Map<?, Optional<T>> to Map<?, T>

Question

I have a Map with Optional values.

I want to filter it to remove empty values.

Map<K, Optional<T>> input;

Map<K, T> result = input.
// here is some code I'm looking for
                   .collect(Collector.toMap(???));

What is easiest way to achieve it?

All I can think of requires Optional.get and I don't like that IDE warns me about it.

Answer 1

The easiest way to do it, especially while avoiding Optional.get(), isn't going to involve streams at all; more like

Map<K, T> result = new LinkedHashMap<>();
input.forEach((k, vOpt) -> {
  vOpt.ifPresent(v -> result.put(k, v));
});

(In general, if there isn't a really obvious way to do something with streams, it's generally going to be simpler and cleaner not to use streams at all.)

If you insist on using a Collector, it's possible but requires a carefully customized one, especially if you won't tolerate Optional.get().

map.entrySet().stream()
  .collect(Collector.of(
    LinkedHashMap::new,
    (map, entry) -> entry.getValue().ifPresent(v -> map.put(entry.getKey(), v)),
    (m1, m2) -> { 
      m1.putAll(m2);
      return m1;
    }));

If you can tolerate Optional.get(), you might consider it simpler to write

map.entrySet().stream()
  .filter(e -> e.getValue().isPresent())
  .collect(Collectors.toMap(Entry::getKey, e -> e.getValue().get()));