Comparator for sorting by frequency without Creating a Comparator implementation class

Question

Just curious to know on whether we can sort a List according to the frequency of repeated numbers using Java 8 without writing a Custom Comparator class. I need to sort the given integers based on its frequency and then by the natural numerical order. I'm getting error at Comparator.naturalOrder(); Here is the code which I tried: Integer[] given = new Integer[]{0,0,1,22,11,22,22,11,44,555,55,66,77,88,99}; List intList = Arrays.asList(given); Map frequencyMap = intList.stream().collect(Collectors.groupingBy(Function.identity(), Collectors.counting())); List newList = intList.stream().sorted(Comparator.comparing(frequencyMap::get).thenComparing(Comparator.naturalOrder())).collect(Collectors.toList()); System.out.println(newList.toString()); The expected output is [1, 44, 55, 66, 77, 88, 99, 555, 0, 0, 11, 11, 22, 22, 22] PS: Used arrays in first line in order to avoid list.add() in multiple lines and for clear understanding. JavaScript questions and answers, JavaScript questions pdf, JavaScript question bank, JavaScript questions and answers pdf, mcq on JavaScript pdf, JavaScript questions and solutions, JavaScript mcq Test , Interview JavaScript questions, JavaScript Questions for Interview, JavaScript MCQ (Multiple Choice Questions)

Answer 1

Unfortunately, Java’s type inference can’t recognize the type of the compared object when chaining Comparator.comparing(frequencyMap::get) with thenComparing(Comparator.naturalOrder()). Since the method signature of Map.get is get(Object), the compiler infers Comparator as result type of Comparator.comparing(frequencyMap::get). You can fix this by inserting an explicit type. But note that you are not using the result of collect(Collectors.toList()) but just printing the original, unaffected List. On the other hand, you don’t need the List when the array is given: Integer[] given = {0,0,1,22,11,22,22,11,44,555,55,66,77,88,99}; Map frequencyMap = Arrays.stream(given) .collect(Collectors.groupingBy(Function.identity(), Collectors.counting())); Arrays.sort(given, Comparator.comparingLong(frequencyMap::get) .thenComparing(Comparator.naturalOrder())); System.out.println(Arrays.toString(given)); For printing without changing the array you can also use the following alternative Arrays.stream(given) .collect(Collectors.groupingBy(Function.identity(), Collectors.counting())) .entrySet().stream() .sorted(Map.Entry.comparingByValue() .thenComparing(Map.Entry.comparingByKey())) .flatMap(e -> LongStream.range(0, e.getValue()).mapToObj(l -> e.getKey())) .forEach(System.out::println); This sorts the groups instead of the individual values and prints identical values as often as they’ve been counted.