How to get the first element of the List or Set?

Question

I'd like to know if I can get the first element of a list or set. Which method to use?

Answer 1

See the javadoc

of List

list.get(0);

or Set

set.iterator().next();

and check the size before using the above methods by invoking isEmpty()

!list_or_set.isEmpty()

Answer 2

Collection c;

Iterator iter = c.iterator();

Object first = iter.next();

(This is the closest you'll get to having the "first" element of a Set. You should realize that it has absolutely no meaning for most implementations of Set. This may have meaning for LinkedHashSet and TreeSet, but not for HashSet.)

Answer 3

In Java >=8 you could also use the Streaming API:

Optional<String> first = set.stream().findFirst();

(Useful if the Set/List may be empty.)

Answer 4

Let's assume that you have a List<String> strings that you want the first item from.

There are several ways to do that:

Java (pre-8):

String firstElement = null;
if (!strings.isEmpty() && strings.size() > 0) {
    firstElement = strings.get(0);
}

Java 8:

Optional<String> firstElement = strings.stream().findFirst();

Guava

String firstElement = Iterables.getFirst(strings, null);

Apache commons (4+)

String firstElement = (String) IteratorUtils.get(strings, 0);

Apache commons (before 4)

String firstElement = (String) CollectionUtils.get(strings, 0);

Followed by or encapsulated within the appropriate checks or try-catch blocks.

Kotlin:

In Kotlin both Arrays and most of the Collections (eg: List) have a first method call. So your code would look something like this

for a List:

val stringsList: List<String?> = listOf("a", "b", null)
val first: String? = stringsList.first()

for an Array:

val stringArray: Array<String?> = arrayOf("a", "b", null)
val first: String? = stringArray.first()

Followed by or encapsulated within the appropriate checks or try-catch blocks.

Kotlin also includes safer ways to do that for kotlin.collections, for example firstOrNull or getOrElse, or getOrDefault when using JRE8

Answer 5

I'm surprised that nobody suggested guava solution yet:

com.google.common.collect.Iterables.get(collection, 0)
// or
com.google.common.collect.Iterables.get(collection, 0, defaultValue)
// or
com.google.common.collect.Iterables.getFirst(collection, defaultValue)

or if you expect single element:

com.google.common.collect.Iterables.getOnlyElement(collection, defaultValue)
// or
com.google.common.collect.Iterables.getOnlyElement(collection)

Answer 6

java8 and further

Set<String> set = new TreeSet<>();
    set.add("2");
    set.add("1");
    set.add("3");
    String first = set.stream().findFirst().get();

This will help you retrieve the first element of the list or set. Given that the set or list is not empty (get() on empty optional will throw java.util.NoSuchElementException)

orElse() can be used as: (this is just a work around - not recommended)

String first = set.stream().findFirst().orElse("");
set.removeIf(String::isEmpty);

Below is the appropriate approach :

Optional<String> firstString = set.stream().findFirst();
if(firstString.isPresent()){
    String first = firstString.get();
}

Similarly first element of the list can be retrieved.

Hope this helps.

Answer 7

Set

set.toArray()[0];

List

list.get(0);

Answer 8

This is not an exact answer to this question, but in case the objects should be sorted SortedSet has a first() method:

SortedSet<String> sortedSet = new TreeSet<String>();
sortedSet.add("2");
sortedSet.add("1");
sortedSet.add("3");
String first = sortedSet.first(); //first="1"

The sorted objects must implement the Comparable interface (like String does)

Answer 9

You can use the get(index) method to access an element from a List.

Sets, by definition, simply contain elements and have no particular order. Therefore, there is no "first" element you can get, but it is possible to iterate through it using iterator (using the for each loop) or convert it to an array using the toArray() method.