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.

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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();