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up vote 178 down vote favorite
28

I'd like to convert an array to a set in Java. There are some obvious ways of doing this (i.e. with a loop) but I would like something a bit neater, something like:

java.util.Arrays.asList(Object[] a);

Any ideas?

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2  
What, you think the Set constructor that takes a Collection or .addAll method doesn't do a loop? –  Paul Tomblin Jun 17 '10 at 18:29
15  
No, it's just that I would prefer not to write out the whole loop in my code (it's really ugly!). –  Peter Jun 17 '10 at 18:32
1  
Write your own:) –  Petar Minchev Jun 17 '10 at 18:38
    
possible duplicate of initialize java HashSet values by construction –  datakey May 22 at 10:57

8 Answers 8

up vote 316 down vote accepted

Like this:

Set<T> mySet = new HashSet<T>(Arrays.asList(someArray));
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8  
+1 for generics. –  JavadocMD Jun 17 '10 at 18:29
4  
I would leave out the last <T>, otherwise nice oneliner! –  despot Jul 6 '11 at 10:06
2  
this is highly innefient. Why create a List to iterate over it to create a Set? Just iterate over the array yourself. –  dataoz Mar 2 '12 at 17:56
67  
@dataoz: Wrong; Arrays.asList is O(1). –  SLaks Jun 8 '12 at 18:34
13  
Note that if you use this method on an array of primitives such as int[] it will return a List<int[]> so you should use wrapper classes to get the intended behavior. –  T. Markle Sep 22 '12 at 4:53
up vote 60 down vote 
Set<T> mySet = new HashSet<T>();
Collections.addAll(mySet, myArray);

That's Collections.addAll(java.util.Collection, T...) from jdk1.6.

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3  
You can do that with java.util.Collections.addAll. Plus, I wouldn't recommend Commons Collections anymore, what with it not being generified and Guava existing . –  ColinD Jun 17 '10 at 18:40
    
Yeah I noticed that and altered the above before I read your comment. :) –  JavadocMD Jun 17 '10 at 18:42
11  
+1 for being more efficient than SLaks's answer, even though it's not a one-liner. –  Adrian Apr 10 '12 at 22:17
    
@Adrian I question that. I think addAll will be O(n). –  Steve Powell Oct 22 '13 at 15:35
3  
Per the Collections.addAll() javadoc (Java 6): "The behavior of this convenience method is identical to that of c.addAll(Arrays.asList(elements)), but this method is likely to run significantly faster under most implementations." –  Bert F Feb 3 at 22:50
up vote 47 down vote

With Guava you can do:

T[] array = ...
Set<T> set = Sets.newHashSet(array);
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8  
also ImmutableSet.copyOf(array). (I like to point out also's, I guess.) –  Kevin Bourrillion Jun 17 '10 at 21:26
    
I love Guava. It's API is concise and epressive. –  oschrenk Dec 11 '11 at 20:41
    
+1 / makes for more readable and succinct code than java's collections. –  Jubbat Jan 7 '12 at 21:22
    
+1. I agree with Jubbat. The accepted answer is correct if limited to standard library, but this is far more elegant. Worth including Guava in your project. –  Chet Mar 27 '13 at 14:59
up vote 10 down vote

After you do Arrays.asList(array) you can execute Set set = new HashSet(list);

Here is a sample method, you can write:

public <T> Set<T> GetSetFromArray(T[] array) {
    return new HashSet<T>(Arrays.asList(array));
}
share|improve this answer
    
I was hoping for a method that returns a set directly from an array, does one exist? –  Peter Jun 17 '10 at 18:31
    
Honestly, I doubt:) –  Petar Minchev Jun 17 '10 at 18:31
1  
You can write your own, if you are so eager:) –  Petar Minchev Jun 17 '10 at 18:34
    
I know you meant Arrays.asList(array) instead of Arrays.List(array) in the first line –  dimitrisli Oct 25 '11 at 21:24
    
@dimitrisli - Thanks, of course:) –  Petar Minchev Oct 26 '11 at 6:20
up vote 2 down vote

Quickly : you can do :

// Fixed-size list
List list = Arrays.asList(array);

// Growable list
list = new LinkedList(Arrays.asList(array));

// Duplicate elements are discarded
Set set = new HashSet(Arrays.asList(array));

and to reverse

// Create an array containing the elements in a list
Object[] objectArray = list.toArray();
MyClass[] array = (MyClass[])list.toArray(new MyClass[list.size()]);

// Create an array containing the elements in a set
objectArray = set.toArray();
array = (MyClass[])set.toArray(new MyClass[set.size()]);
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the most comprehensive answer! –  davidgale Oct 2 '13 at 18:37
up vote 2 down vote

In GS Collections, the following will work:

Set<Integer> set1 = Sets.mutable.of(1, 2, 3, 4, 5);
Set<Integer> set2 = Sets.mutable.of(new Integer[]{1, 2, 3, 4, 5});
MutableSet<Integer> mutableSet = Sets.mutable.of(1, 2, 3, 4, 5);
ImmutableSet<Integer> immutableSet = Sets.immutable.of(1, 2, 3, 4, 5);

Set<Integer> unmodifiableSet = Sets.mutable.of(1, 2, 3, 4, 5).asUnmodifiable();
Set<Integer> synchronizedSet = Sets.mutable.of(1, 2, 3, 4, 5).asSynchronized();

Note: I am a developer on GS Collections

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up vote 1 down vote

Sometime using some standard libraries helps a lot. Try to look at the Apache Commons Collections. In this case your problems is simply transformed to something like this

String[] keys = {"blah", "blahblah"}
Set<String> myEmptySet = new HashSet<String>();
CollectionUtils.addAll(pythonKeywordSet, keys);

And here is the CollectionsUtils javadoc

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2  
user might not use apache commons –  Adrian Jan 4 '12 at 17:17
    
if the user does not use apache commons, then that is his first mistake. –  Jeryl Cook Oct 4 '13 at 13:46
up vote -3 down vote

new HashSet<Object>(Arrays.asList(Object[] a));

But I think this would be more efficient:

final Set s = new HashSet<Object>();
for (Object o : a) { s.add(o); }
share|improve this answer
    
That wouldn't really be more efficient (at least not worth thinking about). –  ColinD Jun 17 '10 at 18:46
3  
With the constructor version, the initial capacity of the HashSet is set based on the size of the array, for instance. –  ColinD Jun 17 '10 at 18:55

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