I need to remove all empty/null values from List<Optional<String>>
.
Example:
List<Optional<String>> list = new ArrayList<>();
list.add(Optional.empty());
list.add(Optional.of("Str1"));
list.add(Optional.of("Str2"));
list.add(Optional.of("Str3"));
list.add(Optional.of("Str4"));
list.add(Optional.of("Str5"));
list.add(Optional.empty());
list.add(Optional.ofNullable(null));
Currently, I'm using one of the below approaches:
Way 1:
List<String> collect = list.stream()
.filter(Optional::isPresent)
.map(obj ->obj.get())
.collect(Collectors.toList());
Way 2:
List<Optional<String>> emptlist = new ArrayList<>();
emptlist.add(Optional.empty());
list.removeAll(emptlist);
Is there any other better way?
An Optional object in Java is a container object that can hold both empty and a non-null values. If an Optional object does contain a value, we say that it is present; if it does not contain a value, we say that it is empty.
An Optional always contains a non-null value or is empty, yes, but you don't have an Optional, you have a reference of type Optional pointing to null.
A HashSet , being a set, only contains one "copy" of any object, which also means that it can only contain one instance of null . Thus, you can just use HashSet. remove(null) .
With Java9, you can do this using the newly added Optional::stream
API :
List<String> collect = list.stream()
.flatMap(Optional::stream)
.collect(Collectors.toList());
This method can be used to transform a
Stream
of optional elements to aStream
of present value elements.
Sticking with Java8, the Way1 in the question is good enough IMHO -
List<String> collect = list.stream()
.filter(Optional::isPresent)
.map(Optional::get) // just a small update of using reference
.collect(Collectors.toList());
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