Transform and filter a Java Map with streams

Question

I have a Java Map that I'd like to transform and filter. As a trivial example, suppose I want to convert all values to Integers then remove the odd entries.

Map<String, String> input = new HashMap<>();
input.put("a", "1234");
input.put("b", "2345");
input.put("c", "3456");
input.put("d", "4567");

Map<String, Integer> output = input.entrySet().stream()
        .collect(Collectors.toMap(
                Map.Entry::getKey,
                e -> Integer.parseInt(e.getValue())
        ))
        .entrySet().stream()
        .filter(e -> e.getValue() % 2 == 0)
        .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));


System.out.println(output.toString());

This is correct and yields: {a=1234, c=3456}

However, I can't help but wonder if there's a way to avoid calling .entrySet().stream() twice.

Is there a way I can perform both transform and filter operations and call .collect() only once at the end?

Answer 1

Yes, you can map each entry to another temporary entry that will hold the key and the parsed integer value. Then you can filter each entry based on their value.

Map<String, Integer> output =
    input.entrySet()
         .stream()
         .map(e -> new AbstractMap.SimpleEntry<>(e.getKey(), Integer.valueOf(e.getValue())))
         .filter(e -> e.getValue() % 2 == 0)
         .collect(Collectors.toMap(
             Map.Entry::getKey,
             Map.Entry::getValue
         ));

Note that I used Integer.valueOf instead of parseInt since we actually want a boxed int.

---

If you have the luxury to use the StreamEx library, you can do it quite simply:

Map<String, Integer> output =
    EntryStream.of(input).mapValues(Integer::valueOf).filterValues(v -> v % 2 == 0).toMap();