Count Int Occurrences with Java8

Count int occurrences with Java8

Try:

 Map<Integer, Long> counters = persons.stream()
     .collect(Collectors.groupingBy(p -> p.getBirthday().getMonthValue(), 
         Collectors.counting()));

Java 8 : Count the occurrence of digit 4 in the given int array

First, you need to obtain a stream from the given array of int[]. You can do it either by using the static method IntStream.of, or with Arrays utility class and it's method stream().

Both will give you an IntStream - a stream of int primitives. In order to collect stream elements into a list you need to convert it to a stream of objects. For that you can apply method boxed() on the stream pipeline and each int element will get wrapped with Integer object.

In order to find all elements in the given list that contain a target digit (4), you can turn the target digit into a string and then apply a filter based on it.

public static List<Integer> findOccurrencesOfDigit(List<Integer> source, 
                                                   int digit) {
    String target = String.valueOf(digit);

    return source.stream()
                    .filter(n -> String.valueOf(n).contains(target))
                    .collect(Collectors.toList());
}

main()

public static void main(String[] args) {
    int[] nums = {48,44,4,88,84,16,12,13};

    System.out.println(findOccurrencesOfDigit(nums, 4));
}

Output

[48, 44, 4, 84]

Note: method filter() expects a Predicate (a function represented by a boolean condition, that takes an element and return true of false)

Your attempt to create a predicate filter(n -> n % 10) is incorrect syntactically and logically. n % 10 doesn't produce a boolean value, it'll give the right most digit of n.

If you want to use modulus (%) operator to create a predicate, you have to divide the element by 10 in a loop until it'll not become equal to 0. And check before every division weather remainder is equal to the target digit. It could be done like that:

public static boolean containsDigit(int candidate, int digit) {
    boolean flag = false;
    while (candidate != 0) {
        if (candidate % 10 == digit) {
            flag = true;
            break;
        }
        candidate /= 10; // same as candidate = candidate / 10;
    }
    return flag;
}

You can utilize this method inside a lambda expression like that (note: it's a preferred way to avoid multiline lambda expressions by extracting them into separate methods)

filter(n -> containsDigit(n, digit))

Java count occurrence of each element in an integer array

For the latter question, you have to change

.forEach((k, v) -> if(v.size() > 1) System.out.println(k+" "+v.size()));

to

.forEach((k, v) -> {if(v.size() > 1) System.out.println(k+" "+v.size());});

For the first part, it's not clear why you need the first collect followed by a second Stream pipeline.
If the purpose was to convert an IntStream to a Stream<Integer>, use boxed():

Arrays.stream(arr)
      .boxed()
      .collect(Collectors.groupingBy(s -> s))
      .forEach((k, v) -> System.out.println(k+" "+v.size()));

As Dici suggested, you can also chain Collectors to group each number with its number of occurrences :

Map<Integer,Integer> occurrences = 
    Arrays.stream(arr)
          .boxed()
          .collect(Collectors.groupingBy(s -> s, Collectors.counting()));

Counting occurrences in a list with Java 8

You can use a Stream<PDFDataItem> with Collectors.groupingBy() Collector to group the PDFDataItem instances by the value property and then count the number of elements in each group using the Collectors.counting() Collector.

Map<Double,Long> valueCounts = 
    pdfList.stream()
           .collect(Collectors.groupingBy(PDFDataItem::getValue,Collectors.counting()));

create map of int occurrences using Java 8

IntStream has one method collect where the second argument operates on an int not an Object. Using boxed() turns an IntStream into a Stream<Integer>

Also counting() returns a long.

Map<Integer, Long> result = IntStream.range(0, 100).boxed()
        .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

Counting word occurence with arrays stream in Java 8

    String sentence = "The cat has black fur and black eyes";
    String[] bites = sentence.trim().split("\\s+");
    String in = "black cat";
    long i = Stream.of(bites).filter(e->(Arrays.asList(in.split("\\s")).contains(e))).count();
    System.out.println(i);

Count occurrences of a String in a List, then sort the result - Java 8

You can get the desired output if you use a TreeMap instead of the default HashMap when you create the Map like this:

Map<String, Long> employeeFirstNameCount = employees.stream()
            .collect(Collectors.groupingBy(Employee::getFirstName,
                                           TreeMap::new,
                                           Collectors.counting()));

The java.util.TreeMap uses the natural ordering of its keys (that is enough for what you need) or you can provide a custom Comparator

Note I use the lambda expression Employee::getFirstName instead of p -> p.getFirstName(), but both produce the same result.

How can I count occurrences with groupBy?

I think you're just looking for the overload which takes another Collector to specify what to do with each group... and then Collectors.counting() to do the counting:

import java.util.*;
import java.util.stream.*;

class Test {
    public static void main(String[] args) {
        List<String> list = new ArrayList<>();

        list.add("Hello");
        list.add("Hello");
        list.add("World");

        Map<String, Long> counted = list.stream()
            .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

        System.out.println(counted);
    }
}

Result:

{Hello=2, World=1}

(There's also the possibility of using groupingByConcurrent for more efficiency. Something to bear in mind for your real code, if it would be safe in your context.)

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