n.getValue()))
        .collect(Collectors.joining());

Results in ab2c2d .

Answer 2

Unfortunately, the other two Answers both fail with most characters.

Avoid legacy type char

The char type is legacy, essentially broken since Java 2, legacy since Java 5. As a 16-bit value, char is physically incapable of representing most of the 144,697 characters defined in Unicode .

See one Answer's code break:

String input = "😷😷abbccd";
String output =
        input
                .chars()
                .mapToObj( s -> Character.toLowerCase( ( char ) s ) ) // notice here Character.valueOf was redundant, we're already dealing with a char
                .collect( Collectors.groupingBy( Function.identity() , LinkedHashMap :: new , Collectors.counting() ) )
                .entrySet().stream()
                .map( n -> n.getKey() + "" + ( n.getValue() == 1 ? "" : n.getValue() ) )
                .collect( Collectors.joining() );

System.out.println( "output = " + output );

output =?2?2ab2c2d

Code point

Use code point integer numbers instead, when working with individual characters. A code point is the number permanently assigned to each character in Unicode. They range from zero to just over a million.

You will find code point related method scattered around the Java classes. These include String , StringBuilder , Character , etc.

The String#codePoints method returns an IntStream of code points, the code point number for each character in the string.

Here is a re-worked version of the clever code from Answer by Federico klez Culloca . Cudos to him, as I would not have come up with that approach.

String input = "😷😷abbccd";
String output =
        input
                .codePoints()
                .map( Character :: toLowerCase )
                .mapToObj( codePoint -> Character.toString( codePoint ) )
                .collect( Collectors.groupingBy( Function.identity() , LinkedHashMap :: new , Collectors.counting() ) )
                .entrySet().stream()
                .map( n -> n.getKey() + "" + ( n.getValue() == 1 ? "" : n.getValue() ) )
                .collect( Collectors.joining() );
System.out.println( "output = " + output );

output = 2ab2c2d

Answer 3

For the sake of variety, a different approach using regex:

String input1 = "abbccd";

String output = Pattern.compile("(?<=(.))(?!\\1)")
                       .splitAsStream(input1)
                       .map(str -> str.length() == 1 ? str : "" + str.charAt(0) + str.length())
                       .collect(Collectors.joining());

System.out.println(output);