Business
Jobs
  • About Us
  • Solutions
    • Job Postings
      Post your job and receive qualified candidates in 48h.
    • Candidate Assessments
      500+ technical and psychological tests, plus anti-fraud.
    • Headhunting
      Tailor-made executive search from start to finish.
    • Payroll + EOR
      Payroll dispersal and EOR across 15+ LATAM countries.
  • Pricing
  • Jobs

0

259
Views
Processing a list of maps using Java 8 streams

How can I simplify this code into a single lambda expression? The idea is that there is a list of maps and I would like to create a new list of maps, using a filter on the key. In this example, I want to remap it so that it only keeps the keys "x" and "z".

    Map<String, String> m0 = new LinkedHashMap<>();
    m0.put("x", "123");
    m0.put("y", "456");
    m0.put("z", "789");

    Map<String, String> m1 = new LinkedHashMap<>();
    m1.put("x", "000");
    m1.put("y", "111");
    m1.put("z", "222");

    List<Map> l = new ArrayList<>(Arrays.asList(m0, m1));
    List<Map> tx = new ArrayList<>();
    for(Map<String, String> m : l) {
        Map<String, String> filtered = m.entrySet()
                .stream()
                .filter(map -> map.getKey().equals("x") || map.getKey().equals("z"))
                .collect(Collectors.toMap(p -> p.getKey(), p -> p.getValue()));
        tx.add(filtered);
    }
    System.err.println("l: " + l);
    System.err.println("tx: " + tx);

Output:

    l: [{x=123, y=456, z=789}, {x=000, y=111, z=222}]
    tx: [{x=123, z=789}, {x=000, z=222}]
about 4 years ago · Santiago Trujillo
3 answers
Answer question

0

Of course, you can convert your entire operation into one Stream operation.

// no need to copy a List (result of Array.asList) to an ArrayList, by the way
List<Map<String, String>> l = Arrays.asList(m0, m1);

List<Map<String, String>> tx = l.stream().map(m -> m.entrySet().stream()
        .filter(map -> map.getKey().equals("x") || map.getKey().equals("z"))
        .collect(Collectors.toMap(p -> p.getKey(), p -> p.getValue())))
    .collect(Collectors.toList());

But note that streaming over a Map and filtering is an operation with a linear time complexity, as it will check each key of each map against the filter, while you have only a very small number of actual keys you want to retain. So here, it is much simpler and more efficient (for larger maps) to use

List<Map<String, String>> tx = l.stream()
    .map(m -> Stream.of("x", "y")
                    .filter(m::containsKey).collect(Collectors.toMap(key->key, m::get)))
    .collect(Collectors.toList());

which will only perform four lookups per map. If it bothers you, you could even reduce it to two lookups, however, the constant factor is irrelevant for the overall time complexity, which will be constant time, if the map has a constant time lookup, like HashMap. Even for map’s with O(log(n)) lookup time complexity, like TreeMap, this will be more efficient than the linear scan, if the maps are larger than the three mappings of the example code.

about 4 years ago · Santiago Trujillo Report

0

You can try something like this:

List<Map<String, String>> l = Arrays.asList(m0, m1);

l.forEach(map -> {
    map.entrySet().removeIf(e -> !e.getKey().equals("x") && !e.getKey().equals("z"));
});

It simply removes all of the mappings in every Map<String, String> if the entry key is not x or z.

Edit: You should utilize Radiodef's equivalent, but shorter method!

List<Map<String, String>> l = Arrays.asList(m0, m1);

l.forEach(map -> map.keySet().retainAll(Arrays.asList("x", "z"));
about 4 years ago · Santiago Trujillo Report

0

Try the following code(I declared a list for desiredKeys):

public class Main {
    public static void main(String[] args) {
        Map<String, String> m0 = new HashMap<>();
        m0.put("x", "123");
        m0.put("y", "456");
        m0.put("z", "789");

        Map<String, String> m1 = new HashMap<>();
        m1.put("x", "000");
        m1.put("y", "111");
        m1.put("z", "222");

        List<Map<String, String>> l = new ArrayList<>(Arrays.asList(m0, m1));

        List<String> desiredKeys = Lists.newArrayList("x", "z");

        List<Map<String, String>> transformed = l.stream().map(map -> map.entrySet().stream()
                .filter(e -> desiredKeys.stream().anyMatch(k -> k.equals(e.getKey())))
                .collect(Collectors.toMap(e -> e.getKey(), p -> p.getValue()))).filter(m -> !m.isEmpty()).collect(Collectors.toList());

        System.err.println(l);
        System.err.println(transformed);
    }
}
about 4 years ago · Santiago Trujillo Report
Answer question
Find remote jobs

Discover the new way to find a job!

Top jobs
Top job categories
Business
Post vacancy Pricing Sales
Legal
Terms and conditions Privacy policy
© 2026 PeakU Inc. All Rights Reserved.
Andres GPT
Show me some job opportunities
There's an error!