How to Iterate over a Set/HashSet without an Itera

问题:

How can I iterate over a Set/HashSet without the following?

Iterator iter = set.iterator();
while (iter.hasNext()) {
    System.out.println(iter.next());
}

回答1:

You can use an enhanced for loop:

Set<String> set = new HashSet<String>();

//populate set

for (String s : set) {
    System.out.println(s);
}

Or with Java 8:

set.forEach(System.out::println);

回答2:

There are at least six additional ways to iterate over a set. The following are known to me:

Method 1

// Obsolete Collection
Enumeration e = new Vector(movies).elements();
while (e.hasMoreElements()) {
  System.out.println(e.nextElement());
}

Method 2

for (String movie : movies) {
  System.out.println(movie);
}

Method 3

String[] movieArray = movies.toArray(new String[movies.size()]);
for (int i = 0; i < movieArray.length; i++) {
  System.out.println(movieArray[i]);
}

Method 4

// Supported in Java 8 and above
movies.stream().forEach((movie) -> {
  System.out.println(movie);
});

Method 5

// Supported in Java 8 and above
movies.stream().forEach(movie -> System.out.println(movie));

Method 6

// Supported in Java 8 and above
movies.stream().forEach(System.out::println);

This is the HashSet which I used for my examples:

Set<String> movies = new HashSet<>();
movies.add("Avatar");
movies.add("The Lord of the Rings");
movies.add("Titanic");

回答3:

Converting your set into an array may also help you for iterating over the elements:

Object[] array = set.toArray();

for(int i=0; i<array.length; i++)
   Object o = array[i];

回答4:

To demonstrate, consider the following set, which holds different Person objects:

Set<Person> people = new HashSet<Person>();
people.add(new Person("Tharindu", 10));
people.add(new Person("Martin", 20));
people.add(new Person("Fowler", 30));

Person Model Class

public class Person {
    private String name;
    private int age;

    public Person(String name, int age) {
        this.name = name;
        this.age = age;
    }

    //TODO - getters,setters ,overridden toString & compareTo methods

}

1. The for statement has a form designed for iteration through Collections and arrays .This form is sometimes referred to as the enhanced for statement, and can be used to make your loops more compact and easy to read.

for(Person p:people){
  System.out.println(p.getName());
}

2. Java 8 - java.lang.Iterable.forEach(Consumer)

people.forEach(p -> System.out.println(p.getName()));

default void forEach(Consumer<? super T> action)

Performs the given action for each element of the Iterable until all elements have been processed or the action throws an exception. Unless otherwise specified by the implementing class, actions are performed in the order of iteration (if an iteration order is specified). Exceptions thrown by the action are relayed to the caller. Implementation Requirements:

The default implementation behaves as if: 

for (T t : this)
     action.accept(t);

Parameters: action - The action to be performed for each element

Throws: NullPointerException - if the specified action is null

Since: 1.8

回答5:

Here are few tips on how to iterate a Set along with their performances:

public class IterateSet {

    public static void main(String[] args) {

        //example Set
        Set<String> set = new HashSet<>();

        set.add("Jack");
        set.add("John");
        set.add("Joe");
        set.add("Josh");

        long startTime = System.nanoTime();
        long endTime = System.nanoTime();

        //using iterator
        System.out.println("Using Iterator");
        startTime = System.nanoTime();
        Iterator<String> setIterator = set.iterator();
        while(setIterator.hasNext()){
            System.out.println(setIterator.next());
        }
        endTime = System.nanoTime();
        long durationIterator = (endTime - startTime);


        //using lambda
        System.out.println("Using Lambda");
        startTime = System.nanoTime();
        set.forEach((s) -> System.out.println(s));
        endTime = System.nanoTime();
        long durationLambda = (endTime - startTime);


        //using Stream API
        System.out.println("Using Stream API");
        startTime = System.nanoTime();
        set.stream().forEach((s) -> System.out.println(s));
        endTime = System.nanoTime();
        long durationStreamAPI = (endTime - startTime);


        //using Split Iterator (not recommended)
        System.out.println("Using Split Iterator");
        startTime = System.nanoTime();
        Spliterator<String> splitIterator = set.spliterator();
        splitIterator.forEachRemaining((s) -> System.out.println(s));
        endTime = System.nanoTime();
        long durationSplitIterator = (endTime - startTime);


        //time calculations
        System.out.println("Iterator Duration:" + durationIterator);
        System.out.println("Lamda Duration:" + durationLambda);
        System.out.println("Stream API:" + durationStreamAPI);
        System.out.println("Split Iterator:"+ durationSplitIterator);
    }
}

The code is self explanatory.

The result of the durations are:

Iterator Duration: 495287
Lambda Duration: 50207470
Stream Api:       2427392
Split Iterator:    567294

We can see the Lambda takes the longest while Iterator is the fastest.

回答6:

You can use functional operation for a more neat code

Set<String> set = new HashSet<String>();

set.forEach((s) -> {
     System.out.println(s);
});

回答7:

Enumeration(?):

Enumeration e = new Vector(set).elements();
while (e.hasMoreElements())
    {
        System.out.println(e.nextElement());
    }

Another way (java.util.Collections.enumeration()):

for (Enumeration e1 = Collections.enumeration(set); e1.hasMoreElements();)
    {
        System.out.println(e1.nextElement());
    }

Java 8:

set.forEach(element -> System.out.println(element));

or

set.stream().forEach((elem) -> {
    System.out.println(elem);
});