How can I check if an element exists in a Set of items?

Question

In an if statement in Java how can I check whether an object exists in a set of items. E.g. In this scenario i need to validate that the fruit will be an apple, orange or banana.

if (fruitname in ["APPLE", "ORANGES", "GRAPES"]) {
    //Do something
}

It's a very trivial thing but I couldn't figure out a short and concise way to accomplish this.

score 69 · Accepted Answer · edited May 16 '19 at 09:29

69

static final List<String> fruits = Arrays.asList("APPLE", "ORANGES", "GRAPES");

if (fruits.contains(fruitname))

If your list was much larger, a set would be more efficient.

static final Set<String> fruits = new HashSet<String>(
       Arrays.asList("APPLE", "ORANGES", "GRAPES", /*many more*/));

edited May 16 '19 at 09:29

T04435

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answered Jan 14 '11 at 08:51

Peter Lawrey

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Might as well make it a `Set`. Seems more appropriate. – wds Jan 14 '11 at 08:53
@wds There isn't really an Arrays.asSet method, and if this is the only use for the collection the extra conversion (to Set) step wouldn't seem worth the effort. – extraneon Jan 14 '11 at 10:02
Not for a list with three entries. For more entries it might be worth it. – Peter Lawrey Jan 14 '11 at 14:53
I'm guessing having to hash the key actually takes longer than doing a lookup in a table of a few dozen elements. Still, `Set` is the right type for the job, even if the particular implementation might not be. – wds Jan 17 '11 at 09:08
@wds, true, for a small number of entries, it hardly matters which is faster. You can scan a List of 100K entries in about a milli-second which might be fast enough. – Peter Lawrey Jan 17 '11 at 10:51
@PeterLawrey If the list is not a collection of Strings, lets say just a custom object, do we need to override equal method? – WowBow May 29 '12 at 19:50
@WowBow It depends on the class, however you may need to override equals/hashCode or compareTo depending on the collection used. – Peter Lawrey May 29 '12 at 20:03

score 9 · Answer 2 · answered Jan 14 '11 at 09:09

for completeness using google-collections/guava:

import com.google.common.collect.Sets;

static final Set<String> fruit = Sets.newHashSet("APPLE", "ORANGES", "GRAPES");

if (fruit.contains(fruitname))

or using the plane old jdk classes:

static final Set<String> fruit = new HashSet<String>(Arrays.asList("APPLE", "ORANGES", "GRAPES"));

score 2 · Answer 3 · edited Jun 02 '14 at 17:18

2

Is Arrays.binarySearch what you are looking for?

String [] fruits = new String[]{"APPLE", "ORANGES", "GRAPES"};
Arrays.sort(fruits); // binarySearch requires that the array is sorted

if (Arrays.binarySearch(fruits), fruitname) >= 0) {
  // found!
}

And of course the trusted Apache Commons ArrayUtils:

if (ArrayUtils.contains(new String[]{"APPLE", "ORANGES", "GRAPES"}, fruitname){
  // found
}

I knew there would be something in Apache Commons :)

edited Jun 02 '14 at 17:18

Gareth Davis

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answered Jan 14 '11 at 08:51

extraneon

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binary search assumes the array is sorted, I remember O coming after G ;) – Peter Lawrey Jan 14 '11 at 08:52
I noticed :) So I added a sort. I like your solution better though (that's the upvote:) – extraneon Jan 14 '11 at 08:53
@peter: there's an Array.sort() in there ;) – fasseg Jan 14 '11 at 08:55
k thx for the info! sorry peter I upvoted your comment then ;) – fasseg Jan 14 '11 at 08:59
`Arrays.sort` is a void method, so this doesn't really compile. – Natix Dec 05 '13 at 14:35

score 1 · Answer 4 · edited Dec 30 '18 at 06:54

1

If you have Set, List, Map of fruits which all have the same parent: Collection, you can try this example.

String fruitName = "Orange";
Collection<String> fruits = ... // set of fruits
if (fruits.contains(fruitName)) {
    ...
}

(For Java 8/9/10 ways of creating literal Set please see this SO answer.)

Be careful with case sensitivity (Orange != orange).

edited Dec 30 '18 at 06:54

Greg Dubicki

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answered Jan 14 '11 at 10:04

lukastymo

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How can I check if an element exists in a Set of items?

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